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

Optimal. Leaf size=223 \[ -\frac{3 i \text{PolyLog}\left (2,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{16 a^{5/2} \sqrt{b} \log ^2(f)}+\frac{3 i \text{PolyLog}\left (2,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{16 a^{5/2} \sqrt{b} \log ^2(f)}-\frac{f^x}{8 a^2 \log ^2(f) \left (a+b f^{2 x}\right )}+\frac{3 x f^x}{8 a^2 \log (f) \left (a+b f^{2 x}\right )}-\frac{\tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{2 a^{5/2} \sqrt{b} \log ^2(f)}+\frac{3 x \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{8 a^{5/2} \sqrt{b} \log (f)}+\frac{x f^x}{4 a \log (f) \left (a+b f^{2 x}\right )^2} \]

[Out]

-f^x/(8*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)/Sqrt[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)

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Rubi [A]  time = 0.217052, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {2249, 199, 205, 2245, 2282, 4848, 2391} \[ -\frac{3 i \text{PolyLog}\left (2,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{16 a^{5/2} \sqrt{b} \log ^2(f)}+\frac{3 i \text{PolyLog}\left (2,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{16 a^{5/2} \sqrt{b} \log ^2(f)}-\frac{f^x}{8 a^2 \log ^2(f) \left (a+b f^{2 x}\right )}+\frac{3 x f^x}{8 a^2 \log (f) \left (a+b f^{2 x}\right )}-\frac{\tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{2 a^{5/2} \sqrt{b} \log ^2(f)}+\frac{3 x \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{8 a^{5/2} \sqrt{b} \log (f)}+\frac{x f^x}{4 a \log (f) \left (a+b f^{2 x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(f^x*x)/(a + b*f^(2*x))^3,x]

[Out]

-f^x/(8*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)/Sqrt[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)

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

Mathematica [A]  time = 0.305697, size = 184, normalized size = 0.83 \[ \frac{\frac{6 i \left (-\text{PolyLog}\left (2,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )+\text{PolyLog}\left (2,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )+x \log (f) \left (\log \left (1-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )-\log \left (1+\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )\right )\right )}{\sqrt{a} \sqrt{b}}+\frac{8 a x f^x \log (f)}{\left (a+b f^{2 x}\right )^2}+\frac{4 f^x (3 x \log (f)-1)}{a+b f^{2 x}}-\frac{16 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b}}}{32 a^2 \log ^2(f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-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)/Sqr
t[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)

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Maple [A]  time = 0.069, size = 223, normalized size = 1. \begin{align*}{\frac{{f}^{x} \left ( 3\,\ln \left ( f \right ) bx \left ({f}^{x} \right ) ^{2}+5\,\ln \left ( f \right ) ax-b \left ({f}^{x} \right ) ^{2}-a \right ) }{8\, \left ( \ln \left ( f \right ) \right ) ^{2}{a}^{2} \left ( a+b \left ({f}^{x} \right ) ^{2} \right ) ^{2}}}-{\frac{1}{2\, \left ( \ln \left ( f \right ) \right ) ^{2}{a}^{2}}\arctan \left ({b{f}^{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,x}{16\,\ln \left ( f \right ){a}^{2}}\ln \left ({ \left ( -b{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}-{\frac{3\,x}{16\,\ln \left ( f \right ){a}^{2}}\ln \left ({ \left ( b{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}+{\frac{3}{16\, \left ( \ln \left ( f \right ) \right ) ^{2}{a}^{2}}{\it dilog} \left ({ \left ( -b{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}-{\frac{3}{16\, \left ( \ln \left ( f \right ) \right ) ^{2}{a}^{2}}{\it dilog} \left ({ \left ( b{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(f^x*x/(a+b*f^(2*x))^3,x)

[Out]

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/ln(f)^2/a^2/(a*b)^(1/2)*
arctan(b*f^x/(a*b)^(1/2))+3/16/ln(f)/a^2*x/(-a*b)^(1/2)*ln((-b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-3/16/ln(f)/a^2*
x/(-a*b)^(1/2)*ln((b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))+3/16/ln(f)^2/a^2/(-a*b)^(1/2)*dilog((-b*f^x+(-a*b)^(1/2))
/(-a*b)^(1/2))-3/16/ln(f)^2/a^2/(-a*b)^(1/2)*dilog((b*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(f^x*x/(a+b*f^(2*x))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.60431, size = 1068, normalized size = 4.79 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*sq
rt(-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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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) + (Integral(-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))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{x} x}{{\left (b f^{2 \, x} + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^x*x/(b*f^(2*x) + a)^3, x)