\(\int \frac {f^x x^2}{(a+b f^{2 x})^3} \, dx\) [109]

Optimal result

Integrand size = 18, antiderivative size = 420 \[ \int \frac {f^x x^2}{\left (a+b f^{2 x}\right )^3} \, dx=\frac {\arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{4 a^{5/2} \sqrt {b} \log ^3(f)}-\frac {f^x x}{4 a^2 \left (a+b f^{2 x}\right ) \log ^2(f)}-\frac {x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {b} \log ^2(f)}+\frac {f^x x^2}{4 a \left (a+b f^{2 x}\right )^2 \log (f)}+\frac {3 f^x x^2}{8 a^2 \left (a+b f^{2 x}\right ) \log (f)}+\frac {3 x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log (f)}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b} \log ^3(f)}-\frac {3 i x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log ^2(f)}-\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b} \log ^3(f)}+\frac {3 i x \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log ^2(f)}+\frac {3 i \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log ^3(f)}-\frac {3 i \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log ^3(f)} \] Output:

1/4*arctan(b^(1/2)*f^x/a^(1/2))/a^(5/2)/b^(1/2)/ln(f)^3-1/4*f^x*x/a^2/(a+b 
*f^(2*x))/ln(f)^2-x*arctan(b^(1/2)*f^x/a^(1/2))/a^(5/2)/b^(1/2)/ln(f)^2+1/ 
4*f^x*x^2/a/(a+b*f^(2*x))^2/ln(f)+3/8*f^x*x^2/a^2/(a+b*f^(2*x))/ln(f)+3/8* 
x^2*arctan(b^(1/2)*f^x/a^(1/2))/a^(5/2)/b^(1/2)/ln(f)+1/2*I*polylog(2,-I*b 
^(1/2)*f^x/a^(1/2))/a^(5/2)/b^(1/2)/ln(f)^3-3/8*I*x*polylog(2,-I*b^(1/2)*f 
^x/a^(1/2))/a^(5/2)/b^(1/2)/ln(f)^2-1/2*I*polylog(2,I*b^(1/2)*f^x/a^(1/2)) 
/a^(5/2)/b^(1/2)/ln(f)^3+3/8*I*x*polylog(2,I*b^(1/2)*f^x/a^(1/2))/a^(5/2)/ 
b^(1/2)/ln(f)^2+3/8*I*polylog(3,-I*b^(1/2)*f^x/a^(1/2))/a^(5/2)/b^(1/2)/ln 
(f)^3-3/8*I*polylog(3,I*b^(1/2)*f^x/a^(1/2))/a^(5/2)/b^(1/2)/ln(f)^3
 

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 353, normalized size of antiderivative = 0.84 \[ \int \frac {f^x x^2}{\left (a+b f^{2 x}\right )^3} \, dx=\frac {\frac {4 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}}+\frac {4 a f^x x^2 \log ^2(f)}{\left (a+b f^{2 x}\right )^2}+\frac {2 f^x x \log (f) (-2+3 x \log (f))}{a+b f^{2 x}}-\frac {8 i \left (x \log (f) \left (\log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-\log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )\right )-\operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )\right )}{\sqrt {a} \sqrt {b}}+\frac {3 i \left (x^2 \log ^2(f) \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-x^2 \log ^2(f) \log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-2 x \log (f) \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+2 x \log (f) \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+2 \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-2 \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )\right )}{\sqrt {a} \sqrt {b}}}{16 a^2 \log ^3(f)} \] Input:

Integrate[(f^x*x^2)/(a + b*f^(2*x))^3,x]
 

Output:

((4*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]) + (4*a*f^x*x^2*Log[f] 
^2)/(a + b*f^(2*x))^2 + (2*f^x*x*Log[f]*(-2 + 3*x*Log[f]))/(a + b*f^(2*x)) 
 - ((8*I)*(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*S 
qrt[b]*f^x)/Sqrt[a]]))/(Sqrt[a]*Sqrt[b]) + ((3*I)*(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)/Sqrt[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*Pol 
yLog[3, (I*Sqrt[b]*f^x)/Sqrt[a]]))/(Sqrt[a]*Sqrt[b]))/(16*a^2*Log[f]^3)
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 416, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2675, 27, 2010, 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.

