3.7.6 \(\int \frac {F^{f (a+b \log (c (d+e x)^n))^2}}{d g+e g x} \, dx\) [606]

Optimal. Leaf size=70 \[ \frac {\sqrt {\pi } \text {erfi}\left (a \sqrt {f} \sqrt {\log (F)}+b \sqrt {f} \sqrt {\log (F)} \log \left (c (d+e x)^n\right )\right )}{2 b e \sqrt {f} g n \sqrt {\log (F)}} \]

[Out]

1/2*erfi(a*f^(1/2)*ln(F)^(1/2)+b*ln(c*(e*x+d)^n)*f^(1/2)*ln(F)^(1/2))*Pi^(1/2)/b/e/g/n/f^(1/2)/ln(F)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2314, 2308, 2266, 2235} \begin {gather*} \frac {\sqrt {\pi } \text {Erfi}\left (a \sqrt {f} \sqrt {\log (F)}+b \sqrt {f} \sqrt {\log (F)} \log \left (c (d+e x)^n\right )\right )}{2 b e \sqrt {f} g n \sqrt {\log (F)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2235

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]

Rule 2266

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 2308

Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]^2*(b_.))*(f_.))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol]
:> Dist[(g + h*x)^(m + 1)/(h*n*(c*(d + e*x)^n)^((m + 1)/n)), Subst[Int[E^(a*f*Log[F] + ((m + 1)*x)/n + b*f*Lo
g[F]*x^2), x], x, Log[c*(d + e*x)^n]], x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m, n}, x] && EqQ[e*g - d*h, 0]

Rule 2314

Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]*(b_.))^2*(f_.))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol]
:> Dist[(g + h*x)^m*((c*(d + e*x)^n)^(2*a*b*f*Log[F])/(d + e*x)^(m + 2*a*b*f*n*Log[F])), Int[(d + e*x)^(m + 2
*a*b*f*n*Log[F])*F^(a^2*f + b^2*f*Log[c*(d + e*x)^n]^2), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m, n}, x]
&& EqQ[e*g - d*h, 0]

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

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Mathematica [A]
time = 0.06, size = 59, normalized size = 0.84 \begin {gather*} \frac {\sqrt {\pi } \text {erfi}\left (\sqrt {f} \sqrt {\log (F)} \left (a+b \log \left (c (d+e x)^n\right )\right )\right )}{2 b e \sqrt {f} g n \sqrt {\log (F)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.28, size = 125, normalized size = 1.79 \[-\frac {\sqrt {\pi }\, \erf \left (-b \sqrt {-f \ln \left (F \right )}\, \ln \left (\left (e x +d \right )^{n}\right )+\frac {f \left (a +b \left (\ln \left (c \right )-\frac {i \pi \,\mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) \left (-\mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )+\mathrm {csgn}\left (i c \right )\right ) \left (-\mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )+\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right )\right )}{2}\right )\right ) \ln \left (F \right )}{\sqrt {-f \ln \left (F \right )}}\right )}{2 g e n b \sqrt {-f \ln \left (F \right )}}\]

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),x)

[Out]

-1/2/g/e/n*Pi^(1/2)/b/(-f*ln(F))^(1/2)*erf(-b*(-f*ln(F))^(1/2)*ln((e*x+d)^n)+f*(a+b*(ln(c)-1/2*I*Pi*csgn(I*c*(
e*x+d)^n)*(-csgn(I*c*(e*x+d)^n)+csgn(I*c))*(-csgn(I*c*(e*x+d)^n)+csgn(I*(e*x+d)^n))))*ln(F)/(-f*ln(F))^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 66, normalized size = 0.94 \begin {gather*} -\frac {\sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \left (F\right )} \operatorname {erf}\left (\frac {\sqrt {-b^{2} f n^{2} \log \left (F\right )} {\left (b n \log \left (x e + d\right ) + b \log \left (c\right ) + a\right )}}{b n}\right ) e^{\left (-1\right )}}{2 \, b g n} \end {gather*}

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),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {F^{a^{2} f} F^{b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} F^{2 a b f \log {\left (c \left (d + e x\right )^{n} \right )}}}{d + e x}\, dx}{g} \end {gather*}

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),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 3.69, size = 63, normalized size = 0.90 \begin {gather*} -\frac {\sqrt {\pi }\,\mathrm {erf}\left (\frac {1{}\mathrm {i}\,f\,\ln \left (F\right )\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\,b^2+1{}\mathrm {i}\,a\,f\,\ln \left (F\right )\,b}{\sqrt {b^2\,f\,\ln \left (F\right )}}\right )\,1{}\mathrm {i}}{2\,e\,g\,n\,\sqrt {b^2\,f\,\ln \left (F\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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