Optimal. Leaf size=101 \[ \frac {a q^2 x^{q+1} (d x)^m \, _2F_1\left (1,\frac {m+q+1}{q};\frac {m+2 q+1}{q};a x^q\right )}{(m+1)^2 (m+q+1)}+\frac {(d x)^{m+1} \text {Li}_2\left (a x^q\right )}{d (m+1)}+\frac {q (d x)^{m+1} \log \left (1-a x^q\right )}{d (m+1)^2} \]
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Rubi [A] time = 0.06, antiderivative size = 101, 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, 20, 364} \[ \frac {(d x)^{m+1} \text {PolyLog}\left (2,a x^q\right )}{d (m+1)}+\frac {a q^2 x^{q+1} (d x)^m \, _2F_1\left (1,\frac {m+q+1}{q};\frac {m+2 q+1}{q};a x^q\right )}{(m+1)^2 (m+q+1)}+\frac {q (d x)^{m+1} \log \left (1-a x^q\right )}{d (m+1)^2} \]
Antiderivative was successfully verified.
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Rule 20
Rule 364
Rule 2455
Rule 6591
Rubi steps
\begin {align*} \int (d x)^m \text {Li}_2\left (a x^q\right ) \, dx &=\frac {(d x)^{1+m} \text {Li}_2\left (a x^q\right )}{d (1+m)}+\frac {q \int (d x)^m \log \left (1-a x^q\right ) \, dx}{1+m}\\ &=\frac {q (d x)^{1+m} \log \left (1-a x^q\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_2\left (a x^q\right )}{d (1+m)}+\frac {\left (a q^2\right ) \int \frac {x^{-1+q} (d x)^{1+m}}{1-a x^q} \, dx}{d (1+m)^2}\\ &=\frac {q (d x)^{1+m} \log \left (1-a x^q\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_2\left (a x^q\right )}{d (1+m)}+\frac {\left (a q^2 x^{-m} (d x)^m\right ) \int \frac {x^{m+q}}{1-a x^q} \, dx}{(1+m)^2}\\ &=\frac {a q^2 x^{1+q} (d x)^m \, _2F_1\left (1,\frac {1+m+q}{q};\frac {1+m+2 q}{q};a x^q\right )}{(1+m)^2 (1+m+q)}+\frac {q (d x)^{1+m} \log \left (1-a x^q\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_2\left (a x^q\right )}{d (1+m)}\\ \end {align*}
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Mathematica [A] time = 0.07, size = 80, normalized size = 0.79 \[ \frac {x (d x)^m \left (a q^2 x^q \, _2F_1\left (1,\frac {m+q+1}{q};\frac {m+2 q+1}{q};a x^q\right )+(m+q+1) \left ((m+1) \text {Li}_2\left (a x^q\right )+q \log \left (1-a x^q\right )\right )\right )}{(m+1)^2 (m+q+1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (d x\right )^{m} {\rm Li}_2\left (a x^{q}\right ), x\right ) \]
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 \[ \int \left (d x\right )^{m} {\rm Li}_2\left (a x^{q}\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.14, size = 148, normalized size = 1.47 \[ -\frac {\left (d x \right )^{m} x^{-m} \left (-a \right )^{-\frac {m}{q}-\frac {1}{q}} \left (-\frac {q^{2} x^{1+m} \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \ln \left (1-a \,x^{q}\right )}{\left (1+m \right )^{2}}-\frac {q \,x^{1+m} \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \polylog \left (2, a \,x^{q}\right )}{1+m}-\frac {q^{2} x^{1+m +q} a \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \Phi \left (a \,x^{q}, 1, \frac {1+m +q}{q}\right )}{\left (1+m \right )^{2}}\right )}{q} \]
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 \[ -d^{m} q^{2} \int -\frac {x^{m}}{m^{2} - {\left (a m^{2} + 2 \, a m + a\right )} x^{q} + 2 \, m + 1}\,{d x} - \frac {d^{m} q^{2} x x^{m} - {\left (d^{m} m + d^{m}\right )} q x x^{m} \log \left (-a x^{q} + 1\right ) - {\left (d^{m} m^{2} + 2 \, d^{m} m + d^{m}\right )} x x^{m} {\rm Li}_2\left (a x^{q}\right )}{m^{3} + 3 \, m^{2} + 3 \, m + 1} \]
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 \[ \int {\left (d\,x\right )}^m\,\mathrm {polylog}\left (2,a\,x^q\right ) \,d x \]
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 \[ \int \left (d x\right )^{m} \operatorname {Li}_{2}\left (a x^{q}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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