3.597 \(\int F^{f (a+b \log ^2(c (d+e x)^n))} (g+h x) \, dx\)

Optimal. Leaf size=242 \[ \frac{\sqrt{\pi } F^{a f} (d+e x) (e g-d h) e^{-\frac{1}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-1/n} \text{Erfi}\left (\frac{2 b f n \log (F) \log \left (c (d+e x)^n\right )+1}{2 \sqrt{b} \sqrt{f} n \sqrt{\log (F)}}\right )}{2 \sqrt{b} e^2 \sqrt{f} n \sqrt{\log (F)}}+\frac{\sqrt{\pi } h F^{a f} (d+e x)^2 e^{-\frac{1}{b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-2/n} \text{Erfi}\left (\frac{b f n \log (F) \log \left (c (d+e x)^n\right )+1}{\sqrt{b} \sqrt{f} n \sqrt{\log (F)}}\right )}{2 \sqrt{b} e^2 \sqrt{f} n \sqrt{\log (F)}} \]

[Out]

(F^(a*f)*h*Sqrt[Pi]*(d + e*x)^2*Erfi[(1 + b*f*n*Log[F]*Log[c*(d + e*x)^n])/(Sqrt[b]*Sqrt[f]*n*Sqrt[Log[F]])])/
(2*Sqrt[b]*e^2*E^(1/(b*f*n^2*Log[F]))*Sqrt[f]*n*(c*(d + e*x)^n)^(2/n)*Sqrt[Log[F]]) + (F^(a*f)*(e*g - d*h)*Sqr
t[Pi]*(d + e*x)*Erfi[(1 + 2*b*f*n*Log[F]*Log[c*(d + e*x)^n])/(2*Sqrt[b]*Sqrt[f]*n*Sqrt[Log[F]])])/(2*Sqrt[b]*e
^2*E^(1/(4*b*f*n^2*Log[F]))*Sqrt[f]*n*(c*(d + e*x)^n)^n^(-1)*Sqrt[Log[F]])

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Rubi [F]  time = 0.22953, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (g+h x) \, dx \]

Verification is Not applicable to the result.

[In]

Int[F^(f*(a + b*Log[c*(d + e*x)^n]^2))*(g + h*x),x]

[Out]

(F^(a*f)*g*Sqrt[Pi]*(d + e*x)*Erfi[(1 + 2*b*f*n*Log[F]*Log[c*(d + e*x)^n])/(2*Sqrt[b]*Sqrt[f]*n*Sqrt[Log[F]])]
)/(2*Sqrt[b]*e*E^(1/(4*b*f*n^2*Log[F]))*Sqrt[f]*n*(c*(d + e*x)^n)^n^(-1)*Sqrt[Log[F]]) + h*Defer[Int][F^(f*(a
+ b*Log[c*(d + e*x)^n]^2))*x, x]

Rubi steps

\begin{align*} \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (g+h x) \, dx &=\int \left (F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} g+F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} h x\right ) \, dx\\ &=g \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} \, dx+h \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} x \, dx\\ &=\frac{g \operatorname{Subst}\left (\int F^{f \left (a+b \log ^2\left (c x^n\right )\right )} \, dx,x,d+e x\right )}{e}+h \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} x \, dx\\ &=h \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} x \, dx+\frac{\left (g (d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \operatorname{Subst}\left (\int e^{\frac{x}{n}+a f \log (F)+b f x^2 \log (F)} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e n}\\ &=h \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} x \, dx+\frac{\left (e^{-\frac{1}{4 b f n^2 \log (F)}} F^{a f} g (d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \operatorname{Subst}\left (\int e^{\frac{\left (\frac{1}{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 n}\\ &=\frac{e^{-\frac{1}{4 b f n^2 \log (F)}} F^{a f} g \sqrt{\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text{erfi}\left (\frac{1+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} n \sqrt{\log (F)}}+h \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} x \, dx\\ \end{align*}

Mathematica [A]  time = 0.374243, size = 204, normalized size = 0.84 \[ \frac{\sqrt{\pi } F^{a f} (d+e x) e^{-\frac{1}{b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-2/n} \left ((e g-d h) e^{\frac{3}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{\frac{1}{n}} \text{Erfi}\left (\frac{2 b f n \log (F) \log \left (c (d+e x)^n\right )+1}{2 \sqrt{b} \sqrt{f} n \sqrt{\log (F)}}\right )+h (d+e x) \text{Erfi}\left (\frac{b f n \log (F) \log \left (c (d+e x)^n\right )+1}{\sqrt{b} \sqrt{f} n \sqrt{\log (F)}}\right )\right )}{2 \sqrt{b} e^2 \sqrt{f} n \sqrt{\log (F)}} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(f*(a + b*Log[c*(d + e*x)^n]^2))*(g + h*x),x]

