3.42 \(\int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 \, dx\)

Optimal. Leaf size=103 \[ a^3 x+\frac{3 a^2 b \left (F^{g (e+f x)}\right )^n}{f g n \log (F)}+\frac{3 a b^2 \left (F^{g (e+f x)}\right )^{2 n}}{2 f g n \log (F)}+\frac{b^3 \left (F^{g (e+f x)}\right )^{3 n}}{3 f g n \log (F)} \]

[Out]

a^3*x + (3*a^2*b*(F^(g*(e + f*x)))^n)/(f*g*n*Log[F]) + (3*a*b^2*(F^(g*(e + f*x))
)^(2*n))/(2*f*g*n*Log[F]) + (b^3*(F^(g*(e + f*x)))^(3*n))/(3*f*g*n*Log[F])

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Rubi [A]  time = 0.112289, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176 \[ a^3 x+\frac{3 a^2 b \left (F^{g (e+f x)}\right )^n}{f g n \log (F)}+\frac{3 a b^2 \left (F^{g (e+f x)}\right )^{2 n}}{2 f g n \log (F)}+\frac{b^3 \left (F^{g (e+f x)}\right )^{3 n}}{3 f g n \log (F)} \]

Antiderivative was successfully verified.

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

[Out]

a^3*x + (3*a^2*b*(F^(g*(e + f*x)))^n)/(f*g*n*Log[F]) + (3*a*b^2*(F^(g*(e + f*x))
)^(2*n))/(2*f*g*n*Log[F]) + (b^3*(F^(g*(e + f*x)))^(3*n))/(3*f*g*n*Log[F])

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Rubi in Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \frac{a^{3} \log{\left (\left (F^{g \left (e + f x\right )}\right )^{n} \right )}}{f g n \log{\left (F \right )}} + \frac{3 a^{2} b \left (F^{g \left (e + f x\right )}\right )^{n}}{f g n \log{\left (F \right )}} + \frac{3 a b^{2} \int ^{\left (F^{g \left (e + f x\right )}\right )^{n}} x\, dx}{f g n \log{\left (F \right )}} + \frac{b^{3} \left (F^{g \left (e + f x\right )}\right )^{3 n}}{3 f g n \log{\left (F \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((a+b*(F**(g*(f*x+e)))**n)**3,x)

[Out]

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

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Mathematica [A]  time = 0.136143, size = 74, normalized size = 0.72 \[ a^3 x+\frac{b \left (F^{g (e+f x)}\right )^n \left (18 a^2+9 a b \left (F^{g (e+f x)}\right )^n+2 b^2 \left (F^{g (e+f x)}\right )^{2 n}\right )}{6 f g n \log (F)} \]

Antiderivative was successfully verified.

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

[Out]

a^3*x + (b*(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)))/(6*f*g*n*Log[F])

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Maple [A]  time = 0.005, size = 124, normalized size = 1.2 \[{\frac{{b}^{3} \left ( \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) ^{3}}{3\,ngf\ln \left ( F \right ) }}+{\frac{3\,a{b}^{2} \left ( \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) ^{2}}{2\,ngf\ln \left ( F \right ) }}+3\,{\frac{{a}^{2}b \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n}}{ngf\ln \left ( F \right ) }}+{\frac{{a}^{3}\ln \left ( \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) }{ngf\ln \left ( F \right ) }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((a+b*(F^(g*(f*x+e)))^n)^3,x)

[Out]

1/3/g/f/ln(F)/n*b^3*((F^(g*(f*x+e)))^n)^3+3/2/g/f/ln(F)/n*a*b^2*((F^(g*(f*x+e)))
^n)^2+3*a^2*b*(F^(g*(f*x+e)))^n/f/g/n/ln(F)+1/g/f/ln(F)/n*a^3*ln((F^(g*(f*x+e)))
^n)

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Maxima [A]  time = 0.817554, size = 155, normalized size = 1.5 \[ a^{3} x + \frac{3 \,{\left (F^{f g x}\right )}^{n}{\left (F^{e g}\right )}^{n} a^{2} b}{f g n \log \left (F\right )} + \frac{3 \,{\left (F^{f g x}\right )}^{2 \, n}{\left (F^{e g}\right )}^{2 \, n} a b^{2}}{2 \, f g n \log \left (F\right )} + \frac{{\left (F^{f g x}\right )}^{3 \, n}{\left (F^{e g}\right )}^{3 \, n} b^{3}}{3 \, f g n \log \left (F\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(((F^((f*x + e)*g))^n*b + a)^3,x, algorithm="maxima")

