Integrand size = 17, antiderivative size = 110 \[ \int \frac {x}{b f^{-x}+a f^x} \, dx=\frac {x \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)} \] Output:
x*arctan(a^(1/2)*f^x/b^(1/2))/a^(1/2)/b^(1/2)/ln(f)-1/2*I*polylog(2,-I*a^( 1/2)*f^x/b^(1/2))/a^(1/2)/b^(1/2)/ln(f)^2+1/2*I*polylog(2,I*a^(1/2)*f^x/b^ (1/2))/a^(1/2)/b^(1/2)/ln(f)^2
Time = 0.12 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.98 \[ \int \frac {x}{b f^{-x}+a f^x} \, dx=\frac {i \left (x \log (f) \left (\log \left (1-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )-\log \left (1+\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )\right )-\operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)} \] Input:
Integrate[x/(b/f^x + a*f^x),x]
Output:
((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)
Time = 0.58 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2696, 27, 2720, 5355, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x}{a f^x+b f^{-x}} \, dx\) |
\(\Big \downarrow \) 2696 |
\(\displaystyle \frac {x \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\int \frac {\arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {\int \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )dx}{\sqrt {a} \sqrt {b} \log (f)}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {x \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {\int f^{-x} \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )df^x}{\sqrt {a} \sqrt {b} \log ^2(f)}\) |
\(\Big \downarrow \) 5355 |
\(\displaystyle \frac {x \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {\frac {1}{2} i \int f^{-x} \log \left (1-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )df^x-\frac {1}{2} i \int f^{-x} \log \left (\frac {i \sqrt {a} f^x}{\sqrt {b}}+1\right )df^x}{\sqrt {a} \sqrt {b} \log ^2(f)}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {x \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {\frac {1}{2} i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)}\) |
Input:
Int[x/(b/f^x + a*f^x),x]
Output:
(x*ArcTan[(Sqrt[a]*f^x)/Sqrt[b]])/(Sqrt[a]*Sqrt[b]*Log[f]) - ((I/2)*PolyLo g[2, ((-I)*Sqrt[a]*f^x)/Sqrt[b]] - (I/2)*PolyLog[2, (I*Sqrt[a]*f^x)/Sqrt[b ]])/(Sqrt[a]*Sqrt[b]*Log[f]^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)^(m_.)/((b_.)*(F_)^(v_) + (a_.)*(F_)^((c_.) + (d_.)*(x_))), x_Symbo l] :> With[{u = IntHide[1/(a*F^(c + d*x) + b*F^v), x]}, Simp[x^m*u, x] - Si mp[m Int[x^(m - 1)*u, x], x]] /; FreeQ[{F, a, b, c, d}, x] && EqQ[v, -(c + d*x)] && GtQ[m, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
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]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Simp[I*(b/2) Int[Log[1 - I*c*x]/x, x], x] - Simp[I*(b/2) Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]
Time = 0.06 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.22
method | result | size |
risch | \(\frac {x \ln \left (\frac {-a \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 \ln \left (f \right ) \sqrt {-a b}}-\frac {x \ln \left (\frac {a \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 \ln \left (f \right ) \sqrt {-a b}}+\frac {\operatorname {dilog}\left (\frac {-a \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 \ln \left (f \right )^{2} \sqrt {-a b}}-\frac {\operatorname {dilog}\left (\frac {a \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 \ln \left (f \right )^{2} \sqrt {-a b}}\) | \(134\) |
Input:
int(x/(b/(f^x)+a*f^x),x,method=_RETURNVERBOSE)
Output:
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)*dil og((a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))
Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.02 \[ \int \frac {x}{b f^{-x}+a f^x} \, dx=-\frac {x \sqrt {-\frac {a}{b}} \log \left (f^{x} \sqrt {-\frac {a}{b}} + 1\right ) \log \left (f\right ) - x \sqrt {-\frac {a}{b}} \log \left (-f^{x} \sqrt {-\frac {a}{b}} + 1\right ) \log \left (f\right ) - \sqrt {-\frac {a}{b}} {\rm Li}_2\left (f^{x} \sqrt {-\frac {a}{b}}\right ) + \sqrt {-\frac {a}{b}} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {a}{b}}\right )}{2 \, a \log \left (f\right )^{2}} \] Input:
integrate(x/(b/(f^x)+a*f^x),x, algorithm="fricas")
Output:
-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)*dil og(-f^x*sqrt(-a/b)))/(a*log(f)^2)
\[ \int \frac {x}{b f^{-x}+a f^x} \, dx=\int \frac {f^{x} x}{a f^{2 x} + b}\, dx \] Input:
integrate(x/(b/(f**x)+a*f**x),x)
Output:
Integral(f**x*x/(a*f**(2*x) + b), x)
\[ \int \frac {x}{b f^{-x}+a f^x} \, dx=\int { \frac {x}{a f^{x} + \frac {b}{f^{x}}} \,d x } \] Input:
integrate(x/(b/(f^x)+a*f^x),x, algorithm="maxima")
Output:
integrate(x/(a*f^x + b/f^x), x)
\[ \int \frac {x}{b f^{-x}+a f^x} \, dx=\int { \frac {x}{a f^{x} + \frac {b}{f^{x}}} \,d x } \] Input:
integrate(x/(b/(f^x)+a*f^x),x, algorithm="giac")
Output:
integrate(x/(a*f^x + b/f^x), x)
Timed out. \[ \int \frac {x}{b f^{-x}+a f^x} \, dx=\int \frac {x}{\frac {b}{f^x}+a\,f^x} \,d x \] Input:
int(x/(b/f^x + a*f^x),x)
Output:
int(x/(b/f^x + a*f^x), x)
\[ \int \frac {x}{b f^{-x}+a f^x} \, dx=\int \frac {f^{x} x}{f^{2 x} a +b}d x \] Input:
int(x/(b/(f^x)+a*f^x),x)
Output:
int((f**x*x)/(f**(2*x)*a + b),x)