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

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

[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])

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Rubi [A]  time = 0.063461, antiderivative size = 111, 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, 44} \[ -\frac{\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^3 f g n \log (F)}+\frac{1}{a^2 f g n \log (F) \left (a+b \left (F^{g (e+f x)}\right )^n\right )}+\frac{x}{a^3}+\frac{1}{2 a f g n \log (F) \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2} \]

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 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 44

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] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3} \, dx &=\frac{\operatorname{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{\operatorname{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{\operatorname{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*}

Mathematica [A]  time = 0.10563, size = 84, normalized size = 0.76 \[ \frac{\frac{a \left (3 a+2 b \left (F^{g (e+f x)}\right )^n\right )}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2}-2 \log \left (a+b \left (F^{g (e+f x)}\right )^n\right )+2 f g n x \log (F)}{2 a^3 f g n \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 134, normalized size = 1.2 \begin{align*}{\frac{\ln \left ( \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) }{ngf\ln \left ( F \right ){a}^{3}}}-{\frac{\ln \left ( a+b \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) }{ngf\ln \left ( F \right ){a}^{3}}}+{\frac{1}{{a}^{2}f \left ( a+b \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) gn\ln \left ( F \right ) }}+{\frac{1}{2\,af \left ( a+b \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) ^{2}gn\ln \left ( F \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.16756, size = 196, normalized size = 1.77 \begin{align*} \frac{2 \,{\left (F^{f g x + e g}\right )}^{n} b + 3 \, a}{2 \,{\left (2 \,{\left (F^{f g x + e g}\right )}^{n} a^{3} b n +{\left (F^{f g x + e g}\right )}^{2 \, n} a^{2} b^{2} n + a^{4} n\right )} f g \log \left (F\right )} + \frac{\log \left (F^{f g x + e g}\right )}{a^{3} f g \log \left (F\right )} - \frac{\log \left (\frac{{\left (F^{f g x + e g}\right )}^{n} b + a}{b}\right )}{a^{3} f g n \log \left (F\right )} \end{align*}

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

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Fricas [A]  time = 1.59137, size = 464, normalized size = 4.18 \begin{align*} \frac{2 \, F^{2 \, f g n x + 2 \, e g n} 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 + e g n} + 3 \, a^{2} - 2 \,{\left (2 \, F^{f g n x + e g n} a b + F^{2 \, f g n x + 2 \, e g n} b^{2} + a^{2}\right )} \log \left (F^{f g n x + e g n} b + a\right )}{2 \,{\left (2 \, F^{f g n x + e g n} a^{4} b f g n \log \left (F\right ) + F^{2 \, f g n x + 2 \, e g n} a^{3} b^{2} f g n \log \left (F\right ) + a^{5} f g n \log \left (F\right )\right )}} \end{align*}

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

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Sympy [A]  time = 0.226713, size = 116, normalized size = 1.05 \begin{align*} \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{align*}

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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left ({\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a\right )}^{3}}\,{d x} \end{align*}

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]

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