3.10 \(\int x^4 \text{sech}^{-1}(a x)^3 \, dx\)

Optimal. Leaf size=297 \[ \frac{9 i \text{sech}^{-1}(a x) \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}-\frac{9 i \text{sech}^{-1}(a x) \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}-\frac{9 i \text{PolyLog}\left (3,-i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}+\frac{9 i \text{PolyLog}\left (3,i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}-\frac{3 x^3 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2}{20 a^2}-\frac{x^3 \text{sech}^{-1}(a x)}{10 a^2}+\frac{x \sqrt{\frac{1-a x}{a x+1}} (a x+1)}{20 a^4}+\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1)}{a x}\right )}{2 a^5}-\frac{9 x \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2}{40 a^4}-\frac{9 x \text{sech}^{-1}(a x)}{20 a^4}-\frac{9 \text{sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}+\frac{1}{5} x^5 \text{sech}^{-1}(a x)^3 \]

[Out]

(x*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x))/(20*a^4) - (9*x*ArcSech[a*x])/(20*a^4) - (x^3*ArcSech[a*x])/(10*a^2) -
(9*x*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x]^2)/(40*a^4) - (3*x^3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x
)*ArcSech[a*x]^2)/(20*a^2) + (x^5*ArcSech[a*x]^3)/5 - (9*ArcSech[a*x]^2*ArcTan[E^ArcSech[a*x]])/(20*a^5) + Arc
Tan[(Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x))/(a*x)]/(2*a^5) + (((9*I)/20)*ArcSech[a*x]*PolyLog[2, (-I)*E^ArcSech[
a*x]])/a^5 - (((9*I)/20)*ArcSech[a*x]*PolyLog[2, I*E^ArcSech[a*x]])/a^5 - (((9*I)/20)*PolyLog[3, (-I)*E^ArcSec
h[a*x]])/a^5 + (((9*I)/20)*PolyLog[3, I*E^ArcSech[a*x]])/a^5

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Rubi [A]  time = 0.203194, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.9, Rules used = {6285, 5418, 4186, 3768, 3770, 4180, 2531, 2282, 6589} \[ \frac{9 i \text{sech}^{-1}(a x) \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}-\frac{9 i \text{sech}^{-1}(a x) \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}-\frac{9 i \text{PolyLog}\left (3,-i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}+\frac{9 i \text{PolyLog}\left (3,i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}-\frac{3 x^3 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2}{20 a^2}-\frac{x^3 \text{sech}^{-1}(a x)}{10 a^2}+\frac{x \sqrt{\frac{1-a x}{a x+1}} (a x+1)}{20 a^4}+\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1)}{a x}\right )}{2 a^5}-\frac{9 x \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2}{40 a^4}-\frac{9 x \text{sech}^{-1}(a x)}{20 a^4}-\frac{9 \text{sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}+\frac{1}{5} x^5 \text{sech}^{-1}(a x)^3 \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcSech[a*x]^3,x]

[Out]

(x*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x))/(20*a^4) - (9*x*ArcSech[a*x])/(20*a^4) - (x^3*ArcSech[a*x])/(10*a^2) -
(9*x*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x]^2)/(40*a^4) - (3*x^3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x
)*ArcSech[a*x]^2)/(20*a^2) + (x^5*ArcSech[a*x]^3)/5 - (9*ArcSech[a*x]^2*ArcTan[E^ArcSech[a*x]])/(20*a^5) + Arc
Tan[(Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x))/(a*x)]/(2*a^5) + (((9*I)/20)*ArcSech[a*x]*PolyLog[2, (-I)*E^ArcSech[
a*x]])/a^5 - (((9*I)/20)*ArcSech[a*x]*PolyLog[2, I*E^ArcSech[a*x]])/a^5 - (((9*I)/20)*PolyLog[3, (-I)*E^ArcSec
h[a*x]])/a^5 + (((9*I)/20)*PolyLog[3, I*E^ArcSech[a*x]])/a^5

Rule 6285

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b
*x)^n*Sech[x]^(m + 1)*Tanh[x], x], x, ArcSech[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] &
& (GtQ[n, 0] || LtQ[m, -1])

Rule 5418

Int[(x_)^(m_.)*Sech[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tanh[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> -Simp[(
x^(m - n + 1)*Sech[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sech[a + b*x^n]^p, x],
x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

