3.1.41 \(\int (a+b (F^{g (e+f x)})^n)^3 (c+d x) \, dx\) [41]

Optimal. Leaf size=236 \[ \frac {a^3 (c+d x)^2}{2 d}-\frac {3 a^2 b d \left (F^{e g+f g x}\right )^n}{f^2 g^2 n^2 \log ^2(F)}-\frac {3 a b^2 d \left (F^{e g+f g x}\right )^{2 n}}{4 f^2 g^2 n^2 \log ^2(F)}-\frac {b^3 d \left (F^{e g+f g x}\right )^{3 n}}{9 f^2 g^2 n^2 \log ^2(F)}+\frac {3 a^2 b \left (F^{e g+f g x}\right )^n (c+d x)}{f g n \log (F)}+\frac {3 a b^2 \left (F^{e g+f g x}\right )^{2 n} (c+d x)}{2 f g n \log (F)}+\frac {b^3 \left (F^{e g+f g x}\right )^{3 n} (c+d x)}{3 f g n \log (F)} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2214, 2207, 2225} \begin {gather*} \frac {a^3 (c+d x)^2}{2 d}+\frac {3 a^2 b (c+d x) \left (F^{e g+f g x}\right )^n}{f g n \log (F)}-\frac {3 a^2 b d \left (F^{e g+f g x}\right )^n}{f^2 g^2 n^2 \log ^2(F)}+\frac {3 a b^2 (c+d x) \left (F^{e g+f g x}\right )^{2 n}}{2 f g n \log (F)}-\frac {3 a b^2 d \left (F^{e g+f g x}\right )^{2 n}}{4 f^2 g^2 n^2 \log ^2(F)}+\frac {b^3 (c+d x) \left (F^{e g+f g x}\right )^{3 n}}{3 f g n \log (F)}-\frac {b^3 d \left (F^{e g+f g x}\right )^{3 n}}{9 f^2 g^2 n^2 \log ^2(F)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(c + d*x)^2)/(2*d) - (3*a^2*b*d*(F^(e*g + f*g*x))^n)/(f^2*g^2*n^2*Log[F]^2) - (3*a*b^2*d*(F^(e*g + f*g*x)
)^(2*n))/(4*f^2*g^2*n^2*Log[F]^2) - (b^3*d*(F^(e*g + f*g*x))^(3*n))/(9*f^2*g^2*n^2*Log[F]^2) + (3*a^2*b*(F^(e*
g + f*g*x))^n*(c + d*x))/(f*g*n*Log[F]) + (3*a*b^2*(F^(e*g + f*g*x))^(2*n)*(c + d*x))/(2*f*g*n*Log[F]) + (b^3*
(F^(e*g + f*g*x))^(3*n)*(c + d*x))/(3*f*g*n*Log[F])

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

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]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin {align*} \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 (c+d x) \, dx &=\int \left (a^3 (c+d x)+3 a^2 b \left (F^{e g+f g x}\right )^n (c+d x)+3 a b^2 \left (F^{e g+f g x}\right )^{2 n} (c+d x)+b^3 \left (F^{e g+f g x}\right )^{3 n} (c+d x)\right ) \, dx\\ &=\frac {a^3 (c+d x)^2}{2 d}+\left (3 a^2 b\right ) \int \left (F^{e g+f g x}\right )^n (c+d x) \, dx+\left (3 a b^2\right ) \int \left (F^{e g+f g x}\right )^{2 n} (c+d x) \, dx+b^3 \int \left (F^{e g+f g x}\right )^{3 n} (c+d x) \, dx\\ &=\frac {a^3 (c+d x)^2}{2 d}+\frac {3 a^2 b \left (F^{e g+f g x}\right )^n (c+d x)}{f g n \log (F)}+\frac {3 a b^2 \left (F^{e g+f g x}\right )^{2 n} (c+d x)}{2 f g n \log (F)}+\frac {b^3 \left (F^{e g+f g x}\right )^{3 n} (c+d x)}{3 f g n \log (F)}-\frac {\left (3 a^2 b d\right ) \int \left (F^{e g+f g x}\right )^n \, dx}{f g n \log (F)}-\frac {\left (3 a b^2 d\right ) \int \left (F^{e g+f g x}\right )^{2 n} \, dx}{2 f g n \log (F)}-\frac {\left (b^3 d\right ) \int \left (F^{e g+f g x}\right )^{3 n} \, dx}{3 f g n \log (F)}\\ &=\frac {a^3 (c+d x)^2}{2 d}-\frac {3 a^2 b d \left (F^{e g+f g x}\right )^n}{f^2 g^2 n^2 \log ^2(F)}-\frac {3 a b^2 d \left (F^{e g+f g x}\right )^{2 n}}{4 f^2 g^2 n^2 \log ^2(F)}-\frac {b^3 d \left (F^{e g+f g x}\right )^{3 n}}{9 f^2 g^2 n^2 \log ^2(F)}+\frac {3 a^2 b \left (F^{e g+f g x}\right )^n (c+d x)}{f g n \log (F)}+\frac {3 a b^2 \left (F^{e g+f g x}\right )^{2 n} (c+d x)}{2 f g n \log (F)}+\frac {b^3 \left (F^{e g+f g x}\right )^{3 n} (c+d x)}{3 f g n \log (F)}\\ \end {align*}

