Integrand size = 31, antiderivative size = 137 \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx=\frac {e^{-\frac {(1+m)^2}{4 b f n^2 \log (F)}} F^{a f} \sqrt {\pi } \left (c (d+e x)^n\right )^{-\frac {1+m}{n}} (d g+e g x)^{1+m} \text {erfi}\left (\frac {1+m+2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g n \sqrt {\log (F)}} \]
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Time = 0.17 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2308, 2266, 2235} \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx=\frac {\sqrt {\pi } F^{a f} (d g+e g x)^{m+1} e^{-\frac {(m+1)^2}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-\frac {m+1}{n}} \text {erfi}\left (\frac {2 b f n \log (F) \log \left (c (d+e x)^n\right )+m+1}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g n \sqrt {\log (F)}} \]
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Rule 2235
Rule 2266
Rule 2308
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\left (c (d+e x)^n\right )^{-\frac {1+m}{n}} (d g+e g x)^{1+m}\right ) \text {Subst}\left (\int e^{\frac {(1+m) x}{n}+a f \log (F)+b f x^2 \log (F)} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e g n} \\ & = \frac {\left (e^{-\frac {(1+m)^2}{4 b f n^2 \log (F)}} F^{a f} \left (c (d+e x)^n\right )^{-\frac {1+m}{n}} (d g+e g x)^{1+m}\right ) \text {Subst}\left (\int e^{\frac {\left (\frac {1+m}{n}+2 b f x \log (F)\right )^2}{4 b f \log (F)}} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e g n} \\ & = \frac {e^{-\frac {(1+m)^2}{4 b f n^2 \log (F)}} F^{a f} \sqrt {\pi } \left (c (d+e x)^n\right )^{-\frac {1+m}{n}} (d g+e g x)^{1+m} \text {erfi}\left (\frac {1+m+2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g n \sqrt {\log (F)}} \\ \end{align*}
\[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx=\int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx \]
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\[\int F^{f \left (a +b \ln \left (c \left (e x +d \right )^{n}\right )^{2}\right )} \left (e g x +d g \right )^{m}d x\]
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Time = 0.35 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04 \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx=-\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \left (F\right )} \operatorname {erf}\left (\frac {{\left (2 \, b f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + 2 \, b f n \log \left (F\right ) \log \left (c\right ) + m + 1\right )} \sqrt {-b f n^{2} \log \left (F\right )}}{2 \, b f n^{2} \log \left (F\right )}\right ) e^{\left (\frac {4 \, a b f^{2} n^{2} \log \left (F\right )^{2} + 4 \, b f m n^{2} \log \left (F\right ) \log \left (g\right ) - 4 \, {\left (b f m + b f\right )} n \log \left (F\right ) \log \left (c\right ) - m^{2} - 2 \, m - 1}{4 \, b f n^{2} \log \left (F\right )}\right )}}{2 \, e n} \]
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\[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx=\int F^{f \left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}^{2}\right )} \left (g \left (d + e x\right )\right )^{m}\, dx \]
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\[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx=\int { {\left (e g x + d g\right )}^{m} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + a\right )} f} \,d x } \]
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\[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx=\int { {\left (e g x + d g\right )}^{m} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + a\right )} f} \,d x } \]
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Timed out. \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx=\int {\mathrm {e}}^{f\,\ln \left (F\right )\,\left (b\,{\ln \left (c\,{\left (d+e\,x\right )}^n\right )}^2+a\right )}\,{\left (d\,g+e\,g\,x\right )}^m \,d x \]
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