Rule 4186

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(b^2*(c + d*x)^m*Cot[e
+ f*x]*(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*d^2*m*(m - 1))/(f^2*(n - 1)*(n - 2)), Int[(c + d
*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)^m*(b*Csc[e + f*x])^(n
- 2), x], x] - Simp[(b^2*d*m*(c + d*x)^(m - 1)*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[
{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int x^4 \text{sech}^{-1}(a x)^3 \, dx &=-\frac{\operatorname{Subst}\left (\int x^3 \text{sech}^5(x) \tanh (x) \, dx,x,\text{sech}^{-1}(a x)\right )}{a^5}\\ &=\frac{1}{5} x^5 \text{sech}^{-1}(a x)^3-\frac{3 \operatorname{Subst}\left (\int x^2 \text{sech}^5(x) \, dx,x,\text{sech}^{-1}(a x)\right )}{5 a^5}\\ &=-\frac{x^3 \text{sech}^{-1}(a x)}{10 a^2}-\frac{3 x^3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{20 a^2}+\frac{1}{5} x^5 \text{sech}^{-1}(a x)^3+\frac{\operatorname{Subst}\left (\int \text{sech}^3(x) \, dx,x,\text{sech}^{-1}(a x)\right )}{10 a^5}-\frac{9 \operatorname{Subst}\left (\int x^2 \text{sech}^3(x) \, dx,x,\text{sech}^{-1}(a x)\right )}{20 a^5}\\ &=\frac{x \sqrt{\frac{1-a x}{1+a x}} (1+a x)}{20 a^4}-\frac{9 x \text{sech}^{-1}(a x)}{20 a^4}-\frac{x^3 \text{sech}^{-1}(a x)}{10 a^2}-\frac{9 x \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{40 a^4}-\frac{3 x^3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{20 a^2}+\frac{1}{5} x^5 \text{sech}^{-1}(a x)^3+\frac{\operatorname{Subst}\left (\int \text{sech}(x) \, dx,x,\text{sech}^{-1}(a x)\right )}{20 a^5}-\frac{9 \operatorname{Subst}\left (\int x^2 \text{sech}(x) \, dx,x,\text{sech}^{-1}(a x)\right )}{40 a^5}+\frac{9 \operatorname{Subst}\left (\int \text{sech}(x) \, dx,x,\text{sech}^{-1}(a x)\right )}{20 a^5}\\ &=\frac{x \sqrt{\frac{1-a x}{1+a x}} (1+a x)}{20 a^4}-\frac{9 x \text{sech}^{-1}(a x)}{20 a^4}-\frac{x^3 \text{sech}^{-1}(a x)}{10 a^2}-\frac{9 x \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{40 a^4}-\frac{3 x^3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{20 a^2}+\frac{1}{5} x^5 \text{sech}^{-1}(a x)^3-\frac{9 \text{sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}+\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)}{a x}\right )}{2 a^5}+\frac{(9 i) \operatorname{Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\text{sech}^{-1}(a x)\right )}{20 a^5}-\frac{(9 i) \operatorname{Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\text{sech}^{-1}(a x)\right )}{20 a^5}\\ &=\frac{x \sqrt{\frac{1-a x}{1+a x}} (1+a x)}{20 a^4}-\frac{9 x \text{sech}^{-1}(a x)}{20 a^4}-\frac{x^3 \text{sech}^{-1}(a x)}{10 a^2}-\frac{9 x \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{40 a^4}-\frac{3 x^3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{20 a^2}+\frac{1}{5} x^5 \text{sech}^{-1}(a x)^3-\frac{9 \text{sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}+\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)}{a x}\right )}{2 a^5}+\frac{9 i \text{sech}^{-1}(a x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}-\frac{9 i \text{sech}^{-1}(a x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}-\frac{(9 i) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\text{sech}^{-1}(a x)\right )}{20 a^5}+\frac{(9 i) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\text{sech}^{-1}(a x)\right )}{20 a^5}\\ &=\frac{x \sqrt{\frac{1-a x}{1+a x}} (1+a x)}{20 a^4}-\frac{9 x \text{sech}^{-1}(a x)}{20 a^4}-\frac{x^3 \text{sech}^{-1}(a x)}{10 a^2}-\frac{9 x \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{40 a^4}-\frac{3 x^3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{20 a^2}+\frac{1}{5} x^5 \text{sech}^{-1}(a x)^3-\frac{9 \text{sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}+\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)}{a x}\right )}{2 a^5}+\frac{9 i \text{sech}^{-1}(a x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}-\frac{9 i \text{sech}^{-1}(a x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}-\frac{(9 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}+\frac{(9 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}\\ &=\frac{x \sqrt{\frac{1-a x}{1+a x}} (1+a x)}{20 a^4}-\frac{9 x \text{sech}^{-1}(a x)}{20 a^4}-\frac{x^3 \text{sech}^{-1}(a x)}{10 a^2}-\frac{9 x \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{40 a^4}-\frac{3 x^3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{20 a^2}+\frac{1}{5} x^5 \text{sech}^{-1}(a x)^3-\frac{9 \text{sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}+\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)}{a x}\right )}{2 a^5}+\frac{9 i \text{sech}^{-1}(a x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}-\frac{9 i \text{sech}^{-1}(a x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}-\frac{9 i \text{Li}_3\left (-i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}+\frac{9 i \text{Li}_3\left (i e^{\text{sech}^{-1}(a x)}\right )}{20 a^5}\\ \end{align*}