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Mathematica [A]
time = 0.40, size = 161, normalized size = 0.68 \begin {gather*} \frac {-b d \left (F^{g (e+f x)}\right )^n \left (108 a^2+27 a b \left (F^{g (e+f x)}\right )^n+4 b^2 \left (F^{g (e+f x)}\right )^{2 n}\right )+6 b f \left (F^{g (e+f x)}\right )^n \left (18 a^2+9 a b \left (F^{g (e+f x)}\right )^n+2 b^2 \left (F^{g (e+f x)}\right )^{2 n}\right ) g n (c+d x) \log (F)+18 a^3 f^2 g^2 n^2 x (2 c+d x) \log ^2(F)}{36 f^2 g^2 n^2 \log ^2(F)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*d*(F^(g*(e + f*x)))^n*(108*a^2 + 27*a*b*(F^(g*(e + f*x)))^n + 4*b^2*(F^(g*(e + f*x)))^(2*n))) + 6*b*f*(F^
(g*(e + f*x)))^n*(18*a^2 + 9*a*b*(F^(g*(e + f*x)))^n + 2*b^2*(F^(g*(e + f*x)))^(2*n))*g*n*(c + d*x)*Log[F] + 1
8*a^3*f^2*g^2*n^2*x*(2*c + d*x)*Log[F]^2)/(36*f^2*g^2*n^2*Log[F]^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 278, normalized size = 1.18 \begin {gather*} \frac {1}{2} \, a^{3} d x^{2} + a^{3} c x + \frac {3 \, F^{f g n x + g n e} a^{2} b c}{f g n \log \left (F\right )} + \frac {3 \, F^{2 \, f g n x + 2 \, g n e} a b^{2} c}{2 \, f g n \log \left (F\right )} + \frac {F^{3 \, f g n x + 3 \, g n e} b^{3} c}{3 \, f g n \log \left (F\right )} + \frac {3 \, {\left (F^{g n e} f g n x \log \left (F\right ) - F^{g n e}\right )} F^{f g n x} a^{2} b d}{f^{2} g^{2} n^{2} \log \left (F\right )^{2}} + \frac {3 \, {\left (2 \, F^{2 \, g n e} f g n x \log \left (F\right ) - F^{2 \, g n e}\right )} F^{2 \, f g n x} a b^{2} d}{4 \, f^{2} g^{2} n^{2} \log \left (F\right )^{2}} + \frac {{\left (3 \, F^{3 \, g n e} f g n x \log \left (F\right ) - F^{3 \, g n e}\right )} F^{3 \, f g n x} b^{3} d}{9 \, f^{2} g^{2} n^{2} \log \left (F\right )^{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),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.45, size = 195, normalized size = 0.83 \begin {gather*} \frac {18 \, {\left (a^{3} d f^{2} g^{2} n^{2} x^{2} + 2 \, a^{3} c f^{2} g^{2} n^{2} x\right )} \log \left (F\right )^{2} - 4 \, {\left (b^{3} d - 3 \, {\left (b^{3} d f g n x + b^{3} c f g n\right )} \log \left (F\right )\right )} F^{3 \, f g n x + 3 \, g n e} - 27 \, {\left (a b^{2} d - 2 \, {\left (a b^{2} d f g n x + a b^{2} c f g n\right )} \log \left (F\right )\right )} F^{2 \, f g n x + 2 \, g n e} - 108 \, {\left (a^{2} b d - {\left (a^{2} b d f g n x + a^{2} b c f g n\right )} \log \left (F\right )\right )} F^{f g n x + g n e}}{36 \, f^{2} g^{2} n^{2} \log \left (F\right )^{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),x, algorithm="fricas")

