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

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 = 0.27, antiderivative size = 70, normalized size of antiderivative = 1.00, 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} \[ \frac {\sqrt {\pi } \text {Erfi}\left (a \sqrt {f} \sqrt {\log (F)}+b \sqrt {f} \sqrt {\log (F)} \log \left (c (d+e x)^n\right )\right )}{2 b e \sqrt {f} g n \sqrt {\log (F)}} \]

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 12

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

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 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 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 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]

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 {\operatorname {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 {\operatorname {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 {\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} \, dx,x,d+e x\right )}{e g}\\ &=\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} \, 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 ) \operatorname {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 {\operatorname {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 {\operatorname {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 = 0.05, size = 59, normalized size = 0.84 \[ \frac {\sqrt {\pi } \text {erfi}\left (\sqrt {f} \sqrt {\log (F)} \left (a+b \log \left (c (d+e x)^n\right )\right )\right )}{2 b e \sqrt {f} g n \sqrt {\log (F)}} \]

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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fricas [A]  time = 0.42, size = 66, normalized size = 0.94 \[ -\frac {\sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \relax (F)} \operatorname {erf}\left (\frac {\sqrt {-b^{2} f n^{2} \log \relax (F)} {\left (b n \log \left (e x + d\right ) + b \log \relax (c) + a\right )}}{b n}\right )}{2 \, b e g n} \]

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(e*x + d) + b*log(c) + a)/(b*n))/(b*
e*g*n)

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

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

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{\left (b \ln \left (c \left (e x +d \right )^{n}\right )+a \right )^{2} f}}{e g x +d g}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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

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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