3.6.90 \(\int F^{f (a+b \log ^2(c (d+e x)^n))} \, dx\) [590]

Optimal. Leaf size=118 \[ \frac {e^{-\frac {1}{4 b f n^2 \log (F)}} F^{a f} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {1+2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} n \sqrt {\log (F)}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2307, 2266, 2235} \begin {gather*} \frac {\sqrt {\pi } F^{a f} (d+e x) e^{-\frac {1}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-1/n} \text {Erfi}\left (\frac {2 b f n \log (F) \log \left (c (d+e x)^n\right )+1}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {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)),x]

[Out]

(F^(a*f)*Sqrt[Pi]*(d + e*x)*Erfi[(1 + 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/(4*b*f*n^2*Log[F]))*Sqrt[f]*n*(c*(d + e*x)^n)^n^(-1)*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 2307

Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]^2*(b_.))*(f_.)), x_Symbol] :> Dist[(d + e*x)/(e*n*(c*
(d + e*x)^n)^(1/n)), Subst[Int[E^(a*f*Log[F] + x/n + b*f*Log[F]*x^2), x], x, Log[c*(d + e*x)^n]], x] /; FreeQ[
{F, a, b, c, d, e, f, n}, x]

Rubi steps

\begin {align*} \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} \, dx &=\frac {\text {Subst}\left (\int F^{f \left (a+b \log ^2\left (c x^n\right )\right )} \, dx,x,d+e x\right )}{e}\\ &=\frac {\left ((d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \text {Subst}\left (\int e^{\frac {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 {1}{4 b f n^2 \log (F)}} F^{a f} (d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \text {Subst}\left (\int e^{\frac {\left (\frac {1}{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 {1}{4 b f n^2 \log (F)}} F^{a f} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {1+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 = 0.07, size = 118, normalized size = 1.00 \begin {gather*} \frac {e^{-\frac {1}{4 b f n^2 \log (F)}} F^{a f} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {1+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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 116, normalized size = 0.98 \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 ) + 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 n \log \left (F\right ) \log \left (c\right ) - 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)),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) + 1)*sqrt(-
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*n*log(F)*log(c) - 1)/(b*f*n^2*log(F))
- 1)/n

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (105) = 210\).
time = 7.98, size = 228, normalized size = 1.93 \begin {gather*} \begin {cases} - \frac {2 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d f n^{2} \log {\left (F \right )}}{e} - \frac {2 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d f n \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{e} + 2 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b f n^{2} x \log {\left (F \right )} - 2 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b f n x \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )} + \frac {F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} d}{e} + F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} x & \text {for}\: e \neq 0 \\F^{f \left (a + b \log {\left (c d^{n} \right )}^{2}\right )} x & \text {otherwise} \end {cases} \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)),x)

[Out]

Piecewise((-2*F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*b*d*f*n**2*log(F)/e - 2*F**(a*f)*F**(b*f*log(c*(d + e*x
)**n)**2)*b*d*f*n*log(F)*log(c*(d + e*x)**n)/e + 2*F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*b*f*n**2*x*log(F)
- 2*F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*b*f*n*x*log(F)*log(c*(d + e*x)**n) + F**(a*f)*F**(b*f*log(c*(d +
e*x)**n)**2)*d/e + F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*x, Ne(e, 0)), (F**(f*(a + b*log(c*d**n)**2))*x, Tr
ue))

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Giac [A]
time = 5.12, size = 101, normalized size = 0.86 \begin {gather*} -\frac {\sqrt {\pi } F^{a f} \operatorname {erf}\left (-\sqrt {-b f \log \left (F\right )} n \log \left (x e + d\right ) - \sqrt {-b f \log \left (F\right )} \log \left (c\right ) - \frac {\sqrt {-b f \log \left (F\right )}}{2 \, b f n \log \left (F\right )}\right ) e^{\left (-\frac {1}{4 \, b f n^{2} \log \left (F\right )} - 1\right )}}{2 \, \sqrt {-b f \log \left (F\right )} c^{\left (\frac {1}{n}\right )} 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)),x, algorithm="giac")

[Out]

-1/2*sqrt(pi)*F^(a*f)*erf(-sqrt(-b*f*log(F))*n*log(x*e + d) - sqrt(-b*f*log(F))*log(c) - 1/2*sqrt(-b*f*log(F))
/(b*f*n*log(F)))*e^(-1/4/(b*f*n^2*log(F)) - 1)/(sqrt(-b*f*log(F))*c^(1/n)*n)

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

[Out]

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

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