Optimal. Leaf size=302 \[ -\frac {3 f^3 \text {Li}_5\left (-i e^{2 i (a+b x)}\right )}{16 b^4}+\frac {3 f^3 \text {Li}_5\left (i e^{2 i (a+b x)}\right )}{16 b^4}+\frac {3 i f^2 (e+f x) \text {Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 i f^2 (e+f x) \text {Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}+\frac {3 f (e+f x)^2 \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f (e+f x)^2 \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {i (e+f x)^3 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}+\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f} \]
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Rubi [A] time = 0.24, antiderivative size = 302, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6252, 4181, 2531, 6609, 2282, 6589} \[ \frac {3 i f^2 (e+f x) \text {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 i f^2 (e+f x) \text {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{8 b^3}+\frac {3 f (e+f x)^2 \text {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f (e+f x)^2 \text {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f^3 \text {PolyLog}\left (5,-i e^{2 i (a+b x)}\right )}{16 b^4}+\frac {3 f^3 \text {PolyLog}\left (5,i e^{2 i (a+b x)}\right )}{16 b^4}-\frac {i (e+f x)^3 \text {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}+\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 4181
Rule 6252
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int (e+f x)^3 \coth ^{-1}(\tan (a+b x)) \, dx &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}-\frac {b \int (e+f x)^4 \sec (2 a+2 b x) \, dx}{4 f}\\ &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}+\frac {1}{2} \int (e+f x)^3 \log \left (1-i e^{i (2 a+2 b x)}\right ) \, dx-\frac {1}{2} \int (e+f x)^3 \log \left (1+i e^{i (2 a+2 b x)}\right ) \, dx\\ &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac {i (e+f x)^3 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {(3 i f) \int (e+f x)^2 \text {Li}_2\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{4 b}-\frac {(3 i f) \int (e+f x)^2 \text {Li}_2\left (i e^{i (2 a+2 b x)}\right ) \, dx}{4 b}\\ &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac {i (e+f x)^3 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {3 f (e+f x)^2 \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f (e+f x)^2 \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {\left (3 f^2\right ) \int (e+f x) \text {Li}_3\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{4 b^2}+\frac {\left (3 f^2\right ) \int (e+f x) \text {Li}_3\left (i e^{i (2 a+2 b x)}\right ) \, dx}{4 b^2}\\ &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac {i (e+f x)^3 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {3 f (e+f x)^2 \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f (e+f x)^2 \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}+\frac {3 i f^2 (e+f x) \text {Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 i f^2 (e+f x) \text {Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {\left (3 i f^3\right ) \int \text {Li}_4\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{8 b^3}+\frac {\left (3 i f^3\right ) \int \text {Li}_4\left (i e^{i (2 a+2 b x)}\right ) \, dx}{8 b^3}\\ &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac {i (e+f x)^3 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {3 f (e+f x)^2 \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f (e+f x)^2 \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}+\frac {3 i f^2 (e+f x) \text {Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 i f^2 (e+f x) \text {Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {\left (3 f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_4(-i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{16 b^4}+\frac {\left (3 f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_4(i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{16 b^4}\\ &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac {i (e+f x)^3 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {3 f (e+f x)^2 \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f (e+f x)^2 \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}+\frac {3 i f^2 (e+f x) \text {Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 i f^2 (e+f x) \text {Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 f^3 \text {Li}_5\left (-i e^{2 i (a+b x)}\right )}{16 b^4}+\frac {3 f^3 \text {Li}_5\left (i e^{2 i (a+b x)}\right )}{16 b^4}\\ \end {align*}
