\(\int \frac {f^x x^2}{a+b f^{2 x}} \, dx\) [101]

Optimal result

Integrand size = 18, antiderivative size = 184 \[ \int \frac {f^x x^2}{a+b f^{2 x}} \, dx=\frac {x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {i x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i x \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {i \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)} \] Output:

x^2*arctan(b^(1/2)*f^x/a^(1/2))/a^(1/2)/b^(1/2)/ln(f)-I*x*polylog(2,-I*b^( 
1/2)*f^x/a^(1/2))/a^(1/2)/b^(1/2)/ln(f)^2+I*x*polylog(2,I*b^(1/2)*f^x/a^(1 
/2))/a^(1/2)/b^(1/2)/ln(f)^2+I*polylog(3,-I*b^(1/2)*f^x/a^(1/2))/a^(1/2)/b 
^(1/2)/ln(f)^3-I*polylog(3,I*b^(1/2)*f^x/a^(1/2))/a^(1/2)/b^(1/2)/ln(f)^3
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.91 \[ \int \frac {f^x x^2}{a+b f^{2 x}} \, dx=\frac {i \left (x^2 \log ^2(f) \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-x^2 \log ^2(f) \log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-2 x \log (f) \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+2 x \log (f) \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+2 \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-2 \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )\right )}{2 \sqrt {a} \sqrt {b} \log ^3(f)} \] Input:

Integrate[(f^x*x^2)/(a + b*f^(2*x)),x]
 

Output:

((I/2)*(x^2*Log[f]^2*Log[1 - (I*Sqrt[b]*f^x)/Sqrt[a]] - x^2*Log[f]^2*Log[1 
 + (I*Sqrt[b]*f^x)/Sqrt[a]] - 2*x*Log[f]*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sqr 
t[a]] + 2*x*Log[f]*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[a]] + 2*PolyLog[3, ((-I 
)*Sqrt[b]*f^x)/Sqrt[a]] - 2*PolyLog[3, (I*Sqrt[b]*f^x)/Sqrt[a]]))/(Sqrt[a] 
*Sqrt[b]*Log[f]^3)
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2675, 27, 5666, 3011, 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.

\(\displaystyle \int \frac {x^2 f^x}{a+b f^{2 x}} \, dx\)

\(\Big \downarrow \) 2675

\(\displaystyle \frac {x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-2 \int \frac {x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {2 \int x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )dx}{\sqrt {a} \sqrt {b} \log (f)}\)

\(\Big \downarrow \) 5666

\(\displaystyle \frac {x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {2 \left (\frac {1}{2} i \int x \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )dx-\frac {1}{2} i \int x \log \left (\frac {i \sqrt {b} f^x}{\sqrt {a}}+1\right )dx\right )}{\sqrt {a} \sqrt {b} \log (f)}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {2 \left (\frac {1}{2} i \left (\frac {\int \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )dx}{\log (f)}-\frac {x \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )-\frac {1}{2} i \left (\frac {\int \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )dx}{\log (f)}-\frac {x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )\right )}{\sqrt {a} \sqrt {b} \log (f)}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {2 \left (\frac {1}{2} i \left (\frac {\int f^{-x} \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )df^x}{\log ^2(f)}-\frac {x \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )-\frac {1}{2} i \left (\frac {\int f^{-x} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )df^x}{\log ^2(f)}-\frac {x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )\right )}{\sqrt {a} \sqrt {b} \log (f)}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {2 \left (\frac {1}{2} i \left (\frac {\operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log ^2(f)}-\frac {x \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )-\frac {1}{2} i \left (\frac {\operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log ^2(f)}-\frac {x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\log (f)}\right )\right )}{\sqrt {a} \sqrt {b} \log (f)}\)

Input:

Int[(f^x*x^2)/(a + b*f^(2*x)),x]
 

Output:

(x^2*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]) - (2*((-1/2*I 
)*(-((x*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/Log[f]) + PolyLog[3, ((-I) 
*Sqrt[b]*f^x)/Sqrt[a]]/Log[f]^2) + (I/2)*(-((x*PolyLog[2, (I*Sqrt[b]*f^x)/ 
Sqrt[a]])/Log[f]) + PolyLog[3, (I*Sqrt[b]*f^x)/Sqrt[a]]/Log[f]^2)))/(Sqrt[ 
a]*Sqrt[b]*Log[f])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2675

