3.7.4 \(\int F^{f (a+b \log (c (d+e x)^n))^2} (d g+e g x) \, dx\) [604]

Optimal. Leaf size=122 \[ \frac {e^{-\frac {1+2 a b f n \log (F)}{b^2 f n^2 \log (F)}} g \sqrt {\pi } (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {erfi}\left (\frac {\frac {1}{n}+a b f \log (F)+b^2 f \log (F) \log \left (c (d+e x)^n\right )}{b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2314, 2308, 2266, 2235} \begin {gather*} \frac {\sqrt {\pi } g (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} e^{-\frac {2 a b f n \log (F)+1}{b^2 f n^2 \log (F)}} \text {Erfi}\left (\frac {a b f \log (F)+b^2 f \log (F) \log \left (c (d+e x)^n\right )+\frac {1}{n}}{b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} 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),x]

[Out]

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

Rule 2314

Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]*(b_.))^2*(f_.))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol]
:> Dist[(g + h*x)^m*((c*(d + e*x)^n)^(2*a*b*f*Log[F])/(d + e*x)^(m + 2*a*b*f*n*Log[F])), Int[(d + e*x)^(m + 2
*a*b*f*n*Log[F])*F^(a^2*f + b^2*f*Log[c*(d + e*x)^n]^2), x], 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 \left (c (d+e x)^n\right )\right )^2} (d g+e g x) \, dx &=\frac {\text {Subst}\left (\int F^{f \left (a+b \log \left (c x^n\right )\right )^2} g x \, dx,x,d+e x\right )}{e}\\ &=\frac {g \text {Subst}\left (\int F^{f \left (a+b \log \left (c x^n\right )\right )^2} x \, dx,x,d+e x\right )}{e}\\ &=\frac {g \text {Subst}\left (\int F^{a^2 f+2 a b f \log \left (c x^n\right )+b^2 f \log ^2\left (c x^n\right )} x \, dx,x,d+e x\right )}{e}\\ &=\frac {g \text {Subst}\left (\int F^{a^2 f+b^2 f \log ^2\left (c x^n\right )} x \left (c x^n\right )^{2 a b f \log (F)} \, dx,x,d+e x\right )}{e}\\ &=\frac {\left (g (d+e x)^{-2 a b f n \log (F)} \left (c (d+e x)^n\right )^{2 a b f \log (F)}\right ) \text {Subst}\left (\int F^{a^2 f+b^2 f \log ^2\left (c x^n\right )} x^{1+2 a b f n \log (F)} \, dx,x,d+e x\right )}{e}\\ &=\frac {\left (g (d+e x)^2 \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {2+2 a b f n \log (F)}{n}}\right ) \text {Subst}\left (\int \exp \left (a^2 f \log (F)+b^2 f x^2 \log (F)+\frac {x (2+2 a b f n \log (F))}{n}\right ) \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e n}\\ &=\frac {\left (\exp \left (a^2 f \log (F)-\frac {(2+2 a b f n \log (F))^2}{4 b^2 f n^2 \log (F)}\right ) g (d+e x)^2 \left (c (d+e x)^n\right )^{2 a b f \log (F)-\frac {2+2 a b f n \log (F)}{n}}\right ) \text {Subst}\left (\int \exp \left (\frac {\left (2 b^2 f x \log (F)+\frac {2+2 a b f n \log (F)}{n}\right )^2}{4 b^2 f \log (F)}\right ) \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e n}\\ &=\frac {e^{-\frac {1+2 a b f n \log (F)}{b^2 f n^2 \log (F)}} g \sqrt {\pi } (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {erfi}\left (\frac {\frac {1}{n}+a b f \log (F)+b^2 f \log (F) \log \left (c (d+e x)^n\right )}{b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}}\\ \end {align*}

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Mathematica [A]
time = 0.60, size = 120, normalized size = 0.98 \begin {gather*} \frac {e^{-\frac {1+2 a b f n \log (F)}{b^2 f n^2 \log (F)}} g \sqrt {\pi } (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {erfi}\left (\frac {1+b f n \log (F) \left (a+b \log \left (c (d+e x)^n\right )\right )}{b \sqrt {f} n \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \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),x]

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int F^{f \left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )^{2}} \left (e g x +d g \right )\, dx\]

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),x)

[Out]

int(F^(f*(a+b*ln(c*(e*x+d)^n))^2)*(e*g*x+d*g),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),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.41, size = 128, normalized size = 1.05 \begin {gather*} -\frac {\sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \left (F\right )} g \operatorname {erf}\left (\frac {{\left (b^{2} f n^{2} \log \left (x e + d\right ) \log \left (F\right ) + b^{2} f n \log \left (F\right ) \log \left (c\right ) + a b f n \log \left (F\right ) + 1\right )} \sqrt {-b^{2} f n^{2} \log \left (F\right )}}{b^{2} f n^{2} \log \left (F\right )}\right ) e^{\left (-\frac {2 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 2 \, a b f n \log \left (F\right ) + 1}{b^{2} f n^{2} \log \left (F\right )} - 1\right )}}{2 \, b 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),x, algorithm="fricas")

[Out]

-1/2*sqrt(pi)*sqrt(-b^2*f*n^2*log(F))*g*erf((b^2*f*n^2*log(x*e + d)*log(F) + b^2*f*n*log(F)*log(c) + a*b*f*n*l
og(F) + 1)*sqrt(-b^2*f*n^2*log(F))/(b^2*f*n^2*log(F)))*e^(-(2*b^2*f*n*log(F)*log(c) + 2*a*b*f*n*log(F) + 1)/(b
^2*f*n^2*log(F)) - 1)/(b*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),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),x, algorithm="giac")

[Out]

integrate((g*x*e + d*g)*F^((b*log((x*e + d)^n*c) + a)^2*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 (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2}\,\left (d\,g+e\,g\,x\right ) \,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),x)

[Out]

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

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