3.1.61 \(\int \frac {1}{(a+b (F^{g (e+f x)})^n)^3} \, dx\) [61]

Optimal. Leaf size=111 \[ \frac {x}{a^3}+\frac {1}{2 a f \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 g n \log (F)}+\frac {1}{a^2 f \left (a+b \left (F^{g (e+f x)}\right )^n\right ) g n \log (F)}-\frac {\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^3 f g n \log (F)} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 111, 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 = {2320, 272, 46} \begin {gather*} -\frac {\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^3 f g n \log (F)}+\frac {x}{a^3}+\frac {1}{a^2 f g n \log (F) \left (a+b \left (F^{g (e+f x)}\right )^n\right )}+\frac {1}{2 a f g n \log (F) \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 272

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 2320

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 \frac {1}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{x \left (a+b x^n\right )^3} \, dx,x,F^{g (e+f x)}\right )}{f g \log (F)}\\ &=\frac {\text {Subst}\left (\int \frac {1}{x (a+b x)^3} \, dx,x,\left (F^{g (e+f x)}\right )^n\right )}{f g n \log (F)}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{a^3 x}-\frac {b}{a (a+b x)^3}-\frac {b}{a^2 (a+b x)^2}-\frac {b}{a^3 (a+b x)}\right ) \, dx,x,\left (F^{g (e+f x)}\right )^n\right )}{f g n \log (F)}\\ &=\frac {x}{a^3}+\frac {1}{2 a f \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 g n \log (F)}+\frac {1}{a^2 f \left (a+b \left (F^{g (e+f x)}\right )^n\right ) g n \log (F)}-\frac {\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^3 f g n \log (F)}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 128, normalized size = 1.15 \begin {gather*} \frac {\frac {3 a+2 b \left (F^{g (e+f x)}\right )^n}{2 a^2 f \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 g n}+\frac {\log \left (\left (F^{g (e+f x)}\right )^n\right )}{a^3 f g n}-\frac {\log \left (a^4 f g n \log (F)+a^3 b f \left (F^{g (e+f x)}\right )^n g n \log (F)\right )}{a^3 f g n}}{\log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 96, normalized size = 0.86

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a^{3}}-\frac {\ln \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a^{3}}+\frac {1}{a^{2} \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}+\frac {1}{2 a \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )^{2}}}{g f \ln \left (F \right ) n}\)	\(96\)
default	\(\frac {\frac {\ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a^{3}}-\frac {\ln \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a^{3}}+\frac {1}{a^{2} \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}+\frac {1}{2 a \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )^{2}}}{g f \ln \left (F \right ) n}\)	\(96\)
risch	\(\frac {\ln \left (F^{g \left (f x +e \right )}\right )}{\ln \left (F \right ) a^{3} f g}+\frac {2 b \left (F^{g \left (f x +e \right )}\right )^{n}+3 a}{2 \ln \left (F \right ) f g n \,a^{2} \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )^{2}}-\frac {\ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}+\frac {a}{b}\right )}{\ln \left (F \right ) a^{3} f g n}\)	\(115\)
norman	\(\frac {\frac {b \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right ) f g n \,a^{2}}+\frac {b^{2} x \,{\mathrm e}^{2 n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{a^{3}}+\frac {x}{a}+\frac {2 b x \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{a^{2}}+\frac {3}{2 \ln \left (F \right ) a f g n}}{\left (a +b \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}\right )^{2}}-\frac {\ln \left (a +b \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}\right )}{\ln \left (F \right ) a^{3} f g n}\)	\(161\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*(F^(g*(f*x+e)))^n)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.35, size = 143, normalized size = 1.29 \begin {gather*} \frac {2 \, F^{f g n x + g n e} b + 3 \, a}{2 \, {\left (2 \, F^{f g n x + g n e} a^{3} b + F^{2 \, f g n x + 2 \, g n e} a^{2} b^{2} + a^{4}\right )} f g n \log \left (F\right )} + \frac {f g n x + g n e}{a^{3} f g n} - \frac {\log \left (F^{f g n x + g n e} b + a\right )}{a^{3} f g n \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*(F^(g*(f*x+e)))^n)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.45, size = 197, normalized size = 1.77 \begin {gather*} \frac {2 \, F^{2 \, f g n x + 2 \, g n e} b^{2} f g n x \log \left (F\right ) + 2 \, a^{2} f g n x \log \left (F\right ) + 2 \, {\left (2 \, a b f g n x \log \left (F\right ) + a b\right )} F^{f g n x + g n e} + 3 \, a^{2} - 2 \, {\left (2 \, F^{f g n x + g n e} a b + F^{2 \, f g n x + 2 \, g n e} b^{2} + a^{2}\right )} \log \left (F^{f g n x + g n e} b + a\right )}{2 \, {\left (2 \, F^{f g n x + g n e} a^{4} b f g n \log \left (F\right ) + F^{2 \, f g n x + 2 \, g n e} a^{3} b^{2} f g n \log \left (F\right ) + a^{5} f g n \log \left (F\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*(F^(g*(f*x+e)))^n)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.09, size = 116, normalized size = 1.05 \begin {gather*} \frac {3 a + 2 b \left (F^{g \left (e + f x\right )}\right )^{n}}{2 a^{4} f g n \log {\left (F \right )} + 4 a^{3} b f g n \left (F^{g \left (e + f x\right )}\right )^{n} \log {\left (F \right )} + 2 a^{2} b^{2} f g n \left (F^{g \left (e + f x\right )}\right )^{2 n} \log {\left (F \right )}} + \frac {x}{a^{3}} - \frac {\log {\left (\frac {a}{b} + \left (F^{g \left (e + f x\right )}\right )^{n} \right )}}{a^{3} f g n \log {\left (F \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*(F**(g*(f*x+e)))**n)**3,x)

[Out]

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

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Giac [A]
time = 1.97, size = 136, normalized size = 1.23 \begin {gather*} \frac {\log \left ({\left | F \right |}^{f g n x} {\left | F \right |}^{g n e}\right )}{a^{3} f g n \log \left (F\right )} - \frac {\log \left ({\left | F^{f g n x} F^{g n e} b + a \right |}\right )}{a^{3} f g n \log \left (F\right )} + \frac {2 \, F^{f g n x} F^{g n e} a b + 3 \, a^{2}}{2 \, {\left (F^{f g n x} F^{g n e} b + a\right )}^{2} a^{3} f g n \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*(F^(g*(f*x+e)))^n)^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 3.50, size = 142, normalized size = 1.28 \begin {gather*} \frac {x}{a^3}+\frac {1}{2\,a\,f\,g\,n\,\ln \left (F\right )\,\left (a^2+b^2\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^{2\,n}+2\,a\,b\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n\right )}+\frac {1}{a^2\,f\,g\,n\,\ln \left (F\right )\,\left (a+b\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n\right )}-\frac {\ln \left (a+b\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n\right )}{a^3\,f\,g\,n\,\ln \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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