Optimal. Leaf size=77 \[ \frac {i x^3}{3}-\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i x \text {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac {\text {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3} \]
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Rubi [A]
time = 0.09, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3800, 2221,
2611, 2320, 6724} \begin {gather*} -\frac {\text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {i x \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i x^3}{3} \end {gather*}
Antiderivative was successfully verified.
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Rule 2221
Rule 2320
Rule 2611
Rule 3800
Rule 6724
Rubi steps
\begin {align*} \int x^2 \tan (a+b x) \, dx &=\frac {i x^3}{3}-2 i \int \frac {e^{2 i (a+b x)} x^2}{1+e^{2 i (a+b x)}} \, dx\\ &=\frac {i x^3}{3}-\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {2 \int x \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac {i x^3}{3}-\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i x \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {i \int \text {Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=\frac {i x^3}{3}-\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i x \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {\text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^3}\\ &=\frac {i x^3}{3}-\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i x \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {\text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}\\ \end {align*}
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Mathematica [A]
time = 0.01, size = 77, normalized size = 1.00 \begin {gather*} \frac {i x^3}{3}-\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i x \text {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac {\text {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.06, size = 103, normalized size = 1.34
method | result | size |
risch | \(\frac {i x^{3}}{3}-\frac {2 i a^{2} x}{b^{2}}-\frac {4 i a^{3}}{3 b^{3}}-\frac {x^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}+\frac {i x \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{2}}-\frac {\polylog \left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}\) | \(103\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than
twice the leaf count of optimal. 163 vs. \(2 (62) = 124\).
time = 0.55, size = 163, normalized size = 2.12 \begin {gather*} -\frac {-2 i \, {\left (b x + a\right )}^{3} + 6 i \, {\left (b x + a\right )}^{2} a - 6 i \, b x {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 6 \, a^{2} \log \left (\sec \left (b x + a\right )\right ) - 6 \, {\left (-i \, {\left (b x + a\right )}^{2} + 2 i \, {\left (b x + a\right )} a\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\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 ) + 3 \, {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )})}{6 \, b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 200 vs. \(2 (62) = 124\).
time = 0.37, size = 200, normalized size = 2.60 \begin {gather*} -\frac {2 \, b^{2} x^{2} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, b^{2} x^{2} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 i \, b x {\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 ) - 2 i \, b x {\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 ) + {\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 ) + {\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 )}{4 \, b^{3}} \end {gather*}
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 \begin {gather*} \int x^{2} \tan {\left (a + b x \right )}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x^2\,\mathrm {tan}\left (a+b\,x\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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