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Mathematica [B] time = 0.35, size = 654, normalized size = 2.17 \[ \frac {1}{4} x \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right ) \coth ^{-1}(\tan (a+b x))+\frac {-8 b^4 e^3 x \log \left (1-i e^{2 i (a+b x)}\right )+8 b^4 e^3 x \log \left (1+i e^{2 i (a+b x)}\right )-12 b^4 e^2 f x^2 \log \left (1-i e^{2 i (a+b x)}\right )+12 b^4 e^2 f x^2 \log \left (1+i e^{2 i (a+b x)}\right )-8 b^4 e f^2 x^3 \log \left (1-i e^{2 i (a+b x)}\right )+8 b^4 e f^2 x^3 \log \left (1+i e^{2 i (a+b x)}\right )-2 b^4 f^3 x^4 \log \left (1-i e^{2 i (a+b x)}\right )+2 b^4 f^3 x^4 \log \left (1+i e^{2 i (a+b x)}\right )-4 i b^3 (e+f x)^3 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )+4 i b^3 (e+f x)^3 \text {Li}_2\left (i e^{2 i (a+b x)}\right )+6 b^2 e^2 f \text {Li}_3\left (-i e^{2 i (a+b x)}\right )-6 b^2 e^2 f \text {Li}_3\left (i e^{2 i (a+b x)}\right )+12 b^2 e f^2 x \text {Li}_3\left (-i e^{2 i (a+b x)}\right )-12 b^2 e f^2 x \text {Li}_3\left (i e^{2 i (a+b x)}\right )+6 b^2 f^3 x^2 \text {Li}_3\left (-i e^{2 i (a+b x)}\right )-6 b^2 f^3 x^2 \text {Li}_3\left (i e^{2 i (a+b x)}\right )+6 i b e f^2 \text {Li}_4\left (-i e^{2 i (a+b x)}\right )-6 i b e f^2 \text {Li}_4\left (i e^{2 i (a+b x)}\right )+6 i b f^3 x \text {Li}_4\left (-i e^{2 i (a+b x)}\right )-6 i b f^3 x \text {Li}_4\left (i e^{2 i (a+b x)}\right )-3 f^3 \text {Li}_5\left (-i e^{2 i (a+b x)}\right )+3 f^3 \text {Li}_5\left (i e^{2 i (a+b x)}\right )}{16 b^4} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.77, size = 1808, normalized size = 5.99 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (f x + e\right )}^{3} \operatorname {arcoth}\left (\tan \left (b x + a\right )\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 50.40, size = 7429, normalized size = 24.60 \[ \text {output too large to display} \]
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 \[ \frac {1}{16} \, {\left (f^{3} x^{4} + 4 \, e f^{2} x^{3} + 6 \, e^{2} f x^{2} + 4 \, e^{3} x\right )} \log \left (2 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \sin \left (2 \, b x + 2 \, a\right )^{2} + 4 \, \sin \left (2 \, b x + 2 \, a\right ) + 2\right ) - \frac {1}{16} \, {\left (f^{3} x^{4} + 4 \, e f^{2} x^{3} + 6 \, e^{2} f x^{2} + 4 \, e^{3} x\right )} \log \left (2 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \sin \left (2 \, b x + 2 \, a\right )^{2} - 4 \, \sin \left (2 \, b x + 2 \, a\right ) + 2\right ) - \int \frac {{\left (b f^{3} x^{4} + 4 \, b e f^{2} x^{3} + 6 \, b e^{2} f x^{2} + 4 \, b e^{3} x\right )} \cos \left (4 \, b x + 4 \, a\right ) \cos \left (2 \, b x + 2 \, a\right ) + {\left (b f^{3} x^{4} + 4 \, b e f^{2} x^{3} + 6 \, b e^{2} f x^{2} + 4 \, b e^{3} x\right )} \sin \left (4 \, b x + 4 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + {\left (b f^{3} x^{4} + 4 \, b e f^{2} x^{3} + 6 \, b e^{2} f x^{2} + 4 \, b e^{3} x\right )} \cos \left (2 \, b x + 2 \, a\right )}{2 \, {\left (\cos \left (4 \, b x + 4 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 2 \, \cos \left (4 \, b x + 4 \, a\right ) + 1\right )}}\,{d x} \]
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 \[ \int \mathrm {acoth}\left (\mathrm {tan}\left (a+b\,x\right )\right )\,{\left (e+f\,x\right )}^3 \,d x \]
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 \[ \int \left (e + f x\right )^{3} \operatorname {acoth}{\left (\tan {\left (a + b x \right )} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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