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

Optimal. Leaf size=137 \[ \frac{\sqrt{\pi } F^{a f} (d g+e g x)^{m+1} e^{-\frac{(m+1)^2}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-\frac{m+1}{n}} \text{Erfi}\left (\frac{2 b f n \log (F) \log \left (c (d+e x)^n\right )+m+1}{2 \sqrt{b} \sqrt{f} n \sqrt{\log (F)}}\right )}{2 \sqrt{b} e \sqrt{f} g n \sqrt{\log (F)}} \]

[Out]

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

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Rubi [A]  time = 0.389683, antiderivative size = 136, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {2276, 2234, 2204} \[ \frac{\sqrt{\pi } F^{a f} (g (d+e x))^{m+1} e^{-\frac{(m+1)^2}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-\frac{m+1}{n}} \text{Erfi}\left (\frac{2 b f n \log (F) \log \left (c (d+e x)^n\right )+m+1}{2 \sqrt{b} \sqrt{f} n \sqrt{\log (F)}}\right )}{2 \sqrt{b} e \sqrt{f} g n \sqrt{\log (F)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2276

Int[(F_)^(((a_.) + Log[(c_.)*(x_)^(n_.)]^2*(b_.))*(d_.))*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(e*x)^(m + 1)/(
e*n*(c*x^n)^((m + 1)/n)), Subst[Int[E^(a*d*Log[F] + ((m + 1)*x)/n + b*d*Log[F]*x^2), x], x, Log[c*x^n]], x] /;
FreeQ[{F, a, b, c, d, e, m, n}, x]

Rule 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rubi steps

\begin{align*} \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx &=\frac{\operatorname{Subst}\left (\int F^{f \left (a+b \log ^2\left (c x^n\right )\right )} (g x)^m \, dx,x,d+e x\right )}{e}\\ &=\frac{\left ((g (d+e x))^{1+m} \left (c (d+e x)^n\right )^{-\frac{1+m}{n}}\right ) \operatorname{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} (g (d+e x))^{1+m} \left (c (d+e x)^n\right )^{-\frac{1+m}{n}}\right ) \operatorname{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 } (g (d+e x))^{1+m} \left (c (d+e x)^n\right )^{-\frac{1+m}{n}} \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*}

Mathematica [F]  time = 0.157323, size = 0, normalized size = 0. \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 80.773, 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 ( egx+dg \right ) ^{m}\, 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))*(e*g*x+d*g)^m,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \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))*(e*g*x+d*g)^m,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.02729, size = 382, normalized size = 2.79 \begin{align*} -\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} \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))*(e*g*x+d*g)^m,x, algorithm="fricas")

[Out]

-1/2*sqrt(pi)*sqrt(-b*f*n^2*log(F))*erf(1/2*(2*b*f*n^2*log(e*x + d)*log(F) + 2*b*f*n*log(F)*log(c) + m + 1)*sq
rt(-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*m*n^2*log(F)*log(g) - 4*(b*f*m +
b*f)*n*log(F)*log(c) - m^2 - 2*m - 1)/(b*f*n^2*log(F)))/(e*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))*(e*g*x+d*g)**m,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \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))*(e*g*x+d*g)^m,x, algorithm="giac")

[Out]

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