3.6.88 \(\int F^{f (a+b \log ^2(c (d+e x)^n))} (d g+e g x)^2 \, dx\) [588]

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

[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 = 0.09, antiderivative size = 123, 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 } 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)}} \end {gather*}

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 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)^2 \, dx &=\frac {\text {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 \text {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 ) \text {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 ) \text {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 = 0.19, size = 123, normalized size = 1.00 \begin {gather*} \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 {gather*}

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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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int F^{f \left (a +b \ln \left (c \left (e x +d \right )^{n}\right )^{2}\right )} \left (e g x +d g \right )^{2}\, dx\]

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)^2,x)

[Out]

int(F^(f*(a+b*ln(c*(e*x+d)^n)^2))*(e*g*x+d*g)^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))*(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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Fricas [A]
time = 0.36, size = 119, normalized size = 0.97 \begin {gather*} -\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \left (F\right )} g^{2} \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 ) + 3\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} - 12 \, b f n \log \left (F\right ) \log \left (c\right ) - 9}{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)^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(x*e + 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)) - 1)/n

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 546 vs. \(2 (110) = 220\).
time = 114.99, size = 546, normalized size = 4.44 \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^{3} f g^{2} n^{2} \log {\left (F \right )}}{9 e} - \frac {2 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d^{3} f g^{2} n \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{9 e} + \frac {2 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d^{2} f g^{2} n^{2} x \log {\left (F \right )}}{9} - \frac {2 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d^{2} f g^{2} n x \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{3} + \frac {2 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d e f g^{2} n^{2} x^{2} \log {\left (F \right )}}{9} - \frac {2 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d e f g^{2} n x^{2} \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{3} + \frac {2 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b e^{2} f g^{2} n^{2} x^{3} \log {\left (F \right )}}{27} - \frac {2 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b e^{2} f g^{2} n x^{3} \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{9} + \frac {F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} d^{3} g^{2}}{3 e} + F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} d^{2} g^{2} x + F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} d e g^{2} x^{2} + \frac {F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} e^{2} g^{2} x^{3}}{3} & \text {for}\: e \neq 0 \\F^{f \left (a + b \log {\left (c d^{n} \right )}^{2}\right )} d^{2} g^{2} 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))*(e*g*x+d*g)**2,x)

[Out]

Piecewise((-2*F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*b*d**3*f*g**2*n**2*log(F)/(9*e) - 2*F**(a*f)*F**(b*f*lo
g(c*(d + e*x)**n)**2)*b*d**3*f*g**2*n*log(F)*log(c*(d + e*x)**n)/(9*e) + 2*F**(a*f)*F**(b*f*log(c*(d + e*x)**n
)**2)*b*d**2*f*g**2*n**2*x*log(F)/9 - 2*F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*b*d**2*f*g**2*n*x*log(F)*log(
c*(d + e*x)**n)/3 + 2*F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*b*d*e*f*g**2*n**2*x**2*log(F)/9 - 2*F**(a*f)*F*
*(b*f*log(c*(d + e*x)**n)**2)*b*d*e*f*g**2*n*x**2*log(F)*log(c*(d + e*x)**n)/3 + 2*F**(a*f)*F**(b*f*log(c*(d +
e*x)**n)**2)*b*e**2*f*g**2*n**2*x**3*log(F)/27 - 2*F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*b*e**2*f*g**2*n*x
**3*log(F)*log(c*(d + e*x)**n)/9 + F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*d**3*g**2/(3*e) + F**(a*f)*F**(b*f
*log(c*(d + e*x)**n)**2)*d**2*g**2*x + F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*d*e*g**2*x**2 + F**(a*f)*F**(b
*f*log(c*(d + e*x)**n)**2)*e**2*g**2*x**3/3, Ne(e, 0)), (F**(f*(a + b*log(c*d**n)**2))*d**2*g**2*x, True))

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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)^2,x, algorithm="giac")

[Out]

integrate((g*x*e + d*g)^2*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 )}^2 \,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)^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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