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

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

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Rubi [A]  time = 0.05, antiderivative size = 103, normalized size of antiderivative = 1.00, 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 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])

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 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]]]

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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.43, size = 84, normalized size = 0.82 \[ \frac {6 \, a^{3} f g n x \log \relax (F) + 18 \, F^{f g n x + e g n} a^{2} b + 9 \, F^{2 \, f g n x + 2 \, e g n} a b^{2} + 2 \, F^{3 \, f g n x + 3 \, e g n} b^{3}}{6 \, f g n \log \relax (F)} \]

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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giac [A]  time = 0.44, size = 102, normalized size = 0.99 \[ \frac {18 \, F^{f g n x} F^{g n e} a^{2} b + 9 \, F^{2 \, f g n x} F^{2 \, g n e} a b^{2} + 2 \, F^{3 \, f g n x} F^{3 \, g n e} b^{3} + 6 \, a^{3} \log \left ({\left | F \right |}^{f g n x} {\left | F \right |}^{g n e}\right )}{6 \, f g n \log \relax (F)} \]

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]

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

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maple [A]  time = 0.01, size = 124, normalized size = 1.20 \[ \frac {a^{3} \ln \left (\left (F^{\left (f x +e \right ) g}\right )^{n}\right )}{f g n \ln \relax (F )}+\frac {3 a^{2} b \left (F^{\left (f x +e \right ) g}\right )^{n}}{f g n \ln \relax (F )}+\frac {3 a \,b^{2} \left (F^{\left (f x +e \right ) g}\right )^{2 n}}{2 f g n \ln \relax (F )}+\frac {b^{3} \left (F^{\left (f x +e \right ) g}\right )^{3 n}}{3 f g n \ln \relax (F )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.56, size = 115, normalized size = 1.12 \[ a^{3} x + \frac {3 \, {\left (F^{f g x}\right )}^{n} {\left (F^{e g}\right )}^{n} a^{2} b}{f g n \log \relax (F)} + \frac {3 \, {\left (F^{f g x}\right )}^{2 \, n} {\left (F^{e g}\right )}^{2 \, n} a b^{2}}{2 \, f g n \log \relax (F)} + \frac {{\left (F^{f g x}\right )}^{3 \, n} {\left (F^{e g}\right )}^{3 \, n} b^{3}}{3 \, f g n \log \relax (F)} \]

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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mupad [B]  time = 3.78, size = 124, normalized size = 1.20 \[ \frac {a^3\,\ln \left (F^{f\,g\,x}\right )}{f\,g\,\ln \relax (F)}+\frac {b^3\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^{3\,n}}{3\,f\,g\,n\,\ln \relax (F)}+\frac {3\,a^2\,b\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n}{f\,g\,n\,\ln \relax (F)}+\frac {3\,a\,b^2\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^{2\,n}}{2\,f\,g\,n\,\ln \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(F^(g*(e + f*x)))^n)^3,x)

[Out]

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

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sympy [A]  time = 0.21, size = 153, normalized size = 1.49 \[ a^{3} x + \begin {cases} \frac {18 a^{2} b f^{2} g^{2} n^{2} \left (F^{g \left (e + f x\right )}\right )^{n} \log {\relax (F )}^{2} + 9 a b^{2} f^{2} g^{2} n^{2} \left (F^{g \left (e + f x\right )}\right )^{2 n} \log {\relax (F )}^{2} + 2 b^{3} f^{2} g^{2} n^{2} \left (F^{g \left (e + f x\right )}\right )^{3 n} \log {\relax (F )}^{2}}{6 f^{3} g^{3} n^{3} \log {\relax (F )}^{3}} & \text {for}\: 6 f^{3} g^{3} n^{3} \log {\relax (F )}^{3} \neq 0 \\x \left (3 a^{2} b + 3 a b^{2} + b^{3}\right ) & \text {otherwise} \end {cases} \]

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*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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