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

Optimal. Leaf size=153 \[ \frac{\sqrt{\pi } F^{a^2 f} (d+e x) (d g+e g x)^m \left (c (d+e x)^n\right )^{-\frac{m+1}{n}} \exp \left (-\frac{(2 a b f n \log (F)+m+1)^2}{4 b^2 f n^2 \log (F)}\right ) \text{Erfi}\left (\frac{2 a b f n \log (F)+2 b^2 f n \log (F) \log \left (c (d+e x)^n\right )+m+1}{2 b \sqrt{f} n \sqrt{\log (F)}}\right )}{2 b e \sqrt{f} n \sqrt{\log (F)}} \]

[Out]

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

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Rubi [A]  time = 0.757005, 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.

[In]

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

[Out]

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

Rule 2278

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

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

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 \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*}

Mathematica [F]  time = 0.238361, size = 0, normalized size = 0. \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (d g+e g x)^m \, dx \]

Verification is Not applicable to the result.

[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 = 180., size = 0, normalized size = 0. \begin{align*} \int{F}^{f \left ( a+b\ln \left ( c \left ( ex+d \right ) ^{n} \right ) \right ) ^{2}} \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 ) + a\right )}^{2} 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) + a)^2*f), x)

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Fricas [A]  time = 1.02914, size = 443, normalized size = 2.9 \begin{align*} -\frac{\sqrt{\pi } \sqrt{-b^{2} f n^{2} \log \left (F\right )} \operatorname{erf}\left (\frac{{\left (2 \, b^{2} f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + 2 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 2 \, a b f n \log \left (F\right ) + m + 1\right )} \sqrt{-b^{2} f n^{2} \log \left (F\right )}}{2 \, b^{2} f n^{2} \log \left (F\right )}\right ) e^{\left (\frac{4 \, b^{2} f m n^{2} \log \left (F\right ) \log \left (g\right ) - 4 \,{\left (b^{2} f m + b^{2} f\right )} n \log \left (F\right ) \log \left (c\right ) - 4 \,{\left (a b f m + a b f\right )} n \log \left (F\right ) - m^{2} - 2 \, m - 1}{4 \, b^{2} f n^{2} \log \left (F\right )}\right )}}{2 \, b 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^2*f*n^2*log(F))*erf(1/2*(2*b^2*f*n^2*log(e*x + d)*log(F) + 2*b^2*f*n*log(F)*log(c) + 2*a
*b*f*n*log(F) + m + 1)*sqrt(-b^2*f*n^2*log(F))/(b^2*f*n^2*log(F)))*e^(1/4*(4*b^2*f*m*n^2*log(F)*log(g) - 4*(b^
2*f*m + b^2*f)*n*log(F)*log(c) - 4*(a*b*f*m + a*b*f)*n*log(F) - m^2 - 2*m - 1)/(b^2*f*n^2*log(F)))/(b*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 ) + a\right )}^{2} 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) + a)^2*f), x)