Optimal. Leaf size=124 \[ \frac{i \text{PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )}{8 b^2}+\frac{x \text{PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac{1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac{1}{2} x^2 \tan ^{-1}(c-(1-i c) \tan (a+b x))+\frac{b x^3}{6} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.207879, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {5171, 2184, 2190, 2531, 2282, 6589} \[ \frac{i \text{PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )}{8 b^2}+\frac{x \text{PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac{1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac{1}{2} x^2 \tan ^{-1}(c-(1-i c) \tan (a+b x))+\frac{b x^3}{6} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 5171
Rule 2184
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x \tan ^{-1}(c+(-1+i c) \tan (a+b x)) \, dx &=\frac{1}{2} x^2 \tan ^{-1}(c-(1-i c) \tan (a+b x))-\frac{1}{2} (i b) \int \frac{x^2}{i (-1+i c)+c+c e^{2 i a+2 i b x}} \, dx\\ &=\frac{b x^3}{6}+\frac{1}{2} x^2 \tan ^{-1}(c-(1-i c) \tan (a+b x))-\frac{1}{2} (b c) \int \frac{e^{2 i a+2 i b x} x^2}{i (-1+i c)+c+c e^{2 i a+2 i b x}} \, dx\\ &=\frac{b x^3}{6}+\frac{1}{2} x^2 \tan ^{-1}(c-(1-i c) \tan (a+b x))+\frac{1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )-\frac{1}{2} i \int x \log \left (1+\frac{c e^{2 i a+2 i b x}}{i (-1+i c)+c}\right ) \, dx\\ &=\frac{b x^3}{6}+\frac{1}{2} x^2 \tan ^{-1}(c-(1-i c) \tan (a+b x))+\frac{1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac{x \text{Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac{\int \text{Li}_2\left (-\frac{c e^{2 i a+2 i b x}}{i (-1+i c)+c}\right ) \, dx}{4 b}\\ &=\frac{b x^3}{6}+\frac{1}{2} x^2 \tan ^{-1}(c-(1-i c) \tan (a+b x))+\frac{1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac{x \text{Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i c x)}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^2}\\ &=\frac{b x^3}{6}+\frac{1}{2} x^2 \tan ^{-1}(c-(1-i c) \tan (a+b x))+\frac{1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac{x \text{Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac{i \text{Li}_3\left (-i c e^{2 i a+2 i b x}\right )}{8 b^2}\\ \end{align*}
Mathematica [A] time = 0.306181, size = 111, normalized size = 0.9 \[ \frac{i \left (2 i b x \text{PolyLog}\left (2,\frac{i e^{-2 i (a+b x)}}{c}\right )+\text{PolyLog}\left (3,\frac{i e^{-2 i (a+b x)}}{c}\right )+2 b^2 x^2 \log \left (1-\frac{i e^{-2 i (a+b x)}}{c}\right )\right )}{8 b^2}+\frac{1}{2} x^2 \tan ^{-1}(c+i (c+i) \tan (a+b x)) \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [C] time = 8.505, size = 1498, normalized size = 12.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [B] time = 1.06539, size = 296, normalized size = 2.39 \begin{align*} \frac{\frac{{\left ({\left (b x + a\right )}^{2} - 2 \,{\left (b x + a\right )} a\right )} \arctan \left ({\left (i \, c - 1\right )} \tan \left (b x + a\right ) + c\right )}{b} + \frac{2 \,{\left (-4 i \,{\left (b x + a\right )}^{3} + 12 i \,{\left (b x + a\right )}^{2} a - 6 i \, b x{\rm Li}_2\left (-i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) +{\left (6 i \,{\left (b x + a\right )}^{2} - 12 i \,{\left (b x + a\right )} a\right )} \arctan \left (c \cos \left (2 \, b x + 2 \, a\right ), -c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \,{\left ({\left (b x + a\right )}^{2} - 2 \,{\left (b x + a\right )} a\right )} \log \left (c^{2} \cos \left (2 \, b x + 2 \, a\right )^{2} + c^{2} \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \,{\rm Li}_{3}(-i \, c e^{\left (2 i \, b x + 2 i \, a\right )})\right )}{\left (i \, c - 1\right )}}{b{\left (12 \, c + 12 i\right )}}}{2 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [C] time = 1.94291, size = 760, normalized size = 6.13 \begin{align*} \frac{2 \, b^{3} x^{3} + 3 i \, b^{2} x^{2} \log \left (-\frac{{\left (c + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )}}{c e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) + 2 \, a^{3} + 6 \, b x{\rm Li}_2\left (\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 \, b x{\rm Li}_2\left (-\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 3 i \, a^{2} \log \left (\frac{2 \, c e^{\left (i \, b x + i \, a\right )} + i \, \sqrt{-4 i \, c}}{2 \, c}\right ) + 3 i \, a^{2} \log \left (\frac{2 \, c e^{\left (i \, b x + i \, a\right )} - i \, \sqrt{-4 i \, c}}{2 \, c}\right ) +{\left (3 i \, b^{2} x^{2} - 3 i \, a^{2}\right )} \log \left (\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) +{\left (3 i \, b^{2} x^{2} - 3 i \, a^{2}\right )} \log \left (-\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) + 6 i \,{\rm polylog}\left (3, \frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \,{\rm polylog}\left (3, -\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{b \int \frac{x^{2}}{i c e^{2 i a} e^{2 i b x} + 1}\, dx}{2} + \frac{i x^{2} \log{\left (- i c + \frac{i c}{e^{2 i a} e^{2 i b x} + 1} - \frac{i c e^{i a} e^{i b x}}{e^{i a} e^{i b x} + e^{- i a} e^{- i b x}} + 1 - \frac{1}{e^{2 i a} e^{2 i b x} + 1} + \frac{e^{i a} e^{i b x}}{e^{i a} e^{i b x} + e^{- i a} e^{- i b x}} \right )}}{4} - \frac{i x^{2} \log{\left (i c - \frac{i c}{e^{2 i a} e^{2 i b x} + 1} + \frac{i c e^{i a} e^{i b x}}{e^{i a} e^{i b x} + e^{- i a} e^{- i b x}} + 1 + \frac{1}{e^{2 i a} e^{2 i b x} + 1} - \frac{e^{i a} e^{i b x}}{e^{i a} e^{i b x} + e^{- i a} e^{- i b x}} \right )}}{4} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \arctan \left ({\left (i \, c - 1\right )} \tan \left (b x + a\right ) + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]