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

Optimal. Leaf size=123 \[ \frac {\sqrt {\pi } g^2 F^{a f} (d+e x)^3 e^{-\frac {9}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-3/n} \text {erfi}\left (\frac {2 b f n \log (F) \log \left (c (d+e x)^n\right )+3}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} n \sqrt {\log (F)}} \]

[Out]

1/2*F^(a*f)*g^2*(e*x+d)^3*erfi(1/2*(3+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(9/4/b/f/n^2/ln(F))/n/((c*(e*x+d)^n)^(3/n))/b^(1/2)/f^(1/2)/ln(F)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 123, 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} \[ \frac {\sqrt {\pi } g^2 F^{a f} (d+e x)^3 e^{-\frac {9}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-3/n} \text {Erfi}\left (\frac {2 b f n \log (F) \log \left (c (d+e x)^n\right )+3}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} n \sqrt {\log (F)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^(a*f)*g^2*Sqrt[Pi]*(d + e*x)^3*Erfi[(3 + 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^(9/(4*b*f*n^2*Log[F]))*Sqrt[f]*n*(c*(d + e*x)^n)^(3/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 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)^2 \, dx &=\frac {\operatorname {Subst}\left (\int 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 {g^2 \operatorname {Subst}\left (\int F^{f \left (a+b \log ^2\left (c x^n\right )\right )} x^2 \, dx,x,d+e x\right )}{e}\\ &=\frac {\left (g^2 (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n}\right ) \operatorname {Subst}\left (\int e^{\frac {3 x}{n}+a f \log (F)+b f x^2 \log (F)} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e n}\\ &=\frac {\left (e^{-\frac {9}{4 b f n^2 \log (F)}} F^{a f} g^2 (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n}\right ) \operatorname {Subst}\left (\int e^{\frac {\left (\frac {3}{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 n}\\ &=\frac {e^{-\frac {9}{4 b f n^2 \log (F)}} F^{a f} g^2 \sqrt {\pi } (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n} \text {erfi}\left (\frac {3+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} n \sqrt {\log (F)}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 123, normalized size = 1.00 \[ \frac {\sqrt {\pi } g^2 F^{a f} (d+e x)^3 e^{-\frac {9}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-3/n} \text {erfi}\left (\frac {2 b f n \log (F) \log \left (c (d+e x)^n\right )+3}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} n \sqrt {\log (F)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^(a*f)*g^2*Sqrt[Pi]*(d + e*x)^3*Erfi[(3 + 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^(9/(4*b*f*n^2*Log[F]))*Sqrt[f]*n*(c*(d + e*x)^n)^(3/n)*Sqrt[Log[F]])

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fricas [A]  time = 0.43, size = 119, normalized size = 0.97 \[ -\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \relax (F)} g^{2} \operatorname {erf}\left (\frac {{\left (2 \, b f n^{2} \log \left (e x + d\right ) \log \relax (F) + 2 \, b f n \log \relax (F) \log \relax (c) + 3\right )} \sqrt {-b f n^{2} \log \relax (F)}}{2 \, b f n^{2} \log \relax (F)}\right ) e^{\left (\frac {4 \, a b f^{2} n^{2} \log \relax (F)^{2} - 12 \, b f n \log \relax (F) \log \relax (c) - 9}{4 \, b f n^{2} \log \relax (F)}\right )}}{2 \, e n} \]

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)^2,x, algorithm="fricas")

[Out]

-1/2*sqrt(pi)*sqrt(-b*f*n^2*log(F))*g^2*erf(1/2*(2*b*f*n^2*log(e*x + d)*log(F) + 2*b*f*n*log(F)*log(c) + 3)*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 - 12*b*f*n*log(F)*log(c) - 9)/(b*f*n^2*lo
g(F)))/(e*n)

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

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)^2,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \left (e g x +d g \right )^{2} F^{\left (b \ln \left (c \left (e x +d \right )^{n}\right )^{2}+a \right ) f}\, dx \]

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

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)^2,x, algorithm="maxima")

[Out]

integrate((e*g*x + d*g)^2*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 \[ \int {\mathrm {e}}^{f\,\ln \relax (F)\,\left (b\,{\ln \left (c\,{\left (d+e\,x\right )}^n\right )}^2+a\right )}\,{\left (d\,g+e\,g\,x\right )}^2 \,d x \]

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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