Integrand size = 16, antiderivative size = 110 \[ \int \frac {f^x x}{a+b f^{2 x}} \, dx=\frac {x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)} \]
x*arctan(f^x*b^(1/2)/a^(1/2))/ln(f)/a^(1/2)/b^(1/2)-1/2*I*polylog(2,-I*f^x *b^(1/2)/a^(1/2))/ln(f)^2/a^(1/2)/b^(1/2)+1/2*I*polylog(2,I*f^x*b^(1/2)/a^ (1/2))/ln(f)^2/a^(1/2)/b^(1/2)
Time = 0.07 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.98 \[ \int \frac {f^x x}{a+b f^{2 x}} \, dx=\frac {i \left (x \log (f) \left (\log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-\log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )\right )-\operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)} \]
((I/2)*(x*Log[f]*(Log[1 - (I*Sqrt[b]*f^x)/Sqrt[a]] - Log[1 + (I*Sqrt[b]*f^ x)/Sqrt[a]]) - PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sqrt[a]] + PolyLog[2, (I*Sqrt [b]*f^x)/Sqrt[a]]))/(Sqrt[a]*Sqrt[b]*Log[f]^2)
Time = 0.38 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2675, 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.
\(\displaystyle \int \frac {x f^x}{a+b f^{2 x}} \, dx\) |
\(\Big \downarrow \) 2675 |
\(\displaystyle \frac {x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\int \frac {\arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {\int \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )dx}{\sqrt {a} \sqrt {b} \log (f)}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {\int f^{-x} \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )df^x}{\sqrt {a} \sqrt {b} \log ^2(f)}\) |
\(\Big \downarrow \) 5355 |
\(\displaystyle \frac {x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {\frac {1}{2} i \int f^{-x} \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )df^x-\frac {1}{2} i \int f^{-x} \log \left (\frac {i \sqrt {b} f^x}{\sqrt {a}}+1\right )df^x}{\sqrt {a} \sqrt {b} \log ^2(f)}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {\frac {1}{2} i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)}\) |
(x*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]) - ((I/2)*PolyLo g[2, ((-I)*Sqrt[b]*f^x)/Sqrt[a]] - (I/2)*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[a ]])/(Sqrt[a]*Sqrt[b]*Log[f]^2)
3.1.44.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(F_)^((e_.)*((c_.) + (d_.)*(x_)))*((a_.) + (b_.)*(F_)^(v_))^(p_)*(x_)^( m_.), x_Symbol] :> With[{u = IntHide[F^(e*(c + d*x))*(a + b*F^v)^p, x]}, Si mp[x^m u, x] - Simp[m Int[x^(m - 1)*u, x], x]] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[v, 2*e*(c + d*x)] && GtQ[m, 0] && ILtQ[p, 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 | \(\frac {x \ln \left (\frac {-b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 \ln \left (f \right ) \sqrt {-a b}}-\frac {x \ln \left (\frac {b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 \ln \left (f \right ) \sqrt {-a b}}+\frac {\operatorname {dilog}\left (\frac {-b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 \ln \left (f \right )^{2} \sqrt {-a b}}-\frac {\operatorname {dilog}\left (\frac {b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 \ln \left (f \right )^{2} \sqrt {-a b}}\) | \(134\) |
1/2/ln(f)*x/(-a*b)^(1/2)*ln((-b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1/2/ln(f)* x/(-a*b)^(1/2)*ln((b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))+1/2/ln(f)^2/(-a*b)^(1 /2)*dilog((-b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1/2/ln(f)^2/(-a*b)^(1/2)*dil og((b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))
Time = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.02 \[ \int \frac {f^x x}{a+b f^{2 x}} \, dx=-\frac {x \sqrt {-\frac {b}{a}} \log \left (f^{x} \sqrt {-\frac {b}{a}} + 1\right ) \log \left (f\right ) - x \sqrt {-\frac {b}{a}} \log \left (-f^{x} \sqrt {-\frac {b}{a}} + 1\right ) \log \left (f\right ) - \sqrt {-\frac {b}{a}} {\rm Li}_2\left (f^{x} \sqrt {-\frac {b}{a}}\right ) + \sqrt {-\frac {b}{a}} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {b}{a}}\right )}{2 \, b \log \left (f\right )^{2}} \]
-1/2*(x*sqrt(-b/a)*log(f^x*sqrt(-b/a) + 1)*log(f) - x*sqrt(-b/a)*log(-f^x* sqrt(-b/a) + 1)*log(f) - sqrt(-b/a)*dilog(f^x*sqrt(-b/a)) + sqrt(-b/a)*dil og(-f^x*sqrt(-b/a)))/(b*log(f)^2)
\[ \int \frac {f^x x}{a+b f^{2 x}} \, dx=\int \frac {f^{x} x}{a + b f^{2 x}}\, dx \]
\[ \int \frac {f^x x}{a+b f^{2 x}} \, dx=\int { \frac {f^{x} x}{b f^{2 \, x} + a} \,d x } \]
\[ \int \frac {f^x x}{a+b f^{2 x}} \, dx=\int { \frac {f^{x} x}{b f^{2 \, x} + a} \,d x } \]
Timed out. \[ \int \frac {f^x x}{a+b f^{2 x}} \, dx=\int \frac {f^x\,x}{a+b\,f^{2\,x}} \,d x \]