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]
 

rule 2720

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]]]
 

rule 3011

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]
 

rule 5666

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
 

rule 7143

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]
 

Maple [F]

\[\int \frac {f^{x} x^{2}}{a +b \,f^{2 x}}d x\]

Input:

int(f^x*x^2/(a+b*f^(2*x)),x)
 

Output:

int(f^x*x^2/(a+b*f^(2*x)),x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.96 \[ \int \frac {f^x x^2}{a+b f^{2 x}} \, dx=-\frac {x^{2} \sqrt {-\frac {b}{a}} \log \left (f^{x} \sqrt {-\frac {b}{a}} + 1\right ) \log \left (f\right )^{2} - x^{2} \sqrt {-\frac {b}{a}} \log \left (-f^{x} \sqrt {-\frac {b}{a}} + 1\right ) \log \left (f\right )^{2} - 2 \, x \sqrt {-\frac {b}{a}} {\rm Li}_2\left (f^{x} \sqrt {-\frac {b}{a}}\right ) \log \left (f\right ) + 2 \, x \sqrt {-\frac {b}{a}} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {b}{a}}\right ) \log \left (f\right ) + 2 \, \sqrt {-\frac {b}{a}} {\rm polylog}\left (3, f^{x} \sqrt {-\frac {b}{a}}\right ) - 2 \, \sqrt {-\frac {b}{a}} {\rm polylog}\left (3, -f^{x} \sqrt {-\frac {b}{a}}\right )}{2 \, b \log \left (f\right )^{3}} \] Input:

integrate(f^x*x^2/(a+b*f^(2*x)),x, algorithm="fricas")
 

Output:

-1/2*(x^2*sqrt(-b/a)*log(f^x*sqrt(-b/a) + 1)*log(f)^2 - x^2*sqrt(-b/a)*log 
(-f^x*sqrt(-b/a) + 1)*log(f)^2 - 2*x*sqrt(-b/a)*dilog(f^x*sqrt(-b/a))*log( 
f) + 2*x*sqrt(-b/a)*dilog(-f^x*sqrt(-b/a))*log(f) + 2*sqrt(-b/a)*polylog(3 
, f^x*sqrt(-b/a)) - 2*sqrt(-b/a)*polylog(3, -f^x*sqrt(-b/a)))/(b*log(f)^3)
 

Sympy [F]

\[ \int \frac {f^x x^2}{a+b f^{2 x}} \, dx=\int \frac {f^{x} x^{2}}{a + b f^{2 x}}\, dx \] Input:

integrate(f**x*x**2/(a+b*f**(2*x)),x)
 

Output:

Integral(f**x*x**2/(a + b*f**(2*x)), x)
 

Maxima [F]

\[ \int \frac {f^x x^2}{a+b f^{2 x}} \, dx=\int { \frac {f^{x} x^{2}}{b f^{2 \, x} + a} \,d x } \] Input:

integrate(f^x*x^2/(a+b*f^(2*x)),x, algorithm="maxima")
 

Output:

integrate(f^x*x^2/(b*f^(2*x) + a), x)
 

Giac [F]

\[ \int \frac {f^x x^2}{a+b f^{2 x}} \, dx=\int { \frac {f^{x} x^{2}}{b f^{2 \, x} + a} \,d x } \] Input:

integrate(f^x*x^2/(a+b*f^(2*x)),x, algorithm="giac")
 

Output:

integrate(f^x*x^2/(b*f^(2*x) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {f^x x^2}{a+b f^{2 x}} \, dx=\int \frac {f^x\,x^2}{a+b\,f^{2\,x}} \,d x \] Input:

int((f^x*x^2)/(a + b*f^(2*x)),x)
 

Output:

int((f^x*x^2)/(a + b*f^(2*x)), x)
 

Reduce [F]

\[ \int \frac {f^x x^2}{a+b f^{2 x}} \, dx=\int \frac {f^{x} x^{2}}{f^{2 x} b +a}d x \] Input:

int(f^x*x^2/(a+b*f^(2*x)),x)
                                                                                    
                                                                                    
 

Output:

int((f**x*x**2)/(f**(2*x)*b + a),x)