[Out]

1/36*(18*(a^3*d*f^2*g^2*n^2*x^2 + 2*a^3*c*f^2*g^2*n^2*x)*log(F)^2 - 4*(b^3*d - 3*(b^3*d*f*g*n*x + b^3*c*f*g*n)
*log(F))*F^(3*f*g*n*x + 3*g*n*e) - 27*(a*b^2*d - 2*(a*b^2*d*f*g*n*x + a*b^2*c*f*g*n)*log(F))*F^(2*f*g*n*x + 2*
g*n*e) - 108*(a^2*b*d - (a^2*b*d*f*g*n*x + a^2*b*c*f*g*n)*log(F))*F^(f*g*n*x + g*n*e))/(f^2*g^2*n^2*log(F)^2)

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Sympy [A]
time = 0.15, size = 348, normalized size = 1.47 \begin {gather*} a^{3} c x + \frac {a^{3} d x^{2}}{2} + \begin {cases} \frac {\left (12 b^{3} c f^{5} g^{5} n^{5} \log {\left (F \right )}^{5} + 12 b^{3} d f^{5} g^{5} n^{5} x \log {\left (F \right )}^{5} - 4 b^{3} d f^{4} g^{4} n^{4} \log {\left (F \right )}^{4}\right ) \left (F^{g \left (e + f x\right )}\right )^{3 n} + \left (54 a b^{2} c f^{5} g^{5} n^{5} \log {\left (F \right )}^{5} + 54 a b^{2} d f^{5} g^{5} n^{5} x \log {\left (F \right )}^{5} - 27 a b^{2} d f^{4} g^{4} n^{4} \log {\left (F \right )}^{4}\right ) \left (F^{g \left (e + f x\right )}\right )^{2 n} + \left (108 a^{2} b c f^{5} g^{5} n^{5} \log {\left (F \right )}^{5} + 108 a^{2} b d f^{5} g^{5} n^{5} x \log {\left (F \right )}^{5} - 108 a^{2} b d f^{4} g^{4} n^{4} \log {\left (F \right )}^{4}\right ) \left (F^{g \left (e + f x\right )}\right )^{n}}{36 f^{6} g^{6} n^{6} \log {\left (F \right )}^{6}} & \text {for}\: f^{6} g^{6} n^{6} \log {\left (F \right )}^{6} \neq 0 \\x^{2} \cdot \left (\frac {3 a^{2} b d}{2} + \frac {3 a b^{2} d}{2} + \frac {b^{3} d}{2}\right ) + x \left (3 a^{2} b c + 3 a b^{2} c + b^{3} c\right ) & \text {otherwise} \end {cases} \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),x)

[Out]

a**3*c*x + a**3*d*x**2/2 + Piecewise((((12*b**3*c*f**5*g**5*n**5*log(F)**5 + 12*b**3*d*f**5*g**5*n**5*x*log(F)
**5 - 4*b**3*d*f**4*g**4*n**4*log(F)**4)*(F**(g*(e + f*x)))**(3*n) + (54*a*b**2*c*f**5*g**5*n**5*log(F)**5 + 5
4*a*b**2*d*f**5*g**5*n**5*x*log(F)**5 - 27*a*b**2*d*f**4*g**4*n**4*log(F)**4)*(F**(g*(e + f*x)))**(2*n) + (108
*a**2*b*c*f**5*g**5*n**5*log(F)**5 + 108*a**2*b*d*f**5*g**5*n**5*x*log(F)**5 - 108*a**2*b*d*f**4*g**4*n**4*log
(F)**4)*(F**(g*(e + f*x)))**n)/(36*f**6*g**6*n**6*log(F)**6), Ne(f**6*g**6*n**6*log(F)**6, 0)), (x**2*(3*a**2*
b*d/2 + 3*a*b**2*d/2 + b**3*d/2) + x*(3*a**2*b*c + 3*a*b**2*c + b**3*c), True))

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Giac [C] Result contains complex when optimal does not.
time = 2.86, size = 3558, normalized size = 15.08 \begin {gather*} \text {Too large to display} \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),x, algorithm="giac")

[Out]

