3.6.93 \(\int \frac {F^{f (a+b \log ^2(c (d+e x)^n))}}{(d g+e g x)^3} \, dx\) [593]

Optimal. Leaf size=118 \[ -\frac {e^{-\frac {1}{b f n^2 \log (F)}} F^{a f} \sqrt {\pi } \left (c (d+e x)^n\right )^{2/n} \text {erfi}\left (\frac {1-b f n \log (F) \log \left (c (d+e x)^n\right )}{\sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^3 n (d+e x)^2 \sqrt {\log (F)}} \]

[Out]

1/2*F^(a*f)*(c*(e*x+d)^n)^(2/n)*erfi((-1+b*f*n*ln(F)*ln(c*(e*x+d)^n))/n/b^(1/2)/f^(1/2)/ln(F)^(1/2))*Pi^(1/2)/
e/exp(1/b/f/n^2/ln(F))/g^3/n/(e*x+d)^2/b^(1/2)/f^(1/2)/ln(F)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2308, 2266, 2235} \begin {gather*} -\frac {\sqrt {\pi } F^{a f} e^{-\frac {1}{b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{2/n} \text {Erfi}\left (\frac {1-b f n \log (F) \log \left (c (d+e x)^n\right )}{\sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^3 n \sqrt {\log (F)} (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[F^(f*(a + b*Log[c*(d + e*x)^n]^2))/(d*g + e*g*x)^3,x]

[Out]

-1/2*(F^(a*f)*Sqrt[Pi]*(c*(d + e*x)^n)^(2/n)*Erfi[(1 - b*f*n*Log[F]*Log[c*(d + e*x)^n])/(Sqrt[b]*Sqrt[f]*n*Sqr
t[Log[F]])])/(Sqrt[b]*e*E^(1/(b*f*n^2*Log[F]))*Sqrt[f]*g^3*n*(d + e*x)^2*Sqrt[Log[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]

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)^3} \, dx &=\frac {\text {Subst}\left (\int \frac {F^{f \left (a+b \log ^2\left (c x^n\right )\right )}}{g^3 x^3} \, dx,x,d+e x\right )}{e}\\ &=\frac {\text {Subst}\left (\int \frac {F^{f \left (a+b \log ^2\left (c x^n\right )\right )}}{x^3} \, dx,x,d+e x\right )}{e g^3}\\ &=\frac {\left (c (d+e x)^n\right )^{2/n} \text {Subst}\left (\int e^{-\frac {2 x}{n}+a f \log (F)+b f x^2 \log (F)} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e g^3 n (d+e x)^2}\\ &=\frac {\left (e^{-\frac {1}{b f n^2 \log (F)}} F^{a f} \left (c (d+e x)^n\right )^{2/n}\right ) \text {Subst}\left (\int e^{\frac {\left (-\frac {2}{n}+2 b f x \log (F)\right )^2}{4 b f \log (F)}} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e g^3 n (d+e x)^2}\\ &=-\frac {e^{-\frac {1}{b f n^2 \log (F)}} F^{a f} \sqrt {\pi } \left (c (d+e x)^n\right )^{2/n} \text {erfi}\left (\frac {1-b f n \log (F) \log \left (c (d+e x)^n\right )}{\sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^3 n (d+e x)^2 \sqrt {\log (F)}}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 117, normalized size = 0.99 \begin {gather*} \frac {e^{-\frac {1}{b f n^2 \log (F)}} F^{a f} \sqrt {\pi } \left (c (d+e x)^n\right )^{2/n} \text {erfi}\left (\frac {-1+b f n \log (F) \log \left (c (d+e x)^n\right )}{\sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^3 n (d+e x)^2 \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)^3,x]

[Out]

(F^(a*f)*Sqrt[Pi]*(c*(d + e*x)^n)^(2/n)*Erfi[(-1 + b*f*n*Log[F]*Log[c*(d + e*x)^n])/(Sqrt[b]*Sqrt[f]*n*Sqrt[Lo
g[F]])])/(2*Sqrt[b]*e*E^(1/(b*f*n^2*Log[F]))*Sqrt[f]*g^3*n*(d + e*x)^2*Sqrt[Log[F]])

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {F^{f \left (a +b \ln \left (c \left (e x +d \right )^{n}\right )^{2}\right )}}{\left (e g x +d g \right )^{3}}\, dx\]

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)^3,x)

[Out]

int(F^(f*(a+b*ln(c*(e*x+d)^n)^2))/(e*g*x+d*g)^3,x)

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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)^3,x, algorithm="maxima")

[Out]

integrate(F^((b*log((e*x + d)^n*c)^2 + a)*f)/(e*g*x + d*g)^3, x)

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Fricas [A]
time = 0.37, size = 114, normalized size = 0.97 \begin {gather*} -\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \left (F\right )} \operatorname {erf}\left (\frac {{\left (b f n^{2} \log \left (x e + d\right ) \log \left (F\right ) + b f n \log \left (F\right ) \log \left (c\right ) - 1\right )} \sqrt {-b f n^{2} \log \left (F\right )}}{b f n^{2} \log \left (F\right )}\right ) e^{\left (\frac {a b f^{2} n^{2} \log \left (F\right )^{2} + 2 \, b f n \log \left (F\right ) \log \left (c\right ) - 1}{b f n^{2} \log \left (F\right )} - 1\right )}}{2 \, g^{3} 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)^3,x, algorithm="fricas")

[Out]

-1/2*sqrt(pi)*sqrt(-b*f*n^2*log(F))*erf((b*f*n^2*log(x*e + d)*log(F) + b*f*n*log(F)*log(c) - 1)*sqrt(-b*f*n^2*
log(F))/(b*f*n^2*log(F)))*e^((a*b*f^2*n^2*log(F)^2 + 2*b*f*n*log(F)*log(c) - 1)/(b*f*n^2*log(F)) - 1)/(g^3*n)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**3,x)

[Out]

Timed out

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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)^3,x, algorithm="giac")

[Out]

integrate(F^((b*log((x*e + d)^n*c)^2 + a)*f)/(g*x*e + d*g)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {e}}^{f\,\ln \left (F\right )\,\left (b\,{\ln \left (c\,{\left (d+e\,x\right )}^n\right )}^2+a\right )}}{{\left (d\,g+e\,g\,x\right )}^3} \,d x \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)^3,x)

[Out]

int(exp(f*log(F)*(a + b*log(c*(d + e*x)^n)^2))/(d*g + e*g*x)^3, x)

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