Integrand size = 19, antiderivative size = 268 \[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=\frac {x^3 \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 i x^2 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x \operatorname {PolyLog}\left (3,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {3 i x \operatorname {PolyLog}\left (3,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {3 i \operatorname {PolyLog}\left (4,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^4(f)}+\frac {3 i \operatorname {PolyLog}\left (4,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^4(f)} \]
x^3*arctan(f^x*a^(1/2)/b^(1/2))/ln(f)/a^(1/2)/b^(1/2)-3/2*I*x^2*polylog(2, -I*f^x*a^(1/2)/b^(1/2))/ln(f)^2/a^(1/2)/b^(1/2)+3/2*I*x^2*polylog(2,I*f^x* a^(1/2)/b^(1/2))/ln(f)^2/a^(1/2)/b^(1/2)+3*I*x*polylog(3,-I*f^x*a^(1/2)/b^ (1/2))/ln(f)^3/a^(1/2)/b^(1/2)-3*I*x*polylog(3,I*f^x*a^(1/2)/b^(1/2))/ln(f )^3/a^(1/2)/b^(1/2)-3*I*polylog(4,-I*f^x*a^(1/2)/b^(1/2))/ln(f)^4/a^(1/2)/ b^(1/2)+3*I*polylog(4,I*f^x*a^(1/2)/b^(1/2))/ln(f)^4/a^(1/2)/b^(1/2)
Time = 0.09 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.84 \[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=\frac {i \left (x^3 \log ^3(f) \log \left (1-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )-x^3 \log ^3(f) \log \left (1+\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )-3 x^2 \log ^2(f) \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )+3 x^2 \log ^2(f) \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )+6 x \log (f) \operatorname {PolyLog}\left (3,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )-6 x \log (f) \operatorname {PolyLog}\left (3,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )-6 \operatorname {PolyLog}\left (4,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )+6 \operatorname {PolyLog}\left (4,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )\right )}{2 \sqrt {a} \sqrt {b} \log ^4(f)} \]
((I/2)*(x^3*Log[f]^3*Log[1 - (I*Sqrt[a]*f^x)/Sqrt[b]] - x^3*Log[f]^3*Log[1 + (I*Sqrt[a]*f^x)/Sqrt[b]] - 3*x^2*Log[f]^2*PolyLog[2, ((-I)*Sqrt[a]*f^x) /Sqrt[b]] + 3*x^2*Log[f]^2*PolyLog[2, (I*Sqrt[a]*f^x)/Sqrt[b]] + 6*x*Log[f ]*PolyLog[3, ((-I)*Sqrt[a]*f^x)/Sqrt[b]] - 6*x*Log[f]*PolyLog[3, (I*Sqrt[a ]*f^x)/Sqrt[b]] - 6*PolyLog[4, ((-I)*Sqrt[a]*f^x)/Sqrt[b]] + 6*PolyLog[4, (I*Sqrt[a]*f^x)/Sqrt[b]]))/(Sqrt[a]*Sqrt[b]*Log[f]^4)
Time = 0.76 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2696, 27, 5666, 3011, 7163, 2720, 7143}
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.
| \(\displaystyle \int \frac {x^3}{a f^x+b f^{-x}} \, dx\) |
| \(\Big \downarrow \) 2696 |
| \(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-3 \int \frac {x^2 \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 \int x^2 \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )dx}{\sqrt {a} \sqrt {b} \log (f)}\) |
| \(\Big \downarrow \) 5666 |
| \(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 \left (\frac {1}{2} i \int x^2 \log \left (1-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )dx-\frac {1}{2} i \int x^2 \log \left (\frac {i \sqrt {a} f^x}{\sqrt {b}}+1\right )dx\right )}{\sqrt {a} \sqrt {b} \log (f)}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 \left (\frac {1}{2} i \left (\frac {2 \int x \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )dx}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}\right )-\frac {1}{2} i \left (\frac {2 \int x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )dx}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}\right )\right )}{\sqrt {a} \sqrt {b} \log (f)}\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 \left (\frac {1}{2} i \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}-\frac {\int \operatorname {PolyLog}\left (3,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )dx}{\log (f)}\right )}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}\right )-\frac {1}{2} i \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}-\frac {\int \operatorname {PolyLog}\left (3,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )dx}{\log (f)}\right )}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}\right )\right )}{\sqrt {a} \sqrt {b} \log (f)}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 \left (\frac {1}{2} i \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}-\frac {\int f^{-x} \operatorname {PolyLog}\left (3,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )df^x}{\log ^2(f)}\right )}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}\right )-\frac {1}{2} i \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}-\frac {\int f^{-x} \operatorname {PolyLog}\left (3,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )df^x}{\log ^2(f)}\right )}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}\right )\right )}{\sqrt {a} \sqrt {b} \log (f)}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 \left (\frac {1}{2} i \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}-\frac {\operatorname {PolyLog}\left (4,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log ^2(f)}\right )}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}\right )-\frac {1}{2} i \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}-\frac {\operatorname {PolyLog}\left (4,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log ^2(f)}\right )}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\log (f)}\right )\right )}{\sqrt {a} \sqrt {b} \log (f)}\) |
(x^3*ArcTan[(Sqrt[a]*f^x)/Sqrt[b]])/(Sqrt[a]*Sqrt[b]*Log[f]) - (3*((-1/2*I )*(-((x^2*PolyLog[2, ((-I)*Sqrt[a]*f^x)/Sqrt[b]])/Log[f]) + (2*((x*PolyLog [3, ((-I)*Sqrt[a]*f^x)/Sqrt[b]])/Log[f] - PolyLog[4, ((-I)*Sqrt[a]*f^x)/Sq rt[b]]/Log[f]^2))/Log[f]) + (I/2)*(-((x^2*PolyLog[2, (I*Sqrt[a]*f^x)/Sqrt[ b]])/Log[f]) + (2*((x*PolyLog[3, (I*Sqrt[a]*f^x)/Sqrt[b]])/Log[f] - PolyLo g[4, (I*Sqrt[a]*f^x)/Sqrt[b]]/Log[f]^2))/Log[f])))/(Sqrt[a]*Sqrt[b]*Log[f] )
3.1.57.3.1 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[(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[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[ArcTan[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] : > Simp[I/2 Int[x^m*Log[1 - I*a - I*b*f^(c + d*x)], x], x] - Simp[I/2 In t[x^m*Log[1 + I*a + I*b*f^(c + d*x)], x], x] /; FreeQ[{a, b, c, d, f}, x] & & IntegerQ[m] && m > 0
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. )*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F])) Int[(e + f*x) ^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c , d, e, f, n, p}, x] && GtQ[m, 0]
\[\int \frac {x^{3}}{b \,f^{-x}+a \,f^{x}}d x\]
Time = 0.30 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.89 \[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=-\frac {x^{3} \sqrt {-\frac {a}{b}} \log \left (f^{x} \sqrt {-\frac {a}{b}} + 1\right ) \log \left (f\right )^{3} - x^{3} \sqrt {-\frac {a}{b}} \log \left (-f^{x} \sqrt {-\frac {a}{b}} + 1\right ) \log \left (f\right )^{3} - 3 \, x^{2} \sqrt {-\frac {a}{b}} {\rm Li}_2\left (f^{x} \sqrt {-\frac {a}{b}}\right ) \log \left (f\right )^{2} + 3 \, x^{2} \sqrt {-\frac {a}{b}} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {a}{b}}\right ) \log \left (f\right )^{2} + 6 \, x \sqrt {-\frac {a}{b}} \log \left (f\right ) {\rm polylog}\left (3, f^{x} \sqrt {-\frac {a}{b}}\right ) - 6 \, x \sqrt {-\frac {a}{b}} \log \left (f\right ) {\rm polylog}\left (3, -f^{x} \sqrt {-\frac {a}{b}}\right ) - 6 \, \sqrt {-\frac {a}{b}} {\rm polylog}\left (4, f^{x} \sqrt {-\frac {a}{b}}\right ) + 6 \, \sqrt {-\frac {a}{b}} {\rm polylog}\left (4, -f^{x} \sqrt {-\frac {a}{b}}\right )}{2 \, a \log \left (f\right )^{4}} \]
-1/2*(x^3*sqrt(-a/b)*log(f^x*sqrt(-a/b) + 1)*log(f)^3 - x^3*sqrt(-a/b)*log (-f^x*sqrt(-a/b) + 1)*log(f)^3 - 3*x^2*sqrt(-a/b)*dilog(f^x*sqrt(-a/b))*lo g(f)^2 + 3*x^2*sqrt(-a/b)*dilog(-f^x*sqrt(-a/b))*log(f)^2 + 6*x*sqrt(-a/b) *log(f)*polylog(3, f^x*sqrt(-a/b)) - 6*x*sqrt(-a/b)*log(f)*polylog(3, -f^x *sqrt(-a/b)) - 6*sqrt(-a/b)*polylog(4, f^x*sqrt(-a/b)) + 6*sqrt(-a/b)*poly log(4, -f^x*sqrt(-a/b)))/(a*log(f)^4)
\[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=\int \frac {f^{x} x^{3}}{a f^{2 x} + b}\, dx \]
\[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=\int { \frac {x^{3}}{a f^{x} + \frac {b}{f^{x}}} \,d x } \]
\[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=\int { \frac {x^{3}}{a f^{x} + \frac {b}{f^{x}}} \,d x } \]
Timed out. \[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=\int \frac {x^3}{\frac {b}{f^x}+a\,f^x} \,d x \]