Mathematica [A]  time = 0.554282, size = 281, normalized size = 0.95 \[ \frac{18 i \text{sech}^{-1}(a x) \text{PolyLog}\left (2,-i e^{-\text{sech}^{-1}(a x)}\right )-18 i \text{sech}^{-1}(a x) \text{PolyLog}\left (2,i e^{-\text{sech}^{-1}(a x)}\right )+18 i \text{PolyLog}\left (3,-i e^{-\text{sech}^{-1}(a x)}\right )-18 i \text{PolyLog}\left (3,i e^{-\text{sech}^{-1}(a x)}\right )+8 a^5 x^5 \text{sech}^{-1}(a x)^3-6 a^3 x^3 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2-4 a^3 x^3 \text{sech}^{-1}(a x)+2 a x \sqrt{\frac{1-a x}{a x+1}} (a x+1)-9 a x \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2-18 a x \text{sech}^{-1}(a x)+9 i \text{sech}^{-1}(a x)^2 \log \left (1-i e^{-\text{sech}^{-1}(a x)}\right )-9 i \text{sech}^{-1}(a x)^2 \log \left (1+i e^{-\text{sech}^{-1}(a x)}\right )+40 \tan ^{-1}\left (\tanh \left (\frac{1}{2} \text{sech}^{-1}(a x)\right )\right )}{40 a^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4*ArcSech[a*x]^3,x]

[Out]

(2*a*x*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x) - 18*a*x*ArcSech[a*x] - 4*a^3*x^3*ArcSech[a*x] - 9*a*x*Sqrt[(1 - a*
x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x]^2 - 6*a^3*x^3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x]^2 + 8*a^5*
x^5*ArcSech[a*x]^3 + 40*ArcTan[Tanh[ArcSech[a*x]/2]] + (9*I)*ArcSech[a*x]^2*Log[1 - I/E^ArcSech[a*x]] - (9*I)*
ArcSech[a*x]^2*Log[1 + I/E^ArcSech[a*x]] + (18*I)*ArcSech[a*x]*PolyLog[2, (-I)/E^ArcSech[a*x]] - (18*I)*ArcSec
h[a*x]*PolyLog[2, I/E^ArcSech[a*x]] + (18*I)*PolyLog[3, (-I)/E^ArcSech[a*x]] - (18*I)*PolyLog[3, I/E^ArcSech[a
*x]])/(40*a^5)

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Maple [F]  time = 0.651, size = 0, normalized size = 0. \begin{align*} \int{x}^{4} \left ({\rm arcsech} \left (ax\right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arcsech(a*x)^3,x)

[Out]

int(x^4*arcsech(a*x)^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \operatorname{arsech}\left (a x\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsech(a*x)^3,x, algorithm="maxima")

[Out]

integrate(x^4*arcsech(a*x)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{4} \operatorname{arsech}\left (a x\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsech(a*x)^3,x, algorithm="fricas")

[Out]

integral(x^4*arcsech(a*x)^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \operatorname{asech}^{3}{\left (a x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*asech(a*x)**3,x)

[Out]

Integral(x**4*asech(a*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \operatorname{arsech}\left (a x\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsech(a*x)^3,x, algorithm="giac")

[Out]

integrate(x^4*arcsech(a*x)^3, x)