\(\int \frac {x^2}{(b f^{-x}+a f^x)^2} \, dx\) [60]

Optimal result

Integrand size = 19, antiderivative size = 98 \[ \int \frac {x^2}{\left (b f^{-x}+a f^x\right )^2} \, dx=\frac {x^2}{2 a b \log (f)}-\frac {x^2}{2 a \left (b+a f^{2 x}\right ) \log (f)}-\frac {x \log \left (1+\frac {a f^{2 x}}{b}\right )}{2 a b \log ^2(f)}-\frac {\operatorname {PolyLog}\left (2,-\frac {a f^{2 x}}{b}\right )}{4 a b \log ^3(f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2321, 2222, 2215, 2221, 2317, 2438} \[ \int \frac {x^2}{\left (b f^{-x}+a f^x\right )^2} \, dx=-\frac {\operatorname {PolyLog}\left (2,-\frac {a f^{2 x}}{b}\right )}{4 a b \log ^3(f)}-\frac {x^2}{2 a \log (f) \left (a f^{2 x}+b\right )}-\frac {x \log \left (\frac {a f^{2 x}}{b}+1\right )}{2 a b \log ^2(f)}+\frac {x^2}{2 a b \log (f)} \]

[In]

Int[x^2/(b/f^x + a*f^x)^2,x]

[Out]

x^2/(2*a*b*Log[f]) - x^2/(2*a*(b + a*f^(2*x))*Log[f]) - (x*Log[1 + (a*f^(2*x))/b])/(2*a*b*Log[f]^2) - PolyLog[
2, -((a*f^(2*x))/b)]/(4*a*b*Log[f]^3)

Rule 2215

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n))
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2222

Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*
((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*((a + b*(F^(g*(e + f*x)))^n)^(p + 1)/(b*f*g*n*(p + 1
)*Log[F])), x] - Dist[d*(m/(b*f*g*n*(p + 1)*Log[F])), Int[(c + d*x)^(m - 1)*(a + b*(F^(g*(e + f*x)))^n)^(p + 1
), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n, p}, x] && NeQ[p, -1]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2321

Int[(u_.)*((a_.)*(F_)^(v_) + (b_.)*(F_)^(w_))^(n_), x_Symbol] :> Int[u*F^(n*v)*(a + b*F^ExpandToSum[w - v, x])
^n, x] /; FreeQ[{F, a, b, n}, x] && ILtQ[n, 0] && LinearQ[{v, w}, x]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps \begin{align*} \text {integral}& = \int \frac {f^{2 x} x^2}{\left (b+a f^{2 x}\right )^2} \, dx \\ & = -\frac {x^2}{2 a \left (b+a f^{2 x}\right ) \log (f)}+\frac {\int \frac {x}{b+a f^{2 x}} \, dx}{a \log (f)} \\ & = \frac {x^2}{2 a b \log (f)}-\frac {x^2}{2 a \left (b+a f^{2 x}\right ) \log (f)}-\frac {\int \frac {f^{2 x} x}{b+a f^{2 x}} \, dx}{b \log (f)} \\ & = \frac {x^2}{2 a b \log (f)}-\frac {x^2}{2 a \left (b+a f^{2 x}\right ) \log (f)}-\frac {x \log \left (1+\frac {a f^{2 x}}{b}\right )}{2 a b \log ^2(f)}+\frac {\int \log \left (1+\frac {a f^{2 x}}{b}\right ) \, dx}{2 a b \log ^2(f)} \\ & = \frac {x^2}{2 a b \log (f)}-\frac {x^2}{2 a \left (b+a f^{2 x}\right ) \log (f)}-\frac {x \log \left (1+\frac {a f^{2 x}}{b}\right )}{2 a b \log ^2(f)}+\frac {\text {Subst}\left (\int \frac {\log \left (1+\frac {a x}{b}\right )}{x} \, dx,x,f^{2 x}\right )}{4 a b \log ^3(f)} \\ & = \frac {x^2}{2 a b \log (f)}-\frac {x^2}{2 a \left (b+a f^{2 x}\right ) \log (f)}-\frac {x \log \left (1+\frac {a f^{2 x}}{b}\right )}{2 a b \log ^2(f)}-\frac {\text {Li}_2\left (-\frac {a f^{2 x}}{b}\right )}{4 a b \log ^3(f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.92 \[ \int \frac {x^2}{\left (b f^{-x}+a f^x\right )^2} \, dx=\frac {2 x \log (f) \left (a f^{2 x} x \log (f)-\left (b+a f^{2 x}\right ) \log \left (1+\frac {a f^{2 x}}{b}\right )\right )-\left (b+a f^{2 x}\right ) \operatorname {PolyLog}\left (2,-\frac {a f^{2 x}}{b}\right )}{4 a b \left (b+a f^{2 x}\right ) \log ^3(f)} \]

