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

Optimal. Leaf size=40 \[ \frac{x}{a}-\frac{\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a f g n \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.0300728, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {2282, 266, 36, 29, 31} \[ \frac{x}{a}-\frac{\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a f g n \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(F^(g*(e + f*x)))^n)^(-1),x]

[Out]

x/a - Log[a + b*(F^(g*(e + f*x)))^n]/(a*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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0061125, size = 40, normalized size = 1. \[ \frac{x}{a}-\frac{\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a f g n \log (F)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(F^(g*(e + f*x)))^n)^(-1),x]

[Out]

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

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Maple [A]  time = 0.005, size = 65, normalized size = 1.6 \begin{align*}{\frac{\ln \left ( \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) }{\ln \left ( F \right ) afgn}}-{\frac{\ln \left ( a+b \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) }{\ln \left ( F \right ) afgn}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06782, size = 89, normalized size = 2.22 \begin{align*} -\frac{\log \left ({\left (F^{f g x + e g}\right )}^{n} b + a\right )}{a f g n \log \left (F\right )} + \frac{\log \left ({\left (F^{f g x + e g}\right )}^{n}\right )}{a 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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.79227, size = 92, normalized size = 2.3 \begin{align*} \frac{f g n x \log \left (F\right ) - \log \left (F^{f g n x + e g n} b + a\right )}{a 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),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.200935, size = 27, normalized size = 0.68 \begin{align*} \frac{x}{a} - \frac{\log{\left (\frac{a}{b} + \left (F^{g \left (e + f x\right )}\right )^{n} \right )}}{a 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),x)

[Out]

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

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

[Out]

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