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

Optimal. Leaf size=121 \[ -\frac {\sqrt {\pi } F^{a f} e^{-\frac {1}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{\frac {1}{n}} \text {erfi}\left (\frac {1-2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^2 n \sqrt {\log (F)} (d+e x)} \]

[Out]

1/2*F^(a*f)*(c*(e*x+d)^n)^(1/n)*erfi(1/2*(-1+2*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(1/4/b/f/n^2/ln(F))/g^2/n/(e*x+d)/b^(1/2)/f^(1/2)/ln(F)^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {12, 2276, 2234, 2204} \[ -\frac {\sqrt {\pi } F^{a f} e^{-\frac {1}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{\frac {1}{n}} \text {Erfi}\left (\frac {1-2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^2 n \sqrt {\log (F)} (d+e x)} \]

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]

-(F^(a*f)*Sqrt[Pi]*(c*(d + e*x)^n)^n^(-1)*Erfi[(1 - 2*b*f*n*Log[F]*Log[c*(d + e*x)^n])/(2*Sqrt[b]*Sqrt[f]*n*Sq
rt[Log[F]])])/(2*Sqrt[b]*e*E^(1/(4*b*f*n^2*Log[F]))*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 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 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)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {F^{f \left (a+b \log ^2\left (c x^n\right )\right )}}{g^2 x^2} \, 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^2} \, dx,x,d+e x\right )}{e g^2}\\ &=\frac {\left (c (d+e x)^n\right )^{\frac {1}{n}} \operatorname {Subst}\left (\int e^{-\frac {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^2 n (d+e x)}\\ &=\frac {\left (e^{-\frac {1}{4 b f n^2 \log (F)}} F^{a f} \left (c (d+e x)^n\right )^{\frac {1}{n}}\right ) \operatorname {Subst}\left (\int e^{\frac {\left (-\frac {1}{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^2 n (d+e x)}\\ &=-\frac {e^{-\frac {1}{4 b f n^2 \log (F)}} F^{a f} \sqrt {\pi } \left (c (d+e x)^n\right )^{\frac {1}{n}} \text {erfi}\left (\frac {1-2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^2 n (d+e x) \sqrt {\log (F)}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 121, normalized size = 1.00 \[ \frac {\sqrt {\pi } F^{a f} e^{-\frac {1}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{\frac {1}{n}} \text {erfi}\left (\frac {2 b f n \log (F) \log \left (c (d+e x)^n\right )-1}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g^2 n \sqrt {\log (F)} (d+e x)} \]

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]

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

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fricas [A]  time = 0.41, size = 119, normalized size = 0.98 \[ -\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \relax (F)} \operatorname {erf}\left (\frac {{\left (2 \, b f n^{2} \log \left (e x + d\right ) \log \relax (F) + 2 \, b f n \log \relax (F) \log \relax (c) - 1\right )} \sqrt {-b f n^{2} \log \relax (F)}}{2 \, b f n^{2} \log \relax (F)}\right ) e^{\left (\frac {4 \, a b f^{2} n^{2} \log \relax (F)^{2} + 4 \, b f n \log \relax (F) \log \relax (c) - 1}{4 \, b f n^{2} \log \relax (F)}\right )}}{2 \, e g^{2} n} \]

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)^2,x, algorithm="fricas")

[Out]

-1/2*sqrt(pi)*sqrt(-b*f*n^2*log(F))*erf(1/2*(2*b*f*n^2*log(e*x + d)*log(F) + 2*b*f*n*log(F)*log(c) - 1)*sqrt(-
b*f*n^2*log(F))/(b*f*n^2*log(F)))*e^(1/4*(4*a*b*f^2*n^2*log(F)^2 + 4*b*f*n*log(F)*log(c) - 1)/(b*f*n^2*log(F))
)/(e*g^2*n)

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

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)^2,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.61, size = 0, normalized size = 0.00 \[ \int \frac {F^{\left (b \ln \left (c \left (e x +d \right )^{n}\right )^{2}+a \right ) f}}{\left (e g x +d g \right )^{2}}\, dx \]

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)^2,x)

[Out]

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

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

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)^2,x, algorithm="maxima")

[Out]

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

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

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)^2,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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