[In]

Integrate[x^2/(b/f^x + a*f^x)^2,x]

[Out]

(2*x*Log[f]*(a*f^(2*x)*x*Log[f] - (b + a*f^(2*x))*Log[1 + (a*f^(2*x))/b]) - (b + a*f^(2*x))*PolyLog[2, -((a*f^
(2*x))/b)])/(4*a*b*(b + a*f^(2*x))*Log[f]^3)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.93

method	result	size
risch	\(\frac {x^{2}}{2 a b \ln \left (f \right )}-\frac {x^{2}}{2 a \left (b +a \,f^{2 x}\right ) \ln \left (f \right )}-\frac {x \ln \left (1+\frac {a \,f^{2 x}}{b}\right )}{2 a b \ln \left (f \right )^{2}}-\frac {\operatorname {Li}_{2}\left (-\frac {a \,f^{2 x}}{b}\right )}{4 a b \ln \left (f \right )^{3}}\)	\(91\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.62 \[ \int \frac {x^2}{\left (b f^{-x}+a f^x\right )^2} \, dx=\frac {a f^{2 \, x} x^{2} \log \left (f\right )^{2} - {\left (a f^{2 \, x} + b\right )} {\rm Li}_2\left (f^{x} \sqrt {-\frac {a}{b}}\right ) - {\left (a f^{2 \, x} + b\right )} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {a}{b}}\right ) - {\left (a f^{2 \, x} x \log \left (f\right ) + b x \log \left (f\right )\right )} \log \left (f^{x} \sqrt {-\frac {a}{b}} + 1\right ) - {\left (a f^{2 \, x} x \log \left (f\right ) + b x \log \left (f\right )\right )} \log \left (-f^{x} \sqrt {-\frac {a}{b}} + 1\right )}{2 \, {\left (a^{2} b f^{2 \, x} \log \left (f\right )^{3} + a b^{2} \log \left (f\right )^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.85 \[ \int \frac {x^2}{\left (b f^{-x}+a f^x\right )^2} \, dx=-\frac {x^{2}}{2 \, {\left (a^{2} f^{2 \, x} \log \left (f\right ) + a b \log \left (f\right )\right )}} + \frac {x^{2}}{2 \, a b \log \left (f\right )} - \frac {2 \, x \log \left (\frac {a f^{2 \, x}}{b} + 1\right ) \log \left (f\right ) + {\rm Li}_2\left (-\frac {a f^{2 \, x}}{b}\right )}{4 \, a b \log \left (f\right )^{3}} \]

[In]

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

[Out]

-1/2*x^2/(a^2*f^(2*x)*log(f) + a*b*log(f)) + 1/2*x^2/(a*b*log(f)) - 1/4*(2*x*log(a*f^(2*x)/b + 1)*log(f) + dil
og(-a*f^(2*x)/b))/(a*b*log(f)^3)

Giac [F]

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

[In]

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

[Out]

integrate(x^2/(a*f^x + b/f^x)^2, x)

Mupad [F(-1)]

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

[In]

int(x^2/(b/f^x + a*f^x)^2,x)

[Out]

int(x^2/(b/f^x + a*f^x)^2, x)