3.55 \(\int \frac{x}{b f^{-x}+a f^x} \, dx\)

Optimal. Leaf size=110 \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{b} \log (f)} \]

[Out]

(x*ArcTan[(Sqrt[a]*f^x)/Sqrt[b]])/(Sqrt[a]*Sqrt[b]*Log[f]) - ((I/2)*PolyLog[2, ((-I)*Sqrt[a]*f^x)/Sqrt[b]])/(S
qrt[a]*Sqrt[b]*Log[f]^2) + ((I/2)*PolyLog[2, (I*Sqrt[a]*f^x)/Sqrt[b]])/(Sqrt[a]*Sqrt[b]*Log[f]^2)

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Rubi [A]  time = 0.0952606, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {2282, 205, 2266, 12, 4848, 2391} \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{b} \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[x/(b/f^x + a*f^x),x]

[Out]

(x*ArcTan[(Sqrt[a]*f^x)/Sqrt[b]])/(Sqrt[a]*Sqrt[b]*Log[f]) - ((I/2)*PolyLog[2, ((-I)*Sqrt[a]*f^x)/Sqrt[b]])/(S
qrt[a]*Sqrt[b]*Log[f]^2) + ((I/2)*PolyLog[2, (I*Sqrt[a]*f^x)/Sqrt[b]])/(Sqrt[a]*Sqrt[b]*Log[f]^2)

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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2266

Int[(x_)^(m_.)/((b_.)*(F_)^(v_) + (a_.)*(F_)^((c_.) + (d_.)*(x_))), x_Symbol] :> With[{u = IntHide[1/(a*F^(c +
d*x) + b*F^v), x]}, Simp[x^m*u, x] - Dist[m, Int[x^(m - 1)*u, x], x]] /; FreeQ[{F, a, b, c, d}, x] && EqQ[v,
-(c + d*x)] && GtQ[m, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{x}{b f^{-x}+a f^x} \, dx &=\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{b} \log (f)}-\int \frac{\tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{b} \log (f)} \, dx\\ &=\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{b} \log (f)}-\frac{\int \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right ) \, dx}{\sqrt{a} \sqrt{b} \log (f)}\\ &=\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{b} \log (f)}-\frac{\operatorname{Subst}\left (\int \frac{\tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{x} \, dx,x,f^x\right )}{\sqrt{a} \sqrt{b} \log ^2(f)}\\ &=\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{b} \log (f)}-\frac{i \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{i \sqrt{a} x}{\sqrt{b}}\right )}{x} \, dx,x,f^x\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{i \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{i \sqrt{a} x}{\sqrt{b}}\right )}{x} \, dx,x,f^x\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}\\ &=\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{b} \log (f)}-\frac{i \text{Li}_2\left (-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{i \text{Li}_2\left (\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}\\ \end{align*}

Mathematica [A]  time = 0.062469, size = 108, normalized size = 0.98 \[ \frac{i \left (-\text{PolyLog}\left (2,-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )+\text{PolyLog}\left (2,\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )+x \log (f) \left (\log \left (1-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )-\log \left (1+\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )\right )\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(b/f^x + a*f^x),x]

[Out]

((I/2)*(x*Log[f]*(Log[1 - (I*Sqrt[a]*f^x)/Sqrt[b]] - Log[1 + (I*Sqrt[a]*f^x)/Sqrt[b]]) - PolyLog[2, ((-I)*Sqrt
[a]*f^x)/Sqrt[b]] + PolyLog[2, (I*Sqrt[a]*f^x)/Sqrt[b]]))/(Sqrt[a]*Sqrt[b]*Log[f]^2)

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Maple [A]  time = 0.041, size = 134, normalized size = 1.2 \begin{align*}{\frac{x}{2\,\ln \left ( f \right ) }\ln \left ({ \left ( -a{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}-{\frac{x}{2\,\ln \left ( f \right ) }\ln \left ({ \left ( a{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}+{\frac{1}{2\, \left ( \ln \left ( f \right ) \right ) ^{2}}{\it dilog} \left ({ \left ( -a{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}-{\frac{1}{2\, \left ( \ln \left ( f \right ) \right ) ^{2}}{\it dilog} \left ({ \left ( a{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b/(f^x)+a*f^x),x)

[Out]

1/2/ln(f)*x/(-a*b)^(1/2)*ln((-a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1/2/ln(f)*x/(-a*b)^(1/2)*ln((a*f^x+(-a*b)^(1/2
))/(-a*b)^(1/2))+1/2/ln(f)^2/(-a*b)^(1/2)*dilog((-a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1/2/ln(f)^2/(-a*b)^(1/2)*d
ilog((a*f^x+(-a*b)^(1/2))/(-a*b)^(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(x/(b/(f^x)+a*f^x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72942, size = 247, normalized size = 2.25 \begin{align*} -\frac{x \sqrt{-\frac{a}{b}} \log \left (f^{x} \sqrt{-\frac{a}{b}} + 1\right ) \log \left (f\right ) - x \sqrt{-\frac{a}{b}} \log \left (-f^{x} \sqrt{-\frac{a}{b}} + 1\right ) \log \left (f\right ) - \sqrt{-\frac{a}{b}}{\rm Li}_2\left (f^{x} \sqrt{-\frac{a}{b}}\right ) + \sqrt{-\frac{a}{b}}{\rm Li}_2\left (-f^{x} \sqrt{-\frac{a}{b}}\right )}{2 \, a \log \left (f\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(x*sqrt(-a/b)*log(f^x*sqrt(-a/b) + 1)*log(f) - x*sqrt(-a/b)*log(-f^x*sqrt(-a/b) + 1)*log(f) - sqrt(-a/b)*
dilog(f^x*sqrt(-a/b)) + sqrt(-a/b)*dilog(-f^x*sqrt(-a/b)))/(a*log(f)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{x} x}{a f^{2 x} + b}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b/(f**x)+a*f**x),x)

[Out]

Integral(f**x*x/(a*f**(2*x) + b), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(a*f^x + b/f^x), x)