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

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

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Rubi [A]
time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.00, 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 = {2320, 272, 36, 29, 31} \begin {gather*} \frac {x}{a}-\frac {\log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a f g n \log (F)} \end {gather*}

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

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Mathematica [A]
time = 0.10, size = 55, normalized size = 1.38 \begin {gather*} \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)} \end {gather*}

Antiderivative was successfully verified.

[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])

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Maple [A]
time = 0.03, size = 53, normalized size = 1.32

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 61, normalized size = 1.52 \begin {gather*} \frac {f g n x + g n e}{a f g n} - \frac {\log \left (F^{f g n x + g n e} b + a\right )}{a 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),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 45, normalized size = 1.12 \begin {gather*} \frac {f g n x \log \left (F\right ) - \log \left (F^{f g n x + g n e} b + a\right )}{a 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),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.06, size = 27, normalized size = 0.68 \begin {gather*} \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 {gather*}

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 [A]
time = 2.75, size = 74, normalized size = 1.85 \begin {gather*} \frac {\log \left ({\left | F \right |}^{f g n x} {\left | F \right |}^{g n e}\right )}{a 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 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),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 3.45, size = 44, normalized size = 1.10 \begin {gather*} -\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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