Optimal. Leaf size=234 \[ \frac{f (e+f x) \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac{f (e+f x) \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{4 b^2}+\frac{i f^2 \text{PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{i f^2 \text{PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{i (e+f x)^2 \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac{(e+f x)^3 \tanh ^{-1}(\tan (a+b x))}{3 f} \]
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Rubi [A] time = 0.165241, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6251, 4181, 2531, 6609, 2282, 6589} \[ \frac{f (e+f x) \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac{f (e+f x) \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{4 b^2}+\frac{i f^2 \text{PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{i f^2 \text{PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{i (e+f x)^2 \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac{(e+f x)^3 \tanh ^{-1}(\tan (a+b x))}{3 f} \]
Antiderivative was successfully verified.
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Rule 6251
Rule 4181
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int (e+f x)^2 \tanh ^{-1}(\tan (a+b x)) \, dx &=\frac{(e+f x)^3 \tanh ^{-1}(\tan (a+b x))}{3 f}-\frac{b \int (e+f x)^3 \sec (2 a+2 b x) \, dx}{3 f}\\ &=\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac{(e+f x)^3 \tanh ^{-1}(\tan (a+b x))}{3 f}+\frac{1}{2} \int (e+f x)^2 \log \left (1-i e^{i (2 a+2 b x)}\right ) \, dx-\frac{1}{2} \int (e+f x)^2 \log \left (1+i e^{i (2 a+2 b x)}\right ) \, dx\\ &=\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac{(e+f x)^3 \tanh ^{-1}(\tan (a+b x))}{3 f}-\frac{i (e+f x)^2 \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{(i f) \int (e+f x) \text{Li}_2\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{2 b}-\frac{(i f) \int (e+f x) \text{Li}_2\left (i e^{i (2 a+2 b x)}\right ) \, dx}{2 b}\\ &=\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac{(e+f x)^3 \tanh ^{-1}(\tan (a+b x))}{3 f}-\frac{i (e+f x)^2 \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{f (e+f x) \text{Li}_3\left (-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac{f (e+f x) \text{Li}_3\left (i e^{2 i (a+b x)}\right )}{4 b^2}-\frac{f^2 \int \text{Li}_3\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{4 b^2}+\frac{f^2 \int \text{Li}_3\left (i e^{i (2 a+2 b x)}\right ) \, dx}{4 b^2}\\ &=\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac{(e+f x)^3 \tanh ^{-1}(\tan (a+b x))}{3 f}-\frac{i (e+f x)^2 \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{f (e+f x) \text{Li}_3\left (-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac{f (e+f x) \text{Li}_3\left (i e^{2 i (a+b x)}\right )}{4 b^2}+\frac{\left (i f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{8 b^3}-\frac{\left (i f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{8 b^3}\\ &=\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac{(e+f x)^3 \tanh ^{-1}(\tan (a+b x))}{3 f}-\frac{i (e+f x)^2 \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{f (e+f x) \text{Li}_3\left (-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac{f (e+f x) \text{Li}_3\left (i e^{2 i (a+b x)}\right )}{4 b^2}+\frac{i f^2 \text{Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{i f^2 \text{Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}\\ \end{align*}
Mathematica [A] time = 0.64761, size = 409, normalized size = 1.75 \[ \frac{1}{3} x \left (3 e^2+3 e f x+f^2 x^2\right ) \tanh ^{-1}(\tan (a+b x))+\frac{-6 i b^2 (e+f x)^2 \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )+6 i b^2 (e+f x)^2 \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )+6 b e f \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )-6 b e f \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )+6 b f^2 x \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )-6 b f^2 x \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )+3 i f^2 \text{PolyLog}\left (4,-i e^{2 i (a+b x)}\right )-3 i f^2 \text{PolyLog}\left (4,i e^{2 i (a+b x)}\right )-12 b^3 e^2 x \log \left (1-i e^{2 i (a+b x)}\right )+12 b^3 e^2 x \log \left (1+i e^{2 i (a+b x)}\right )-12 b^3 e f x^2 \log \left (1-i e^{2 i (a+b x)}\right )+12 b^3 e f x^2 \log \left (1+i e^{2 i (a+b x)}\right )-4 b^3 f^2 x^3 \log \left (1-i e^{2 i (a+b x)}\right )+4 b^3 f^2 x^3 \log \left (1+i e^{2 i (a+b x)}\right )}{24 b^3} \]
Antiderivative was successfully verified.
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Maple [C] time = 8.575, size = 5543, normalized size = 23.7 \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}{12} \,{\left (f^{2} x^{3} + 3 \, e f x^{2} + 3 \, e^{2} x\right )} \log \left (2 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \sin \left (2 \, b x + 2 \, a\right )^{2} + 4 \, \sin \left (2 \, b x + 2 \, a\right ) + 2\right ) - \frac{1}{12} \,{\left (f^{2} x^{3} + 3 \, e f x^{2} + 3 \, e^{2} x\right )} \log \left (2 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \sin \left (2 \, b x + 2 \, a\right )^{2} - 4 \, \sin \left (2 \, b x + 2 \, a\right ) + 2\right ) - \int \frac{2 \,{\left ({\left (b f^{2} x^{3} + 3 \, b e f x^{2} + 3 \, b e^{2} x\right )} \cos \left (4 \, b x + 4 \, a\right ) \cos \left (2 \, b x + 2 \, a\right ) +{\left (b f^{2} x^{3} + 3 \, b e f x^{2} + 3 \, b e^{2} x\right )} \sin \left (4 \, b x + 4 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) +{\left (b f^{2} x^{3} + 3 \, b e f x^{2} + 3 \, b e^{2} x\right )} \cos \left (2 \, b x + 2 \, a\right )\right )}}{3 \,{\left (\cos \left (4 \, b x + 4 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 2 \, \cos \left (4 \, b x + 4 \, a\right ) + 1\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.4866, size = 3359, normalized size = 14.35 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e + f x\right )^{2} \operatorname{atanh}{\left (\tan{\left (a + b x \right )} \right )}\, dx \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{\left (f x + e\right )}^{2} \operatorname{artanh}\left (\tan \left (b x + a\right )\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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