3.1.49 \(\int \frac {f^x x^2}{(a+b f^{2 x})^2} \, dx\) [49]

3.1.49.1 Optimal result

Integrand size = 18, antiderivative size = 333 \[ \int \frac {f^x x^2}{\left (a+b f^{2 x}\right )^2} \, dx=-\frac {x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b} \log ^2(f)}+\frac {f^x x^2}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac {x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}-\frac {i x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}-\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}+\frac {i x \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {i \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}-\frac {i \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)} \]

output

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

3.1.49.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 477, normalized size of antiderivative = 1.43 \[ \int \frac {f^x x^2}{\left (a+b f^{2 x}\right )^2} \, dx=\frac {f^x x^2}{2 a \left (a+b f^{2 x}\right ) \log (f)}-\frac {\frac {\frac {i x^2}{2 \sqrt {a}}-\frac {i x \log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log (f)}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^2(f)}}{2 \sqrt {b}}+\frac {-\frac {i x^2}{2 \sqrt {a}}+\frac {i x \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log (f)}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^2(f)}}{2 \sqrt {b}}}{a \log (f)}+\frac {\frac {\frac {i x^3}{3 \sqrt {a}}-\frac {i x^2 \log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log (f)}-\frac {2 i x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^2(f)}+\frac {2 i \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^3(f)}}{2 \sqrt {b}}+\frac {-\frac {i x^3}{3 \sqrt {a}}+\frac {i x^2 \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log (f)}+\frac {2 i x \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^2(f)}-\frac {2 i \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^3(f)}}{2 \sqrt {b}}}{2 a} \]

input

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

output

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

3.1.49.3 Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.00, 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.

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

\(\Big \downarrow \) 2675

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2010

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.1.49.3.1 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]
 

3.1.49.4 Maple [F]

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

input

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

output

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

3.1.49.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.17 \[ \int \frac {f^x x^2}{\left (a+b f^{2 x}\right )^2} \, dx=\frac {2 \, b f^{x} x^{2} \log \left (f\right )^{2} + 2 \, {\left ({\left (b x \log \left (f\right ) - b\right )} f^{2 \, x} \sqrt {-\frac {b}{a}} + {\left (a x \log \left (f\right ) - a\right )} \sqrt {-\frac {b}{a}}\right )} {\rm Li}_2\left (f^{x} \sqrt {-\frac {b}{a}}\right ) - 2 \, {\left ({\left (b x \log \left (f\right ) - b\right )} f^{2 \, x} \sqrt {-\frac {b}{a}} + {\left (a x \log \left (f\right ) - a\right )} \sqrt {-\frac {b}{a}}\right )} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {b}{a}}\right ) - {\left ({\left (b x^{2} \log \left (f\right )^{2} - 2 \, b x \log \left (f\right )\right )} f^{2 \, x} \sqrt {-\frac {b}{a}} + {\left (a x^{2} \log \left (f\right )^{2} - 2 \, a x \log \left (f\right )\right )} \sqrt {-\frac {b}{a}}\right )} \log \left (f^{x} \sqrt {-\frac {b}{a}} + 1\right ) + {\left ({\left (b x^{2} \log \left (f\right )^{2} - 2 \, b x \log \left (f\right )\right )} f^{2 \, x} \sqrt {-\frac {b}{a}} + {\left (a x^{2} \log \left (f\right )^{2} - 2 \, a x \log \left (f\right )\right )} \sqrt {-\frac {b}{a}}\right )} \log \left (-f^{x} \sqrt {-\frac {b}{a}} + 1\right ) - 2 \, {\left (b f^{2 \, x} \sqrt {-\frac {b}{a}} + a \sqrt {-\frac {b}{a}}\right )} {\rm polylog}\left (3, f^{x} \sqrt {-\frac {b}{a}}\right ) + 2 \, {\left (b f^{2 \, x} \sqrt {-\frac {b}{a}} + a \sqrt {-\frac {b}{a}}\right )} {\rm polylog}\left (3, -f^{x} \sqrt {-\frac {b}{a}}\right )}{4 \, {\left (a b^{2} f^{2 \, x} \log \left (f\right )^{3} + a^{2} b \log \left (f\right )^{3}\right )}} \]

input

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

output

1/4*(2*b*f^x*x^2*log(f)^2 + 2*((b*x*log(f) - b)*f^(2*x)*sqrt(-b/a) + (a*x* 
log(f) - a)*sqrt(-b/a))*dilog(f^x*sqrt(-b/a)) - 2*((b*x*log(f) - b)*f^(2*x 
)*sqrt(-b/a) + (a*x*log(f) - a)*sqrt(-b/a))*dilog(-f^x*sqrt(-b/a)) - ((b*x 
^2*log(f)^2 - 2*b*x*log(f))*f^(2*x)*sqrt(-b/a) + (a*x^2*log(f)^2 - 2*a*x*l 
og(f))*sqrt(-b/a))*log(f^x*sqrt(-b/a) + 1) + ((b*x^2*log(f)^2 - 2*b*x*log( 
f))*f^(2*x)*sqrt(-b/a) + (a*x^2*log(f)^2 - 2*a*x*log(f))*sqrt(-b/a))*log(- 
f^x*sqrt(-b/a) + 1) - 2*(b*f^(2*x)*sqrt(-b/a) + a*sqrt(-b/a))*polylog(3, f 
^x*sqrt(-b/a)) + 2*(b*f^(2*x)*sqrt(-b/a) + a*sqrt(-b/a))*polylog(3, -f^x*s 
qrt(-b/a)))/(a*b^2*f^(2*x)*log(f)^3 + a^2*b*log(f)^3)
 

3.1.49.6 Sympy [F]

\[ \int \frac {f^x x^2}{\left (a+b f^{2 x}\right )^2} \, dx=\frac {f^{x} x^{2}}{2 a^{2} \log {\left (f \right )} + 2 a b f^{2 x} \log {\left (f \right )}} + \frac {\int \left (- \frac {2 f^{x} x}{a + b f^{2 x}}\right )\, dx + \int \frac {f^{x} x^{2} \log {\left (f \right )}}{a + b f^{2 x}}\, dx}{2 a \log {\left (f \right )}} \]

input

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

output

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

3.1.49.7 Maxima [F]

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

input

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

output

1/2*f^x*x^2/(a*b*f^(2*x)*log(f) + a^2*log(f)) + integrate(1/2*(x^2*log(f) 
- 2*x)*f^x/(a*b*f^(2*x)*log(f) + a^2*log(f)), x)
 

3.1.49.8 Giac [F]

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

input

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

output

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

3.1.49.9 Mupad [F(-1)]

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

input

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

output

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