3.6.97 \(\int F^{f (a+b \log ^2(c (d+e x)^n))} (g+h x) \, dx\) [597]

Optimal. Leaf size=242 \[ \frac {e^{-\frac {1}{b f n^2 \log (F)}} F^{a f} h \sqrt {\pi } (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {erfi}\left (\frac {1+b f n \log (F) \log \left (c (d+e x)^n\right )}{\sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e^2 \sqrt {f} n \sqrt {\log (F)}}+\frac {e^{-\frac {1}{4 b f n^2 \log (F)}} F^{a f} (e g-d h) \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {1+2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e^2 \sqrt {f} n \sqrt {\log (F)}} \]

[Out]

1/2*F^(a*f)*h*(e*x+d)^2*erfi((1+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^2/exp(1
/b/f/n^2/ln(F))/n/((c*(e*x+d)^n)^(2/n))/b^(1/2)/f^(1/2)/ln(F)^(1/2)+1/2*F^(a*f)*(-d*h+e*g)*(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^2/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 = 0.25, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2309, 2307, 2266, 2235, 2308} \begin {gather*} \frac {\sqrt {\pi } F^{a f} (d+e x) (e g-d h) 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^2 \sqrt {f} n \sqrt {\log (F)}}+\frac {\sqrt {\pi } h F^{a f} (d+e x)^2 e^{-\frac {1}{b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-2/n} \text {Erfi}\left (\frac {b f n \log (F) \log \left (c (d+e x)^n\right )+1}{\sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e^2 \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))*(g + h*x),x]

[Out]

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

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 2309

Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]^2*(b_.))*(f_.))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol]
 :> Dist[1/e^(m + 1), Subst[Int[ExpandIntegrand[F^(f*(a + b*Log[c*x^n]^2)), (e*g - d*h + h*x)^m, x], x], x, d
+ e*x], x] /; FreeQ[{F, a, b, c, d, e, f, g, h, n}, x] && IGtQ[m, 0]

Rubi steps

\begin {align*} \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (g+h x) \, dx &=\int \left (F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} g+F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} h x\right ) \, dx\\ &=g \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} \, dx+h \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} x \, dx\\ &=\frac {g \text {Subst}\left (\int F^{f \left (a+b \log ^2\left (c x^n\right )\right )} \, dx,x,d+e x\right )}{e}+h \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} x \, dx\\ &=h \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} x \, dx+\frac {\left (g (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}\\ &=h \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} x \, dx+\frac {\left (e^{-\frac {1}{4 b f n^2 \log (F)}} F^{a f} g (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} g \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)}}+h \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} x \, dx\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 204, normalized size = 0.84 \begin {gather*} \frac {e^{-\frac {1}{b f n^2 \log (F)}} F^{a f} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-2/n} \left (h (d+e x) \text {erfi}\left (\frac {1+b f n \log (F) \log \left (c (d+e x)^n\right )}{\sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )+e^{\frac {3}{4 b f n^2 \log (F)}} (e g-d h) \left (c (d+e x)^n\right )^{\frac {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 )\right )}{2 \sqrt {b} e^2 \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))*(g + h*x),x]

[Out]

(F^(a*f)*Sqrt[Pi]*(d + e*x)*(h*(d + e*x)*Erfi[(1 + b*f*n*Log[F]*Log[c*(d + e*x)^n])/(Sqrt[b]*Sqrt[f]*n*Sqrt[Lo
g[F]])] + E^(3/(4*b*f*n^2*Log[F]))*(e*g - d*h)*(c*(d + e*x)^n)^n^(-1)*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^2*E^(1/(b*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 = 0.03, size = 0, normalized size = 0.00 \[\int F^{f \left (a +b \ln \left (c \left (e x +d \right )^{n}\right )^{2}\right )} \left (h x +g \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 234, normalized size = 0.97 \begin {gather*} \frac {{\left (\sqrt {\pi } \sqrt {-b f n^{2} \log \left (F\right )} {\left (d h - g e\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 )}\right )} - \sqrt {\pi } \sqrt {-b f n^{2} \log \left (F\right )} h \operatorname {erf}\left (\frac {{\left (b f n^{2} \log \left (x e + d\right ) \log \left (F\right ) + b f n \log \left (F\right ) \log \left (c\right ) + 1\right )} \sqrt {-b f n^{2} \log \left (F\right )}}{b f n^{2} \log \left (F\right )}\right ) e^{\left (\frac {a b f^{2} n^{2} \log \left (F\right )^{2} - 2 \, b f n \log \left (F\right ) \log \left (c\right ) - 1}{b f n^{2} \log \left (F\right )}\right )}\right )} e^{\left (-2\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))*(h*x+g),x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 581 vs. \(2 (223) = 446\).
time = 27.69, size = 581, normalized size = 2.40 \begin {gather*} \begin {cases} \frac {3 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d^{2} f h n^{2} \log {\left (F \right )}}{2 e^{2}} + \frac {3 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d^{2} f h n \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{2 e^{2}} - \frac {2 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d f g 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 g n \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{e} - \frac {3 F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d f h n^{2} x \log {\left (F \right )}}{2 e} + \frac {F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d f h n x \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 g 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 g 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}} b f h n^{2} x^{2} \log {\left (F \right )}}{4} - \frac {F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b f h n x^{2} \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{2} - \frac {F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} d^{2} h}{2 e^{2}} + \frac {F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} d g}{e} + F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} g x + \frac {F^{a f} F^{b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} h x^{2}}{2} & \text {for}\: e \neq 0 \\F^{f \left (a + b \log {\left (c d^{n} \right )}^{2}\right )} \left (g x + \frac {h x^{2}}{2}\right ) & \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))*(h*x+g),x)

[Out]

Piecewise((3*F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*b*d**2*f*h*n**2*log(F)/(2*e**2) + 3*F**(a*f)*F**(b*f*log
(c*(d + e*x)**n)**2)*b*d**2*f*h*n*log(F)*log(c*(d + e*x)**n)/(2*e**2) - 2*F**(a*f)*F**(b*f*log(c*(d + e*x)**n)
**2)*b*d*f*g*n**2*log(F)/e - 2*F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*b*d*f*g*n*log(F)*log(c*(d + e*x)**n)/e
 - 3*F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*b*d*f*h*n**2*x*log(F)/(2*e) + F**(a*f)*F**(b*f*log(c*(d + e*x)**
n)**2)*b*d*f*h*n*x*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*g*n**2*x*log(
F) - 2*F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*b*f*g*n*x*log(F)*log(c*(d + e*x)**n) + F**(a*f)*F**(b*f*log(c*
(d + e*x)**n)**2)*b*f*h*n**2*x**2*log(F)/4 - F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*b*f*h*n*x**2*log(F)*log(
c*(d + e*x)**n)/2 - F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*d**2*h/(2*e**2) + F**(a*f)*F**(b*f*log(c*(d + e*x
)**n)**2)*d*g/e + F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*g*x + F**(a*f)*F**(b*f*log(c*(d + e*x)**n)**2)*h*x*
*2/2, Ne(e, 0)), (F**(f*(a + b*log(c*d**n)**2))*(g*x + h*x**2/2), 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))*(h*x+g),x, algorithm="giac")

[Out]

integrate((h*x + g)*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 = -0.00 \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 (g+h\,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))*(g + h*x),x)

[Out]

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

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