Optimal. Leaf size=128 \[ \frac {x^2}{2 b}-\frac {i x^3}{3}+\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {\log (\cos (a+b x))}{b^3}-\frac {i x \text {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac {\text {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {x \tan (a+b x)}{b^2}+\frac {x^2 \tan ^2(a+b x)}{2 b} \]
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Rubi [A]
time = 0.13, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3801, 3556, 30,
3800, 2221, 2611, 2320, 6724} \begin {gather*} \frac {\text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {\log (\cos (a+b x))}{b^3}-\frac {i x \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {x \tan (a+b x)}{b^2}+\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {x^2 \tan ^2(a+b x)}{2 b}+\frac {x^2}{2 b}-\frac {i x^3}{3} \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2221
Rule 2320
Rule 2611
Rule 3556
Rule 3800
Rule 3801
Rule 6724
Rubi steps
\begin {align*} \int x^2 \tan ^3(a+b x) \, dx &=\frac {x^2 \tan ^2(a+b x)}{2 b}-\frac {\int x \tan ^2(a+b x) \, dx}{b}-\int x^2 \tan (a+b x) \, dx\\ &=-\frac {i x^3}{3}-\frac {x \tan (a+b x)}{b^2}+\frac {x^2 \tan ^2(a+b x)}{2 b}+2 i \int \frac {e^{2 i (a+b x)} x^2}{1+e^{2 i (a+b x)}} \, dx+\frac {\int \tan (a+b x) \, dx}{b^2}+\frac {\int x \, dx}{b}\\ &=\frac {x^2}{2 b}-\frac {i x^3}{3}+\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {\log (\cos (a+b x))}{b^3}-\frac {x \tan (a+b x)}{b^2}+\frac {x^2 \tan ^2(a+b x)}{2 b}-\frac {2 \int x \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac {x^2}{2 b}-\frac {i x^3}{3}+\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {\log (\cos (a+b x))}{b^3}-\frac {i x \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {x \tan (a+b x)}{b^2}+\frac {x^2 \tan ^2(a+b x)}{2 b}+\frac {i \int \text {Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=\frac {x^2}{2 b}-\frac {i x^3}{3}+\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {\log (\cos (a+b x))}{b^3}-\frac {i x \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {x \tan (a+b x)}{b^2}+\frac {x^2 \tan ^2(a+b x)}{2 b}+\frac {\text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^3}\\ &=\frac {x^2}{2 b}-\frac {i x^3}{3}+\frac {x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {\log (\cos (a+b x))}{b^3}-\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}-\frac {x \tan (a+b x)}{b^2}+\frac {x^2 \tan ^2(a+b x)}{2 b}\\ \end {align*}
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Mathematica [A]
time = 3.75, size = 179, normalized size = 1.40 \begin {gather*} \frac {e^{-i a} \left (2 b^2 x^2 \left (-2 i b e^{2 i a} x+3 \left (1+e^{2 i a}\right ) \log \left (1+e^{2 i (a+b x)}\right )\right )-6 i b \left (1+e^{2 i a}\right ) x \text {PolyLog}\left (2,-e^{2 i (a+b x)}\right )+3 \left (1+e^{2 i a}\right ) \text {PolyLog}\left (3,-e^{2 i (a+b x)}\right )\right ) \sec (a)+6 b^2 x^2 \sec ^2(a+b x)-12 b x \sec (a) \sec (a+b x) \sin (b x)-4 b^3 x^3 \tan (a)-12 (\log (\cos (a+b x))+b x \tan (a))}{12 b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.09, size = 180, normalized size = 1.41
method | result | size |
risch | \(-\frac {i x^{3}}{3}+\frac {2 x \left (b x \,{\mathrm e}^{2 i \left (b x +a \right )}-i {\mathrm e}^{2 i \left (b x +a \right )}-i\right )}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 i a^{2} x}{b^{2}}+\frac {4 i a^{3}}{3 b^{3}}+\frac {x^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}-\frac {i x \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{2}}+\frac {\polylog \left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}-\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b^{3}}+\frac {2 \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}\) | \(180\) |
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. 736 vs. \(2 (109) = 218\).
