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

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

[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 = 0.0493079, 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, Rules used = {2282, 266, 43} \[ \frac{3 a^2 b \left (F^{g (e+f x)}\right )^n}{f g n \log (F)}+a^3 x+\frac{3 a b^2 \left (F^{g (e+f x)}\right )^{2 n}}{2 f g n \log (F)}+\frac{b^3 \left (F^{g (e+f x)}\right )^{3 n}}{3 f g n \log (F)} \]

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])

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0487694, size = 74, normalized size = 0.72 \[ \frac{b \left (F^{g (e+f x)}\right )^n \left (18 a^2+9 a b \left (F^{g (e+f x)}\right )^n+2 b^2 \left (F^{g (e+f x)}\right )^{2 n}\right )}{6 f g n \log (F)}+a^3 x \]

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 = 0.002, size = 124, normalized size = 1.2 \begin{align*}{\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 ) }} \end{align*}

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 = 1.22977, size = 155, normalized size = 1.5 \begin{align*} 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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F^(g*(f*x+e)))^n)^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 = 1.51905, size = 189, normalized size = 1.83 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F^(g*(f*x+e)))^n)^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 = 0.291525, size = 153, normalized size = 1.49 \begin{align*} 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} \end{align*}

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*l
og(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 [C]  time = 1.37772, size = 1393, normalized size = 13.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x + 2/3*(2*b^3*f*g*n*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)*log
(abs(F))/(4*f^2*g^2*n^2*log(abs(F))^2 + (pi*f*g*n*sgn(F) - pi*f*g*n)^2) - (pi*f*g*n*sgn(F) - pi*f*g*n)*b^3*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)/(4*f^2*g^2*n^2*log(abs(F))^2 +
(pi*f*g*n*sgn(F) - pi*f*g*n)^2))*e^(3*f*g*n*x*log(abs(F)) + 3*g*n*e*log(abs(F))) - 1/2*I*(-2*I*b^3*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)/(3*I*pi*f*g*n*sgn(F) - 3*I*pi*f*g
*n + 6*f*g*n*log(abs(F))) + 2*I*b^3*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)/(-3*I*pi*f*g*n*sgn(F) + 3*I*pi*f*g*n + 6*f*g*n*log(abs(F))))*e^(3*f*g*n*x*log(abs(F)) + 3*g*n*e
*log(abs(F))) + 3*(2*a*b^2*f*g*n*cos(-pi*f*g*n*x*sgn(F) + pi*f*g*n*x - pi*g*n*e*sgn(F) + pi*g*n*e)*log(abs(F))
/(4*f^2*g^2*n^2*log(abs(F))^2 + (pi*f*g*n*sgn(F) - pi*f*g*n)^2) - (pi*f*g*n*sgn(F) - pi*f*g*n)*a*b^2*sin(-pi*f
*g*n*x*sgn(F) + pi*f*g*n*x - pi*g*n*e*sgn(F) + pi*g*n*e)/(4*f^2*g^2*n^2*log(abs(F))^2 + (pi*f*g*n*sgn(F) - pi*
f*g*n)^2))*e^(2*f*g*n*x*log(abs(F)) + 2*g*n*e*log(abs(F))) - 1/2*I*(-3*I*a*b^2*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)/(I*pi*f*g*n*sgn(F) - I*pi*f*g*n + 2*f*g*n*log(abs(F))) + 3*I*a*b^2*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)/(-I*pi*f*g*n*sgn(F) + I*pi*f*g*n + 2*f*
g*n*log(abs(F))))*e^(2*f*g*n*x*log(abs(F)) + 2*g*n*e*log(abs(F))) + 6*(2*a^2*b*f*g*n*cos(-1/2*pi*f*g*n*x*sgn(F
) + 1/2*pi*f*g*n*x - 1/2*pi*g*n*e*sgn(F) + 1/2*pi*g*n*e)*log(abs(F))/(4*f^2*g^2*n^2*log(abs(F))^2 + (pi*f*g*n*
sgn(F) - pi*f*g*n)^2) - (pi*f*g*n*sgn(F) - pi*f*g*n)*a^2*b*sin(-1/2*pi*f*g*n*x*sgn(F) + 1/2*pi*f*g*n*x - 1/2*p
i*g*n*e*sgn(F) + 1/2*pi*g*n*e)/(4*f^2*g^2*n^2*log(abs(F))^2 + (pi*f*g*n*sgn(F) - pi*f*g*n)^2))*e^(f*g*n*x*log(
abs(F)) + g*n*e*log(abs(F))) - 1/2*I*(-6*I*a^2*b*e^(1/2*I*pi*f*g*n*x*sgn(F) - 1/2*I*pi*f*g*n*x + 1/2*I*pi*g*n*
e*sgn(F) - 1/2*I*pi*g*n*e)/(I*pi*f*g*n*sgn(F) - I*pi*f*g*n + 2*f*g*n*log(abs(F))) + 6*I*a^2*b*e^(-1/2*I*pi*f*g
*n*x*sgn(F) + 1/2*I*pi*f*g*n*x - 1/2*I*pi*g*n*e*sgn(F) + 1/2*I*pi*g*n*e)/(-I*pi*f*g*n*sgn(F) + I*pi*f*g*n + 2*
f*g*n*log(abs(F))))*e^(f*g*n*x*log(abs(F)) + g*n*e*log(abs(F)))