3.1.45 \(\int \frac {(a+b (F^{g (e+f x)})^n)^3}{(c+d x)^3} \, dx\) [45]

Optimal. Leaf size=447 \[ -\frac {a^3}{2 d (c+d x)^2}-\frac {3 a^2 b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2}-\frac {3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{2 d (c+d x)^2}-\frac {b^3 \left (F^{e g+f g x}\right )^{3 n}}{2 d (c+d x)^2}-\frac {3 a^2 b f \left (F^{e g+f g x}\right )^n g n \log (F)}{2 d^2 (c+d x)}-\frac {3 a b^2 f \left (F^{e g+f g x}\right )^{2 n} g n \log (F)}{d^2 (c+d x)}-\frac {3 b^3 f \left (F^{e g+f g x}\right )^{3 n} g n \log (F)}{2 d^2 (c+d x)}+\frac {3 a^2 b f^2 F^{\left (e-\frac {c f}{d}\right ) g n-g n (e+f x)} \left (F^{e g+f g x}\right )^n g^2 n^2 \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)}{2 d^3}+\frac {6 a b^2 f^2 F^{2 \left (e-\frac {c f}{d}\right ) g n-2 g n (e+f x)} \left (F^{e g+f g x}\right )^{2 n} g^2 n^2 \text {Ei}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)}{d^3}+\frac {9 b^3 f^2 F^{3 \left (e-\frac {c f}{d}\right ) g n-3 g n (e+f x)} \left (F^{e g+f g x}\right )^{3 n} g^2 n^2 \text {Ei}\left (\frac {3 f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)}{2 d^3} \]

[Out]

-1/2*a^3/d/(d*x+c)^2-3/2*a^2*b*(F^(f*g*x+e*g))^n/d/(d*x+c)^2-3/2*a*b^2*(F^(f*g*x+e*g))^(2*n)/d/(d*x+c)^2-1/2*b
^3*(F^(f*g*x+e*g))^(3*n)/d/(d*x+c)^2-3/2*a^2*b*f*(F^(f*g*x+e*g))^n*g*n*ln(F)/d^2/(d*x+c)-3*a*b^2*f*(F^(f*g*x+e
*g))^(2*n)*g*n*ln(F)/d^2/(d*x+c)-3/2*b^3*f*(F^(f*g*x+e*g))^(3*n)*g*n*ln(F)/d^2/(d*x+c)+3/2*a^2*b*f^2*F^((e-c*f
/d)*g*n-g*n*(f*x+e))*(F^(f*g*x+e*g))^n*g^2*n^2*Ei(f*g*n*(d*x+c)*ln(F)/d)*ln(F)^2/d^3+6*a*b^2*f^2*F^(2*(e-c*f/d
)*g*n-2*g*n*(f*x+e))*(F^(f*g*x+e*g))^(2*n)*g^2*n^2*Ei(2*f*g*n*(d*x+c)*ln(F)/d)*ln(F)^2/d^3+9/2*b^3*f^2*F^(3*(e
-c*f/d)*g*n-3*g*n*(f*x+e))*(F^(f*g*x+e*g))^(3*n)*g^2*n^2*Ei(3*f*g*n*(d*x+c)*ln(F)/d)*ln(F)^2/d^3