\(\displaystyle \int \frac {x^2 f^x}{\left (a+b f^{2 x}\right )^3} \, dx\)

\(\Big \downarrow \) 2675

\(\displaystyle -2 \int \frac {1}{8} x \left (\frac {3 f^x}{a^2 \left (b f^{2 x}+a\right ) \log (f)}+\frac {2 f^x}{a \left (b f^{2 x}+a\right )^2 \log (f)}+\frac {3 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {b} \log (f)}\right )dx+\frac {3 x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log (f)}+\frac {3 x^2 f^x}{8 a^2 \log (f) \left (a+b f^{2 x}\right )}+\frac {x^2 f^x}{4 a \log (f) \left (a+b f^{2 x}\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{4} \int x \left (\frac {3 f^x}{a^2 \left (b f^{2 x}+a\right ) \log (f)}+\frac {2 f^x}{a \left (b f^{2 x}+a\right )^2 \log (f)}+\frac {3 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {b} \log (f)}\right )dx+\frac {3 x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log (f)}+\frac {3 x^2 f^x}{8 a^2 \log (f) \left (a+b f^{2 x}\right )}+\frac {x^2 f^x}{4 a \log (f) \left (a+b f^{2 x}\right )^2}\)

\(\Big \downarrow \) 2010

\(\displaystyle -\frac {1}{4} \int \left (\frac {3 x f^x}{a^2 \left (b f^{2 x}+a\right ) \log (f)}+\frac {2 x f^x}{a \left (b f^{2 x}+a\right )^2 \log (f)}+\frac {3 x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {b} \log (f)}\right )dx+\frac {3 x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log (f)}+\frac {3 x^2 f^x}{8 a^2 \log (f) \left (a+b f^{2 x}\right )}+\frac {x^2 f^x}{4 a \log (f) \left (a+b f^{2 x}\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log (f)}+\frac {3 x^2 f^x}{8 a^2 \log (f) \left (a+b f^{2 x}\right )}+\frac {1}{4} \left (\frac {\arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {b} \log ^3(f)}-\frac {4 x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {b} \log ^2(f)}+\frac {2 i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {b} \log ^3(f)}-\frac {2 i \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {b} \log ^3(f)}+\frac {3 i \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b} \log ^3(f)}-\frac {3 i \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b} \log ^3(f)}-\frac {3 i x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b} \log ^2(f)}+\frac {3 i x \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b} \log ^2(f)}-\frac {x f^x}{a^2 \log ^2(f) \left (a+b f^{2 x}\right )}\right )+\frac {x^2 f^x}{4 a \log (f) \left (a+b f^{2 x}\right )^2}\)

Input:

Int[(f^x*x^2)/(a + b*f^(2*x))^3,x]
 

Output:

(f^x*x^2)/(4*a*(a + b*f^(2*x))^2*Log[f]) + (3*f^x*x^2)/(8*a^2*(a + b*f^(2* 
x))*Log[f]) + (3*x^2*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(8*a^(5/2)*Sqrt[b]*Log 
[f]) + (ArcTan[(Sqrt[b]*f^x)/Sqrt[a]]/(a^(5/2)*Sqrt[b]*Log[f]^3) - (f^x*x) 
/(a^2*(a + b*f^(2*x))*Log[f]^2) - (4*x*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(a^( 
5/2)*Sqrt[b]*Log[f]^2) + ((2*I)*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/(a 
^(5/2)*Sqrt[b]*Log[f]^3) - (((3*I)/2)*x*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sqrt 
[a]])/(a^(5/2)*Sqrt[b]*Log[f]^2) - ((2*I)*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[ 
a]])/(a^(5/2)*Sqrt[b]*Log[f]^3) + (((3*I)/2)*x*PolyLog[2, (I*Sqrt[b]*f^x)/ 
Sqrt[a]])/(a^(5/2)*Sqrt[b]*Log[f]^2) + (((3*I)/2)*PolyLog[3, ((-I)*Sqrt[b] 
*f^x)/Sqrt[a]])/(a^(5/2)*Sqrt[b]*Log[f]^3) - (((3*I)/2)*PolyLog[3, (I*Sqrt 
[b]*f^x)/Sqrt[a]])/(a^(5/2)*Sqrt[b]*Log[f]^3))/4
 

