Optimal. Leaf size=229 \[ \frac {i f^2 \text {Li}_4\left (-i e^{2 a+2 b x}\right )}{8 b^3}-\frac {i f^2 \text {Li}_4\left (i e^{2 a+2 b x}\right )}{8 b^3}-\frac {i f (e+f x) \text {Li}_3\left (-i e^{2 a+2 b x}\right )}{4 b^2}+\frac {i f (e+f x) \text {Li}_3\left (i e^{2 a+2 b x}\right )}{4 b^2}+\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac {(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f} \]
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Rubi [A] time = 0.15, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5186, 4180, 2531, 6609, 2282, 6589} \[ -\frac {i f (e+f x) \text {PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{4 b^2}+\frac {i f (e+f x) \text {PolyLog}\left (3,i e^{2 a+2 b x}\right )}{4 b^2}+\frac {i f^2 \text {PolyLog}\left (4,-i e^{2 a+2 b x}\right )}{8 b^3}-\frac {i f^2 \text {PolyLog}\left (4,i e^{2 a+2 b x}\right )}{8 b^3}+\frac {i (e+f x)^2 \text {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}-\frac {i (e+f x)^2 \text {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}-\frac {(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 4180
Rule 5186
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int (e+f x)^2 \cot ^{-1}(\coth (a+b x)) \, dx &=\frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {b \int (e+f x)^3 \text {sech}(2 a+2 b x) \, dx}{3 f}\\ &=\frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac {1}{2} i \int (e+f x)^2 \log \left (1-i e^{2 a+2 b x}\right ) \, dx-\frac {1}{2} i \int (e+f x)^2 \log \left (1+i e^{2 a+2 b x}\right ) \, dx\\ &=\frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac {(i f) \int (e+f x) \text {Li}_2\left (-i e^{2 a+2 b x}\right ) \, dx}{2 b}+\frac {(i f) \int (e+f x) \text {Li}_2\left (i e^{2 a+2 b x}\right ) \, dx}{2 b}\\ &=\frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac {i f (e+f x) \text {Li}_3\left (-i e^{2 a+2 b x}\right )}{4 b^2}+\frac {i f (e+f x) \text {Li}_3\left (i e^{2 a+2 b x}\right )}{4 b^2}+\frac {\left (i f^2\right ) \int \text {Li}_3\left (-i e^{2 a+2 b x}\right ) \, dx}{4 b^2}-\frac {\left (i f^2\right ) \int \text {Li}_3\left (i e^{2 a+2 b x}\right ) \, dx}{4 b^2}\\ &=\frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac {i f (e+f x) \text {Li}_3\left (-i e^{2 a+2 b x}\right )}{4 b^2}+\frac {i f (e+f x) \text {Li}_3\left (i e^{2 a+2 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^{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^{2 a+2 b x}\right )}{8 b^3}\\ &=\frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac {i f (e+f x) \text {Li}_3\left (-i e^{2 a+2 b x}\right )}{4 b^2}+\frac {i f (e+f x) \text {Li}_3\left (i e^{2 a+2 b x}\right )}{4 b^2}+\frac {i f^2 \text {Li}_4\left (-i e^{2 a+2 b x}\right )}{8 b^3}-\frac {i f^2 \text {Li}_4\left (i e^{2 a+2 b x}\right )}{8 b^3}\\ \end {align*}
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Mathematica [A] time = 0.22, size = 375, normalized size = 1.64 \[ \frac {1}{3} x \left (3 e^2+3 e f x+f^2 x^2\right ) \cot ^{-1}(\coth (a+b x))-\frac {i \left (12 b^3 e^2 x \log \left (1-i e^{2 (a+b x)}\right )-12 b^3 e^2 x \log \left (1+i e^{2 (a+b x)}\right )+12 b^3 e f x^2 \log \left (1-i e^{2 (a+b x)}\right )-12 b^3 e f x^2 \log \left (1+i e^{2 (a+b x)}\right )+4 b^3 f^2 x^3 \log \left (1-i e^{2 (a+b x)}\right )-4 b^3 f^2 x^3 \log \left (1+i e^{2 (a+b x)}\right )-6 b^2 (e+f x)^2 \text {Li}_2\left (-i e^{2 (a+b x)}\right )+6 b^2 (e+f x)^2 \text {Li}_2\left (i e^{2 (a+b x)}\right )+6 b e f \text {Li}_3\left (-i e^{2 (a+b x)}\right )-6 b e f \text {Li}_3\left (i e^{2 (a+b x)}\right )+6 b f^2 x \text {Li}_3\left (-i e^{2 (a+b x)}\right )-6 b f^2 x \text {Li}_3\left (i e^{2 (a+b x)}\right )-3 f^2 \text {Li}_4\left (-i e^{2 (a+b x)}\right )+3 f^2 \text {Li}_4\left (i e^{2 (a+b x)}\right )\right )}{24 b^3} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.84, size = 994, normalized size = 4.34 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 41.30, size = 5425, normalized size = 23.69 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, {\left (f^{2} x^{3} + 3 \, e f x^{2} + 3 \, e^{2} x\right )} \arctan \left (\frac {e^{\left (2 \, b x + 2 \, a\right )} - 1}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - \int \frac {2 \, {\left (b f^{2} x^{3} e^{\left (2 \, a\right )} + 3 \, b e f x^{2} e^{\left (2 \, a\right )} + 3 \, b e^{2} x e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{3 \, {\left (e^{\left (4 \, b x + 4 \, a\right )} + 1\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \mathrm {acot}\left (\mathrm {coth}\left (a+b\,x\right )\right )\,{\left (e+f\,x\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e + f x\right )^{2} \operatorname {acot}{\left (\coth {\left (a + b x \right )} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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