3.6.87 \(\int F^{f (a+b \log ^2(c (d+e x)^n))} (d g+e g x)^m \, dx\) [587]

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

[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 = 0.17, antiderivative size = 137, 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} (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)}} \end {gather*}

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]*(d*g + e*g*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 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 F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx &=\frac {\text {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 ) \text {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 ) \text {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 = 0.11, size = 0, normalized size = 0.00 \begin {gather*} \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (d g+e g x)^m \, dx \end {gather*}

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

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 \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)^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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Fricas [A]
time = 0.38, size = 143, normalized size = 1.04 \begin {gather*} -\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \left (F\right )} \operatorname {erf}\left (\frac {{\left (2 \, b f n^{2} \log \left (x e + d\right ) \log \left (F\right ) + 2 \, b f n \log \left (F\right ) \log \left (c\right ) + m + 1\right )} \sqrt {-b f n^{2} \log \left (F\right )}}{2 \, b f n^{2} \log \left (F\right )}\right ) e^{\left (\frac {4 \, a b f^{2} n^{2} \log \left (F\right )^{2} + 4 \, b f m n^{2} \log \left (F\right ) \log \left (g\right ) - 4 \, {\left (b f m + b f\right )} n \log \left (F\right ) \log \left (c\right ) - m^{2} - 2 \, m - 1}{4 \, b f n^{2} \log \left (F\right )} - 1\right )}}{2 \, 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)^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(x*e + 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)) - 1)/n

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \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)**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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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)^m,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\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 )}^m \,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)^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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