[Out]

(F^(a*f)*Sqrt[Pi]*(d + e*x)*(h*(d + e*x)*Erfi[(1 + b*f*n*Log[F]*Log[c*(d + e*x)^n])/(Sqrt[b]*Sqrt[f]*n*Sqrt[Lo
g[F]])] + E^(3/(4*b*f*n^2*Log[F]))*(e*g - d*h)*(c*(d + e*x)^n)^n^(-1)*Erfi[(1 + 2*b*f*n*Log[F]*Log[c*(d + e*x)
^n])/(2*Sqrt[b]*Sqrt[f]*n*Sqrt[Log[F]])]))/(2*Sqrt[b]*e^2*E^(1/(b*f*n^2*Log[F]))*Sqrt[f]*n*(c*(d + e*x)^n)^(2/
n)*Sqrt[Log[F]])

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Maple [F]  time = 0.372, size = 0, normalized size = 0. \begin{align*} \int{F}^{f \left ( a+b \left ( \ln \left ( c \left ( ex+d \right ) ^{n} \right ) \right ) ^{2} \right ) } \left ( hx+g \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(f*(a+b*ln(c*(e*x+d)^n)^2))*(h*x+g),x)

[Out]

int(F^(f*(a+b*ln(c*(e*x+d)^n)^2))*(h*x+g),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (h x + g\right )} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + a\right )} f}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(f*(a+b*log(c*(e*x+d)^n)^2))*(h*x+g),x, algorithm="maxima")

[Out]

integrate((h*x + g)*F^((b*log((e*x + d)^n*c)^2 + a)*f), x)

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Fricas [A]  time = 1.05963, size = 610, normalized size = 2.52 \begin{align*} -\frac{\sqrt{\pi } \sqrt{-b f n^{2} \log \left (F\right )}{\left (e g - d h\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 ) + 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 n \log \left (F\right ) \log \left (c\right ) - 1}{4 \, b f n^{2} \log \left (F\right )}\right )} + \sqrt{\pi } \sqrt{-b f n^{2} \log \left (F\right )} h \operatorname{erf}\left (\frac{{\left (b f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + b f n \log \left (F\right ) \log \left (c\right ) + 1\right )} \sqrt{-b f n^{2} \log \left (F\right )}}{b f n^{2} \log \left (F\right )}\right ) e^{\left (\frac{a b f^{2} n^{2} \log \left (F\right )^{2} - 2 \, b f n \log \left (F\right ) \log \left (c\right ) - 1}{b f n^{2} \log \left (F\right )}\right )}}{2 \, e^{2} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(f*(a+b*log(c*(e*x+d)^n)^2))*(h*x+g),x, algorithm="fricas")

[Out]

-1/2*(sqrt(pi)*sqrt(-b*f*n^2*log(F))*(e*g - d*h)*erf(1/2*(2*b*f*n^2*log(e*x + d)*log(F) + 2*b*f*n*log(F)*log(c
) + 1)*sqrt(-b*f*n^2*log(F))/(b*f*n^2*log(F)))*e^(1/4*(4*a*b*f^2*n^2*log(F)^2 - 4*b*f*n*log(F)*log(c) - 1)/(b*
f*n^2*log(F))) + sqrt(pi)*sqrt(-b*f*n^2*log(F))*h*erf((b*f*n^2*log(e*x + d)*log(F) + b*f*n*log(F)*log(c) + 1)*
sqrt(-b*f*n^2*log(F))/(b*f*n^2*log(F)))*e^((a*b*f^2*n^2*log(F)^2 - 2*b*f*n*log(F)*log(c) - 1)/(b*f*n^2*log(F))
))/(e^2*n)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(f*(a+b*ln(c*(e*x+d)**n)**2))*(h*x+g),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (h x + g\right )} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + a\right )} f}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(f*(a+b*log(c*(e*x+d)^n)^2))*(h*x+g),x, algorithm="giac")

[Out]

integrate((h*x + g)*F^((b*log((e*x + d)^n*c)^2 + a)*f), x)