Optimal. Leaf size=355 \[ -\frac{i x \text{PolyLog}\left (3,-\frac{(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b^2}+\frac{i x \text{PolyLog}\left (3,-\frac{(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b^2}+\frac{i \text{PolyLog}\left (4,-\frac{(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{8 b^3}-\frac{i \text{PolyLog}\left (4,-\frac{(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{8 b^3}+\frac{i x^2 \text{PolyLog}\left (2,-\frac{(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b}-\frac{i x^2 \text{PolyLog}\left (2,-\frac{(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b}+\frac{1}{6} i x^3 \log \left (1+\frac{(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )-\frac{1}{6} i x^3 \log \left (1+\frac{(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )+\frac{1}{3} x^3 \tan ^{-1}(d \tanh (a+b x)+c) \]
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Rubi [A] time = 0.461361, antiderivative size = 355, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5199, 2190, 2531, 6609, 2282, 6589} \[ -\frac{i x \text{PolyLog}\left (3,-\frac{(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b^2}+\frac{i x \text{PolyLog}\left (3,-\frac{(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b^2}+\frac{i \text{PolyLog}\left (4,-\frac{(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{8 b^3}-\frac{i \text{PolyLog}\left (4,-\frac{(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{8 b^3}+\frac{i x^2 \text{PolyLog}\left (2,-\frac{(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b}-\frac{i x^2 \text{PolyLog}\left (2,-\frac{(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b}+\frac{1}{6} i x^3 \log \left (1+\frac{(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )-\frac{1}{6} i x^3 \log \left (1+\frac{(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )+\frac{1}{3} x^3 \tan ^{-1}(d \tanh (a+b x)+c) \]
Antiderivative was successfully verified.
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Rule 5199
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^2 \tan ^{-1}(c+d \tanh (a+b x)) \, dx &=\frac{1}{3} x^3 \tan ^{-1}(c+d \tanh (a+b x))+\frac{1}{3} (b (1-i (c+d))) \int \frac{e^{2 a+2 b x} x^3}{i+c-d+(i+c+d) e^{2 a+2 b x}} \, dx-\frac{1}{3} (b (1+i (c+d))) \int \frac{e^{2 a+2 b x} x^3}{i-c+d+(i-c-d) e^{2 a+2 b x}} \, dx\\ &=\frac{1}{3} x^3 \tan ^{-1}(c+d \tanh (a+b x))+\frac{1}{6} i x^3 \log \left (1+\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )-\frac{1}{6} i x^3 \log \left (1+\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac{1}{2} i \int x^2 \log \left (1+\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right ) \, dx+\frac{1}{2} i \int x^2 \log \left (1+\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right ) \, dx\\ &=\frac{1}{3} x^3 \tan ^{-1}(c+d \tanh (a+b x))+\frac{1}{6} i x^3 \log \left (1+\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )-\frac{1}{6} i x^3 \log \left (1+\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )+\frac{i x^2 \text{Li}_2\left (-\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}-\frac{i x^2 \text{Li}_2\left (-\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}-\frac{i \int x \text{Li}_2\left (-\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right ) \, dx}{2 b}+\frac{i \int x \text{Li}_2\left (-\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right ) \, dx}{2 b}\\ &=\frac{1}{3} x^3 \tan ^{-1}(c+d \tanh (a+b x))+\frac{1}{6} i x^3 \log \left (1+\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )-\frac{1}{6} i x^3 \log \left (1+\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )+\frac{i x^2 \text{Li}_2\left (-\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}-\frac{i x^2 \text{Li}_2\left (-\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}-\frac{i x \text{Li}_3\left (-\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b^2}+\frac{i x \text{Li}_3\left (-\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b^2}+\frac{i \int \text{Li}_3\left (-\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right ) \, dx}{4 b^2}-\frac{i \int \text{Li}_3\left (-\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right ) \, dx}{4 b^2}\\ &=\frac{1}{3} x^3 \tan ^{-1}(c+d \tanh (a+b x))+\frac{1}{6} i x^3 \log \left (1+\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )-\frac{1}{6} i x^3 \log \left (1+\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )+\frac{i x^2 \text{Li}_2\left (-\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}-\frac{i x^2 \text{Li}_2\left (-\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}-\frac{i x \text{Li}_3\left (-\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b^2}+\frac{i x \text{Li}_3\left (-\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b^2}+\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (-\frac{(-i+c+d) x}{-i+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}-\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (-\frac{(i+c+d) x}{i+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}\\ &=\frac{1}{3} x^3 \tan ^{-1}(c+d \tanh (a+b x))+\frac{1}{6} i x^3 \log \left (1+\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )-\frac{1}{6} i x^3 \log \left (1+\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )+\frac{i x^2 \text{Li}_2\left (-\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}-\frac{i x^2 \text{Li}_2\left (-\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}-\frac{i x \text{Li}_3\left (-\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b^2}+\frac{i x \text{Li}_3\left (-\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b^2}+\frac{i \text{Li}_4\left (-\frac{(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{8 b^3}-\frac{i \text{Li}_4\left (-\frac{(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{8 b^3}\\ \end{align*}
Mathematica [A] time = 5.225, size = 305, normalized size = 0.86 \[ \frac{1}{3} x^3 \tan ^{-1}(d \tanh (a+b x)+c)+\frac{i \left (6 b^2 x^2 \text{PolyLog}\left (2,-\frac{(c+d-i) e^{2 (a+b x)}}{c-d-i}\right )-6 b^2 x^2 \text{PolyLog}\left (2,-\frac{(c+d+i) e^{2 (a+b x)}}{c-d+i}\right )-6 b x \text{PolyLog}\left (3,-\frac{(c+d-i) e^{2 (a+b x)}}{c-d-i}\right )+6 b x \text{PolyLog}\left (3,-\frac{(c+d+i) e^{2 (a+b x)}}{c-d+i}\right )+3 \text{PolyLog}\left (4,-\frac{(c+d-i) e^{2 (a+b x)}}{c-d-i}\right )-3 \text{PolyLog}\left (4,-\frac{(c+d+i) e^{2 (a+b x)}}{c-d+i}\right )+4 b^3 x^3 \log \left (1+\frac{(c+d-i) e^{2 (a+b x)}}{c-d-i}\right )-4 b^3 x^3 \log \left (1+\frac{(c+d+i) e^{2 (a+b x)}}{c-d+i}\right )\right )}{24 b^3} \]
Antiderivative was successfully verified.
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Maple [C] time = 6.736, size = 6981, normalized size = 19.7 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \, x^{3} \arctan \left (\frac{{\left (c e^{\left (2 \, a\right )} + d e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + c - d}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - 4 \, b d \int \frac{x^{3} e^{\left (2 \, b x + 2 \, a\right )}}{3 \,{\left (c^{2} - 2 \, c d + d^{2} +{\left (c^{2} e^{\left (4 \, a\right )} + 2 \, c d e^{\left (4 \, a\right )} + d^{2} e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )} + 2 \,{\left (c^{2} e^{\left (2 \, a\right )} - d^{2} e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + 1\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.92378, size = 3641, normalized size = 10.26 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \arctan \left (d \tanh \left (b x + a\right ) + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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