Integrand size = 17, antiderivative size = 196 \[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^3} \, dx=\frac {f^x}{8 a b \left (b+a f^{2 x}\right ) \log ^2(f)}-\frac {f^x x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)} \]
1/8*f^x/a/b/(b+a*f^(2*x))/ln(f)^2-1/4*f^x*x/a/(b+a*f^(2*x))^2/ln(f)+1/8*f^ x*x/a/b/(b+a*f^(2*x))/ln(f)+1/8*x*arctan(f^x*a^(1/2)/b^(1/2))/a^(3/2)/b^(3 /2)/ln(f)-1/16*I*polylog(2,-I*f^x*a^(1/2)/b^(1/2))/a^(3/2)/b^(3/2)/ln(f)^2 +1/16*I*polylog(2,I*f^x*a^(1/2)/b^(1/2))/a^(3/2)/b^(3/2)/ln(f)^2
Time = 0.19 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.07 \[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^3} \, dx=\frac {\frac {2 \sqrt {a} f^x}{b^2+a b f^{2 x}}-\frac {4 \sqrt {a} f^x x \log (f)}{\left (b+a f^{2 x}\right )^2}+\frac {2 \sqrt {a} f^x x \log (f)}{b^2+a b f^{2 x}}+\frac {i x \log (f) \log \left (1-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{b^{3/2}}-\frac {i x \log (f) \log \left (1+\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{b^{3/2}}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{b^{3/2}}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{b^{3/2}}}{16 a^{3/2} \log ^2(f)} \]
((2*Sqrt[a]*f^x)/(b^2 + a*b*f^(2*x)) - (4*Sqrt[a]*f^x*x*Log[f])/(b + a*f^( 2*x))^2 + (2*Sqrt[a]*f^x*x*Log[f])/(b^2 + a*b*f^(2*x)) + (I*x*Log[f]*Log[1 - (I*Sqrt[a]*f^x)/Sqrt[b]])/b^(3/2) - (I*x*Log[f]*Log[1 + (I*Sqrt[a]*f^x) /Sqrt[b]])/b^(3/2) - (I*PolyLog[2, ((-I)*Sqrt[a]*f^x)/Sqrt[b]])/b^(3/2) + (I*PolyLog[2, (I*Sqrt[a]*f^x)/Sqrt[b]])/b^(3/2))/(16*a^(3/2)*Log[f]^2)
Time = 0.69 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2721, 2684, 2009}
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.
| \(\displaystyle \int \frac {x}{\left (a f^x+b f^{-x}\right )^3} \, dx\) |
| \(\Big \downarrow \) 2721 |
| \(\displaystyle \int \frac {x f^{3 x}}{\left (a f^{2 x}+b\right )^3}dx\) |
| \(\Big \downarrow \) 2684 |
| \(\displaystyle \int \left (\frac {x f^x}{a \left (a f^{2 x}+b\right )^2}-\frac {b x f^x}{a \left (a f^{2 x}+b\right )^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac {f^x}{8 a b \log ^2(f) \left (a f^{2 x}+b\right )}+\frac {x f^x}{8 a b \log (f) \left (a f^{2 x}+b\right )}-\frac {x f^x}{4 a \log (f) \left (a f^{2 x}+b\right )^2}\) |
f^x/(8*a*b*(b + a*f^(2*x))*Log[f]^2) - (f^x*x)/(4*a*(b + a*f^(2*x))^2*Log[ f]) + (f^x*x)/(8*a*b*(b + a*f^(2*x))*Log[f]) + (x*ArcTan[(Sqrt[a]*f^x)/Sqr t[b]])/(8*a^(3/2)*b^(3/2)*Log[f]) - ((I/16)*PolyLog[2, ((-I)*Sqrt[a]*f^x)/ Sqrt[b]])/(a^(3/2)*b^(3/2)*Log[f]^2) + ((I/16)*PolyLog[2, (I*Sqrt[a]*f^x)/ Sqrt[b]])/(a^(3/2)*b^(3/2)*Log[f]^2)
3.1.63.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(F_)^(u_))^(p_.)*((c_.) + (d_.)*(F_)^(v_))^(q_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> With[{w = ExpandIntegrand[(e + f*x)^m, (a + b*F^u)^p*(c + d*F^v)^q, x]}, Int[w, x] /; SumQ[w]] /; FreeQ[{F, a, b, c, d, e, f, m}, x] && IntegersQ[p, q] && LinearQ[{u, v}, x] && RationalQ[Simp lify[u/v]]
Int[(u_.)*((a_.)*(F_)^(v_) + (b_.)*(F_)^(w_))^(n_), x_Symbol] :> Int[u*F^(n *v)*(a + b*F^ExpandToSum[w - v, x])^n, x] /; FreeQ[{F, a, b, n}, x] && ILtQ [n, 0] && LinearQ[{v, w}, x]
Time = 0.15 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.07
| method | result | size |
| risch | \(\frac {f^{x} \left (f^{2 x} \ln \left (f \right ) a x -\ln \left (f \right ) b x +a \,f^{2 x}+b \right )}{8 \ln \left (f \right )^{2} b a \left (b +a \,f^{2 x}\right )^{2}}+\frac {x \ln \left (\frac {-a \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{16 \ln \left (f \right ) a b \sqrt {-a b}}-\frac {x \ln \left (\frac {a \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{16 \ln \left (f \right ) a b \sqrt {-a b}}+\frac {\operatorname {dilog}\left (\frac {-a \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{16 \ln \left (f \right )^{2} a b \sqrt {-a b}}-\frac {\operatorname {dilog}\left (\frac {a \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{16 \ln \left (f \right )^{2} a b \sqrt {-a b}}\) | \(209\) |
1/8*f^x*((f^x)^2*ln(f)*a*x-ln(f)*b*x+(f^x)^2*a+b)/ln(f)^2/b/a/((f^x)^2*a+b )^2+1/16/ln(f)/a/b*x/(-a*b)^(1/2)*ln((-a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1 /16/ln(f)/a/b*x/(-a*b)^(1/2)*ln((a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))+1/16/ln (f)^2/a/b/(-a*b)^(1/2)*dilog((-a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1/16/ln(f )^2/a/b/(-a*b)^(1/2)*dilog((a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (150) = 300\).
