Optimal. Leaf size=94 \[ \frac {9 a (d x)^{m+4} \, _2F_1\left (1,\frac {m+4}{3};\frac {m+7}{3};a x^3\right )}{d^4 (m+1)^2 (m+4)}+\frac {\text {Li}_2\left (a x^3\right ) (d x)^{m+1}}{d (m+1)}+\frac {3 \log \left (1-a x^3\right ) (d x)^{m+1}}{d (m+1)^2} \]
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Rubi [A] time = 0.05, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {6591, 2455, 16, 364} \[ \frac {(d x)^{m+1} \text {PolyLog}\left (2,a x^3\right )}{d (m+1)}+\frac {9 a (d x)^{m+4} \, _2F_1\left (1,\frac {m+4}{3};\frac {m+7}{3};a x^3\right )}{d^4 (m+1)^2 (m+4)}+\frac {3 \log \left (1-a x^3\right ) (d x)^{m+1}}{d (m+1)^2} \]
Antiderivative was successfully verified.
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Rule 16
Rule 364
Rule 2455
Rule 6591
Rubi steps
\begin {align*} \int (d x)^m \text {Li}_2\left (a x^3\right ) \, dx &=\frac {(d x)^{1+m} \text {Li}_2\left (a x^3\right )}{d (1+m)}+\frac {3 \int (d x)^m \log \left (1-a x^3\right ) \, dx}{1+m}\\ &=\frac {3 (d x)^{1+m} \log \left (1-a x^3\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_2\left (a x^3\right )}{d (1+m)}+\frac {(9 a) \int \frac {x^2 (d x)^{1+m}}{1-a x^3} \, dx}{d (1+m)^2}\\ &=\frac {3 (d x)^{1+m} \log \left (1-a x^3\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_2\left (a x^3\right )}{d (1+m)}+\frac {(9 a) \int \frac {(d x)^{3+m}}{1-a x^3} \, dx}{d^3 (1+m)^2}\\ &=\frac {9 a (d x)^{4+m} \, _2F_1\left (1,\frac {4+m}{3};\frac {7+m}{3};a x^3\right )}{d^4 (1+m)^2 (4+m)}+\frac {3 (d x)^{1+m} \log \left (1-a x^3\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_2\left (a x^3\right )}{d (1+m)}\\ \end {align*}
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Mathematica [A] time = 0.05, size = 72, normalized size = 0.77 \[ \frac {x (d x)^m \left (9 a x^3 \, _2F_1\left (1,\frac {m+4}{3};\frac {m+7}{3};a x^3\right )+(m+4) \left ((m+1) \text {Li}_2\left (a x^3\right )+3 \log \left (1-a x^3\right )\right )\right )}{(m+1)^2 (m+4)} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (d x\right )^{m} {\rm Li}_2\left (a x^{3}\right ), x\right ) \]
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 \[ \int \left (d x\right )^{m} {\rm Li}_2\left (a x^{3}\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.15, size = 177, normalized size = 1.88 \[ -\frac {\left (d x \right )^{m} x^{-m} \left (-a \right )^{-\frac {1}{3}-\frac {m}{3}} \left (\frac {3 x^{1+m} \left (-a \right )^{\frac {4}{3}+\frac {m}{3}} \left (-36-9 m \right )}{\left (4+m \right ) \left (1+m \right )^{3} a}-\frac {3 x^{1+m} \left (-a \right )^{\frac {4}{3}+\frac {m}{3}} \left (-12-3 m \right ) \ln \left (-a \,x^{3}+1\right )}{\left (4+m \right ) \left (1+m \right )^{2} a}+\frac {3 x^{1+m} \left (-a \right )^{\frac {4}{3}+\frac {m}{3}} \polylog \left (2, a \,x^{3}\right )}{\left (1+m \right ) a}+\frac {3 x^{1+m} \left (-a \right )^{\frac {4}{3}+\frac {m}{3}} \left (12+3 m \right ) \Phi \left (a \,x^{3}, 1, \frac {m}{3}+\frac {1}{3}\right )}{\left (4+m \right ) \left (1+m \right )^{2} a}\right )}{3} \]
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 \[ -9 \, a d^{m} \int \frac {x^{3} x^{m}}{{\left (a m^{2} + 2 \, a m + a\right )} x^{3} - m^{2} - 2 \, m - 1}\,{d x} + \frac {{\left (d^{m} m + d^{m}\right )} x x^{m} {\rm Li}_2\left (a x^{3}\right ) + 3 \, d^{m} x x^{m} \log \left (-a x^{3} + 1\right )}{m^{2} + 2 \, m + 1} \]
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 \[ \int \mathrm {polylog}\left (2,a\,x^3\right )\,{\left (d\,x\right )}^m \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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