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Rubi [A]
time = 0.49, antiderivative size = 447, normalized size of antiderivative = 1.00, number of steps used = 14, 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}{2 d (c+d x)^2}+\frac {3 a^2 b f^2 g^2 n^2 \log ^2(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 )}{2 d^3}-\frac {3 a^2 b f g n \log (F) \left (F^{e g+f g x}\right )^n}{2 d^2 (c+d x)}-\frac {3 a^2 b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2}+\frac {6 a b^2 f^2 g^2 n^2 \log ^2(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^3}-\frac {3 a b^2 f g n \log (F) \left (F^{e g+f g x}\right )^{2 n}}{d^2 (c+d x)}-\frac {3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{2 d (c+d x)^2}+\frac {9 b^3 f^2 g^2 n^2 \log ^2(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 )}{2 d^3}-\frac {3 b^3 f g n \log (F) \left (F^{e g+f g x}\right )^{3 n}}{2 d^2 (c+d x)}-\frac {b^3 \left (F^{e g+f g x}\right )^{3 n}}{2 d (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*a^3/(d*(c + d*x)^2) - (3*a^2*b*(F^(e*g + f*g*x))^n)/(2*d*(c + d*x)^2) - (3*a*b^2*(F^(e*g + f*g*x))^(2*n))
/(2*d*(c + d*x)^2) - (b^3*(F^(e*g + f*g*x))^(3*n))/(2*d*(c + d*x)^2) - (3*a^2*b*f*(F^(e*g + f*g*x))^n*g*n*Log[
F])/(2*d^2*(c + d*x)) - (3*a*b^2*f*(F^(e*g + f*g*x))^(2*n)*g*n*Log[F])/(d^2*(c + d*x)) - (3*b^3*f*(F^(e*g + f*
g*x))^(3*n)*g*n*Log[F])/(2*d^2*(c + d*x)) + (3*a^2*b*f^2*F^((e - (c*f)/d)*g*n - g*n*(e + f*x))*(F^(e*g + f*g*x
))^n*g^2*n^2*ExpIntegralEi[(f*g*n*(c + d*x)*Log[F])/d]*Log[F]^2)/(2*d^3) + (6*a*b^2*f^2*F^(2*(e - (c*f)/d)*g*n
- 2*g*n*(e + f*x))*(F^(e*g + f*g*x))^(2*n)*g^2*n^2*ExpIntegralEi[(2*f*g*n*(c + d*x)*Log[F])/d]*Log[F]^2)/d^3
+ (9*b^3*f^2*F^(3*(e - (c*f)/d)*g*n - 3*g*n*(e + f*x))*(F^(e*g + f*g*x))^(3*n)*g^2*n^2*ExpIntegralEi[(3*f*g*n*
(c + d*x)*Log[F])/d]*Log[F]^2)/(2*d^3)

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)^3} \, dx &=\int \left (\frac {a^3}{(c+d x)^3}+\frac {3 a^2 b \left (F^{e g+f g x}\right )^n}{(c+d x)^3}+\frac {3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{(c+d x)^3}+\frac {b^3 \left (F^{e g+f g x}\right )^{3 n}}{(c+d x)^3}\right ) \, dx\\ &=-\frac {a^3}{2 d (c+d x)^2}+\left (3 a^2 b\right ) \int \frac {\left (F^{e g+f g x}\right )^n}{(c+d x)^3} \, dx+\left (3 a b^2\right ) \int \frac {\left (F^{e g+f g x}\right )^{2 n}}{(c+d x)^3} \, dx+b^3 \int \frac {\left (F^{e g+f g x}\right )^{3 n}}{(c+d x)^3} \, dx\\ &=-\frac {a^3}{2 d (c+d x)^2}-\frac {3 a^2 b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2}-\frac {3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{2 d (c+d x)^2}-\frac {b^3 \left (F^{e g+f g x}\right )^{3 n}}{2 d (c+d x)^2}+\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)^2} \, dx}{2 d}+\frac {\left (3 a b^2 f g n \log (F)\right ) \int \frac {\left (F^{e g+f g x}\right )^{2 n}}{(c+d x)^2} \, 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)^2} \, dx}{2 d}\\ &=-\frac {a^3}{2 d (c+d x)^2}-\frac {3 a^2 b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2}-\frac {3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{2 d (c+d x)^2}-\frac {b^3 \left (F^{e g+f g x}\right )^{3 n}}{2 d (c+d x)^2}-\frac {3 a^2 b f \left (F^{e g+f g x}\right )^n g n \log (F)}{2 d^2 (c+d x)}-\frac {3 a b^2 f \left (F^{e g+f g x}\right )^{2 n} g n \log (F)}{d^2 (c+d x)}-\frac {3 b^3 f \left (F^{e g+f g x}\right )^{3 n} g n \log (F)}{2 d^2 (c+d x)}+\frac {\left (3 a^2 b f^2 g^2 n^2 \log ^2(F)\right ) \int \frac {\left (F^{e g+f g x}\right )^n}{c+d x} \, dx}{2 d^2}+\frac {\left (6 a b^2 f^2 g^2 n^2 \log ^2(F)\right ) \int \frac {\left (F^{e g+f g x}\right )^{2 n}}{c+d x} \, dx}{d^2}+\frac {\left (9 b^3 f^2 g^2 n^2 \log ^2(F)\right ) \int \frac {\left (F^{e g+f g x}\right )^{3 n}}{c+d x} \, dx}{2 d^2}\\ &=-\frac {a^3}{2 d (c+d x)^2}-\frac {3 a^2 b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2}-\frac {3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{2 d (c+d x)^2}-\frac {b^3 \left (F^{e g+f g x}\right )^{3 n}}{2 d (c+d x)^2}-\frac {3 a^2 b f \left (F^{e g+f g x}\right )^n g n \log (F)}{2 d^2 (c+d x)}-\frac {3 a b^2 f \left (F^{e g+f g x}\right )^{2 n} g n \log (F)}{d^2 (c+d x)}-\frac {3 b^3 f \left (F^{e g+f g x}\right )^{3 n} g n \log (F)}{2 d^2 (c+d x)}+\frac {\left (3 a^2 b f^2 F^{-n (e g+f g x)} \left (F^{e g+f g x}\right )^n g^2 n^2 \log ^2(F)\right ) \int \frac {F^{n (e g+f g x)}}{c+d x} \, dx}{2 d^2}+\frac {\left (6 a b^2 f^2 F^{-2 n (e g+f g x)} \left (F^{e g+f g x}\right )^{2 n} g^2 n^2 \log ^2(F)\right ) \int \frac {F^{2 n (e g+f g x)}}{c+d x} \, dx}{d^2}+\frac {\left (9 b^3 f^2 F^{-3 n (e g+f g x)} \left (F^{e g+f g x}\right )^{3 n} g^2 n^2 \log ^2(F)\right ) \int \frac {F^{3 n (e g+f g x)}}{c+d x} \, dx}{2 d^2}\\ &=-\frac {a^3}{2 d (c+d x)^2}-\frac {3 a^2 b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2}-\frac {3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{2 d (c+d x)^2}-\frac {b^3 \left (F^{e g+f g x}\right )^{3 n}}{2 d (c+d x)^2}-\frac {3 a^2 b f \left (F^{e g+f g x}\right )^n g n \log (F)}{2 d^2 (c+d x)}-\frac {3 a b^2 f \left (F^{e g+f g x}\right )^{2 n} g n \log (F)}{d^2 (c+d x)}-\frac {3 b^3 f \left (F^{e g+f g x}\right )^{3 n} g n \log (F)}{2 d^2 (c+d x)}+\frac {3 a^2 b f^2 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^2 n^2 \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)}{2 d^3}+\frac {6 a b^2 f^2 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^2 n^2 \text {Ei}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)}{d^3}+\frac {9 b^3 f^2 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^2 n^2 \text {Ei}\left (\frac {3 f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)}{2 d^3}\\ \end {align*}

