Optimal. Leaf size=153 \[ \frac {\sqrt {\pi } F^{a^2 f} (d+e x) (d g+e g x)^m \left (c (d+e x)^n\right )^{-\frac {m+1}{n}} \exp \left (-\frac {(2 a b f n \log (F)+m+1)^2}{4 b^2 f n^2 \log (F)}\right ) \text {erfi}\left (\frac {2 a b f n \log (F)+2 b^2 f n \log (F) \log \left (c (d+e x)^n\right )+m+1}{2 b \sqrt {f} n \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]
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Rubi [A] time = 0.76, antiderivative size = 152, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2278, 2274, 15, 20, 2276, 2234, 2204} \[ \frac {\sqrt {\pi } F^{a^2 f} (d+e x) (g (d+e x))^m \left (c (d+e x)^n\right )^{-\frac {m+1}{n}} \exp \left (-\frac {(2 a b f n \log (F)+m+1)^2}{4 b^2 f n^2 \log (F)}\right ) \text {Erfi}\left (\frac {2 a b f n \log (F)+2 b^2 f n \log (F) \log \left (c (d+e x)^n\right )+m+1}{2 b \sqrt {f} n \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]
Antiderivative was successfully verified.
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Rule 15
Rule 20
Rule 2204
Rule 2234
Rule 2274
Rule 2276
Rule 2278
Rubi steps
\begin {align*} \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx &=\frac {\operatorname {Subst}\left (\int F^{f \left (a+b \log \left (c x^n\right )\right )^2} (g x)^m \, dx,x,d+e x\right )}{e}\\ &=\frac {\operatorname {Subst}\left (\int F^{a^2 f+2 a b f \log \left (c x^n\right )+b^2 f \log ^2\left (c x^n\right )} (g x)^m \, dx,x,d+e x\right )}{e}\\ &=\frac {\operatorname {Subst}\left (\int F^{a^2 f+b^2 f \log ^2\left (c x^n\right )} (g x)^m \left (c x^n\right )^{2 a b f \log (F)} \, dx,x,d+e x\right )}{e}\\ &=\frac {\left ((d+e x)^{-2 a b f n \log (F)} \left (c (d+e x)^n\right )^{2 a b f \log (F)}\right ) \operatorname {Subst}\left (\int F^{a^2 f+b^2 f \log ^2\left (c x^n\right )} x^{2 a b f n \log (F)} (g x)^m \, dx,x,d+e x\right )}{e}\\ &=\frac {\left ((d+e x)^{-m-2 a b f n \log (F)} (g (d+e x))^m \left (c (d+e x)^n\right )^{2 a b f \log (F)}\right ) \operatorname {Subst}\left (\int F^{a^2 f+b^2 f \log ^2\left (c x^n\right )} x^{m+2 a b f n \log (F)} \, dx,x,d+e x\right )}{e}\\ &=\frac {\left ((d+e x) (g (d+e x))^m \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {1+m+2 a b f n \log (F)}{n}}\right ) \operatorname {Subst}\left (\int \exp \left (a^2 f \log (F)+b^2 f x^2 \log (F)+\frac {x (1+m+2 a b f n \log (F))}{n}\right ) \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e n}\\ &=\frac {\left (\exp \left (-\frac {(1+m+2 a b f n \log (F))^2}{4 b^2 f n^2 \log (F)}\right ) F^{a^2 f} (d+e x) (g (d+e x))^m \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {1+m+2 a b f n \log (F)}{n}}\right ) \operatorname {Subst}\left (\int \exp \left (\frac {\left (2 b^2 f x \log (F)+\frac {1+m+2 a b f n \log (F)}{n}\right )^2}{4 b^2 f \log (F)}\right ) \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e n}\\ &=\frac {\exp \left (-\frac {(1+m+2 a b f n \log (F))^2}{4 b^2 f n^2 \log (F)}\right ) F^{a^2 f} \sqrt {\pi } (d+e x) (g (d+e x))^m \left (c (d+e x)^n\right )^{-\frac {1+m}{n}} \text {erfi}\left (\frac {1+m+2 a b f n \log (F)+2 b^2 f n \log (F) \log \left (c (d+e x)^n\right )}{2 b \sqrt {f} n \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}}\\ \end {align*}
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Mathematica [F] time = 0.25, size = 0, normalized size = 0.00 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx \]
Verification is Not applicable to the result.
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fricas [A] time = 0.42, size = 169, normalized size = 1.10 \[ -\frac {\sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \relax (F)} \operatorname {erf}\left (\frac {{\left (2 \, b^{2} f n^{2} \log \left (e x + d\right ) \log \relax (F) + 2 \, b^{2} f n \log \relax (F) \log \relax (c) + 2 \, a b f n \log \relax (F) + m + 1\right )} \sqrt {-b^{2} f n^{2} \log \relax (F)}}{2 \, b^{2} f n^{2} \log \relax (F)}\right ) e^{\left (\frac {4 \, b^{2} f m n^{2} \log \relax (F) \log \relax (g) - 4 \, {\left (b^{2} f m + b^{2} f\right )} n \log \relax (F) \log \relax (c) - 4 \, {\left (a b f m + a b f\right )} n \log \relax (F) - m^{2} - 2 \, m - 1}{4 \, b^{2} f n^{2} \log \relax (F)}\right )}}{2 \, b e n} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e g x + d g\right )}^{m} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F(-1)] time = 180.00, size = 0, normalized size = 0.00 \[ \int F^{\left (b \ln \left (c \left (e x +d \right )^{n}\right )+a \right )^{2} f} \left (e g x +d g \right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e g x + d g\right )}^{m} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {e}}^{f\,\ln \relax (F)\,{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2}\,{\left (d\,g+e\,g\,x\right )}^m \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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