$\int \frac{\tan^2 x}{1+x^2} dx$

Question

Motivated by this integral, which does not evaluate in Mathematica, I'm wondering if there is some trick or substitution that will solve the indefinite integral.

My motivation is to find integrals that Mathematica cannot solve "directly," and learn from this community tricks, insights, and alternate approaches in which the problem can be solved. Yes... this often involves some mathematical (rather than coding) insight, but there too, sometimes the mathematical insight leads to a coding insight, for instance some clever coded variable substitution, or casting in some transform space, or...

---

Integrate[Tan[x]^2/(1+x^2),x]

Answer 1

We can convert the integral into a direct sum by using the series expansion of the tangent: $$ \tan^2(x)=\sum_{k=1}^{\infty} \frac{2\,(2^{2k+2}-1)\, (2k+1)\, \zeta(2k+2)}{\pi^{2k+2}} x^{2 k} $$ in terms of the Riemann zeta function (or alternatively in terms of the Bernoulli numbers through $\zeta(2k+2)=\frac{(-1)^k 2^{2k+1} \pi^{2k+2} B_{2k+2}}{(2 k+2)!}$).

The integral in question becomes $$ \int\frac{\tan^2(x)}{1+x^2}dx =\sum_{k=1}^{\infty} \frac{2\,(2^{2k+2}-1)\, (2k+1)\, \zeta(2k+2)}{\pi^{2k+2}} \int\frac{x^{2 k}}{1+x^2}dx $$

which we can integrate component-by-component,

Integrate[x^(2 k) / (1 + x^2), x]
(*    (x^(1 + 2 k) Hypergeometric2F1[1, 1/2 + k, 3/2 + k, -x^2])/(1 + 2 k)    *)

getting the sum $$ \int\frac{\tan^2(x)}{1+x^2}dx =\sum_{k=1}^{\infty} \frac{2\,(2^{2k+2}-1)\, \zeta(2k+2)}{\pi^{2k+2}} x^{2k+1} {_2}F_1\left(1,k+\frac{1}{2};k+\frac{3}{2};-x^2\right). $$

This sum converges rapidly: For large $k$, $\zeta(2k+2)\approx1$ and the hypergeometric function is ${_2}F_1\left(1,k+\frac12;k+\frac32;-x^2\right)\approx\frac{1}{1+x^2}$, which makes the $k^{\text{th}}$ term approximately $$ \frac{2\,(2^{2k+2}-1)\, \zeta(2k+2)}{\pi^{2k+2}} x^{2k+1} {_2}F_1\left(1,k+\frac{1}{2};k+\frac{3}{2};-x^2\right) \approx \left(\frac{2x}{\pi}\right)^{2k+2}\frac{2}{x(1+x^2)}, $$ which becomes exponentially smaller with $k\to\infty$ as long as $|x|<\frac{\pi}{2}$.

Let's try out the case of definite integration over $[0,\frac{\pi}{4}]$:

NIntegrate[Tan[x]^2/(1 + x^2), {x, 0, π/4}]
(*    0.1565032456995724`    *)

With[{x = π/4},
  Sum[2(2^(2k+2)-1) Zeta[2k+2]/π^(2k+2) x^(2k+1) *
      Hypergeometric2F1[1, k+1/2, k+3/2, -x^2], {k, 20}]] // N
(*    0.1565032456994413`    *)

side note on hypergeometric functions

The hypergeometric function can also be written as a Hurwitz–Lerch transcendent or as an incomplete beta function: $$ {_2}F_1\left(1,k+\frac12;k+\frac32;z\right) =\left(k+\frac12\right) \Phi \left(z,1,k+\frac12\right) =\frac{k+\frac12}{z^{k+1/2}} B_z\left(k+\frac12,0\right) $$ Of these, the Hurwitz–Lerch transcendent is probably the simplest to evaluate in practice because it reduces to a finite sum: $$ \Phi \left(z,1,k+\frac12\right) = \frac{2}{z^{k+1/2}}\tanh^{-1}\sqrt{z}-\sum_{s=1}^k \frac{1}{\left(k+\frac12-s\right)z^s} $$

With this simplification the result can be written in terms of $\arctan$ and $\zeta$ only:

With[{x = π/4},
  Sum[2 (-1)^k (2^(2k+2)-1) (2k+1) π^(-2k-2) Zeta[2k+2] *
      (ArcTan[x] - Sum[((-1)^q x^(2q+1))/(2q+1), {q, 0, k-1}]),
      {k, 20}]] // N
(*    0.1565032456994413`    *)