Integrand size = 24, antiderivative size = 140 \[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {x^3}{a b d \log (F)}-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {3 x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a b d^2 \log ^2(F)}-\frac {6 x \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}+\frac {6 \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{a b d^4 \log ^4(F)} \]
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Time = 0.16 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2222, 2215, 2221, 2611, 2320, 6724} \[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {6 \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{a b d^4 \log ^4(F)}-\frac {6 x \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}-\frac {3 x^2 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{a b d^2 \log ^2(F)}-\frac {x^3}{b d \log (F) \left (a+b F^{c+d x}\right )}+\frac {x^3}{a b d \log (F)} \]
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Rule 2215
Rule 2221
Rule 2222
Rule 2320
Rule 2611
Rule 6724
Rubi steps \begin{align*} \text {integral}& = -\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}+\frac {3 \int \frac {x^2}{a+b F^{c+d x}} \, dx}{b d \log (F)} \\ & = \frac {x^3}{a b d \log (F)}-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {3 \int \frac {F^{c+d x} x^2}{a+b F^{c+d x}} \, dx}{a d \log (F)} \\ & = \frac {x^3}{a b d \log (F)}-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {3 x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a b d^2 \log ^2(F)}+\frac {6 \int x \log \left (1+\frac {b F^{c+d x}}{a}\right ) \, dx}{a b d^2 \log ^2(F)} \\ & = \frac {x^3}{a b d \log (F)}-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {3 x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a b d^2 \log ^2(F)}-\frac {6 x \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}+\frac {6 \int \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right ) \, dx}{a b d^3 \log ^3(F)} \\ & = \frac {x^3}{a b d \log (F)}-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {3 x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a b d^2 \log ^2(F)}-\frac {6 x \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}+\frac {6 \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {b x}{a}\right )}{x} \, dx,x,F^{c+d x}\right )}{a b d^4 \log ^4(F)} \\ & = \frac {x^3}{a b d \log (F)}-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {3 x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a b d^2 \log ^2(F)}-\frac {6 x \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}+\frac {6 \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right )}{a b d^4 \log ^4(F)} \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.98 \[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}+\frac {3 \left (\frac {x^3}{3 a}-\frac {x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a d \log (F)}-\frac {2 x \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{a d^2 \log ^2(F)}+\frac {2 \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{a d^3 \log ^3(F)}\right )}{b d \log (F)} \]
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Time = 0.05 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.96
method | result | size |
risch | \(-\frac {x^{3}}{b d \left (a +b \,F^{d x +c}\right ) \ln \left (F \right )}+\frac {x^{3}}{a b d \ln \left (F \right )}-\frac {3 c^{2} x}{b \ln \left (F \right ) d^{3} a}-\frac {2 c^{3}}{b \ln \left (F \right ) d^{4} a}-\frac {3 \ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) x^{2}}{b \ln \left (F \right )^{2} d^{2} a}+\frac {3 \ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) c^{2}}{b \ln \left (F \right )^{2} d^{4} a}-\frac {6 \,\operatorname {Li}_{2}\left (-\frac {b \,F^{d x} F^{c}}{a}\right ) x}{b \ln \left (F \right )^{3} d^{3} a}+\frac {6 \,\operatorname {Li}_{3}\left (-\frac {b \,F^{d x} F^{c}}{a}\right )}{b \ln \left (F \right )^{4} d^{4} a}-\frac {3 c^{2} \ln \left (F^{c} F^{d x} b +a \right )}{b \ln \left (F \right )^{2} d^{4} a}+\frac {3 c^{2} \ln \left (F^{d x} F^{c}\right )}{b \ln \left (F \right )^{2} d^{4} a}\) | \(274\) |
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Time = 0.24 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.76 \[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {a c^{3} \log \left (F\right )^{3} + {\left (b d^{3} x^{3} + b c^{3}\right )} F^{d x + c} \log \left (F\right )^{3} - 6 \, {\left (F^{d x + c} b d x \log \left (F\right ) + a d x \log \left (F\right )\right )} {\rm Li}_2\left (-\frac {F^{d x + c} b + a}{a} + 1\right ) - 3 \, {\left (F^{d x + c} b c^{2} \log \left (F\right )^{2} + a c^{2} \log \left (F\right )^{2}\right )} \log \left (F^{d x + c} b + a\right ) - 3 \, {\left ({\left (b d^{2} x^{2} - b c^{2}\right )} F^{d x + c} \log \left (F\right )^{2} + {\left (a d^{2} x^{2} - a c^{2}\right )} \log \left (F\right )^{2}\right )} \log \left (\frac {F^{d x + c} b + a}{a}\right ) + 6 \, {\left (F^{d x + c} b + a\right )} {\rm polylog}\left (3, -\frac {F^{d x + c} b}{a}\right )}{F^{d x + c} a b^{2} d^{4} \log \left (F\right )^{4} + a^{2} b d^{4} \log \left (F\right )^{4}} \]
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\[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=- \frac {x^{3}}{F^{c + d x} b^{2} d \log {\left (F \right )} + a b d \log {\left (F \right )}} + \frac {3 \int \frac {x^{2}}{a + b e^{c \log {\left (F \right )}} e^{d x \log {\left (F \right )}}}\, dx}{b d \log {\left (F \right )}} \]
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Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.92 \[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=-\frac {x^{3}}{F^{d x} F^{c} b^{2} d \log \left (F\right ) + a b d \log \left (F\right )} + \frac {x^{3}}{a b d \log \left (F\right )} - \frac {3 \, {\left (d^{2} x^{2} \log \left (\frac {F^{d x} F^{c} b}{a} + 1\right ) \log \left (F\right )^{2} + 2 \, d x {\rm Li}_2\left (-\frac {F^{d x} F^{c} b}{a}\right ) \log \left (F\right ) - 2 \, {\rm Li}_{3}(-\frac {F^{d x} F^{c} b}{a})\right )}}{a b d^{4} \log \left (F\right )^{4}} \]
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\[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=\int { \frac {F^{d x + c} x^{3}}{{\left (F^{d x + c} b + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=\int \frac {F^{c+d\,x}\,x^3}{{\left (a+F^{c+d\,x}\,b\right )}^2} \,d x \]
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