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

Optimal. Leaf size=128 \[ -\frac{\sqrt{\pi } \left (c (d+e x)^n\right )^{\frac{1}{n}} e^{\frac{a}{b n}-\frac{1}{4 b^2 f n^2 \log (F)}} \text{Erfi}\left (\frac{-2 a b f \log (F)-2 b^2 f \log (F) \log \left (c (d+e x)^n\right )+\frac{1}{n}}{2 b \sqrt{f} \sqrt{\log (F)}}\right )}{2 b e \sqrt{f} g^2 n \sqrt{\log (F)} (d+e x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.411869, antiderivative size = 128, normalized size of antiderivative = 1., 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 = {12, 2278, 2274, 15, 2276, 2234, 2204} \[ -\frac{\sqrt{\pi } \left (c (d+e x)^n\right )^{\frac{1}{n}} e^{\frac{a}{b n}-\frac{1}{4 b^2 f n^2 \log (F)}} \text{Erfi}\left (\frac{-2 a b f \log (F)-2 b^2 f \log (F) \log \left (c (d+e x)^n\right )+\frac{1}{n}}{2 b \sqrt{f} \sqrt{\log (F)}}\right )}{2 b e \sqrt{f} g^2 n \sqrt{\log (F)} (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.256672, size = 126, normalized size = 0.98 \[ \frac{\sqrt{\pi } \left (c (d+e x)^n\right )^{\frac{1}{n}} e^{\frac{4 a b f n \log (F)-1}{4 b^2 f n^2 \log (F)}} \text{Erfi}\left (\frac{2 b f n \log (F) \left (a+b \log \left (c (d+e x)^n\right )\right )-1}{2 b \sqrt{f} n \sqrt{\log (F)}}\right )}{2 b e \sqrt{f} g^2 n \sqrt{\log (F)} (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.521, size = 0, normalized size = 0. \begin{align*} \int{\frac{{F}^{f \left ( a+b\ln \left ( c \left ( ex+d \right ) ^{n} \right ) \right ) ^{2}}}{ \left ( egx+dg \right ) ^{2}}}\, 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)^2,x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.01412, size = 354, normalized size = 2.77 \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 ) - 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 n \log \left (F\right ) \log \left (c\right ) + 4 \, a b f n \log \left (F\right ) - 1}{4 \, b^{2} f n^{2} \log \left (F\right )}\right )}}{2 \, b e g^{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)/(e*g*x+d*g)^2,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) - 1)*sqrt(-b^2*f*n^2*log(F))/(b^2*f*n^2*log(F)))*e^(1/4*(4*b^2*f*n*log(F)*log(c) + 4*a*b*f*n*log
(F) - 1)/(b^2*f*n^2*log(F)))/(b*e*g^2*n)

________________________________________________________________________________________

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)**2,x)

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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