3.44 \(\int \frac{(a+b (F^{g (e+f x)})^n)^3}{(c+d x)^2} \, dx\)

Optimal. Leaf size=305 \[ \frac{3 a^2 b f g n \log (F) \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac{c f}{d}\right )-g n (e+f x)} \text{Ei}\left (\frac{f g n (c+d x) \log (F)}{d}\right )}{d^2}-\frac{3 a^2 b \left (F^{e g+f g x}\right )^n}{d (c+d x)}-\frac{a^3}{d (c+d x)}+\frac{6 a b^2 f g n \log (F) \left (F^{e g+f g x}\right )^{2 n} F^{2 g n \left (e-\frac{c f}{d}\right )-2 g n (e+f x)} \text{Ei}\left (\frac{2 f g n (c+d x) \log (F)}{d}\right )}{d^2}-\frac{3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{d (c+d x)}+\frac{3 b^3 f g n \log (F) \left (F^{e g+f g x}\right )^{3 n} F^{3 g n \left (e-\frac{c f}{d}\right )-3 g n (e+f x)} \text{Ei}\left (\frac{3 f g n (c+d x) \log (F)}{d}\right )}{d^2}-\frac{b^3 \left (F^{e g+f g x}\right )^{3 n}}{d (c+d x)} \]

[Out]

-(a^3/(d*(c + d*x))) - (3*a^2*b*(F^(e*g + f*g*x))^n)/(d*(c + d*x)) - (3*a*b^2*(F^(e*g + f*g*x))^(2*n))/(d*(c +
d*x)) - (b^3*(F^(e*g + f*g*x))^(3*n))/(d*(c + d*x)) + (3*a^2*b*f*F^((e - (c*f)/d)*g*n - g*n*(e + f*x))*(F^(e*
g + f*g*x))^n*g*n*ExpIntegralEi[(f*g*n*(c + d*x)*Log[F])/d]*Log[F])/d^2 + (6*a*b^2*f*F^(2*(e - (c*f)/d)*g*n -
2*g*n*(e + f*x))*(F^(e*g + f*g*x))^(2*n)*g*n*ExpIntegralEi[(2*f*g*n*(c + d*x)*Log[F])/d]*Log[F])/d^2 + (3*b^3*
f*F^(3*(e - (c*f)/d)*g*n - 3*g*n*(e + f*x))*(F^(e*g + f*g*x))^(3*n)*g*n*ExpIntegralEi[(3*f*g*n*(c + d*x)*Log[F
])/d]*Log[F])/d^2

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Rubi [A]  time = 0.490896, antiderivative size = 305, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2183, 2177, 2182, 2178} \[ \frac{3 a^2 b f g n \log (F) \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac{c f}{d}\right )-g n (e+f x)} \text{Ei}\left (\frac{f g n (c+d x) \log (F)}{d}\right )}{d^2}-\frac{3 a^2 b \left (F^{e g+f g x}\right )^n}{d (c+d x)}-\frac{a^3}{d (c+d x)}+\frac{6 a b^2 f g n \log (F) \left (F^{e g+f g x}\right )^{2 n} F^{2 g n \left (e-\frac{c f}{d}\right )-2 g n (e+f x)} \text{Ei}\left (\frac{2 f g n (c+d x) \log (F)}{d}\right )}{d^2}-\frac{3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{d (c+d x)}+\frac{3 b^3 f g n \log (F) \left (F^{e g+f g x}\right )^{3 n} F^{3 g n \left (e-\frac{c f}{d}\right )-3 g n (e+f x)} \text{Ei}\left (\frac{3 f g n (c+d x) \log (F)}{d}\right )}{d^2}-\frac{b^3 \left (F^{e g+f g x}\right )^{3 n}}{d (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(F^(g*(e + f*x)))^n)^3/(c + d*x)^2,x]

[Out]

