\(\int \frac {f^x x}{(a+b f^{2 x})^3} \, dx\) [108]

Optimal result

Integrand size = 16, antiderivative size = 223 \[ \int \frac {f^x x}{\left (a+b f^{2 x}\right )^3} \, dx=-\frac {f^x}{8 a^2 \left (a+b f^{2 x}\right ) \log ^2(f)}-\frac {\arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b} \log ^2(f)}+\frac {f^x x}{4 a \left (a+b f^{2 x}\right )^2 \log (f)}+\frac {3 f^x x}{8 a^2 \left (a+b f^{2 x}\right ) \log (f)}+\frac {3 x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log (f)}-\frac {3 i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{16 a^{5/2} \sqrt {b} \log ^2(f)}+\frac {3 i \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{16 a^{5/2} \sqrt {b} \log ^2(f)} \] Output:

-1/8*f^x/a^2/(a+b*f^(2*x))/ln(f)^2-1/2*arctan(b^(1/2)*f^x/a^(1/2))/a^(5/2) 
/b^(1/2)/ln(f)^2+1/4*f^x*x/a/(a+b*f^(2*x))^2/ln(f)+3/8*f^x*x/a^2/(a+b*f^(2 
*x))/ln(f)+3/8*x*arctan(b^(1/2)*f^x/a^(1/2))/a^(5/2)/b^(1/2)/ln(f)-3/16*I* 
polylog(2,-I*b^(1/2)*f^x/a^(1/2))/a^(5/2)/b^(1/2)/ln(f)^2+3/16*I*polylog(2 
,I*b^(1/2)*f^x/a^(1/2))/a^(5/2)/b^(1/2)/ln(f)^2
 

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.83 \[ \int \frac {f^x x}{\left (a+b f^{2 x}\right )^3} \, dx=\frac {-\frac {16 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}}+\frac {8 a f^x x \log (f)}{\left (a+b f^{2 x}\right )^2}+\frac {4 f^x (-1+3 x \log (f))}{a+b f^{2 x}}+\frac {6 i \left (x \log (f) \left (\log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-\log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )\right )-\operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )\right )}{\sqrt {a} \sqrt {b}}}{32 a^2 \log ^2(f)} \] Input:

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

Output:

((-16*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]) + (8*a*f^x*x*Log[f] 
)/(a + b*f^(2*x))^2 + (4*f^x*(-1 + 3*x*Log[f]))/(a + b*f^(2*x)) + ((6*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*Sqrt[b]*f^x) 
/Sqrt[a]]))/(Sqrt[a]*Sqrt[b]))/(32*a^2*Log[f]^2)
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2675, 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.

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

\(\Big \downarrow \) 2675

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b} \log ^2(f)}+\frac {3 x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \log (f)}-\frac {3 i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{16 a^{5/2} \sqrt {b} \log ^2(f)}+\frac {3 i \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{16 a^{5/2} \sqrt {b} \log ^2(f)}-\frac {f^x}{8 a^2 \log ^2(f) \left (a+b f^{2 x}\right )}+\frac {3 x f^x}{8 a^2 \log (f) \left (a+b f^{2 x}\right )}+\frac {x f^x}{4 a \log (f) \left (a+b f^{2 x}\right )^2}\)

Input:

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

Output:

-1/8*f^x/(a^2*(a + b*f^(2*x))*Log[f]^2) - ArcTan[(Sqrt[b]*f^x)/Sqrt[a]]/(2 
*a^(5/2)*Sqrt[b]*Log[f]^2) + (f^x*x)/(4*a*(a + b*f^(2*x))^2*Log[f]) + (3*f 
^x*x)/(8*a^2*(a + b*f^(2*x))*Log[f]) + (3*x*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]]) 
/(8*a^(5/2)*Sqrt[b]*Log[f]) - (((3*I)/16)*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sq 
rt[a]])/(a^(5/2)*Sqrt[b]*Log[f]^2) + (((3*I)/16)*PolyLog[2, (I*Sqrt[b]*f^x 
)/Sqrt[a]])/(a^(5/2)*Sqrt[b]*Log[f]^2)
 

Defintions of rubi rules used

rule 2009

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

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 [A] (verified)

Time = 0.34 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {f^{x} \left (3 x b \,f^{2 x} \ln \left (f \right )+5 \ln \left (f \right ) a x -b \,f^{2 x}-a \right )}{8 \ln \left (f \right )^{2} a^{2} \left (a +b \,f^{2 x}\right )^{2}}-\frac {\arctan \left (\frac {b \,f^{x}}{\sqrt {a b}}\right )}{2 a^{2} \ln \left (f \right )^{2} \sqrt {a b}}+\frac {3 x \ln \left (\frac {-b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{16 a^{2} \ln \left (f \right ) \sqrt {-a b}}-\frac {3 x \ln \left (\frac {b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{16 a^{2} \ln \left (f \right ) \sqrt {-a b}}+\frac {3 \operatorname {dilog}\left (\frac {-b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{16 a^{2} \ln \left (f \right )^{2} \sqrt {-a b}}-\frac {3 \operatorname {dilog}\left (\frac {b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{16 a^{2} \ln \left (f \right )^{2} \sqrt {-a b}}\)	\(223\)

