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

Optimal. Leaf size=98 \[ -\frac{\text{PolyLog}\left (2,-\frac{a f^{2 x}}{b}\right )}{4 a b \log ^3(f)}-\frac{x^2}{2 a \log (f) \left (a f^{2 x}+b\right )}-\frac{x \log \left (\frac{a f^{2 x}}{b}+1\right )}{2 a b \log ^2(f)}+\frac{x^2}{2 a b \log (f)} \]

[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)

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Rubi [A]  time = 0.169159, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {2283, 2191, 2184, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,-\frac{a f^{2 x}}{b}\right )}{4 a b \log ^3(f)}-\frac{x^2}{2 a \log (f) \left (a f^{2 x}+b\right )}-\frac{x \log \left (\frac{a f^{2 x}}{b}+1\right )}{2 a b \log ^2(f)}+\frac{x^2}{2 a b \log (f)} \]

Antiderivative was successfully verified.

[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 2283

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 2191

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 2184

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2279

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 2391

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*} \int \frac{x^2}{\left (b f^{-x}+a f^x\right )^2} \, dx &=\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{\operatorname{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]  time = 0.0624651, size = 90, normalized size = 0.92 \[ \frac{2 x \log (f) \left (a x f^{2 x} \log (f)-\left (a f^{2 x}+b\right ) \log \left (\frac{a f^{2 x}}{b}+1\right )\right )-\left (a f^{2 x}+b\right ) \text{PolyLog}\left (2,-\frac{a f^{2 x}}{b}\right )}{4 a b \log ^3(f) \left (a f^{2 x}+b\right )} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.029, size = 91, normalized size = 0.9 \begin{align*} -{\frac{{x}^{2}}{2\,\ln \left ( f \right ) a \left ( a \left ({f}^{x} \right ) ^{2}+b \right ) }}+{\frac{{x}^{2}}{2\,\ln \left ( f \right ) ab}}-{\frac{x}{2\, \left ( \ln \left ( f \right ) \right ) ^{2}ab}\ln \left ( 1+{\frac{a{f}^{2\,x}}{b}} \right ) }-{\frac{1}{4\,ab \left ( \ln \left ( f \right ) \right ) ^{3}}{\it polylog} \left ( 2,-{\frac{a{f}^{2\,x}}{b}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/ln(f)*x^2/a/(a*(f^x)^2+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

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Maxima [A]  time = 1.10902, size = 117, normalized size = 1.19 \begin{align*} -\frac{x^{2}}{2 \,{\left (a^{2} f^{2 \, x} \log \left (f\right ) + a b \log \left (f\right )\right )}} + \frac{\log \left (f^{x}\right )^{2}}{2 \, a b \log \left (f\right )^{3}} - \frac{2 \, \log \left (f^{x}\right ) \log \left (\frac{a f^{2 \, x}}{b} + 1\right ) +{\rm Li}_2\left (-\frac{a f^{2 \, x}}{b}\right )}{4 \, a b \log \left (f\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Fricas [A]  time = 1.53634, size = 370, normalized size = 3.78 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (a f^{x} + \frac{b}{f^{x}}\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)