-(a^3/(d*(c + d*x))) - (3*a^2*b*(F^(e*g + f*g*x))^n)/(d*(c + d*x)) - (3*a*b^2*(F^(e*g + f*g*x))^(2*n))/(d*(c +
d*x)) - (b^3*(F^(e*g + f*g*x))^(3*n))/(d*(c + d*x)) + (3*a^2*b*f*F^((e - (c*f)/d)*g*n - g*n*(e + f*x))*(F^(e*
g + f*g*x))^n*g*n*ExpIntegralEi[(f*g*n*(c + d*x)*Log[F])/d]*Log[F])/d^2 + (6*a*b^2*f*F^(2*(e - (c*f)/d)*g*n -
2*g*n*(e + f*x))*(F^(e*g + f*g*x))^(2*n)*g*n*ExpIntegralEi[(2*f*g*n*(c + d*x)*Log[F])/d]*Log[F])/d^2 + (3*b^3*
f*F^(3*(e - (c*f)/d)*g*n - 3*g*n*(e + f*x))*(F^(e*g + f*g*x))^(3*n)*g*n*ExpIntegralEi[(3*f*g*n*(c + d*x)*Log[F
])/d]*Log[F])/d^2

Rule 2183

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> In
t[ExpandIntegrand[(c + d*x)^m, (a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n},
x] && IGtQ[p, 0]

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2182

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[(b*F^(g*(e +
f*x)))^n/F^(g*n*(e + f*x)), Int[(c + d*x)^m*F^(g*n*(e + f*x)), x], x] /; FreeQ[{F, b, c, d, e, f, g, m, n}, x]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rubi steps

\begin{align*} \int \frac{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3}{(c+d x)^2} \, dx &=\int \left (\frac{a^3}{(c+d x)^2}+\frac{3 a^2 b \left (F^{e g+f g x}\right )^n}{(c+d x)^2}+\frac{3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{(c+d x)^2}+\frac{b^3 \left (F^{e g+f g x}\right )^{3 n}}{(c+d x)^2}\right ) \, dx\\ &=-\frac{a^3}{d (c+d x)}+\left (3 a^2 b\right ) \int \frac{\left (F^{e g+f g x}\right )^n}{(c+d x)^2} \, dx+\left (3 a b^2\right ) \int \frac{\left (F^{e g+f g x}\right )^{2 n}}{(c+d x)^2} \, dx+b^3 \int \frac{\left (F^{e g+f g x}\right )^{3 n}}{(c+d x)^2} \, dx\\ &=-\frac{a^3}{d (c+d x)}-\frac{3 a^2 b \left (F^{e g+f g x}\right )^n}{d (c+d x)}-\frac{3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{d (c+d x)}-\frac{b^3 \left (F^{e g+f g x}\right )^{3 n}}{d (c+d x)}+\frac{\left (3 a^2 b f g n \log (F)\right ) \int \frac{\left (F^{e g+f g x}\right )^n}{c+d x} \, dx}{d}+\frac{\left (6 a b^2 f g n \log (F)\right ) \int \frac{\left (F^{e g+f g x}\right )^{2 n}}{c+d x} \, dx}{d}+\frac{\left (3 b^3 f g n \log (F)\right ) \int \frac{\left (F^{e g+f g x}\right )^{3 n}}{c+d x} \, dx}{d}\\ &=-\frac{a^3}{d (c+d x)}-\frac{3 a^2 b \left (F^{e g+f g x}\right )^n}{d (c+d x)}-\frac{3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{d (c+d x)}-\frac{b^3 \left (F^{e g+f g x}\right )^{3 n}}{d (c+d x)}+\frac{\left (3 a^2 b f F^{-n (e g+f g x)} \left (F^{e g+f g x}\right )^n g n \log (F)\right ) \int \frac{F^{n (e g+f g x)}}{c+d x} \, dx}{d}+\frac{\left (6 a b^2 f F^{-2 n (e g+f g x)} \left (F^{e g+f g x}\right )^{2 n} g n \log (F)\right ) \int \frac{F^{2 n (e g+f g x)}}{c+d x} \, dx}{d}+\frac{\left (3 b^3 f F^{-3 n (e g+f g x)} \left (F^{e g+f g x}\right )^{3 n} g n \log (F)\right ) \int \frac{F^{3 n (e g+f g x)}}{c+d x} \, dx}{d}\\ &=-\frac{a^3}{d (c+d x)}-\frac{3 a^2 b \left (F^{e g+f g x}\right )^n}{d (c+d x)}-\frac{3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{d (c+d x)}-\frac{b^3 \left (F^{e g+f g x}\right )^{3 n}}{d (c+d x)}+\frac{3 a^2 b f F^{\left (e-\frac{c f}{d}\right ) g n-g n (e+f x)} \left (F^{e g+f g x}\right )^n g n \text{Ei}\left (\frac{f g n (c+d x) \log (F)}{d}\right ) \log (F)}{d^2}+\frac{6 a b^2 f F^{2 \left (e-\frac{c f}{d}\right ) g n-2 g n (e+f x)} \left (F^{e g+f g x}\right )^{2 n} g n \text{Ei}\left (\frac{2 f g n (c+d x) \log (F)}{d}\right ) \log (F)}{d^2}+\frac{3 b^3 f F^{3 \left (e-\frac{c f}{d}\right ) g n-3 g n (e+f x)} \left (F^{e g+f g x}\right )^{3 n} g n \text{Ei}\left (\frac{3 f g n (c+d x) \log (F)}{d}\right ) \log (F)}{d^2}\\ \end{align*}