[Out]

a^3*x + 3*(F^(f*g*x))^n*(F^(e*g))^n*a^2*b/(f*g*n*log(F)) + 3/2*(F^(f*g*x))^(2*n)
*(F^(e*g))^(2*n)*a*b^2/(f*g*n*log(F)) + 1/3*(F^(f*g*x))^(3*n)*(F^(e*g))^(3*n)*b^
3/(f*g*n*log(F))

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Fricas [A]  time = 0.298676, size = 113, normalized size = 1.1 \[ \frac{6 \, a^{3} f g n x \log \left (F\right ) + 18 \, F^{f g n x + e g n} a^{2} b + 9 \, F^{2 \, f g n x + 2 \, e g n} a b^{2} + 2 \, F^{3 \, f g n x + 3 \, e g n} b^{3}}{6 \, f g n \log \left (F\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(((F^((f*x + e)*g))^n*b + a)^3,x, algorithm="fricas")

[Out]

1/6*(6*a^3*f*g*n*x*log(F) + 18*F^(f*g*n*x + e*g*n)*a^2*b + 9*F^(2*f*g*n*x + 2*e*
g*n)*a*b^2 + 2*F^(3*f*g*n*x + 3*e*g*n)*b^3)/(f*g*n*log(F))

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Sympy [A]  time = 0.487415, size = 153, normalized size = 1.49 \[ a^{3} x + \begin{cases} \frac{18 a^{2} b f^{2} g^{2} n^{2} \left (F^{g \left (e + f x\right )}\right )^{n} \log{\left (F \right )}^{2} + 9 a b^{2} f^{2} g^{2} n^{2} \left (F^{g \left (e + f x\right )}\right )^{2 n} \log{\left (F \right )}^{2} + 2 b^{3} f^{2} g^{2} n^{2} \left (F^{g \left (e + f x\right )}\right )^{3 n} \log{\left (F \right )}^{2}}{6 f^{3} g^{3} n^{3} \log{\left (F \right )}^{3}} & \text{for}\: 6 f^{3} g^{3} n^{3} \log{\left (F \right )}^{3} \neq 0 \\x \left (3 a^{2} b + 3 a b^{2} + b^{3}\right ) & \text{otherwise} \end{cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((a+b*(F**(g*(f*x+e)))**n)**3,x)

[Out]

a**3*x + Piecewise(((18*a**2*b*f**2*g**2*n**2*(F**(g*(e + f*x)))**n*log(F)**2 +
9*a*b**2*f**2*g**2*n**2*(F**(g*(e + f*x)))**(2*n)*log(F)**2 + 2*b**3*f**2*g**2*n
**2*(F**(g*(e + f*x)))**(3*n)*log(F)**2)/(6*f**3*g**3*n**3*log(F)**3), Ne(6*f**3
*g**3*n**3*log(F)**3, 0)), (x*(3*a**2*b + 3*a*b**2 + b**3), True))

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GIAC/XCAS [A]  time = 0.304981, size = 1393, normalized size = 13.52 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(((F^((f*x + e)*g))^n*b + a)^3,x, algorithm="giac")

[Out]

