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

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

[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 = 0.37, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {12, 2278, 2274, 15, 2276, 2234, 2204} \[ \frac {\sqrt {\pi } g (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} e^{-\frac {2 a b f n \log (F)+1}{b^2 f n^2 \log (F)}} \text {Erfi}\left (\frac {a b f \log (F)+b^2 f \log (F) \log \left (c (d+e x)^n\right )+\frac {1}{n}}{b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

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 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, 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]

Rule 2278

Int[(F_)^(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^2*(d_.))*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*F^(a^2*d
+ 2*a*b*d*Log[c*x^n] + b^2*d*Log[c*x^n]^2), x] /; FreeQ[{F, a, b, c, d, e, m, n}, x]

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 {\operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 0.73, size = 120, normalized size = 0.98 \[ \frac {\sqrt {\pi } g (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} e^{-\frac {2 a b f n \log (F)+1}{b^2 f n^2 \log (F)}} \text {erfi}\left (\frac {b f n \log (F) \left (a+b \log \left (c (d+e x)^n\right )\right )+1}{b \sqrt {f} n \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]

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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fricas [A]  time = 0.42, size = 128, normalized size = 1.05 \[ -\frac {\sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \relax (F)} g \operatorname {erf}\left (\frac {{\left (b^{2} f n^{2} \log \left (e x + d\right ) \log \relax (F) + b^{2} f n \log \relax (F) \log \relax (c) + a b f n \log \relax (F) + 1\right )} \sqrt {-b^{2} f n^{2} \log \relax (F)}}{b^{2} f n^{2} \log \relax (F)}\right ) e^{\left (-\frac {2 \, b^{2} f n \log \relax (F) \log \relax (c) + 2 \, a b f n \log \relax (F) + 1}{b^{2} f n^{2} \log \relax (F)}\right )}}{2 \, b e n} \]

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),x, algorithm="fricas")

[Out]

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

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

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),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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),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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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),x)

[Out]

Timed out

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