3.1.55 \(\int \frac {1}{(a+b (F^{g (e+f x)})^n)^2} \, dx\) [55]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, 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 = {2320, 272, 46} \begin {gather*} -\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 )} \end {gather*}

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 46

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] && Lt
Q[m + n + 2, 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}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2} \, dx &=\frac {\text {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 {\text {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 {\text {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*}

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Mathematica [A]
time = 0.16, size = 107, normalized size = 1.45 \begin {gather*} \frac {\frac {1}{a f \left (a+b \left (F^{g (e+f x)}\right )^n\right ) g n}+\frac {\log \left (\left (F^{g (e+f x)}\right )^n\right )}{a^2 f g n}-\frac {\log \left (a^3 f g n \log (F)+a^2 b f \left (F^{g (e+f x)}\right )^n g n \log (F)\right )}{a^2 f g n}}{\log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.04, size = 74, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a^{2}}-\frac {\ln \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a^{2}}+\frac {1}{a \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}}{g f \ln \left (F \right ) n}\)	\(74\)
default	\(\frac {\frac {\ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a^{2}}-\frac {\ln \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{a^{2}}+\frac {1}{a \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}}{g f \ln \left (F \right ) n}\)	\(74\)
risch	\(\frac {\ln \left (F^{g \left (f x +e \right )}\right )}{\ln \left (F \right ) f g \,a^{2}}+\frac {1}{a f \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right ) g n \ln \left (F \right )}-\frac {\ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}+\frac {a}{b}\right )}{\ln \left (F \right ) f g n \,a^{2}}\)	\(96\)
norman	\(\frac {-\frac {b \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right ) f g n \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right ) {\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{a^{2} \ln \left (F \right ) f g}+\frac {\ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}{\ln \left (F \right ) a f g}}{a +b \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}-\frac {\ln \left (a +b \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}\right )}{\ln \left (F \right ) f g n \,a^{2}}\)	\(159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*(F^(g*(f*x+e)))^n)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 97, normalized size = 1.31 \begin {gather*} \frac {f g n x + g n e}{a^{2} f g n} + \frac {1}{{\left (F^{f g n x + g n e} a b + a^{2}\right )} f g n \log \left (F\right )} - \frac {\log \left (F^{f g n x + g n e} b + a\right )}{a^{2} 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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 104, normalized size = 1.41 \begin {gather*} \frac {F^{f g n x + g n e} b f g n x \log \left (F\right ) + a f g n x \log \left (F\right ) - {\left (F^{f g n x + g n e} b + a\right )} \log \left (F^{f g n x + g n e} b + a\right ) + a}{F^{f g n x + g n e} a^{2} b f g n \log \left (F\right ) + a^{3} 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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.08, size = 66, normalized size = 0.89 \begin {gather*} \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 {gather*}

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 [A]
time = 2.27, size = 111, normalized size = 1.50 \begin {gather*} \frac {\log \left ({\left | F \right |}^{f g n x} {\left | F \right |}^{g n e}\right )}{a^{2} 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^{2} f g n \log \left (F\right )} + \frac {1}{{\left (F^{f g n x} F^{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)^2,x, algorithm="giac")

[Out]

log(abs(F)^(f*g*n*x)*abs(F)^(g*n*e))/(a^2*f*g*n*log(F)) - log(abs(F^(f*g*n*x)*F^(g*n*e)*b + a))/(a^2*f*g*n*log
(F)) + 1/((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 = 80, normalized size = 1.08 \begin {gather*} \frac {x}{a^2}+\frac {1}{a\,f\,g\,n\,\ln \left (F\right )\,\left (a+b\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n\right )}-\frac {\ln \left (a+b\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n\right )}{a^2\,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)^2,x)

[Out]

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

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