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

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]
 

Maple [F]

\[\int \frac {f^{x} x^{2}}{\left (a +b \,f^{2 x}\right )^{3}}d x\]

Input:

int(f^x*x^2/(a+b*f^(2*x))^3,x)
 

Output:

int(f^x*x^2/(a+b*f^(2*x))^3,x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 786 vs. \(2 (298) = 596\).

Time = 0.09 (sec) , antiderivative size = 786, normalized size of antiderivative = 1.87 \[ \int \frac {f^x x^2}{\left (a+b f^{2 x}\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(f^x*x^2/(a+b*f^(2*x))^3,x, algorithm="fricas")
 

Output:

1/16*(2*(3*b^2*x^2*log(f)^2 - 2*b^2*x*log(f))*f^(3*x) + 2*(5*a*b*x^2*log(f 
)^2 - 2*a*b*x*log(f))*f^x + 2*((3*b^2*x*log(f) - 4*b^2)*f^(4*x)*sqrt(-b/a) 
 + 2*(3*a*b*x*log(f) - 4*a*b)*f^(2*x)*sqrt(-b/a) + (3*a^2*x*log(f) - 4*a^2 
)*sqrt(-b/a))*dilog(f^x*sqrt(-b/a)) - 2*((3*b^2*x*log(f) - 4*b^2)*f^(4*x)* 
sqrt(-b/a) + 2*(3*a*b*x*log(f) - 4*a*b)*f^(2*x)*sqrt(-b/a) + (3*a^2*x*log( 
f) - 4*a^2)*sqrt(-b/a))*dilog(-f^x*sqrt(-b/a)) + 2*(b^2*f^(4*x)*sqrt(-b/a) 
 + 2*a*b*f^(2*x)*sqrt(-b/a) + a^2*sqrt(-b/a))*log(2*b*f^x + 2*a*sqrt(-b/a) 
) - 2*(b^2*f^(4*x)*sqrt(-b/a) + 2*a*b*f^(2*x)*sqrt(-b/a) + a^2*sqrt(-b/a)) 
*log(2*b*f^x - 2*a*sqrt(-b/a)) - ((3*b^2*x^2*log(f)^2 - 8*b^2*x*log(f))*f^ 
(4*x)*sqrt(-b/a) + 2*(3*a*b*x^2*log(f)^2 - 8*a*b*x*log(f))*f^(2*x)*sqrt(-b 
/a) + (3*a^2*x^2*log(f)^2 - 8*a^2*x*log(f))*sqrt(-b/a))*log(f^x*sqrt(-b/a) 
 + 1) + ((3*b^2*x^2*log(f)^2 - 8*b^2*x*log(f))*f^(4*x)*sqrt(-b/a) + 2*(3*a 
*b*x^2*log(f)^2 - 8*a*b*x*log(f))*f^(2*x)*sqrt(-b/a) + (3*a^2*x^2*log(f)^2 
 - 8*a^2*x*log(f))*sqrt(-b/a))*log(-f^x*sqrt(-b/a) + 1) - 6*(b^2*f^(4*x)*s 
qrt(-b/a) + 2*a*b*f^(2*x)*sqrt(-b/a) + a^2*sqrt(-b/a))*polylog(3, f^x*sqrt 
(-b/a)) + 6*(b^2*f^(4*x)*sqrt(-b/a) + 2*a*b*f^(2*x)*sqrt(-b/a) + a^2*sqrt( 
-b/a))*polylog(3, -f^x*sqrt(-b/a)))/(a^2*b^3*f^(4*x)*log(f)^3 + 2*a^3*b^2* 
f^(2*x)*log(f)^3 + a^4*b*log(f)^3)
 

