\(\int \frac {1}{a+b (F^{g (e+f x)})^n} \, dx\) [49]

Optimal result

Integrand size = 17, antiderivative size = 40 \[ \int \frac {1}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {x}{a}-\frac {\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a f g n \log (F)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2320, 272, 36, 29, 31} \[ \int \frac {1}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {x}{a}-\frac {\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a f g n \log (F)} \]

[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 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]

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 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*} \text {integral}& = \frac {\text {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 {\text {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 {\text {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 \text {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] (verified)

Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.38 \[ \int \frac {1}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {\log \left (\left (F^{g (e+f x)}\right )^n\right )-\log \left (a f \left (a+b \left (F^{g (e+f x)}\right )^n\right ) g n \log (F)\right )}{a f g n \log (F)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.10

method	result	size
norman	\(\frac {x}{a}-\frac {\ln \left (a +b \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}\right )}{\ln \left (F \right ) a f g n}\)	\(44\)
parallelrisch	\(\frac {n g f \ln \left (F \right ) x -\ln \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{\ln \left (F \right ) f g a n}\)	\(44\)
derivativedivides	\(\frac {-\frac {\ln \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a}+\frac {\ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a}}{g f \ln \left (F \right ) n}\)	\(53\)
default	\(\frac {-\frac {\ln \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a}+\frac {\ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a}}{g f \ln \left (F \right ) n}\)	\(53\)
risch	\(\frac {\ln \left (F^{g \left (f x +e \right )}\right )}{\ln \left (F \right ) a f g}-\frac {\ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}+\frac {a}{b}\right )}{\ln \left (F \right ) a f g n}\)	\(62\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.10 \[ \int \frac {1}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {f g n x \log \left (F\right ) - \log \left (F^{f g n x + e g n} b + a\right )}{a f g n \log \left (F\right )} \]

[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))

Sympy [A] (verification not implemented)

Time = 1.89 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.28 \[ \int \frac {1}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {2 \operatorname {atan}{\left (\frac {2 \left (\frac {a}{2 b} + \left (F^{g \left (e + f x\right )}\right )^{n}\right )}{\sqrt {- \frac {a^{2}}{b^{2}}}} \right )}}{b f g n \sqrt {- \frac {a^{2}}{b^{2}}} \log {\left (F \right )}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.48 \[ \int \frac {1}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {f g n x + e g n}{a f g n} - \frac {\log \left (F^{f g n x + e g n} b + a\right )}{a f g n \log \left (F\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.80 \[ \int \frac {1}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {\log \left ({\left | F \right |}^{f g n x} {\left | F \right |}^{e g n}\right )}{a f g n \log \left (F\right )} - \frac {\log \left ({\left | F^{f g n x} F^{e g n} b + a \right |}\right )}{a f g n \log \left (F\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.10 \[ \int \frac {1}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=-\frac {\ln \left (a+b\,{\left (F^{e\,g+f\,g\,x}\right )}^n\right )-f\,g\,n\,x\,\ln \left (F\right )}{a\,f\,g\,n\,\ln \left (F\right )} \]

[In]

int(1/(a + b*(F^(g*(e + f*x)))^n),x)

[Out]

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