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Mathematica [A]
time = 0.72, size = 325, normalized size = 0.73 \begin {gather*} -\frac {a^3 d^2-3 a^2 b f^2 F^{-\frac {f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^n g^2 n^2 (c+d x)^2 \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)-12 a b^2 f^2 F^{-\frac {2 f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^{2 n} g^2 n^2 (c+d x)^2 \text {Ei}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)-9 b^3 f^2 F^{-\frac {3 f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^{3 n} g^2 n^2 (c+d x)^2 \text {Ei}\left (\frac {3 f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)+3 a^2 b d \left (F^{g (e+f x)}\right )^n (d+f g n (c+d x) \log (F))+3 a b^2 d \left (F^{g (e+f x)}\right )^{2 n} (d+2 f g n (c+d x) \log (F))+b^3 d \left (F^{g (e+f x)}\right )^{3 n} (d+3 f g n (c+d x) \log (F))}{2 d^3 (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a^3*d^2 - (3*a^2*b*f^2*(F^(g*(e + f*x)))^n*g^2*n^2*(c + d*x)^2*ExpIntegralEi[(f*g*n*(c + d*x)*Log[F])/d]
*Log[F]^2)/F^((f*g*n*(c + d*x))/d) - (12*a*b^2*f^2*(F^(g*(e + f*x)))^(2*n)*g^2*n^2*(c + d*x)^2*ExpIntegralEi[(
2*f*g*n*(c + d*x)*Log[F])/d]*Log[F]^2)/F^((2*f*g*n*(c + d*x))/d) - (9*b^3*f^2*(F^(g*(e + f*x)))^(3*n)*g^2*n^2*
(c + d*x)^2*ExpIntegralEi[(3*f*g*n*(c + d*x)*Log[F])/d]*Log[F]^2)/F^((3*f*g*n*(c + d*x))/d) + 3*a^2*b*d*(F^(g*
(e + f*x)))^n*(d + f*g*n*(c + d*x)*Log[F]) + 3*a*b^2*d*(F^(g*(e + f*x)))^(2*n)*(d + 2*f*g*n*(c + d*x)*Log[F])
+ b^3*d*(F^(g*(e + f*x)))^(3*n)*(d + 3*f*g*n*(c + d*x)*Log[F]))/(d^3*(c + d*x)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

