Integrand size = 22, antiderivative size = 139 \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {a}{b n}} (e f-d g) (d+e x) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e^2 n}+\frac {e^{-\frac {2 a}{b n}} g (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^2 n} \]
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Time = 0.12 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2446, 2436, 2337, 2209, 2437, 2347} \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {a}{b n}} (d+e x) (e f-d g) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e^2 n}+\frac {g e^{-\frac {2 a}{b n}} (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^2 n} \]
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Rule 2209
Rule 2337
Rule 2347
Rule 2436
Rule 2437
Rule 2446
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e f-d g}{e \left (a+b \log \left (c (d+e x)^n\right )\right )}+\frac {g (d+e x)}{e \left (a+b \log \left (c (d+e x)^n\right )\right )}\right ) \, dx \\ & = \frac {g \int \frac {d+e x}{a+b \log \left (c (d+e x)^n\right )} \, dx}{e}+\frac {(e f-d g) \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx}{e} \\ & = \frac {g \text {Subst}\left (\int \frac {x}{a+b \log \left (c x^n\right )} \, dx,x,d+e x\right )}{e^2}+\frac {(e f-d g) \text {Subst}\left (\int \frac {1}{a+b \log \left (c x^n\right )} \, dx,x,d+e x\right )}{e^2} \\ & = \frac {\left (g (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n}\right ) \text {Subst}\left (\int \frac {e^{\frac {2 x}{n}}}{a+b x} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e^2 n}+\frac {\left ((e f-d g) (d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \text {Subst}\left (\int \frac {e^{\frac {x}{n}}}{a+b x} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e^2 n} \\ & = \frac {e^{-\frac {a}{b n}} (e f-d g) (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {Ei}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e^2 n}+\frac {e^{-\frac {2 a}{b n}} g (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {Ei}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^2 n} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.91 \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {2 a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-2/n} \left (e^{\frac {a}{b n}} (e f-d g) \left (c (d+e x)^n\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )+g (d+e x) \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )\right )}{b e^2 n} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.96 (sec) , antiderivative size = 937, normalized size of antiderivative = 6.74
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Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.76 \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {{\left ({\left (e f - d g\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )} \operatorname {log\_integral}\left ({\left (e x + d\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )}\right ) + g \operatorname {log\_integral}\left ({\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} e^{\left (\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}\right )\right )} e^{\left (-\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}}{b e^{2} n} \]
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\[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\int \frac {f + g x}{a + b \log {\left (c \left (d + e x\right )^{n} \right )}}\, dx \]
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\[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\int { \frac {g x + f}{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a} \,d x } \]
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Time = 0.33 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.14 \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {f {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e n} - \frac {d g {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e^{2} n} + \frac {g {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{n} + \frac {2 \, a}{b n} + 2 \, \log \left (e x + d\right )\right ) e^{\left (-\frac {2 \, a}{b n}\right )}}{b c^{\frac {2}{n}} e^{2} n} \]
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Timed out. \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\int \frac {f+g\,x}{a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )} \,d x \]
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