Mathematica [A]  time = 1.33123, size = 250, normalized size = 0.82 \[ -\frac{-3 a^2 b f g n \log (F) (c+d x) \left (F^{g (e+f x)}\right )^n F^{-\frac{f g n (c+d x)}{d}} \text{Ei}\left (\frac{f g n (c+d x) \log (F)}{d}\right )+3 a^2 b d \left (F^{g (e+f x)}\right )^n+a^3 d-6 a b^2 f g n \log (F) (c+d x) \left (F^{g (e+f x)}\right )^{2 n} F^{-\frac{2 f g n (c+d x)}{d}} \text{Ei}\left (\frac{2 f g n (c+d x) \log (F)}{d}\right )+3 a b^2 d \left (F^{g (e+f x)}\right )^{2 n}-3 b^3 f g n \log (F) (c+d x) \left (F^{g (e+f x)}\right )^{3 n} F^{-\frac{3 f g n (c+d x)}{d}} \text{Ei}\left (\frac{3 f g n (c+d x) \log (F)}{d}\right )+b^3 d \left (F^{g (e+f x)}\right )^{3 n}}{d^2 (c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(F^(g*(e + f*x)))^n)^3/(c + d*x)^2,x]

[Out]

-((a^3*d + 3*a^2*b*d*(F^(g*(e + f*x)))^n + 3*a*b^2*d*(F^(g*(e + f*x)))^(2*n) + b^3*d*(F^(g*(e + f*x)))^(3*n) -
(3*a^2*b*f*(F^(g*(e + f*x)))^n*g*n*(c + d*x)*ExpIntegralEi[(f*g*n*(c + d*x)*Log[F])/d]*Log[F])/F^((f*g*n*(c +
d*x))/d) - (6*a*b^2*f*(F^(g*(e + f*x)))^(2*n)*g*n*(c + d*x)*ExpIntegralEi[(2*f*g*n*(c + d*x)*Log[F])/d]*Log[F
])/F^((2*f*g*n*(c + d*x))/d) - (3*b^3*f*(F^(g*(e + f*x)))^(3*n)*g*n*(c + d*x)*ExpIntegralEi[(3*f*g*n*(c + d*x)
*Log[F])/d]*Log[F])/F^((3*f*g*n*(c + d*x))/d))/(d^2*(c + d*x)))

