Optimal. Leaf size=106 \[ -\frac {3 i \text {Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}-\frac {3 x \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 i x^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i x^4}{4} \]
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Rubi [A] time = 0.16, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3719, 2190, 2531, 6609, 2282, 6589} \[ \frac {3 i x^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 x \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 i \text {Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}-\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i x^4}{4} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3719
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^3 \tan (a+b x) \, dx &=\frac {i x^4}{4}-2 i \int \frac {e^{2 i (a+b x)} x^3}{1+e^{2 i (a+b x)}} \, dx\\ &=\frac {i x^4}{4}-\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 \int x^2 \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac {i x^4}{4}-\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i x^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {(3 i) \int x \text {Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=\frac {i x^4}{4}-\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i x^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 x \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 \int \text {Li}_3\left (-e^{2 i (a+b x)}\right ) \, dx}{2 b^3}\\ &=\frac {i x^4}{4}-\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i x^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 x \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{4 b^4}\\ &=\frac {i x^4}{4}-\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i x^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 x \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 i \text {Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 106, normalized size = 1.00 \[ -\frac {3 i \text {Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}-\frac {3 x \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 i x^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i x^4}{4} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.55, size = 286, normalized size = 2.70 \[ -\frac {4 \, b^{3} x^{3} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 4 \, b^{3} x^{3} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 6 \, b x {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 6 \, b x {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 3 i \, {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 3 i \, {\rm polylog}\left (4, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right )}{8 \, b^{4}} \]
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 \[ \int x^{3} \tan \left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.26, size = 125, normalized size = 1.18 \[ \frac {i x^{4}}{4}+\frac {2 i a^{3} x}{b^{3}}+\frac {3 i a^{4}}{2 b^{4}}-\frac {x^{3} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}+\frac {3 i x^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{2}}-\frac {3 x \polylog \left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}-\frac {3 i \polylog \left (4, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{4 b^{4}}-\frac {2 a^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.05, size = 241, normalized size = 2.27 \[ -\frac {-3 i \, {\left (b x + a\right )}^{4} + 12 i \, {\left (b x + a\right )}^{3} a - 18 i \, {\left (b x + a\right )}^{2} a^{2} + 12 \, a^{3} \log \left (\sec \left (b x + a\right )\right ) + {\left (16 i \, {\left (b x + a\right )}^{3} - 36 i \, {\left (b x + a\right )}^{2} a + 36 i \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + {\left (-24 i \, {\left (b x + a\right )}^{2} + 36 i \, {\left (b x + a\right )} a - 18 i \, a^{2}\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 2 \, {\left (4 \, {\left (b x + a\right )}^{3} - 9 \, {\left (b x + a\right )}^{2} a + 9 \, {\left (b x + a\right )} a^{2}\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 6 \, {\left (4 \, b x + a\right )} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + 12 i \, {\rm Li}_{4}(-e^{\left (2 i \, b x + 2 i \, a\right )})}{12 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\mathrm {tan}\left (a+b\,x\right ) \,d x \]
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 \[ \int x^{3} \tan {\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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