1/2*a^3*d*x^2 + a^3*c*x + 1/9*(2*((3*b^3*d*f*g*n*x*log(abs(F)) + 3*b^3*c*f*g*n*log(abs(F)) - b^3*d)*(pi^2*f^2*
g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2
+ 2*f^2*g^2*n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))^2) + 3*(
pi*b^3*d*f*g*n*x*sgn(F) - pi*b^3*d*f*g*n*x + pi*b^3*c*f*g*n*sgn(F) - pi*b^3*c*f*g*n)*(pi*f^2*g^2*n^2*log(abs(F
))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F
))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))^2))*cos(-3/2*pi*f*g*n*x*sgn(F) +
3/2*pi*f*g*n*x - 3/2*pi*g*n*e*sgn(F) + 3/2*pi*g*n*e) + (3*(pi*b^3*d*f*g*n*x*sgn(F) - pi*b^3*d*f*g*n*x + pi*b^3
*c*f*g*n*sgn(F) - pi*b^3*c*f*g*n)*(pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)/(
(pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*s
gn(F) - pi*f^2*g^2*n^2*log(abs(F)))^2) - 4*(3*b^3*d*f*g*n*x*log(abs(F)) + 3*b^3*c*f*g*n*log(abs(F)) - b^3*d)*(
pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 +
2*f^2*g^2*n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))^2))*sin(-
3/2*pi*f*g*n*x*sgn(F) + 3/2*pi*f*g*n*x - 3/2*pi*g*n*e*sgn(F) + 3/2*pi*g*n*e))*e^(3*f*g*n*x*log(abs(F)) + 3*g*n
*e*log(abs(F))) - 1/18*I*((3*pi*b^3*d*f*g*n*x*sgn(F) - 3*pi*b^3*d*f*g*n*x - 6*I*b^3*d*f*g*n*x*log(abs(F)) + 3*
pi*b^3*c*f*g*n*sgn(F) - 3*pi*b^3*c*f*g*n - 6*I*b^3*c*f*g*n*log(abs(F)) + 2*I*b^3*d)*e^(3/2*I*pi*f*g*n*x*sgn(F)
- 3/2*I*pi*f*g*n*x + 3/2*I*pi*g*n*e*sgn(F) - 3/2*I*pi*g*n*e)/(pi^2*f^2*g^2*n^2*sgn(F) + 2*I*pi*f^2*g^2*n^2*lo
g(abs(F))*sgn(F) - pi^2*f^2*g^2*n^2 - 2*I*pi*f^2*g^2*n^2*log(abs(F)) + 2*f^2*g^2*n^2*log(abs(F))^2) + (3*pi*b^
3*d*f*g*n*x*sgn(F) - 3*pi*b^3*d*f*g*n*x + 6*I*b^3*d*f*g*n*x*log(abs(F)) + 3*pi*b^3*c*f*g*n*sgn(F) - 3*pi*b^3*c
*f*g*n + 6*I*b^3*c*f*g*n*log(abs(F)) - 2*I*b^3*d)*e^(-3/2*I*pi*f*g*n*x*sgn(F) + 3/2*I*pi*f*g*n*x - 3/2*I*pi*g*
n*e*sgn(F) + 3/2*I*pi*g*n*e)/(pi^2*f^2*g^2*n^2*sgn(F) - 2*I*pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi^2*f^2*g^2*n
^2 + 2*I*pi*f^2*g^2*n^2*log(abs(F)) + 2*f^2*g^2*n^2*log(abs(F))^2))*e^(3*f*g*n*x*log(abs(F)) + 3*g*n*e*log(abs
(F))) + 3/2*(((2*a*b^2*d*f*g*n*x*log(abs(F)) + 2*a*b^2*c*f*g*n*log(abs(F)) - a*b^2*d)*(pi^2*f^2*g^2*n^2*sgn(F)
- pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^
2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))^2) + 2*(pi*a*b^2*d*f*g
*n*x*sgn(F) - pi*a*b^2*d*f*g*n*x + pi*a*b^2*c*f*g*n*sgn(F) - pi*a*b^2*c*f*g*n)*(pi*f^2*g^2*n^2*log(abs(F))*sgn
(F) - pi*f^2*g^2*n^2*log(abs(F)))/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)^
2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))^2))*cos(-pi*f*g*n*x*sgn(F) + pi*f*g*n*x
- pi*g*n*e*sgn(F) + pi*g*n*e) + ((pi*a*b^2*d*f*g*n*x*sgn(F) - pi*a*b^2*d*f*g*n*x + pi*a*b^2*c*f*g*n*sgn(F) -
pi*a*b^2*c*f*g*n)*(pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)/((pi^2*f^2*g^2*n^