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) ^{3}}{ \left ( dx+c \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*(F^(g*(f*x+e)))^n)^3/(d*x+c)^2,x)

[Out]

int((a+b*(F^(g*(f*x+e)))^n)^3/(d*x+c)^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\left (F^{e g}\right )}^{3 \, n} b^{3} \int \frac{{\left (F^{f g x}\right )}^{3 \, n}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\,{d x} + 3 \,{\left (F^{e g}\right )}^{2 \, n} a b^{2} \int \frac{{\left (F^{f g x}\right )}^{2 \, n}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\,{d x} + 3 \,{\left (F^{e g}\right )}^{n} a^{2} b \int \frac{{\left (F^{f g x}\right )}^{n}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\,{d x} - \frac{a^{3}}{d^{2} x + c d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F^(g*(f*x+e)))^n)^3/(d*x+c)^2,x, algorithm="maxima")

[Out]

(F^(e*g))^(3*n)*b^3*integrate((F^(f*g*x))^(3*n)/(d^2*x^2 + 2*c*d*x + c^2), x) + 3*(F^(e*g))^(2*n)*a*b^2*integr
ate((F^(f*g*x))^(2*n)/(d^2*x^2 + 2*c*d*x + c^2), x) + 3*(F^(e*g))^n*a^2*b*integrate((F^(f*g*x))^n/(d^2*x^2 + 2
*c*d*x + c^2), x) - a^3/(d^2*x + c*d)

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Fricas [A]  time = 1.56975, size = 589, normalized size = 1.93 \begin{align*} -\frac{3 \, F^{f g n x + e g n} a^{2} b d + 3 \, F^{2 \, f g n x + 2 \, e g n} a b^{2} d + F^{3 \, f g n x + 3 \, e g n} b^{3} d + a^{3} d - 3 \,{\left (b^{3} d f g n x + b^{3} c f g n\right )} F^{\frac{3 \,{\left (d e - c f\right )} g n}{d}}{\rm Ei}\left (\frac{3 \,{\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) \log \left (F\right ) - 6 \,{\left (a b^{2} d f g n x + a b^{2} c f g n\right )} F^{\frac{2 \,{\left (d e - c f\right )} g n}{d}}{\rm Ei}\left (\frac{2 \,{\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) \log \left (F\right ) - 3 \,{\left (a^{2} b d f g n x + a^{2} b c f g n\right )} F^{\frac{{\left (d e - c f\right )} g n}{d}}{\rm Ei}\left (\frac{{\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) \log \left (F\right )}{d^{3} x + c d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F^(g*(f*x+e)))^n)^3/(d*x+c)^2,x, algorithm="fricas")

[Out]

-(3*F^(f*g*n*x + e*g*n)*a^2*b*d + 3*F^(2*f*g*n*x + 2*e*g*n)*a*b^2*d + F^(3*f*g*n*x + 3*e*g*n)*b^3*d + a^3*d -
3*(b^3*d*f*g*n*x + b^3*c*f*g*n)*F^(3*(d*e - c*f)*g*n/d)*Ei(3*(d*f*g*n*x + c*f*g*n)*log(F)/d)*log(F) - 6*(a*b^2
*d*f*g*n*x + a*b^2*c*f*g*n)*F^(2*(d*e - c*f)*g*n/d)*Ei(2*(d*f*g*n*x + c*f*g*n)*log(F)/d)*log(F) - 3*(a^2*b*d*f
*g*n*x + a^2*b*c*f*g*n)*F^((d*e - c*f)*g*n/d)*Ei((d*f*g*n*x + c*f*g*n)*log(F)/d)*log(F))/(d^3*x + c*d^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F**(g*(f*x+e)))**n)**3/(d*x+c)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left ({\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a\right )}^{3}}{{\left (d x + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F^(g*(f*x+e)))^n)^3/(d*x+c)^2,x, algorithm="giac")

[Out]

integrate(((F^((f*x + e)*g))^n*b + a)^3/(d*x + c)^2, x)