Time = 0.28 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.80 \[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^3} \, dx=\frac {2 \, {\left (a^{2} x \log \left (f\right ) + a^{2}\right )} f^{3 \, x} - 2 \, {\left (a b x \log \left (f\right ) - a b\right )} f^{x} + {\left (a^{2} f^{4 \, x} \sqrt {-\frac {a}{b}} + 2 \, a b f^{2 \, x} \sqrt {-\frac {a}{b}} + b^{2} \sqrt {-\frac {a}{b}}\right )} {\rm Li}_2\left (f^{x} \sqrt {-\frac {a}{b}}\right ) - {\left (a^{2} f^{4 \, x} \sqrt {-\frac {a}{b}} + 2 \, a b f^{2 \, x} \sqrt {-\frac {a}{b}} + b^{2} \sqrt {-\frac {a}{b}}\right )} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {a}{b}}\right ) - {\left (a^{2} f^{4 \, x} x \sqrt {-\frac {a}{b}} \log \left (f\right ) + 2 \, a b f^{2 \, x} x \sqrt {-\frac {a}{b}} \log \left (f\right ) + b^{2} x \sqrt {-\frac {a}{b}} \log \left (f\right )\right )} \log \left (f^{x} \sqrt {-\frac {a}{b}} + 1\right ) + {\left (a^{2} f^{4 \, x} x \sqrt {-\frac {a}{b}} \log \left (f\right ) + 2 \, a b f^{2 \, x} x \sqrt {-\frac {a}{b}} \log \left (f\right ) + b^{2} x \sqrt {-\frac {a}{b}} \log \left (f\right )\right )} \log \left (-f^{x} \sqrt {-\frac {a}{b}} + 1\right )}{16 \, {\left (a^{4} b f^{4 \, x} \log \left (f\right )^{2} + 2 \, a^{3} b^{2} f^{2 \, x} \log \left (f\right )^{2} + a^{2} b^{3} \log \left (f\right )^{2}\right )}} \]
1/16*(2*(a^2*x*log(f) + a^2)*f^(3*x) - 2*(a*b*x*log(f) - a*b)*f^x + (a^2*f ^(4*x)*sqrt(-a/b) + 2*a*b*f^(2*x)*sqrt(-a/b) + b^2*sqrt(-a/b))*dilog(f^x*s qrt(-a/b)) - (a^2*f^(4*x)*sqrt(-a/b) + 2*a*b*f^(2*x)*sqrt(-a/b) + b^2*sqrt (-a/b))*dilog(-f^x*sqrt(-a/b)) - (a^2*f^(4*x)*x*sqrt(-a/b)*log(f) + 2*a*b* f^(2*x)*x*sqrt(-a/b)*log(f) + b^2*x*sqrt(-a/b)*log(f))*log(f^x*sqrt(-a/b) + 1) + (a^2*f^(4*x)*x*sqrt(-a/b)*log(f) + 2*a*b*f^(2*x)*x*sqrt(-a/b)*log(f ) + b^2*x*sqrt(-a/b)*log(f))*log(-f^x*sqrt(-a/b) + 1))/(a^4*b*f^(4*x)*log( f)^2 + 2*a^3*b^2*f^(2*x)*log(f)^2 + a^2*b^3*log(f)^2)
\[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^3} \, dx=\frac {f^{- x} \left (a x \log {\left (f \right )} + a\right ) + f^{- 3 x} \left (- b x \log {\left (f \right )} + b\right )}{8 a^{3} b \log {\left (f \right )}^{2} + 16 a^{2} b^{2} f^{- 2 x} \log {\left (f \right )}^{2} + 8 a b^{3} f^{- 4 x} \log {\left (f \right )}^{2}} + \frac {\int \frac {f^{x} x}{a f^{2 x} + b}\, dx}{8 a b} \]
((a*x*log(f) + a)/f**x + (-b*x*log(f) + b)/f**(3*x))/(8*a**3*b*log(f)**2 + 16*a**2*b**2*log(f)**2/f**(2*x) + 8*a*b**3*log(f)**2/f**(4*x)) + Integral (f**x*x/(a*f**(2*x) + b), x)/(8*a*b)
\[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^3} \, dx=\int { \frac {x}{{\left (a f^{x} + \frac {b}{f^{x}}\right )}^{3}} \,d x } \]
1/8*((a*x*log(f) + a)*f^(3*x) - (b*x*log(f) - b)*f^x)/(a^3*b*f^(4*x)*log(f )^2 + 2*a^2*b^2*f^(2*x)*log(f)^2 + a*b^3*log(f)^2) + integrate(1/8*f^x*x/( a^2*b*f^(2*x) + a*b^2), x)
\[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^3} \, dx=\int { \frac {x}{{\left (a f^{x} + \frac {b}{f^{x}}\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x}{\left (b f^{-x}+a f^x\right )^3} \, dx=\int \frac {x}{{\left (\frac {b}{f^x}+a\,f^x\right )}^3} \,d x \]