Integrand size = 18, antiderivative size = 268 \[ \int \frac {f^x x^3}{a+b f^{2 x}} \, dx=\frac {x^3 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 i x^2 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {3 i x \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {3 i \operatorname {PolyLog}\left (4,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^4(f)}+\frac {3 i \operatorname {PolyLog}\left (4,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^4(f)} \] Output:
x^3*arctan(b^(1/2)*f^x/a^(1/2))/a^(1/2)/b^(1/2)/ln(f)-3/2*I*x^2*polylog(2, -I*b^(1/2)*f^x/a^(1/2))/a^(1/2)/b^(1/2)/ln(f)^2+3/2*I*x^2*polylog(2,I*b^(1 /2)*f^x/a^(1/2))/a^(1/2)/b^(1/2)/ln(f)^2+3*I*x*polylog(3,-I*b^(1/2)*f^x/a^ (1/2))/a^(1/2)/b^(1/2)/ln(f)^3-3*I*x*polylog(3,I*b^(1/2)*f^x/a^(1/2))/a^(1 /2)/b^(1/2)/ln(f)^3-3*I*polylog(4,-I*b^(1/2)*f^x/a^(1/2))/a^(1/2)/b^(1/2)/ ln(f)^4+3*I*polylog(4,I*b^(1/2)*f^x/a^(1/2))/a^(1/2)/b^(1/2)/ln(f)^4
Time = 0.12 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.84 \[ \int \frac {f^x x^3}{a+b f^{2 x}} \, dx=\frac {i \left (x^3 \log ^3(f) \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-x^3 \log ^3(f) \log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-3 x^2 \log ^2(f) \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+3 x^2 \log ^2(f) \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+6 x \log (f) \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-6 x \log (f) \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-6 \operatorname {PolyLog}\left (4,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+6 \operatorname {PolyLog}\left (4,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )\right )}{2 \sqrt {a} \sqrt {b} \log ^4(f)} \] Input:
Integrate[(f^x*x^3)/(a + b*f^(2*x)),x]
Output:
((I/2)*(x^3*Log[f]^3*Log[1 - (I*Sqrt[b]*f^x)/Sqrt[a]] - x^3*Log[f]^3*Log[1 + (I*Sqrt[b]*f^x)/Sqrt[a]] - 3*x^2*Log[f]^2*PolyLog[2, ((-I)*Sqrt[b]*f^x) /Sqrt[a]] + 3*x^2*Log[f]^2*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[a]] + 6*x*Log[f ]*PolyLog[3, ((-I)*Sqrt[b]*f^x)/Sqrt[a]] - 6*x*Log[f]*PolyLog[3, (I*Sqrt[b ]*f^x)/Sqrt[a]] - 6*PolyLog[4, ((-I)*Sqrt[b]*f^x)/Sqrt[a]] + 6*PolyLog[4, (I*Sqrt[b]*f^x)/Sqrt[a]]))/(Sqrt[a]*Sqrt[b]*Log[f]^4)
Time = 1.30 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2675, 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.
\(\displaystyle \int \frac {x^3 f^x}{a+b f^{2 x}} \, dx\) |
\(\Big \downarrow \) 2675 |
\(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-3 \int \frac {x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 \int x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )dx}{\sqrt {a} \sqrt {b} \log (f)}\) |
\(\Big \downarrow \) 5666 |
\(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 \left (\frac {1}{2} i \int x^2 \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )dx-\frac {1}{2} i \int x^2 \log \left (\frac {i \sqrt {b} f^x}{\sqrt {a}}+1\right )dx\right )}{\sqrt {a} \sqrt {b} \log (f)}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 \left (\frac {1}{2} i \left (\frac {2 \int x \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )dx}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )-\frac {1}{2} i \left (\frac {2 \int x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )dx}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )\right )}{\sqrt {a} \sqrt {b} \log (f)}\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 \left (\frac {1}{2} i \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}-\frac {\int \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )dx}{\log (f)}\right )}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )-\frac {1}{2} i \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}-\frac {\int \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )dx}{\log (f)}\right )}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )\right )}{\sqrt {a} \sqrt {b} \log (f)}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 \left (\frac {1}{2} i \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}-\frac {\int f^{-x} \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )df^x}{\log ^2(f)}\right )}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )-\frac {1}{2} i \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}-\frac {\int f^{-x} \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )df^x}{\log ^2(f)}\right )}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )\right )}{\sqrt {a} \sqrt {b} \log (f)}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {x^3 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 \left (\frac {1}{2} i \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}-\frac {\operatorname {PolyLog}\left (4,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log ^2(f)}\right )}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )-\frac {1}{2} i \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}-\frac {\operatorname {PolyLog}\left (4,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log ^2(f)}\right )}{\log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )\right )}{\sqrt {a} \sqrt {b} \log (f)}\) |
Input:
Int[(f^x*x^3)/(a + b*f^(2*x)),x]
Output:
(x^3*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]) - (3*((-1/2*I )*(-((x^2*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/Log[f]) + (2*((x*PolyLog [3, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/Log[f] - PolyLog[4, ((-I)*Sqrt[b]*f^x)/Sq rt[a]]/Log[f]^2))/Log[f]) + (I/2)*(-((x^2*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[ a]])/Log[f]) + (2*((x*PolyLog[3, (I*Sqrt[b]*f^x)/Sqrt[a]])/Log[f] - PolyLo g[4, (I*Sqrt[b]*f^x)/Sqrt[a]]/Log[f]^2))/Log[f])))/(Sqrt[a]*Sqrt[b]*Log[f] )
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[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]
\[\int \frac {f^{x} x^{3}}{a +b \,f^{2 x}}d x\]
Input:
int(f^x*x^3/(a+b*f^(2*x)),x)
Output:
int(f^x*x^3/(a+b*f^(2*x)),x)
Time = 0.07 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.89 \[ \int \frac {f^x x^3}{a+b f^{2 x}} \, dx=-\frac {x^{3} \sqrt {-\frac {b}{a}} \log \left (f^{x} \sqrt {-\frac {b}{a}} + 1\right ) \log \left (f\right )^{3} - x^{3} \sqrt {-\frac {b}{a}} \log \left (-f^{x} \sqrt {-\frac {b}{a}} + 1\right ) \log \left (f\right )^{3} - 3 \, x^{2} \sqrt {-\frac {b}{a}} {\rm Li}_2\left (f^{x} \sqrt {-\frac {b}{a}}\right ) \log \left (f\right )^{2} + 3 \, x^{2} \sqrt {-\frac {b}{a}} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {b}{a}}\right ) \log \left (f\right )^{2} + 6 \, x \sqrt {-\frac {b}{a}} \log \left (f\right ) {\rm polylog}\left (3, f^{x} \sqrt {-\frac {b}{a}}\right ) - 6 \, x \sqrt {-\frac {b}{a}} \log \left (f\right ) {\rm polylog}\left (3, -f^{x} \sqrt {-\frac {b}{a}}\right ) - 6 \, \sqrt {-\frac {b}{a}} {\rm polylog}\left (4, f^{x} \sqrt {-\frac {b}{a}}\right ) + 6 \, \sqrt {-\frac {b}{a}} {\rm polylog}\left (4, -f^{x} \sqrt {-\frac {b}{a}}\right )}{2 \, b \log \left (f\right )^{4}} \] Input:
integrate(f^x*x^3/(a+b*f^(2*x)),x, algorithm="fricas")
Output:
-1/2*(x^3*sqrt(-b/a)*log(f^x*sqrt(-b/a) + 1)*log(f)^3 - x^3*sqrt(-b/a)*log (-f^x*sqrt(-b/a) + 1)*log(f)^3 - 3*x^2*sqrt(-b/a)*dilog(f^x*sqrt(-b/a))*lo g(f)^2 + 3*x^2*sqrt(-b/a)*dilog(-f^x*sqrt(-b/a))*log(f)^2 + 6*x*sqrt(-b/a) *log(f)*polylog(3, f^x*sqrt(-b/a)) - 6*x*sqrt(-b/a)*log(f)*polylog(3, -f^x *sqrt(-b/a)) - 6*sqrt(-b/a)*polylog(4, f^x*sqrt(-b/a)) + 6*sqrt(-b/a)*poly log(4, -f^x*sqrt(-b/a)))/(b*log(f)^4)
\[ \int \frac {f^x x^3}{a+b f^{2 x}} \, dx=\int \frac {f^{x} x^{3}}{a + b f^{2 x}}\, dx \] Input:
integrate(f**x*x**3/(a+b*f**(2*x)),x)
Output:
Integral(f**x*x**3/(a + b*f**(2*x)), x)
\[ \int \frac {f^x x^3}{a+b f^{2 x}} \, dx=\int { \frac {f^{x} x^{3}}{b f^{2 \, x} + a} \,d x } \] Input:
integrate(f^x*x^3/(a+b*f^(2*x)),x, algorithm="maxima")
Output:
integrate(f^x*x^3/(b*f^(2*x) + a), x)
\[ \int \frac {f^x x^3}{a+b f^{2 x}} \, dx=\int { \frac {f^{x} x^{3}}{b f^{2 \, x} + a} \,d x } \] Input:
integrate(f^x*x^3/(a+b*f^(2*x)),x, algorithm="giac")
Output:
integrate(f^x*x^3/(b*f^(2*x) + a), x)
Timed out. \[ \int \frac {f^x x^3}{a+b f^{2 x}} \, dx=\int \frac {f^x\,x^3}{a+b\,f^{2\,x}} \,d x \] Input:
int((f^x*x^3)/(a + b*f^(2*x)),x)
Output:
int((f^x*x^3)/(a + b*f^(2*x)), x)
\[ \int \frac {f^x x^3}{a+b f^{2 x}} \, dx=\int \frac {f^{x} x^{3}}{f^{2 x} b +a}d x \] Input:
int(f^x*x^3/(a+b*f^(2*x)),x)
Output:
int((f**x*x**3)/(f**(2*x)*b + a),x)