Integrand size = 16, antiderivative size = 172 \[ \int \frac {f^x x}{\left (a+b f^{2 x}\right )^2} \, dx=-\frac {\arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {f^x x}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac {x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} \log ^2(f)} \] Output:
-1/2*arctan(b^(1/2)*f^x/a^(1/2))/a^(3/2)/b^(1/2)/ln(f)^2+1/2*f^x*x/a/(a+b* f^(2*x))/ln(f)+1/2*x*arctan(b^(1/2)*f^x/a^(1/2))/a^(3/2)/b^(1/2)/ln(f)-1/4 *I*polylog(2,-I*b^(1/2)*f^x/a^(1/2))/a^(3/2)/b^(1/2)/ln(f)^2+1/4*I*polylog (2,I*b^(1/2)*f^x/a^(1/2))/a^(3/2)/b^(1/2)/ln(f)^2
Time = 0.20 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.58 \[ \int \frac {f^x x}{\left (a+b f^{2 x}\right )^2} \, dx=-\frac {\left (1+\frac {b f^{2 x}}{a}\right ) \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} \left (a+b f^{2 x}\right ) \log ^2(f)}+\frac {f^x x}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac {\frac {\frac {i x^2}{2 \sqrt {a}}-\frac {i x \log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log (f)}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^2(f)}}{2 \sqrt {b}}+\frac {-\frac {i x^2}{2 \sqrt {a}}+\frac {i x \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log (f)}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^2(f)}}{2 \sqrt {b}}}{2 a} \] Input:
Integrate[(f^x*x)/(a + b*f^(2*x))^2,x]
Output:
-1/2*((1 + (b*f^(2*x))/a)*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]* (a + b*f^(2*x))*Log[f]^2) + (f^x*x)/(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]))/(2*a)
Time = 0.62 (sec) , antiderivative size = 172, 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.
\(\displaystyle \int \frac {x f^x}{\left (a+b f^{2 x}\right )^2} \, dx\) |
\(\Big \downarrow \) 2675 |
\(\displaystyle -\int \left (\frac {f^x}{2 a \left (b f^{2 x}+a\right ) \log (f)}+\frac {\arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}\right )dx+\frac {x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}+\frac {x f^x}{2 a \log (f) \left (a+b f^{2 x}\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {x \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {x f^x}{2 a \log (f) \left (a+b f^{2 x}\right )}\) |
Input:
Int[(f^x*x)/(a + b*f^(2*x))^2,x]
Output:
-1/2*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]]/(a^(3/2)*Sqrt[b]*Log[f]^2) + (f^x*x)/(2 *a*(a + b*f^(2*x))*Log[f]) + (x*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(2*a^(3/2)* Sqrt[b]*Log[f]) - ((I/4)*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/(a^(3/2)* Sqrt[b]*Log[f]^2) + ((I/4)*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[a]])/(a^(3/2)*S qrt[b]*Log[f]^2)
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]
Time = 0.12 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.13
method | result | size |
risch | \(\frac {f^{x} x}{2 a \left (a +b \,f^{2 x}\right ) \ln \left (f \right )}+\frac {x \ln \left (\frac {-b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{4 a \ln \left (f \right ) \sqrt {-a b}}-\frac {x \ln \left (\frac {b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{4 a \ln \left (f \right ) \sqrt {-a b}}+\frac {\operatorname {dilog}\left (\frac {-b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{4 a \ln \left (f \right )^{2} \sqrt {-a b}}-\frac {\operatorname {dilog}\left (\frac {b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{4 a \ln \left (f \right )^{2} \sqrt {-a b}}-\frac {\arctan \left (\frac {b \,f^{x}}{\sqrt {a b}}\right )}{2 a \ln \left (f \right )^{2} \sqrt {a b}}\) | \(195\) |
Input:
int(f^x*x/(a+b*f^(2*x))^2,x,method=_RETURNVERBOSE)
Output:
1/2/ln(f)/a*f^x*x/(a+b*(f^x)^2)+1/4/a/ln(f)*x/(-a*b)^(1/2)*ln((-b*f^x+(-a* b)^(1/2))/(-a*b)^(1/2))-1/4/a/ln(f)*x/(-a*b)^(1/2)*ln((b*f^x+(-a*b)^(1/2)) /(-a*b)^(1/2))+1/4/a/ln(f)^2/(-a*b)^(1/2)*dilog((-b*f^x+(-a*b)^(1/2))/(-a* b)^(1/2))-1/4/a/ln(f)^2/(-a*b)^(1/2)*dilog((b*f^x+(-a*b)^(1/2))/(-a*b)^(1/ 2))-1/2/a/ln(f)^2/(a*b)^(1/2)*arctan(b*f^x/(a*b)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (120) = 240\).
