3.63 \(\int \frac{x}{(b f^{-x}+a f^x)^3} \, dx\)

Optimal. Leaf size=196 \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}+\frac{f^x}{8 a b \log ^2(f) \left (a f^{2 x}+b\right )}+\frac{x f^x}{8 a b \log (f) \left (a f^{2 x}+b\right )}-\frac{x f^x}{4 a \log (f) \left (a f^{2 x}+b\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.502889, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 9, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.529, Rules used = {2283, 2254, 2249, 199, 205, 2245, 2282, 4848, 2391} \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}+\frac{f^x}{8 a b \log ^2(f) \left (a f^{2 x}+b\right )}+\frac{x f^x}{8 a b \log (f) \left (a f^{2 x}+b\right )}-\frac{x f^x}{4 a \log (f) \left (a f^{2 x}+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x/(b/f^x + a*f^x)^3,x]

[Out]

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

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 2254

Int[((a_.) + (b_.)*(F_)^(u_))^(p_.)*((c_.) + (d_.)*(F_)^(v_))^(q_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> W
ith[{w = ExpandIntegrand[(e + f*x)^m, (a + b*F^u)^p*(c + d*F^v)^q, x]}, Int[w, x] /; SumQ[w]] /; FreeQ[{F, a,
b, c, d, e, f, m}, x] && IntegersQ[p, q] && LinearQ[{u, v}, x] && RationalQ[Simplify[u/v]]

Rule 2249

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(d*e*Log[F])/(g*h*Log[G])]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - (d*e*f)/g)*x^Numerator[m])^p, x], x, G^((h*(f + g*x))/Denominator[m])], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2245

Int[(F_)^((e_.)*((c_.) + (d_.)*(x_)))*((a_.) + (b_.)*(F_)^(v_))^(p_)*(x_)^(m_.), x_Symbol] :> With[{u = IntHid
e[F^(e*(c + d*x))*(a + b*F^v)^p, x]}, Dist[x^m, u, x] - Dist[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]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

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}{\left (b f^{-x}+a f^x\right )^3} \, dx &=\int \frac{f^{3 x} x}{\left (b+a f^{2 x}\right )^3} \, dx\\ &=\int \left (-\frac{b f^x x}{a \left (b+a f^{2 x}\right )^3}+\frac{f^x x}{a \left (b+a f^{2 x}\right )^2}\right ) \, dx\\ &=\frac{\int \frac{f^x x}{\left (b+a f^{2 x}\right )^2} \, dx}{a}-\frac{b \int \frac{f^x x}{\left (b+a f^{2 x}\right )^3} \, dx}{a}\\ &=-\frac{f^x x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac{f^x x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac{\int \left (\frac{f^x}{2 b \left (b+a f^{2 x}\right ) \log (f)}+\frac{\tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{2 \sqrt{a} b^{3/2} \log (f)}\right ) \, dx}{a}+\frac{b \int \left (\frac{f^x}{4 b \left (b+a f^{2 x}\right )^2 \log (f)}+\frac{3 f^x}{8 b^2 \left (b+a f^{2 x}\right ) \log (f)}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 \sqrt{a} b^{5/2} \log (f)}\right ) \, dx}{a}\\ &=-\frac{f^x x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac{f^x x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}+\frac{\int \frac{f^x}{\left (b+a f^{2 x}\right )^2} \, dx}{4 a \log (f)}+\frac{3 \int \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right ) \, dx}{8 a^{3/2} b^{3/2} \log (f)}-\frac{\int \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right ) \, dx}{2 a^{3/2} b^{3/2} \log (f)}+\frac{3 \int \frac{f^x}{b+a f^{2 x}} \, dx}{8 a b \log (f)}-\frac{\int \frac{f^x}{b+a f^{2 x}} \, dx}{2 a b \log (f)}\\ &=-\frac{f^x x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac{f^x x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (b+a x^2\right )^2} \, dx,x,f^x\right )}{4 a \log ^2(f)}+\frac{3 \operatorname{Subst}\left (\int \frac{\tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{x} \, dx,x,f^x\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}-\frac{\operatorname{Subst}\left (\int \frac{\tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{x} \, dx,x,f^x\right )}{2 a^{3/2} b^{3/2} \log ^2(f)}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,f^x\right )}{8 a b \log ^2(f)}-\frac{\operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,f^x\right )}{2 a b \log ^2(f)}\\ &=\frac{f^x}{8 a b \left (b+a f^{2 x}\right ) \log ^2(f)}-\frac{\tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}-\frac{f^x x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac{f^x x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{i \sqrt{a} x}{\sqrt{b}}\right )}{x} \, dx,x,f^x\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{i \sqrt{a} x}{\sqrt{b}}\right )}{x} \, dx,x,f^x\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}-\frac{i \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{i \sqrt{a} x}{\sqrt{b}}\right )}{x} \, dx,x,f^x\right )}{4 a^{3/2} b^{3/2} \log ^2(f)}+\frac{i \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{i \sqrt{a} x}{\sqrt{b}}\right )}{x} \, dx,x,f^x\right )}{4 a^{3/2} b^{3/2} \log ^2(f)}+\frac{\operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,f^x\right )}{8 a b \log ^2(f)}\\ &=\frac{f^x}{8 a b \left (b+a f^{2 x}\right ) \log ^2(f)}-\frac{f^x x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac{f^x x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac{i \text{Li}_2\left (-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac{i \text{Li}_2\left (\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}\\ \end{align*}