Input:

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

Output:

1/8*f^x*(3*ln(f)*b*x*(f^x)^2+5*ln(f)*a*x-b*(f^x)^2-a)/ln(f)^2/a^2/(a+b*(f^ 
x)^2)^2-1/2/a^2/ln(f)^2/(a*b)^(1/2)*arctan(b*f^x/(a*b)^(1/2))+3/16/a^2/ln( 
f)*x/(-a*b)^(1/2)*ln((-b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-3/16/a^2/ln(f)*x/ 
(-a*b)^(1/2)*ln((b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))+3/16/a^2/ln(f)^2/(-a*b) 
^(1/2)*dilog((-b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-3/16/a^2/ln(f)^2/(-a*b)^( 
1/2)*dilog((b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 494 vs. \(2 (167) = 334\).

Time = 0.08 (sec) , antiderivative size = 494, normalized size of antiderivative = 2.22 \[ \int \frac {f^x x}{\left (a+b f^{2 x}\right )^3} \, dx=\frac {2 \, {\left (3 \, b^{2} x \log \left (f\right ) - b^{2}\right )} f^{3 \, x} + 2 \, {\left (5 \, a b x \log \left (f\right ) - a b\right )} f^{x} + 3 \, {\left (b^{2} f^{4 \, x} \sqrt {-\frac {b}{a}} + 2 \, a b f^{2 \, x} \sqrt {-\frac {b}{a}} + a^{2} \sqrt {-\frac {b}{a}}\right )} {\rm Li}_2\left (f^{x} \sqrt {-\frac {b}{a}}\right ) - 3 \, {\left (b^{2} f^{4 \, x} \sqrt {-\frac {b}{a}} + 2 \, a b f^{2 \, x} \sqrt {-\frac {b}{a}} + a^{2} \sqrt {-\frac {b}{a}}\right )} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {b}{a}}\right ) - 4 \, {\left (b^{2} f^{4 \, x} \sqrt {-\frac {b}{a}} + 2 \, a b f^{2 \, x} \sqrt {-\frac {b}{a}} + a^{2} \sqrt {-\frac {b}{a}}\right )} \log \left (2 \, b f^{x} + 2 \, a \sqrt {-\frac {b}{a}}\right ) + 4 \, {\left (b^{2} f^{4 \, x} \sqrt {-\frac {b}{a}} + 2 \, a b f^{2 \, x} \sqrt {-\frac {b}{a}} + a^{2} \sqrt {-\frac {b}{a}}\right )} \log \left (2 \, b f^{x} - 2 \, a \sqrt {-\frac {b}{a}}\right ) - 3 \, {\left (b^{2} f^{4 \, x} x \sqrt {-\frac {b}{a}} \log \left (f\right ) + 2 \, a b f^{2 \, x} x \sqrt {-\frac {b}{a}} \log \left (f\right ) + a^{2} x \sqrt {-\frac {b}{a}} \log \left (f\right )\right )} \log \left (f^{x} \sqrt {-\frac {b}{a}} + 1\right ) + 3 \, {\left (b^{2} f^{4 \, x} x \sqrt {-\frac {b}{a}} \log \left (f\right ) + 2 \, a b f^{2 \, x} x \sqrt {-\frac {b}{a}} \log \left (f\right ) + a^{2} x \sqrt {-\frac {b}{a}} \log \left (f\right )\right )} \log \left (-f^{x} \sqrt {-\frac {b}{a}} + 1\right )}{16 \, {\left (a^{2} b^{3} f^{4 \, x} \log \left (f\right )^{2} + 2 \, a^{3} b^{2} f^{2 \, x} \log \left (f\right )^{2} + a^{4} b \log \left (f\right )^{2}\right )}} \] Input:

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

Output:

1/16*(2*(3*b^2*x*log(f) - b^2)*f^(3*x) + 2*(5*a*b*x*log(f) - a*b)*f^x + 3* 
(b^2*f^(4*x)*sqrt(-b/a) + 2*a*b*f^(2*x)*sqrt(-b/a) + a^2*sqrt(-b/a))*dilog 
(f^x*sqrt(-b/a)) - 3*(b^2*f^(4*x)*sqrt(-b/a) + 2*a*b*f^(2*x)*sqrt(-b/a) + 
a^2*sqrt(-b/a))*dilog(-f^x*sqrt(-b/a)) - 4*(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)) + 4*(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*f^(4*x)*x*sqrt(-b/a)*log(f) + 2*a*b*f^(2*x 
)*x*sqrt(-b/a)*log(f) + a^2*x*sqrt(-b/a)*log(f))*log(f^x*sqrt(-b/a) + 1) + 
 3*(b^2*f^(4*x)*x*sqrt(-b/a)*log(f) + 2*a*b*f^(2*x)*x*sqrt(-b/a)*log(f) + 
a^2*x*sqrt(-b/a)*log(f))*log(-f^x*sqrt(-b/a) + 1))/(a^2*b^3*f^(4*x)*log(f) 
^2 + 2*a^3*b^2*f^(2*x)*log(f)^2 + a^4*b*log(f)^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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