Integrand size = 17, antiderivative size = 111 \[ \int \frac {1}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3} \, dx=\frac {x}{a^3}+\frac {1}{2 a f \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 g n \log (F)}+\frac {1}{a^2 f \left (a+b \left (F^{g (e+f x)}\right )^n\right ) g n \log (F)}-\frac {\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^3 f g n \log (F)} \]
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Time = 0.04 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2320, 272, 46} \[ \int \frac {1}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3} \, dx=-\frac {\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^3 f g n \log (F)}+\frac {x}{a^3}+\frac {1}{a^2 f g n \log (F) \left (a+b \left (F^{g (e+f x)}\right )^n\right )}+\frac {1}{2 a f g n \log (F) \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2} \]
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Rule 46
Rule 272
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x \left (a+b x^n\right )^3} \, dx,x,F^{g (e+f x)}\right )}{f g \log (F)} \\ & = \frac {\text {Subst}\left (\int \frac {1}{x (a+b x)^3} \, dx,x,\left (F^{g (e+f x)}\right )^n\right )}{f g n \log (F)} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{a^3 x}-\frac {b}{a (a+b x)^3}-\frac {b}{a^2 (a+b x)^2}-\frac {b}{a^3 (a+b x)}\right ) \, dx,x,\left (F^{g (e+f x)}\right )^n\right )}{f g n \log (F)} \\ & = \frac {x}{a^3}+\frac {1}{2 a f \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 g n \log (F)}+\frac {1}{a^2 f \left (a+b \left (F^{g (e+f x)}\right )^n\right ) g n \log (F)}-\frac {\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^3 f g n \log (F)} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.15 \[ \int \frac {1}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3} \, dx=\frac {\frac {3 a+2 b \left (F^{g (e+f x)}\right )^n}{2 a^2 f \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 g n}+\frac {\log \left (\left (F^{g (e+f x)}\right )^n\right )}{a^3 f g n}-\frac {\log \left (a^4 f g n \log (F)+a^3 b f \left (F^{g (e+f x)}\right )^n g n \log (F)\right )}{a^3 f g n}}{\log (F)} \]
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Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a^{3}}+\frac {1}{a^{2} \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}+\frac {1}{2 a {\left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}^{2}}+\frac {\ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a^{3}}}{g f \ln \left (F \right ) n}\) | \(96\) |
default | \(\frac {-\frac {\ln \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a^{3}}+\frac {1}{a^{2} \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}+\frac {1}{2 a {\left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}^{2}}+\frac {\ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a^{3}}}{g f \ln \left (F \right ) n}\) | \(96\) |
risch | \(\frac {\ln \left (F^{g \left (f x +e \right )}\right )}{\ln \left (F \right ) a^{3} f g}+\frac {2 b \left (F^{g \left (f x +e \right )}\right )^{n}+3 a}{2 \ln \left (F \right ) f g n \,a^{2} {\left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}^{2}}-\frac {\ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}+\frac {a}{b}\right )}{\ln \left (F \right ) a^{3} f g n}\) | \(115\) |
norman | \(\frac {\frac {b \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right ) f g n \,a^{2}}+\frac {b^{2} x \,{\mathrm e}^{2 n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{a^{3}}+\frac {x}{a}+\frac {2 b x \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{a^{2}}+\frac {3}{2 \ln \left (F \right ) a f g n}}{\left (a +b \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}\right )^{2}}-\frac {\ln \left (a +b \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}\right )}{\ln \left (F \right ) a^{3} f g n}\) | \(161\) |
parallelrisch | \(\frac {2 b^{4} \left (F^{g \left (f x +e \right )}\right )^{2 n} x \ln \left (F \right ) f g n +4 b^{3} \left (F^{g \left (f x +e \right )}\right )^{n} x \ln \left (F \right ) a f g n +2 x \ln \left (F \right ) a^{2} b^{2} f g n -2 \ln \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right ) \left (F^{g \left (f x +e \right )}\right )^{2 n} b^{4}-4 \ln \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right ) \left (F^{g \left (f x +e \right )}\right )^{n} a \,b^{3}-2 \ln \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right ) a^{2} b^{2}+2 \left (F^{g \left (f x +e \right )}\right )^{n} a \,b^{3}+3 a^{2} b^{2}}{2 \ln \left (F \right ) a^{3} b^{2} f g n {\left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}^{2}}\) | \(217\) |
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Time = 0.27 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.71 \[ \int \frac {1}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3} \, dx=\frac {2 \, F^{2 \, f g n x + 2 \, e g n} b^{2} f g n x \log \left (F\right ) + 2 \, a^{2} f g n x \log \left (F\right ) + 2 \, {\left (2 \, a b f g n x \log \left (F\right ) + a b\right )} F^{f g n x + e g n} + 3 \, a^{2} - 2 \, {\left (2 \, F^{f g n x + e g n} a b + F^{2 \, f g n x + 2 \, e g n} b^{2} + a^{2}\right )} \log \left (F^{f g n x + e g n} b + a\right )}{2 \, {\left (2 \, F^{f g n x + e g n} a^{4} b f g n \log \left (F\right ) + F^{2 \, f g n x + 2 \, e g n} a^{3} b^{2} f g n \log \left (F\right ) + a^{5} f g n \log \left (F\right )\right )}} \]
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\[ \int \frac {1}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3} \, dx=\int \frac {1}{\left (a + b \left (F^{g \left (e + f x\right )}\right )^{n}\right )^{3}}\, dx \]
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Time = 0.22 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.24 \[ \int \frac {1}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3} \, dx=\frac {2 \, F^{f g n x + e g n} b + 3 \, a}{2 \, {\left (2 \, F^{f g n x + e g n} a^{3} b + F^{2 \, f g n x + 2 \, e g n} a^{2} b^{2} + a^{4}\right )} f g n \log \left (F\right )} + \frac {f g n x + e g n}{a^{3} f g n} - \frac {\log \left (F^{f g n x + e g n} b + a\right )}{a^{3} f g n \log \left (F\right )} \]
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Time = 0.35 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3} \, dx=\frac {\log \left ({\left | F \right |}^{f g n x} {\left | F \right |}^{e g n}\right )}{a^{3} f g n \log \left (F\right )} - \frac {\log \left ({\left | F^{f g n x} F^{e g n} b + a \right |}\right )}{a^{3} f g n \log \left (F\right )} + \frac {2 \, F^{f g n x} F^{e g n} a b + 3 \, a^{2}}{2 \, {\left (F^{f g n x} F^{e g n} b + a\right )}^{2} a^{3} f g n \log \left (F\right )} \]
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Time = 0.12 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3} \, dx=\frac {x}{a^3}+\frac {1}{2\,a\,f\,g\,n\,\ln \left (F\right )\,\left (a^2+b^2\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^{2\,n}+2\,a\,b\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n\right )}+\frac {1}{a^2\,f\,g\,n\,\ln \left (F\right )\,\left (a+b\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n\right )}-\frac {\ln \left (a+b\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n\right )}{a^3\,f\,g\,n\,\ln \left (F\right )} \]
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