a^3*x + 2/3*(2*b^3*f*g*n*cos(-3/2*pi*f*g*n*x*sign(F) + 3/2*pi*f*g*n*x - 3/2*pi*g
*n*e*sign(F) + 3/2*pi*g*n*e)*ln(abs(F))/(4*f^2*g^2*n^2*ln(abs(F))^2 + (pi*f*g*n*
sign(F) - pi*f*g*n)^2) - (pi*f*g*n*sign(F) - pi*f*g*n)*b^3*sin(-3/2*pi*f*g*n*x*s
ign(F) + 3/2*pi*f*g*n*x - 3/2*pi*g*n*e*sign(F) + 3/2*pi*g*n*e)/(4*f^2*g^2*n^2*ln
(abs(F))^2 + (pi*f*g*n*sign(F) - pi*f*g*n)^2))*e^(3*f*g*n*x*ln(abs(F)) + 3*g*n*e
*ln(abs(F))) - 1/2*I*(-2*I*b^3*e^(3/2*I*pi*f*g*n*x*sign(F) - 3/2*I*pi*f*g*n*x +
3/2*I*pi*g*n*e*sign(F) - 3/2*I*pi*g*n*e)/(3*I*pi*f*g*n*sign(F) - 3*I*pi*f*g*n +
6*f*g*n*ln(abs(F))) + 2*I*b^3*e^(-3/2*I*pi*f*g*n*x*sign(F) + 3/2*I*pi*f*g*n*x -
3/2*I*pi*g*n*e*sign(F) + 3/2*I*pi*g*n*e)/(-3*I*pi*f*g*n*sign(F) + 3*I*pi*f*g*n +
6*f*g*n*ln(abs(F))))*e^(3*f*g*n*x*ln(abs(F)) + 3*g*n*e*ln(abs(F))) + 3*(2*a*b^2
*f*g*n*cos(-pi*f*g*n*x*sign(F) + pi*f*g*n*x - pi*g*n*e*sign(F) + pi*g*n*e)*ln(ab
s(F))/(4*f^2*g^2*n^2*ln(abs(F))^2 + (pi*f*g*n*sign(F) - pi*f*g*n)^2) - (pi*f*g*n
*sign(F) - pi*f*g*n)*a*b^2*sin(-pi*f*g*n*x*sign(F) + pi*f*g*n*x - pi*g*n*e*sign(
F) + pi*g*n*e)/(4*f^2*g^2*n^2*ln(abs(F))^2 + (pi*f*g*n*sign(F) - pi*f*g*n)^2))*e
^(2*f*g*n*x*ln(abs(F)) + 2*g*n*e*ln(abs(F))) - 1/2*I*(-3*I*a*b^2*e^(I*pi*f*g*n*x
*sign(F) - I*pi*f*g*n*x + I*pi*g*n*e*sign(F) - I*pi*g*n*e)/(I*pi*f*g*n*sign(F) -
I*pi*f*g*n + 2*f*g*n*ln(abs(F))) + 3*I*a*b^2*e^(-I*pi*f*g*n*x*sign(F) + I*pi*f*
g*n*x - I*pi*g*n*e*sign(F) + I*pi*g*n*e)/(-I*pi*f*g*n*sign(F) + I*pi*f*g*n + 2*f
*g*n*ln(abs(F))))*e^(2*f*g*n*x*ln(abs(F)) + 2*g*n*e*ln(abs(F))) + 6*(2*a^2*b*f*g
*n*cos(-1/2*pi*f*g*n*x*sign(F) + 1/2*pi*f*g*n*x - 1/2*pi*g*n*e*sign(F) + 1/2*pi*
g*n*e)*ln(abs(F))/(4*f^2*g^2*n^2*ln(abs(F))^2 + (pi*f*g*n*sign(F) - pi*f*g*n)^2)
- (pi*f*g*n*sign(F) - pi*f*g*n)*a^2*b*sin(-1/2*pi*f*g*n*x*sign(F) + 1/2*pi*f*g*
n*x - 1/2*pi*g*n*e*sign(F) + 1/2*pi*g*n*e)/(4*f^2*g^2*n^2*ln(abs(F))^2 + (pi*f*g
*n*sign(F) - pi*f*g*n)^2))*e^(f*g*n*x*ln(abs(F)) + g*n*e*ln(abs(F))) - 1/2*I*(-6
*I*a^2*b*e^(1/2*I*pi*f*g*n*x*sign(F) - 1/2*I*pi*f*g*n*x + 1/2*I*pi*g*n*e*sign(F)
- 1/2*I*pi*g*n*e)/(I*pi*f*g*n*sign(F) - I*pi*f*g*n + 2*f*g*n*ln(abs(F))) + 6*I*
a^2*b*e^(-1/2*I*pi*f*g*n*x*sign(F) + 1/2*I*pi*f*g*n*x - 1/2*I*pi*g*n*e*sign(F) +
1/2*I*pi*g*n*e)/(-I*pi*f*g*n*sign(F) + I*pi*f*g*n + 2*f*g*n*ln(abs(F))))*e^(f*g
*n*x*ln(abs(F)) + g*n*e*ln(abs(F)))