time = 0.62, size = 736, normalized size = 5.75 \begin {gather*} -\frac {a^{2} {\left (\frac {1}{\sin \left (b x + a\right )^{2} - 1} - \log \left (\sin \left (b x + a\right )^{2} - 1\right )\right )} + \frac {2 \, {\left (2 \, {\left (b x + a\right )}^{3} - 6 \, {\left (b x + a\right )}^{2} a - 6 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a + {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a - 1\right )} \cos \left (4 \, b x + 4 \, a\right ) + 2 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - {\left (-i \, {\left (b x + a\right )}^{2} + 2 i \, {\left (b x + a\right )} a + i\right )} \sin \left (4 \, b x + 4 \, a\right ) - 2 \, {\left (-i \, {\left (b x + a\right )}^{2} + 2 i \, {\left (b x + a\right )} a + i\right )} \sin \left (2 \, b x + 2 \, a\right ) - 1\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a - 6 \, b x - 6 \, a\right )} \cos \left (4 \, b x + 4 \, a\right ) + 4 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} {\left (a - i\right )} + 3 \, {\left (b x + a\right )} {\left (-2 i \, a - 1\right )} - 3 \, a\right )} \cos \left (2 \, b x + 2 \, a\right ) + 6 \, {\left (b x \cos \left (4 \, b x + 4 \, a\right ) + 2 \, b x \cos \left (2 \, b x + 2 \, a\right ) + i \, b x \sin \left (4 \, b x + 4 \, a\right ) + 2 i \, b x \sin \left (2 \, b x + 2 \, a\right ) + b x\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 3 \, {\left (i \, {\left (b x + a\right )}^{2} - 2 i \, {\left (b x + a\right )} a + {\left (i \, {\left (b x + a\right )}^{2} - 2 i \, {\left (b x + a\right )} a - i\right )} \cos \left (4 \, b x + 4 \, a\right ) + 2 \, {\left (i \, {\left (b x + a\right )}^{2} - 2 i \, {\left (b x + a\right )} a - i\right )} \cos \left (2 \, b x + 2 \, a\right ) - {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a - 1\right )} \sin \left (4 \, b x + 4 \, a\right ) - 2 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a - 1\right )} \sin \left (2 \, b x + 2 \, a\right ) - i\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 \, {\left (i \, \cos \left (4 \, b x + 4 \, a\right ) + 2 i \, \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (4 \, b x + 4 \, a\right ) - 2 \, \sin \left (2 \, b x + 2 \, a\right ) + i\right )} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + 2 \, {\left (i \, {\left (b x + a\right )}^{3} - 3 i \, {\left (b x + a\right )}^{2} a - 6 i \, b x - 6 i \, a\right )} \sin \left (4 \, b x + 4 \, a\right ) + 4 \, {\left (i \, {\left (b x + a\right )}^{3} + 3 \, {\left (b x + a\right )}^{2} {\left (-i \, a - 1\right )} + 3 \, {\left (b x + a\right )} {\left (2 \, a - i\right )} - 3 i \, a\right )} \sin \left (2 \, b x + 2 \, a\right ) - 12 \, a\right )}}{-6 i \, \cos \left (4 \, b x + 4 \, a\right ) - 12 i \, \cos \left (2 \, b x + 2 \, a\right ) + 6 \, \sin \left (4 \, b x + 4 \, a\right ) + 12 \, \sin \left (2 \, b x + 2 \, a\right ) - 6 i}}{2 \, 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. 240 vs. \(2 (109) = 218\).
time = 0.36, size = 240, normalized size = 1.88 \begin {gather*} \frac {2 \, b^{2} x^{2} \tan \left (b x + a\right )^{2} + 2 \, b^{2} x^{2} + 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 ) - 4 \, b x \tan \left (b x + a\right ) + 2 \, {\left (b^{2} x^{2} - 1\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, {\left (b^{2} x^{2} - 1\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\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 ) + {\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 ^{3}{\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 )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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