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

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

[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 [F]  time = 0.23, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} (g+h x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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 \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 0.39, size = 204, normalized size = 0.84 \[ \frac {\sqrt {\pi } F^{a f} (d+e x) e^{-\frac {1}{b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-2/n} \left ((e g-d h) e^{\frac {3}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{\frac {1}{n}} \text {erfi}\left (\frac {2 b f n \log (F) \log \left (c (d+e x)^n\right )+1}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )+h (d+e x) \text {erfi}\left (\frac {b f n \log (F) \log \left (c (d+e x)^n\right )+1}{\sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )\right )}{2 \sqrt {b} e^2 \sqrt {f} n \sqrt {\log (F)}} \]

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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fricas [A]  time = 0.41, size = 231, normalized size = 0.95 \[ -\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \relax (F)} {\left (e g - d h\right )} \operatorname {erf}\left (\frac {{\left (2 \, b f n^{2} \log \left (e x + d\right ) \log \relax (F) + 2 \, b f n \log \relax (F) \log \relax (c) + 1\right )} \sqrt {-b f n^{2} \log \relax (F)}}{2 \, b f n^{2} \log \relax (F)}\right ) e^{\left (\frac {4 \, a b f^{2} n^{2} \log \relax (F)^{2} - 4 \, b f n \log \relax (F) \log \relax (c) - 1}{4 \, b f n^{2} \log \relax (F)}\right )} + \sqrt {\pi } \sqrt {-b f n^{2} \log \relax (F)} h \operatorname {erf}\left (\frac {{\left (b f n^{2} \log \left (e x + d\right ) \log \relax (F) + b f n \log \relax (F) \log \relax (c) + 1\right )} \sqrt {-b f n^{2} \log \relax (F)}}{b f n^{2} \log \relax (F)}\right ) e^{\left (\frac {a b f^{2} n^{2} \log \relax (F)^{2} - 2 \, b f n \log \relax (F) \log \relax (c) - 1}{b f n^{2} \log \relax (F)}\right )}}{2 \, e^{2} n} \]

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))*(e*g - d*h)*erf(1/2*(2*b*f*n^2*log(e*x + 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(e*x + d)*log(F) + b*f*n*log(F)*log(c) + 1)*
sqrt(-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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (h x + g\right )} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + a\right )} f}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.67, size = 0, normalized size = 0.00 \[ \int \left (h x +g \right ) F^{\left (b \ln \left (c \left (e x +d \right )^{n}\right )^{2}+a \right ) f}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(f*(a + b*log(c*(d + e*x)^n)^2))*(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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sympy [A]  time = 113.97, size = 1329, normalized size = 5.49 \[ \text {result too large to display} \]

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)**2)*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*b*d**2
*f*h*n**2*log(F)*log(d + e*x)/(2*e**2) + 3*F**(a*f)*F**(b*f*log(c)**2)*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f
*n*log(c)*log(d + e*x))*b*d**2*f*h*n**2*log(F)/(2*e**2) + 3*F**(a*f)*F**(b*f*log(c)**2)*F**(b*f*n**2*log(d + e
*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*b*d**2*f*h*n*log(F)*log(c)/(2*e**2) - 2*F**(a*f)*F**(b*f*log(c)**2)*F
**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*b*d*f*g*n**2*log(F)*log(d + e*x)/e - 2*F**(a*f)*
F**(b*f*log(c)**2)*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*b*d*f*g*n**2*log(F)/e - 2*F*
*(a*f)*F**(b*f*log(c)**2)*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*b*d*f*g*n*log(F)*log(
c)/e + F**(a*f)*F**(b*f*log(c)**2)*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*b*d*f*h*n**2
*x*log(F)*log(d + e*x)/e - 3*F**(a*f)*F**(b*f*log(c)**2)*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(
d + e*x))*b*d*f*h*n**2*x*log(F)/(2*e) + F**(a*f)*F**(b*f*log(c)**2)*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*
log(c)*log(d + e*x))*b*d*f*h*n*x*log(F)*log(c)/e - 2*F**(a*f)*F**(b*f*log(c)**2)*F**(b*f*n**2*log(d + e*x)**2)
*F**(2*b*f*n*log(c)*log(d + e*x))*b*f*g*n**2*x*log(F)*log(d + e*x) + 2*F**(a*f)*F**(b*f*log(c)**2)*F**(b*f*n**
2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*b*f*g*n**2*x*log(F) - 2*F**(a*f)*F**(b*f*log(c)**2)*F**(b*
f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*b*f*g*n*x*log(F)*log(c) - F**(a*f)*F**(b*f*log(c)**2)
*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*b*f*h*n**2*x**2*log(F)*log(d + e*x)/2 + F**(a*
f)*F**(b*f*log(c)**2)*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*b*f*h*n**2*x**2*log(F)/4
- F**(a*f)*F**(b*f*log(c)**2)*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*b*f*h*n*x**2*log(
F)*log(c)/2 - F**(a*f)*F**(b*f*log(c)**2)*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*d**2*
h/(2*e**2) + F**(a*f)*F**(b*f*log(c)**2)*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*d*g/e
+ F**(a*f)*F**(b*f*log(c)**2)*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*g*x + F**(a*f)*F*
*(b*f*log(c)**2)*F**(b*f*n**2*log(d + e*x)**2)*F**(2*b*f*n*log(c)*log(d + e*x))*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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