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

Optimal. Leaf size=137 \[ \frac {\sqrt {\pi } F^{a f} (d g+e g x)^{m+1} e^{-\frac {(m+1)^2}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-\frac {m+1}{n}} \text {erfi}\left (\frac {2 b f n \log (F) \log \left (c (d+e x)^n\right )+m+1}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g n \sqrt {\log (F)}} \]

[Out]

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

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Rubi [A]  time = 0.39, antiderivative size = 136, normalized size of antiderivative = 0.99, 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 = {2276, 2234, 2204} \[ \frac {\sqrt {\pi } F^{a f} (g (d+e x))^{m+1} e^{-\frac {(m+1)^2}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-\frac {m+1}{n}} \text {Erfi}\left (\frac {2 b f n \log (F) \log \left (c (d+e x)^n\right )+m+1}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} g n \sqrt {\log (F)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 143, normalized size = 1.04 \[ -\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \relax (F)} \operatorname {erf}\left (\frac {{\left (2 \, b f n^{2} \log \left (e x + d\right ) \log \relax (F) + 2 \, b f n \log \relax (F) \log \relax (c) + m + 1\right )} \sqrt {-b f n^{2} \log \relax (F)}}{2 \, b f n^{2} \log \relax (F)}\right ) e^{\left (\frac {4 \, a b f^{2} n^{2} \log \relax (F)^{2} + 4 \, b f m n^{2} \log \relax (F) \log \relax (g) - 4 \, {\left (b f m + b f\right )} n \log \relax (F) \log \relax (c) - m^{2} - 2 \, m - 1}{4 \, b f n^{2} \log \relax (F)}\right )}}{2 \, e n} \]

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)^m,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) + m + 1)*sq
rt(-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*m*n^2*log(F)*log(g) - 4*(b*f*m +
b*f)*n*log(F)*log(c) - m^2 - 2*m - 1)/(b*f*n^2*log(F)))/(e*n)

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

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)^m,x, algorithm="giac")

[Out]

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

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maple [F]  time = 84.22, size = 0, normalized size = 0.00 \[ \int F^{\left (b \ln \left (c \left (e x +d \right )^{n}\right )^{2}+a \right ) f} \left (e g x +d g \right )^{m}\, dx \]

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)^m,x)

[Out]

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

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

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)^m,x, algorithm="maxima")

[Out]

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

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

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)^m,x)

[Out]

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

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

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)**m,x)

[Out]

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

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