Optimal. Leaf size=299 \[ \frac{3 i f^2 (e+f x) \text{PolyLog}\left (4,-i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{PolyLog}\left (4,i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f (e+f x)^2 \text{PolyLog}\left (3,i e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 i f^3 \text{PolyLog}\left (5,-i e^{2 a+2 b x}\right )}{16 b^4}+\frac{3 i f^3 \text{PolyLog}\left (5,i e^{2 a+2 b x}\right )}{16 b^4}+\frac{i (e+f x)^3 \text{PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{(e+f x)^4 \cot ^{-1}(\coth (a+b x))}{4 f} \]
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Rubi [A] time = 0.208478, antiderivative size = 299, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5186, 4180, 2531, 6609, 2282, 6589} \[ \frac{3 i f^2 (e+f x) \text{PolyLog}\left (4,-i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{PolyLog}\left (4,i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f (e+f x)^2 \text{PolyLog}\left (3,i e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 i f^3 \text{PolyLog}\left (5,-i e^{2 a+2 b x}\right )}{16 b^4}+\frac{3 i f^3 \text{PolyLog}\left (5,i e^{2 a+2 b x}\right )}{16 b^4}+\frac{i (e+f x)^3 \text{PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{(e+f x)^4 \cot ^{-1}(\coth (a+b x))}{4 f} \]
Antiderivative was successfully verified.
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Rule 5186
Rule 4180
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int (e+f x)^3 \cot ^{-1}(\coth (a+b x)) \, dx &=\frac{(e+f x)^4 \cot ^{-1}(\coth (a+b x))}{4 f}-\frac{b \int (e+f x)^4 \text{sech}(2 a+2 b x) \, dx}{4 f}\\ &=\frac{(e+f x)^4 \cot ^{-1}(\coth (a+b x))}{4 f}-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{1}{2} i \int (e+f x)^3 \log \left (1-i e^{2 a+2 b x}\right ) \, dx-\frac{1}{2} i \int (e+f x)^3 \log \left (1+i e^{2 a+2 b x}\right ) \, dx\\ &=\frac{(e+f x)^4 \cot ^{-1}(\coth (a+b x))}{4 f}-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac{(3 i f) \int (e+f x)^2 \text{Li}_2\left (-i e^{2 a+2 b x}\right ) \, dx}{4 b}+\frac{(3 i f) \int (e+f x)^2 \text{Li}_2\left (i e^{2 a+2 b x}\right ) \, dx}{4 b}\\ &=\frac{(e+f x)^4 \cot ^{-1}(\coth (a+b x))}{4 f}-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac{3 i f (e+f x)^2 \text{Li}_3\left (-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f (e+f x)^2 \text{Li}_3\left (i e^{2 a+2 b x}\right )}{8 b^2}+\frac{\left (3 i f^2\right ) \int (e+f x) \text{Li}_3\left (-i e^{2 a+2 b x}\right ) \, dx}{4 b^2}-\frac{\left (3 i f^2\right ) \int (e+f x) \text{Li}_3\left (i e^{2 a+2 b x}\right ) \, dx}{4 b^2}\\ &=\frac{(e+f x)^4 \cot ^{-1}(\coth (a+b x))}{4 f}-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac{3 i f (e+f x)^2 \text{Li}_3\left (-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f (e+f x)^2 \text{Li}_3\left (i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f^2 (e+f x) \text{Li}_4\left (-i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{Li}_4\left (i e^{2 a+2 b x}\right )}{8 b^3}-\frac{\left (3 i f^3\right ) \int \text{Li}_4\left (-i e^{2 a+2 b x}\right ) \, dx}{8 b^3}+\frac{\left (3 i f^3\right ) \int \text{Li}_4\left (i e^{2 a+2 b x}\right ) \, dx}{8 b^3}\\ &=\frac{(e+f x)^4 \cot ^{-1}(\coth (a+b x))}{4 f}-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac{3 i f (e+f x)^2 \text{Li}_3\left (-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f (e+f x)^2 \text{Li}_3\left (i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f^2 (e+f x) \text{Li}_4\left (-i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{Li}_4\left (i e^{2 a+2 b x}\right )}{8 b^3}-\frac{\left (3 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_4(-i x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{16 b^4}+\frac{\left (3 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_4(i x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{16 b^4}\\ &=\frac{(e+f x)^4 \cot ^{-1}(\coth (a+b x))}{4 f}-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac{3 i f (e+f x)^2 \text{Li}_3\left (-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f (e+f x)^2 \text{Li}_3\left (i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f^2 (e+f x) \text{Li}_4\left (-i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{Li}_4\left (i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f^3 \text{Li}_5\left (-i e^{2 a+2 b x}\right )}{16 b^4}+\frac{3 i f^3 \text{Li}_5\left (i e^{2 a+2 b x}\right )}{16 b^4}\\ \end{align*}
Mathematica [B] time = 0.336366, size = 600, normalized size = 2.01 \[ \frac{1}{4} x \left (6 e^2 f x+4 e^3+4 e f^2 x^2+f^3 x^3\right ) \cot ^{-1}(\coth (a+b x))-\frac{i \left (6 b^2 e^2 f \text{PolyLog}\left (3,-i e^{2 (a+b x)}\right )-6 b^2 e^2 f \text{PolyLog}\left (3,i e^{2 (a+b x)}\right )+12 b^2 e f^2 x \text{PolyLog}\left (3,-i e^{2 (a+b x)}\right )-12 b^2 e f^2 x \text{PolyLog}\left (3,i e^{2 (a+b x)}\right )-4 b^3 (e+f x)^3 \text{PolyLog}\left (2,-i e^{2 (a+b x)}\right )+4 b^3 (e+f x)^3 \text{PolyLog}\left (2,i e^{2 (a+b x)}\right )+6 b^2 f^3 x^2 \text{PolyLog}\left (3,-i e^{2 (a+b x)}\right )-6 b^2 f^3 x^2 \text{PolyLog}\left (3,i e^{2 (a+b x)}\right )-6 b e f^2 \text{PolyLog}\left (4,-i e^{2 (a+b x)}\right )+6 b e f^2 \text{PolyLog}\left (4,i e^{2 (a+b x)}\right )-6 b f^3 x \text{PolyLog}\left (4,-i e^{2 (a+b x)}\right )+6 b f^3 x \text{PolyLog}\left (4,i e^{2 (a+b x)}\right )+3 f^3 \text{PolyLog}\left (5,-i e^{2 (a+b x)}\right )-3 f^3 \text{PolyLog}\left (5,i e^{2 (a+b x)}\right )+12 b^4 e^2 f x^2 \log \left (1-i e^{2 (a+b x)}\right )-12 b^4 e^2 f x^2 \log \left (1+i e^{2 (a+b x)}\right )+8 b^4 e^3 x \log \left (1-i e^{2 (a+b x)}\right )-8 b^4 e^3 x \log \left (1+i e^{2 (a+b x)}\right )+8 b^4 e f^2 x^3 \log \left (1-i e^{2 (a+b x)}\right )-8 b^4 e f^2 x^3 \log \left (1+i e^{2 (a+b x)}\right )+2 b^4 f^3 x^4 \log \left (1-i e^{2 (a+b x)}\right )-2 b^4 f^3 x^4 \log \left (1+i e^{2 (a+b x)}\right )\right )}{16 b^4} \]
Antiderivative was successfully verified.
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Maple [C] time = 7.395, size = 7275, normalized size = 24.3 \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}{4} \,{\left (f^{3} x^{4} + 4 \, e f^{2} x^{3} + 6 \, e^{2} f x^{2} + 4 \, e^{3} x\right )} \arctan \left (\frac{e^{\left (2 \, b x + 2 \, a\right )} - 1}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - \int \frac{{\left (b f^{3} x^{4} e^{\left (2 \, a\right )} + 4 \, b e f^{2} x^{3} e^{\left (2 \, a\right )} + 6 \, b e^{2} f x^{2} e^{\left (2 \, a\right )} + 4 \, b e^{3} x e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{2 \,{\left (e^{\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 = 3.03046, size = 4070, normalized size = 13.61 \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{\left (f x + e\right )}^{3} \operatorname{arccot}\left (\coth \left (b x + a\right )\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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