3.55 \(\int \frac{1}{(a+b (F^{g (e+f x)})^n)^2} \, dx\)

Optimal. Leaf size=74 \[ -\frac{\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^2 f g n \log (F)}+\frac{x}{a^2}+\frac{1}{a f g n \log (F) \left (a+b \left (F^{g (e+f x)}\right )^n\right )} \]

[Out]

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

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Rubi [A]  time = 0.0494327, antiderivative size = 74, 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} \[ -\frac{\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^2 f g n \log (F)}+\frac{x}{a^2}+\frac{1}{a f g n \log (F) \left (a+b \left (F^{g (e+f x)}\right )^n\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(F^(g*(e + f*x)))^n)^(-2),x]

[Out]

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

Mathematica [A]  time = 0.0588887, size = 62, normalized size = 0.84 \[ \frac{\frac{a}{a+b \left (F^{g (e+f x)}\right )^n}-\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )+f g n x \log (F)}{a^2 f g n \log (F)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(F^(g*(e + f*x)))^n)^(-2),x]

[Out]

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

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Maple [A]  time = 0.003, size = 99, normalized size = 1.3 \begin{align*}{\frac{\ln \left ( \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) }{ngf\ln \left ( F \right ){a}^{2}}}-{\frac{\ln \left ( a+b \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) }{ngf\ln \left ( F \right ){a}^{2}}}+{\frac{1}{af \left ( a+b \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) 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)^2,x)

[Out]

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

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Maxima [A]  time = 1.02679, size = 135, normalized size = 1.82 \begin{align*} \frac{1}{{\left ({\left (F^{f g x + e g}\right )}^{n} a b n + a^{2} n\right )} f g \log \left (F\right )} + \frac{\log \left (F^{f g x + e g}\right )}{a^{2} f g \log \left (F\right )} - \frac{\log \left (\frac{{\left (F^{f g x + e g}\right )}^{n} b + a}{b}\right )}{a^{2} 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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.56327, size = 248, normalized size = 3.35 \begin{align*} \frac{F^{f g n x + e g n} b f g n x \log \left (F\right ) + a f g n x \log \left (F\right ) -{\left (F^{f g n x + e g n} b + a\right )} \log \left (F^{f g n x + e g n} b + a\right ) + a}{F^{f g n x + e g n} a^{2} b f g n \log \left (F\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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.202329, size = 66, normalized size = 0.89 \begin{align*} \frac{1}{a^{2} f g n \log{\left (F \right )} + a b f g n \left (F^{g \left (e + f x\right )}\right )^{n} \log{\left (F \right )}} + \frac{x}{a^{2}} - \frac{\log{\left (\frac{a}{b} + \left (F^{g \left (e + f x\right )}\right )^{n} \right )}}{a^{2} 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)**2,x)

[Out]

1/(a**2*f*g*n*log(F) + a*b*f*g*n*(F**(g*(e + f*x)))**n*log(F)) + x/a**2 - log(a/b + (F**(g*(e + f*x)))**n)/(a*
*2*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 )}^{2}}\,{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)^2,x, algorithm="giac")

[Out]

integrate(((F^((f*x + e)*g))^n*b + a)^(-2), x)