Integrand size = 24, antiderivative size = 115 \[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {3 x^2 \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac {6 x \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {6 \operatorname {PolyLog}\left (4,-\frac {b F^{c+d x}}{a}\right )}{b d^4 \log ^4(F)} \]
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Time = 0.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2221, 2611, 6744, 2320, 6724} \[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\frac {6 \operatorname {PolyLog}\left (4,-\frac {b F^{c+d x}}{a}\right )}{b d^4 \log ^4(F)}-\frac {6 x \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {3 x^2 \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}+\frac {x^3 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)} \]
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Rule 2221
Rule 2320
Rule 2611
Rule 6724
Rule 6744
Rubi steps \begin{align*} \text {integral}& = \frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}-\frac {3 \int x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right ) \, dx}{b d \log (F)} \\ & = \frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {3 x^2 \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac {6 \int x \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right ) \, dx}{b d^2 \log ^2(F)} \\ & = \frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {3 x^2 \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac {6 x \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {6 \int \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right ) \, dx}{b d^3 \log ^3(F)} \\ & = \frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {3 x^2 \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac {6 x \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {6 \text {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {b x}{a}\right )}{x} \, dx,x,F^{c+d x}\right )}{b d^4 \log ^4(F)} \\ & = \frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {3 x^2 \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac {6 x \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {6 \text {Li}_4\left (-\frac {b F^{c+d x}}{a}\right )}{b d^4 \log ^4(F)} \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00 \[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {3 x^2 \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac {6 x \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {6 \operatorname {PolyLog}\left (4,-\frac {b F^{c+d x}}{a}\right )}{b d^4 \log ^4(F)} \]
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Time = 0.06 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.96
method | result | size |
risch | \(\frac {6 \,\operatorname {Li}_{4}\left (-\frac {b \,F^{d x} F^{c}}{a}\right )}{d^{4} \ln \left (F \right )^{4} b}+\frac {\ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) c^{3}}{d^{4} \ln \left (F \right ) b}-\frac {c^{3} \ln \left (F^{c} F^{d x} b +a \right )}{d^{4} \ln \left (F \right ) b}+\frac {c^{3} \ln \left (F^{d x} F^{c}\right )}{d^{4} \ln \left (F \right ) b}-\frac {c^{3} x}{d^{3} b}+\frac {\ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) x^{3}}{d \ln \left (F \right ) b}+\frac {3 \,\operatorname {Li}_{2}\left (-\frac {b \,F^{d x} F^{c}}{a}\right ) x^{2}}{d^{2} \ln \left (F \right )^{2} b}-\frac {6 \,\operatorname {Li}_{3}\left (-\frac {b \,F^{d x} F^{c}}{a}\right ) x}{d^{3} \ln \left (F \right )^{3} b}-\frac {3 c^{4}}{4 d^{4} b}\) | \(225\) |
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Time = 0.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.17 \[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\frac {3 \, d^{2} x^{2} {\rm Li}_2\left (-\frac {F^{d x + c} b + a}{a} + 1\right ) \log \left (F\right )^{2} - c^{3} \log \left (F^{d x + c} b + a\right ) \log \left (F\right )^{3} + {\left (d^{3} x^{3} + c^{3}\right )} \log \left (F\right )^{3} \log \left (\frac {F^{d x + c} b + a}{a}\right ) - 6 \, d x \log \left (F\right ) {\rm polylog}\left (3, -\frac {F^{d x + c} b}{a}\right ) + 6 \, {\rm polylog}\left (4, -\frac {F^{d x + c} b}{a}\right )}{b d^{4} \log \left (F\right )^{4}} \]
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\[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\int \frac {F^{c + d x} x^{3}}{F^{c + d x} b + a}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.92 \[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\frac {d^{3} x^{3} \log \left (\frac {F^{d x} F^{c} b}{a} + 1\right ) \log \left (F\right )^{3} + 3 \, d^{2} x^{2} {\rm Li}_2\left (-\frac {F^{d x} F^{c} b}{a}\right ) \log \left (F\right )^{2} - 6 \, d x \log \left (F\right ) {\rm Li}_{3}(-\frac {F^{d x} F^{c} b}{a}) + 6 \, {\rm Li}_{4}(-\frac {F^{d x} F^{c} b}{a})}{b d^{4} \log \left (F\right )^{4}} \]
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\[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\int { \frac {F^{d x + c} x^{3}}{F^{d x + c} b + a} \,d x } \]
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Timed out. \[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\int \frac {F^{c+d\,x}\,x^3}{a+F^{c+d\,x}\,b} \,d x \]
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