3.540 \(\int \frac{1}{a+b f^{-c-d x}+c f^{c+d x}} \, dx\)

Optimal. Leaf size=47 \[ -\frac{2 \tanh ^{-1}\left (\frac{a+2 c f^{c+d x}}{\sqrt{a^2-4 b c}}\right )}{d \log (f) \sqrt{a^2-4 b c}} \]

[Out]

(-2*ArcTanh[(a + 2*c*f^(c + d*x))/Sqrt[a^2 - 4*b*c]])/(Sqrt[a^2 - 4*b*c]*d*Log[f])

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Rubi [A]  time = 0.0641623, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2282, 1386, 618, 206} \[ -\frac{2 \tanh ^{-1}\left (\frac{a+2 c f^{c+d x}}{\sqrt{a^2-4 b c}}\right )}{d \log (f) \sqrt{a^2-4 b c}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*f^(-c - d*x) + c*f^(c + d*x))^(-1),x]

[Out]

(-2*ArcTanh[(a + 2*c*f^(c + d*x))/Sqrt[a^2 - 4*b*c]])/(Sqrt[a^2 - 4*b*c]*d*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 1386

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n_.) + (b_.)*(x_)^(mn_))^(p_.), x_Symbol] :> Int[x^(m - n*p)*(b + a*x^n + c
*x^(2*n))^p, x] /; FreeQ[{a, b, c, m, n}, x] && EqQ[mn, -n] && IntegerQ[p] && PosQ[n]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{a+b f^{-c-d x}+c f^{c+d x}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x \left (a+\frac{b}{x}+c x\right )} \, dx,x,f^{c+d x}\right )}{d \log (f)}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{b+a x+c x^2} \, dx,x,f^{c+d x}\right )}{d \log (f)}\\ &=-\frac{2 \operatorname{Subst}\left (\int \frac{1}{a^2-4 b c-x^2} \, dx,x,a+2 c f^{c+d x}\right )}{d \log (f)}\\ &=-\frac{2 \tanh ^{-1}\left (\frac{a+2 c f^{c+d x}}{\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c} d \log (f)}\\ \end{align*}

Mathematica [A]  time = 0.0544174, size = 47, normalized size = 1. \[ -\frac{2 \tanh ^{-1}\left (\frac{a+2 c f^{c+d x}}{\sqrt{a^2-4 b c}}\right )}{d \log (f) \sqrt{a^2-4 b c}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*f^(-c - d*x) + c*f^(c + d*x))^(-1),x]

[Out]

(-2*ArcTanh[(a + 2*c*f^(c + d*x))/Sqrt[a^2 - 4*b*c]])/(Sqrt[a^2 - 4*b*c]*d*Log[f])

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Maple [B]  time = 0.029, size = 135, normalized size = 2.9 \begin{align*}{\frac{1}{d\ln \left ( f \right ) }\ln \left ({f}^{-dx-c}+{\frac{1}{2\,b} \left ( a\sqrt{{a}^{2}-4\,bc}+{a}^{2}-4\,bc \right ){\frac{1}{\sqrt{{a}^{2}-4\,bc}}}} \right ){\frac{1}{\sqrt{{a}^{2}-4\,bc}}}}-{\frac{1}{d\ln \left ( f \right ) }\ln \left ({f}^{-dx-c}+{\frac{1}{2\,b} \left ( a\sqrt{{a}^{2}-4\,bc}-{a}^{2}+4\,bc \right ){\frac{1}{\sqrt{{a}^{2}-4\,bc}}}} \right ){\frac{1}{\sqrt{{a}^{2}-4\,bc}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*f^(-d*x-c)+c*f^(d*x+c)),x)

[Out]

1/(a^2-4*b*c)^(1/2)/d/ln(f)*ln(f^(-d*x-c)+1/2*(a*(a^2-4*b*c)^(1/2)+a^2-4*b*c)/b/(a^2-4*b*c)^(1/2))-1/(a^2-4*b*
c)^(1/2)/d/ln(f)*ln(f^(-d*x-c)+1/2*(a*(a^2-4*b*c)^(1/2)-a^2+4*b*c)/b/(a^2-4*b*c)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*f^(-d*x-c)+c*f^(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.37557, size = 424, normalized size = 9.02 \begin{align*} \left [\frac{\log \left (\frac{2 \, c^{2} f^{2 \, d x + 2 \, c} + a^{2} - 2 \, b c + 2 \,{\left (a c - \sqrt{a^{2} - 4 \, b c} c\right )} f^{d x + c} - \sqrt{a^{2} - 4 \, b c} a}{c f^{2 \, d x + 2 \, c} + a f^{d x + c} + b}\right )}{\sqrt{a^{2} - 4 \, b c} d \log \left (f\right )}, -\frac{2 \, \sqrt{-a^{2} + 4 \, b c} \arctan \left (-\frac{2 \, \sqrt{-a^{2} + 4 \, b c} c f^{d x + c} + \sqrt{-a^{2} + 4 \, b c} a}{a^{2} - 4 \, b c}\right )}{{\left (a^{2} - 4 \, b c\right )} d \log \left (f\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*f^(-d*x-c)+c*f^(d*x+c)),x, algorithm="fricas")

[Out]

[log((2*c^2*f^(2*d*x + 2*c) + a^2 - 2*b*c + 2*(a*c - sqrt(a^2 - 4*b*c)*c)*f^(d*x + c) - sqrt(a^2 - 4*b*c)*a)/(
c*f^(2*d*x + 2*c) + a*f^(d*x + c) + b))/(sqrt(a^2 - 4*b*c)*d*log(f)), -2*sqrt(-a^2 + 4*b*c)*arctan(-(2*sqrt(-a
^2 + 4*b*c)*c*f^(d*x + c) + sqrt(-a^2 + 4*b*c)*a)/(a^2 - 4*b*c))/((a^2 - 4*b*c)*d*log(f))]

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Sympy [A]  time = 0.328432, size = 66, normalized size = 1.4 \begin{align*} \operatorname{RootSum}{\left (z^{2} \left (a^{2} d^{2} \log{\left (f \right )}^{2} - 4 b c d^{2} \log{\left (f \right )}^{2}\right ) - 1, \left ( i \mapsto i \log{\left (f^{c + d x} + \frac{- i a^{2} d \log{\left (f \right )} + 4 i b c d \log{\left (f \right )} + a}{2 c} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*f**(-d*x-c)+c*f**(d*x+c)),x)

[Out]

RootSum(_z**2*(a**2*d**2*log(f)**2 - 4*b*c*d**2*log(f)**2) - 1, Lambda(_i, _i*log(f**(c + d*x) + (-_i*a**2*d*l
og(f) + 4*_i*b*c*d*log(f) + a)/(2*c))))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{c f^{d x + c} + b f^{-d x - c} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*f^(-d*x-c)+c*f^(d*x+c)),x, algorithm="giac")

[Out]

integrate(1/(c*f^(d*x + c) + b*f^(-d*x - c) + a), x)