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

Optimal. Leaf size=67 \[ \frac {\sqrt {\pi } F^{a f} \text {erfi}\left (\sqrt {b} \sqrt {f} \sqrt {\log (F)} \log \left (c (d+e x)^n\right )\right )}{2 \sqrt {b} e \sqrt {f} g n \sqrt {\log (F)}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 67, normalized size of antiderivative = 1.00, 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 = {12, 2276, 2204} \[ \frac {\sqrt {\pi } F^{a f} \text {Erfi}\left (\sqrt {b} \sqrt {f} \sqrt {\log (F)} \log \left (c (d+e x)^n\right )\right )}{2 \sqrt {b} e \sqrt {f} g n \sqrt {\log (F)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 67, normalized size = 1.00 \[ \frac {\sqrt {\pi } F^{a f} \text {erfi}\left (\sqrt {b} \sqrt {f} \sqrt {\log (F)} \log \left (c (d+e x)^n\right )\right )}{2 \sqrt {b} e \sqrt {f} g n \sqrt {\log (F)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 57, normalized size = 0.85 \[ -\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \relax (F)} F^{a f} \operatorname {erf}\left (\frac {\sqrt {-b f n^{2} \log \relax (F)} {\left (n \log \left (e x + d\right ) + \log \relax (c)\right )}}{n}\right )}{2 \, e g n} \]

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

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

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)^2 + a)*f)/(e*g*x + d*g), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)^2 + a)*f)/(e*g*x + d*g), x)

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mupad [B]  time = 3.74, size = 49, normalized size = 0.73 \[ \frac {F^{a\,f}\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,f\,\ln \relax (F)\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{\sqrt {b\,f\,\ln \relax (F)}}\right )}{2\,e\,g\,n\,\sqrt {b\,f\,\ln \relax (F)}} \]

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]

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

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

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

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