F^(3*g*n*e)*b^3*integrate(F^(3*f*g*n*x)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x) + 3*F^(2*g*n*e)*a*b^2*in
tegrate(F^(2*f*g*n*x)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x) + 3*F^(g*n*e)*a^2*b*integrate(F^(f*g*n*x)/
(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x) - 1/2*a^3/(d^3*x^2 + 2*c*d^2*x + c^2*d)

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Fricas [A]
time = 0.45, size = 493, normalized size = 1.10 \begin {gather*} -\frac {a^{3} d^{2} - \frac {9 \, {\left (b^{3} d^{2} f^{2} g^{2} n^{2} x^{2} + 2 \, b^{3} c d f^{2} g^{2} n^{2} x + b^{3} c^{2} f^{2} g^{2} n^{2}\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 )^{2}}{F^{\frac {3 \, {\left (c f g n - d g n e\right )}}{d}}} - \frac {12 \, {\left (a b^{2} d^{2} f^{2} g^{2} n^{2} x^{2} + 2 \, a b^{2} c d f^{2} g^{2} n^{2} x + a b^{2} c^{2} f^{2} g^{2} n^{2}\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 )^{2}}{F^{\frac {2 \, {\left (c f g n - d g n e\right )}}{d}}} - \frac {3 \, {\left (a^{2} b d^{2} f^{2} g^{2} n^{2} x^{2} + 2 \, a^{2} b c d f^{2} g^{2} n^{2} x + a^{2} b c^{2} f^{2} g^{2} n^{2}\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 )^{2}}{F^{\frac {c f g n - d g n e}{d}}} + {\left (b^{3} d^{2} + 3 \, {\left (b^{3} d^{2} f g n x + b^{3} c d f g n\right )} \log \left (F\right )\right )} F^{3 \, f g n x + 3 \, g n e} + 3 \, {\left (a b^{2} d^{2} + 2 \, {\left (a b^{2} d^{2} f g n x + a b^{2} c d f g n\right )} \log \left (F\right )\right )} F^{2 \, f g n x + 2 \, g n e} + 3 \, {\left (a^{2} b d^{2} + {\left (a^{2} b d^{2} f g n x + a^{2} b c d f g n\right )} \log \left (F\right )\right )} F^{f g n x + g n e}}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \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)^3,x, algorithm="fricas")

[Out]

-1/2*(a^3*d^2 - 9*(b^3*d^2*f^2*g^2*n^2*x^2 + 2*b^3*c*d*f^2*g^2*n^2*x + b^3*c^2*f^2*g^2*n^2)*Ei(3*(d*f*g*n*x +
c*f*g*n)*log(F)/d)*log(F)^2/F^(3*(c*f*g*n - d*g*n*e)/d) - 12*(a*b^2*d^2*f^2*g^2*n^2*x^2 + 2*a*b^2*c*d*f^2*g^2*
n^2*x + a*b^2*c^2*f^2*g^2*n^2)*Ei(2*(d*f*g*n*x + c*f*g*n)*log(F)/d)*log(F)^2/F^(2*(c*f*g*n - d*g*n*e)/d) - 3*(
a^2*b*d^2*f^2*g^2*n^2*x^2 + 2*a^2*b*c*d*f^2*g^2*n^2*x + a^2*b*c^2*f^2*g^2*n^2)*Ei((d*f*g*n*x + c*f*g*n)*log(F)
/d)*log(F)^2/F^((c*f*g*n - d*g*n*e)/d) + (b^3*d^2 + 3*(b^3*d^2*f*g*n*x + b^3*c*d*f*g*n)*log(F))*F^(3*f*g*n*x +
3*g*n*e) + 3*(a*b^2*d^2 + 2*(a*b^2*d^2*f*g*n*x + a*b^2*c*d*f*g*n)*log(F))*F^(2*f*g*n*x + 2*g*n*e) + 3*(a^2*b*
d^2 + (a^2*b*d^2*f*g*n*x + a^2*b*c*d*f*g*n)*log(F))*F^(f*g*n*x + g*n*e))/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)

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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 )^{3}}\, 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)**3,x)

[Out]

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

[Out]

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

[Out]

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

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