3.1.44 \(\int \frac {(a+b (F^{g (e+f x)})^n)^3}{(c+d x)^2} \, dx\) [44]

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

[Out]

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

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Rubi [A]
time = 0.35, antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2214, 2208, 2213, 2209} \begin {gather*} -\frac {a^3}{d (c+d x)}+\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 {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)} \end {gather*}

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 2208

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] &&  !TrueQ[$UseGamm
a]

Rule 2209

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

Rule 2213

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 2214

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]

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*}

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Mathematica [A]
time = 1.03, size = 250, normalized size = 0.82 \begin {gather*} -\frac {a^3 d+3 a^2 b d \left (F^{g (e+f x)}\right )^n+3 a b^2 d \left (F^{g (e+f x)}\right )^{2 n}+b^3 d \left (F^{g (e+f x)}\right )^{3 n}-3 a^2 b f F^{-\frac {f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^n g n (c+d x) \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right ) \log (F)-6 a b^2 f F^{-\frac {2 f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^{2 n} g n (c+d x) \text {Ei}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right ) \log (F)-3 b^3 f F^{-\frac {3 f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^{3 n} g n (c+d x) \text {Ei}\left (\frac {3 f g n (c+d x) \log (F)}{d}\right ) \log (F)}{d^2 (c+d x)} \end {gather*}

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 = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )^{3}}{\left (d x +c \right )^{2}}\, dx\]

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 = 0.00, 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((a+b*(F^(g*(f*x+e)))^n)^3/(d*x+c)^2,x, algorithm="maxima")

[Out]

F^(3*g*n*e)*b^3*integrate(F^(3*f*g*n*x)/(d^2*x^2 + 2*c*d*x + c^2), x) + 3*F^(2*g*n*e)*a*b^2*integrate(F^(2*f*g
*n*x)/(d^2*x^2 + 2*c*d*x + c^2), x) + 3*F^(g*n*e)*a^2*b*integrate(F^(f*g*n*x)/(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 = 0.40, size = 277, normalized size = 0.91 \begin {gather*} -\frac {3 \, F^{f g n x + g n e} a^{2} b d + 3 \, F^{2 \, f g n x + 2 \, g n e} a b^{2} d + F^{3 \, f g n x + 3 \, g n e} b^{3} d + a^{3} d - \frac {3 \, {\left (b^{3} d f g n x + b^{3} c f g n\right )} {\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 )}{F^{\frac {3 \, {\left (c f g n - d g n e\right )}}{d}}} - \frac {6 \, {\left (a b^{2} d f g n x + a b^{2} c f g n\right )} {\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 )}{F^{\frac {2 \, {\left (c f g n - d g n e\right )}}{d}}} - \frac {3 \, {\left (a^{2} b d f g n x + a^{2} b c f g n\right )} {\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 )}{F^{\frac {c f g n - d g n e}{d}}}}{d^{3} x + c d^{2}} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \left (F^{e g} F^{f g x}\right )^{n}\right )^{3}}{\left (c + d x\right )^{2}}\, dx \end {gather*}

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]

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

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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((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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,{\left (F^{g\,\left (e+f\,x\right )}\right )}^n\right )}^3}{{\left (c+d\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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