2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g
^2*n^2*log(abs(F)))^2) - 2*(2*a*b^2*d*f*g*n*x*log(abs(F)) + 2*a*b^2*c*f*g*n*log(abs(F)) - a*b^2*d)*(pi*f^2*g^2
*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2
*n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))^2))*sin(-pi*f*g*n*x
*sgn(F) + pi*f*g*n*x - pi*g*n*e*sgn(F) + pi*g*n*e))*e^(2*f*g*n*x*log(abs(F)) + 2*g*n*e*log(abs(F))) - 3/4*I*((
pi*a*b^2*d*f*g*n*x*sgn(F) - pi*a*b^2*d*f*g*n*x - 2*I*a*b^2*d*f*g*n*x*log(abs(F)) + pi*a*b^2*c*f*g*n*sgn(F) - p
i*a*b^2*c*f*g*n - 2*I*a*b^2*c*f*g*n*log(abs(F)) + I*a*b^2*d)*e^(I*pi*f*g*n*x*sgn(F) - I*pi*f*g*n*x + I*pi*g*n*
e*sgn(F) - I*pi*g*n*e)/(pi^2*f^2*g^2*n^2*sgn(F) + 2*I*pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi^2*f^2*g^2*n^2 - 2
*I*pi*f^2*g^2*n^2*log(abs(F)) + 2*f^2*g^2*n^2*log(abs(F))^2) + (pi*a*b^2*d*f*g*n*x*sgn(F) - pi*a*b^2*d*f*g*n*x
+ 2*I*a*b^2*d*f*g*n*x*log(abs(F)) + pi*a*b^2*c*f*g*n*sgn(F) - pi*a*b^2*c*f*g*n + 2*I*a*b^2*c*f*g*n*log(abs(F)
) - I*a*b^2*d)*e^(-I*pi*f*g*n*x*sgn(F) + I*pi*f*g*n*x - I*pi*g*n*e*sgn(F) + I*pi*g*n*e)/(pi^2*f^2*g^2*n^2*sgn(
F) - 2*I*pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi^2*f^2*g^2*n^2 + 2*I*pi*f^2*g^2*n^2*log(abs(F)) + 2*f^2*g^2*n^2
*log(abs(F))^2))*e^(2*f*g*n*x*log(abs(F)) + 2*g*n*e*log(abs(F))) + 3*(2*((a^2*b*d*f*g*n*x*log(abs(F)) + a^2*b*
c*f*g*n*log(abs(F)) - a^2*b*d)*(pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)/((pi
^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(
F) - pi*f^2*g^2*n^2*log(abs(F)))^2) + (pi*a^2*b*d*f*g*n*x*sgn(F) - pi*a^2*b*d*f*g*n*x + pi*a^2*b*c*f*g*n*sgn(F
) - pi*a^2*b*c*f*g*n)*(pi*f^2*g^2*n^2*log(abs(F...

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Mupad [B]
time = 3.77, size = 219, normalized size = 0.93 \begin {gather*} \frac {a^3\,d\,x^2}{2}-{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^{2\,n}\,\left (\frac {3\,a\,b^2\,\left (d-2\,c\,f\,g\,n\,\ln \left (F\right )\right )}{4\,f^2\,g^2\,n^2\,{\ln \left (F\right )}^2}-\frac {3\,a\,b^2\,d\,x}{2\,f\,g\,n\,\ln \left (F\right )}\right )-{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^{3\,n}\,\left (\frac {b^3\,\left (d-3\,c\,f\,g\,n\,\ln \left (F\right )\right )}{9\,f^2\,g^2\,n^2\,{\ln \left (F\right )}^2}-\frac {b^3\,d\,x}{3\,f\,g\,n\,\ln \left (F\right )}\right )-{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n\,\left (\frac {3\,a^2\,b\,\left (d-c\,f\,g\,n\,\ln \left (F\right )\right )}{f^2\,g^2\,n^2\,{\ln \left (F\right )}^2}-\frac {3\,a^2\,b\,d\,x}{f\,g\,n\,\ln \left (F\right )}\right )+a^3\,c\,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),x)

[Out]

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

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