Sympy [F]

\[ \int \frac {f^x x^2}{\left (a+b f^{2 x}\right )^3} \, dx=\frac {f^{3 x} \left (3 b x^{2} \log {\left (f \right )} - 2 b x\right ) + f^{x} \left (5 a x^{2} \log {\left (f \right )} - 2 a x\right )}{8 a^{4} \log {\left (f \right )}^{2} + 16 a^{3} b f^{2 x} \log {\left (f \right )}^{2} + 8 a^{2} b^{2} f^{4 x} \log {\left (f \right )}^{2}} + \frac {\int \frac {2 f^{x}}{a + b f^{2 x}}\, dx + \int \left (- \frac {8 f^{x} x \log {\left (f \right )}}{a + b f^{2 x}}\right )\, dx + \int \frac {3 f^{x} x^{2} \log {\left (f \right )}^{2}}{a + b f^{2 x}}\, dx}{8 a^{2} \log {\left (f \right )}^{2}} \] Input:

integrate(f**x*x**2/(a+b*f**(2*x))**3,x)
 

Output:

(f**(3*x)*(3*b*x**2*log(f) - 2*b*x) + f**x*(5*a*x**2*log(f) - 2*a*x))/(8*a 
**4*log(f)**2 + 16*a**3*b*f**(2*x)*log(f)**2 + 8*a**2*b**2*f**(4*x)*log(f) 
**2) + (Integral(2*f**x/(a + b*f**(2*x)), x) + Integral(-8*f**x*x*log(f)/( 
a + b*f**(2*x)), x) + Integral(3*f**x*x**2*log(f)**2/(a + b*f**(2*x)), x)) 
/(8*a**2*log(f)**2)
 

Maxima [F]

\[ \int \frac {f^x x^2}{\left (a+b f^{2 x}\right )^3} \, dx=\int { \frac {f^{x} x^{2}}{{\left (b f^{2 \, x} + a\right )}^{3}} \,d x } \] Input:

integrate(f^x*x^2/(a+b*f^(2*x))^3,x, algorithm="maxima")
 

Output:

1/8*((3*b*x^2*log(f) - 2*b*x)*f^(3*x) + (5*a*x^2*log(f) - 2*a*x)*f^x)/(a^2 
*b^2*f^(4*x)*log(f)^2 + 2*a^3*b*f^(2*x)*log(f)^2 + a^4*log(f)^2) + integra 
te(1/8*(3*x^2*log(f)^2 - 8*x*log(f) + 2)*f^x/(a^2*b*f^(2*x)*log(f)^2 + a^3 
*log(f)^2), x)
 

Giac [F]

\[ \int \frac {f^x x^2}{\left (a+b f^{2 x}\right )^3} \, dx=\int { \frac {f^{x} x^{2}}{{\left (b f^{2 \, x} + a\right )}^{3}} \,d x } \] Input:

integrate(f^x*x^2/(a+b*f^(2*x))^3,x, algorithm="giac")
 

Output:

integrate(f^x*x^2/(b*f^(2*x) + a)^3, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {f^x x^2}{\left (a+b f^{2 x}\right )^3} \, dx=\int \frac {f^x\,x^2}{{\left (a+b\,f^{2\,x}\right )}^3} \,d x \] Input:

int((f^x*x^2)/(a + b*f^(2*x))^3,x)
 

Output:

int((f^x*x^2)/(a + b*f^(2*x))^3, x)
 

Reduce [F]

\[ \int \frac {f^x x^2}{\left (a+b f^{2 x}\right )^3} \, dx=\int \frac {f^{x} x^{2}}{f^{6 x} b^{3}+3 f^{4 x} a \,b^{2}+3 f^{2 x} a^{2} b +a^{3}}d x \] Input:

int(f^x*x^2/(a+b*f^(2*x))^3,x)
 

Output:

int((f**x*x**2)/(f**(6*x)*b**3 + 3*f**(4*x)*a*b**2 + 3*f**(2*x)*a**2*b + a 
**3),x)