Time = 0.10 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.81 \[ \int \frac {f^x x}{\left (a+b f^{2 x}\right )^2} \, dx=\frac {2 \, b f^{x} x \log \left (f\right ) + {\left (b f^{2 \, x} \sqrt {-\frac {b}{a}} + a \sqrt {-\frac {b}{a}}\right )} {\rm Li}_2\left (f^{x} \sqrt {-\frac {b}{a}}\right ) - {\left (b f^{2 \, x} \sqrt {-\frac {b}{a}} + a \sqrt {-\frac {b}{a}}\right )} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {b}{a}}\right ) - {\left (b f^{2 \, x} \sqrt {-\frac {b}{a}} + a \sqrt {-\frac {b}{a}}\right )} \log \left (2 \, b f^{x} + 2 \, a \sqrt {-\frac {b}{a}}\right ) + {\left (b f^{2 \, x} \sqrt {-\frac {b}{a}} + a \sqrt {-\frac {b}{a}}\right )} \log \left (2 \, b f^{x} - 2 \, a \sqrt {-\frac {b}{a}}\right ) - {\left (b f^{2 \, x} x \sqrt {-\frac {b}{a}} \log \left (f\right ) + a x \sqrt {-\frac {b}{a}} \log \left (f\right )\right )} \log \left (f^{x} \sqrt {-\frac {b}{a}} + 1\right ) + {\left (b f^{2 \, x} x \sqrt {-\frac {b}{a}} \log \left (f\right ) + a x \sqrt {-\frac {b}{a}} \log \left (f\right )\right )} \log \left (-f^{x} \sqrt {-\frac {b}{a}} + 1\right )}{4 \, {\left (a b^{2} f^{2 \, x} \log \left (f\right )^{2} + a^{2} b \log \left (f\right )^{2}\right )}} \] Input:
integrate(f^x*x/(a+b*f^(2*x))^2,x, algorithm="fricas")
Output:
1/4*(2*b*f^x*x*log(f) + (b*f^(2*x)*sqrt(-b/a) + a*sqrt(-b/a))*dilog(f^x*sq rt(-b/a)) - (b*f^(2*x)*sqrt(-b/a) + a*sqrt(-b/a))*dilog(-f^x*sqrt(-b/a)) - (b*f^(2*x)*sqrt(-b/a) + a*sqrt(-b/a))*log(2*b*f^x + 2*a*sqrt(-b/a)) + (b* f^(2*x)*sqrt(-b/a) + a*sqrt(-b/a))*log(2*b*f^x - 2*a*sqrt(-b/a)) - (b*f^(2 *x)*x*sqrt(-b/a)*log(f) + a*x*sqrt(-b/a)*log(f))*log(f^x*sqrt(-b/a) + 1) + (b*f^(2*x)*x*sqrt(-b/a)*log(f) + a*x*sqrt(-b/a)*log(f))*log(-f^x*sqrt(-b/ a) + 1))/(a*b^2*f^(2*x)*log(f)^2 + a^2*b*log(f)^2)
\[ \int \frac {f^x x}{\left (a+b f^{2 x}\right )^2} \, dx=\frac {f^{x} x}{2 a^{2} \log {\left (f \right )} + 2 a b f^{2 x} \log {\left (f \right )}} + \frac {\int \left (- \frac {f^{x}}{a + b f^{2 x}}\right )\, dx + \int \frac {f^{x} x \log {\left (f \right )}}{a + b f^{2 x}}\, dx}{2 a \log {\left (f \right )}} \] Input:
integrate(f**x*x/(a+b*f**(2*x))**2,x)
Output:
f**x*x/(2*a**2*log(f) + 2*a*b*f**(2*x)*log(f)) + (Integral(-f**x/(a + b*f* *(2*x)), x) + Integral(f**x*x*log(f)/(a + b*f**(2*x)), x))/(2*a*log(f))
\[ \int \frac {f^x x}{\left (a+b f^{2 x}\right )^2} \, dx=\int { \frac {f^{x} x}{{\left (b f^{2 \, x} + a\right )}^{2}} \,d x } \] Input:
integrate(f^x*x/(a+b*f^(2*x))^2,x, algorithm="maxima")
Output:
1/2*f^x*x/(a*b*f^(2*x)*log(f) + a^2*log(f)) + integrate(1/2*(x*log(f) - 1) *f^x/(a*b*f^(2*x)*log(f) + a^2*log(f)), x)
\[ \int \frac {f^x x}{\left (a+b f^{2 x}\right )^2} \, dx=\int { \frac {f^{x} x}{{\left (b f^{2 \, x} + a\right )}^{2}} \,d x } \] Input:
integrate(f^x*x/(a+b*f^(2*x))^2,x, algorithm="giac")
Output:
integrate(f^x*x/(b*f^(2*x) + a)^2, x)
Timed out. \[ \int \frac {f^x x}{\left (a+b f^{2 x}\right )^2} \, dx=\int \frac {f^x\,x}{{\left (a+b\,f^{2\,x}\right )}^2} \,d x \] Input:
int((f^x*x)/(a + b*f^(2*x))^2,x)
Output:
int((f^x*x)/(a + b*f^(2*x))^2, x)
\[ \int \frac {f^x x}{\left (a+b f^{2 x}\right )^2} \, dx=\int \frac {f^{x} x}{f^{4 x} b^{2}+2 f^{2 x} a b +a^{2}}d x \] Input:
int(f^x*x/(a+b*f^(2*x))^2,x)
Output:
int((f**x*x)/(f**(4*x)*b**2 + 2*f**(2*x)*a*b + a**2),x)