Mathematica [A]  time = 0.208664, size = 209, normalized size = 1.07 \[ \frac{-\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{b^{3/2}}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{b^{3/2}}+\frac{2 \sqrt{a} f^x}{a b f^{2 x}+b^2}+\frac{2 \sqrt{a} x f^x \log (f)}{a b f^{2 x}+b^2}+\frac{i x \log (f) \log \left (1-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{b^{3/2}}-\frac{i x \log (f) \log \left (1+\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{b^{3/2}}-\frac{4 \sqrt{a} x f^x \log (f)}{\left (a f^{2 x}+b\right )^2}}{16 a^{3/2} \log ^2(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(b/f^x + a*f^x)^3,x]

[Out]

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

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Maple [A]  time = 0.062, size = 209, normalized size = 1.1 \begin{align*}{\frac{{f}^{x} \left ( \left ({f}^{x} \right ) ^{2}\ln \left ( f \right ) ax-\ln \left ( f \right ) bx+a \left ({f}^{x} \right ) ^{2}+b \right ) }{8\, \left ( \ln \left ( f \right ) \right ) ^{2}ab \left ( a \left ({f}^{x} \right ) ^{2}+b \right ) ^{2}}}+{\frac{x}{16\,b\ln \left ( f \right ) a}\ln \left ({ \left ( -a{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}-{\frac{x}{16\,b\ln \left ( f \right ) a}\ln \left ({ \left ( a{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}+{\frac{1}{16\, \left ( \ln \left ( f \right ) \right ) ^{2}ab}{\it dilog} \left ({ \left ( -a{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}-{\frac{1}{16\, \left ( \ln \left ( f \right ) \right ) ^{2}ab}{\it dilog} \left ({ \left ( a{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.60115, size = 774, normalized size = 3.95 \begin{align*} \frac{2 \,{\left (a^{2} x \log \left (f\right ) + a^{2}\right )} f^{3 \, x} - 2 \,{\left (a b x \log \left (f\right ) - a b\right )} f^{x} +{\left (a^{2} f^{4 \, x} \sqrt{-\frac{a}{b}} + 2 \, a b f^{2 \, x} \sqrt{-\frac{a}{b}} + b^{2} \sqrt{-\frac{a}{b}}\right )}{\rm Li}_2\left (f^{x} \sqrt{-\frac{a}{b}}\right ) -{\left (a^{2} f^{4 \, x} \sqrt{-\frac{a}{b}} + 2 \, a b f^{2 \, x} \sqrt{-\frac{a}{b}} + b^{2} \sqrt{-\frac{a}{b}}\right )}{\rm Li}_2\left (-f^{x} \sqrt{-\frac{a}{b}}\right ) -{\left (a^{2} f^{4 \, x} x \sqrt{-\frac{a}{b}} \log \left (f\right ) + 2 \, a b f^{2 \, x} x \sqrt{-\frac{a}{b}} \log \left (f\right ) + b^{2} x \sqrt{-\frac{a}{b}} \log \left (f\right )\right )} \log \left (f^{x} \sqrt{-\frac{a}{b}} + 1\right ) +{\left (a^{2} f^{4 \, x} x \sqrt{-\frac{a}{b}} \log \left (f\right ) + 2 \, a b f^{2 \, x} x \sqrt{-\frac{a}{b}} \log \left (f\right ) + b^{2} x \sqrt{-\frac{a}{b}} \log \left (f\right )\right )} \log \left (-f^{x} \sqrt{-\frac{a}{b}} + 1\right )}{16 \,{\left (a^{4} b f^{4 \, x} \log \left (f\right )^{2} + 2 \, a^{3} b^{2} f^{2 \, x} \log \left (f\right )^{2} + a^{2} b^{3} \log \left (f\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*(a^2*x*log(f) + a^2)*f^(3*x) - 2*(a*b*x*log(f) - a*b)*f^x + (a^2*f^(4*x)*sqrt(-a/b) + 2*a*b*f^(2*x)*sq
rt(-a/b) + b^2*sqrt(-a/b))*dilog(f^x*sqrt(-a/b)) - (a^2*f^(4*x)*sqrt(-a/b) + 2*a*b*f^(2*x)*sqrt(-a/b) + b^2*sq
rt(-a/b))*dilog(-f^x*sqrt(-a/b)) - (a^2*f^(4*x)*x*sqrt(-a/b)*log(f) + 2*a*b*f^(2*x)*x*sqrt(-a/b)*log(f) + b^2*
x*sqrt(-a/b)*log(f))*log(f^x*sqrt(-a/b) + 1) + (a^2*f^(4*x)*x*sqrt(-a/b)*log(f) + 2*a*b*f^(2*x)*x*sqrt(-a/b)*l
og(f) + b^2*x*sqrt(-a/b)*log(f))*log(-f^x*sqrt(-a/b) + 1))/(a^4*b*f^(4*x)*log(f)^2 + 2*a^3*b^2*f^(2*x)*log(f)^
2 + a^2*b^3*log(f)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{f^{- x} \left (a x \log{\left (f \right )} + a\right ) + f^{- 3 x} \left (- b x \log{\left (f \right )} + b\right )}{8 a^{3} b \log{\left (f \right )}^{2} + 16 a^{2} b^{2} f^{- 2 x} \log{\left (f \right )}^{2} + 8 a b^{3} f^{- 4 x} \log{\left (f \right )}^{2}} + \frac{\int \frac{f^{x} x}{a f^{2 x} + b}\, dx}{8 a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(f**(-x)*(a*x*log(f) + a) + f**(-3*x)*(-b*x*log(f) + b))/(8*a**3*b*log(f)**2 + 16*a**2*b**2*f**(-2*x)*log(f)**
2 + 8*a*b**3*f**(-4*x)*log(f)**2) + Integral(f**x*x/(a*f**(2*x) + b), x)/(8*a*b)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (a f^{x} + \frac{b}{f^{x}}\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(a*f^x + b/f^x)^3, x)