\(\int (f x)^m (d+e x^2)^2 (a+b \sec ^{-1}(c x)) \, dx\) [162]

Optimal result

Integrand size = 23, antiderivative size = 374 \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=-\frac {b e \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) x (f x)^{1+m} \sqrt {-1+c^2 x^2}}{c^3 f (2+m) (3+m) (4+m) (5+m) \sqrt {c^2 x^2}}-\frac {b e^2 x (f x)^{3+m} \sqrt {-1+c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {b \left (c^4 d^2 (2+m) (3+m) (4+m) (5+m)+e (1+m)^2 \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right )\right ) x (f x)^{1+m} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{c^3 f (1+m)^2 (2+m) (3+m) (4+m) (5+m) \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2}} \]

[Out]

d^2*(f*x)^(1+m)*(a+b*arcsec(c*x))/f/(1+m)+2*d*e*(f*x)^(3+m)*(a+b*arcsec(c*x))/f^3/(3+m)+e^2*(f*x)^(5+m)*(a+b*a
rcsec(c*x))/f^5/(5+m)-b*(c^4*d^2*(2+m)*(3+m)*(4+m)*(5+m)+e*(1+m)^2*(e*(3+m)^2+2*c^2*d*(m^2+9*m+20)))*x*(f*x)^(
1+m)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],c^2*x^2)*(-c^2*x^2+1)^(1/2)/c^3/f/(1+m)^2/(2+m)/(3+m)/(4+m)/(5+m)/
(c^2*x^2)^(1/2)/(c^2*x^2-1)^(1/2)-b*e*(e*(3+m)^2+2*c^2*d*(m^2+9*m+20))*x*(f*x)^(1+m)*(c^2*x^2-1)^(1/2)/c^3/f/(
4+m)/(5+m)/(m^2+5*m+6)/(c^2*x^2)^(1/2)-b*e^2*x*(f*x)^(3+m)*(c^2*x^2-1)^(1/2)/c/f^3/(4+m)/(5+m)/(c^2*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 355, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {276, 5346, 12, 1281, 470, 372, 371} \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {d^2 (f x)^{m+1} \left (a+b \sec ^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (m+5)}-\frac {b e^2 x \sqrt {c^2 x^2-1} (f x)^{m+3}}{c f^3 (m+4) (m+5) \sqrt {c^2 x^2}}-\frac {b c x \sqrt {1-c^2 x^2} (f x)^{m+1} \left (\frac {e \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^4 (m+2) (m+3) (m+4) (m+5)}+\frac {d^2}{(m+1)^2}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},c^2 x^2\right )}{f \sqrt {c^2 x^2} \sqrt {c^2 x^2-1}}-\frac {b e x \sqrt {c^2 x^2-1} (f x)^{m+1} \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^3 f (m+2) (m+3) (m+4) (m+5) \sqrt {c^2 x^2}} \]

[In]

Int[(f*x)^m*(d + e*x^2)^2*(a + b*ArcSec[c*x]),x]

[Out]

-((b*e*(e*(3 + m)^2 + 2*c^2*d*(20 + 9*m + m^2))*x*(f*x)^(1 + m)*Sqrt[-1 + c^2*x^2])/(c^3*f*(2 + m)*(3 + m)*(4
+ m)*(5 + m)*Sqrt[c^2*x^2])) - (b*e^2*x*(f*x)^(3 + m)*Sqrt[-1 + c^2*x^2])/(c*f^3*(4 + m)*(5 + m)*Sqrt[c^2*x^2]
) + (d^2*(f*x)^(1 + m)*(a + b*ArcSec[c*x]))/(f*(1 + m)) + (2*d*e*(f*x)^(3 + m)*(a + b*ArcSec[c*x]))/(f^3*(3 +
m)) + (e^2*(f*x)^(5 + m)*(a + b*ArcSec[c*x]))/(f^5*(5 + m)) - (b*c*(d^2/(1 + m)^2 + (e*(e*(3 + m)^2 + 2*c^2*d*
(20 + 9*m + m^2)))/(c^4*(2 + m)*(3 + m)*(4 + m)*(5 + m)))*x*(f*x)^(1 + m)*Sqrt[1 - c^2*x^2]*Hypergeometric2F1[
1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(f*Sqrt[c^2*x^2]*Sqrt[-1 + c^2*x^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 470

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

Rule 1281

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Si
mp[c^p*(f*x)^(m + 4*p - 1)*((d + e*x^2)^(q + 1)/(e*f^(4*p - 1)*(m + 4*p + 2*q + 1))), x] + Dist[1/(e*(m + 4*p
+ 2*q + 1)), Int[(f*x)^m*(d + e*x^2)^q*ExpandToSum[e*(m + 4*p + 2*q + 1)*((a + b*x^2 + c*x^4)^p - c^p*x^(4*p))
- d*c^p*(m + 4*p - 1)*x^(4*p - 2), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c, 0] &&
IGtQ[p, 0] &&  !IntegerQ[q] && NeQ[m + 4*p + 2*q + 1, 0]

Rule 5346

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSec[c*x], u, x] - Dist[b*c*(x/Sqrt[c^2*x^2]), Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&  !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {(b c x) \int \frac {(f x)^m \left (d^2 \left (15+8 m+m^2\right )+2 d e \left (5+6 m+m^2\right ) x^2+e^2 \left (3+4 m+m^2\right ) x^4\right )}{(1+m) (3+m) (5+m) \sqrt {-1+c^2 x^2}} \, dx}{\sqrt {c^2 x^2}} \\ & = \frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {(b c x) \int \frac {(f x)^m \left (d^2 \left (15+8 m+m^2\right )+2 d e \left (5+6 m+m^2\right ) x^2+e^2 \left (3+4 m+m^2\right ) x^4\right )}{\sqrt {-1+c^2 x^2}} \, dx}{\left (15+23 m+9 m^2+m^3\right ) \sqrt {c^2 x^2}} \\ & = -\frac {b e^2 x (f x)^{3+m} \sqrt {-1+c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {(b x) \int \frac {(f x)^m \left (c^2 d^2 (3+m) (4+m) (5+m)+e (1+m) \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) x^2\right )}{\sqrt {-1+c^2 x^2}} \, dx}{c (4+m) \left (15+23 m+9 m^2+m^3\right ) \sqrt {c^2 x^2}} \\ & = -\frac {b e \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) x (f x)^{1+m} \sqrt {-1+c^2 x^2}}{c^3 f (2+m) (4+m) \left (15+8 m+m^2\right ) \sqrt {c^2 x^2}}-\frac {b e^2 x (f x)^{3+m} \sqrt {-1+c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}--\frac {\left (b \left (-c^4 d^2 (2+m) (3+m) (4+m) (5+m)-e (1+m)^2 \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right )\right ) x\right ) \int \frac {(f x)^m}{\sqrt {-1+c^2 x^2}} \, dx}{c^3 (2+m) (4+m) \left (15+23 m+9 m^2+m^3\right ) \sqrt {c^2 x^2}} \\ & = -\frac {b e \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) x (f x)^{1+m} \sqrt {-1+c^2 x^2}}{c^3 f (2+m) (4+m) \left (15+8 m+m^2\right ) \sqrt {c^2 x^2}}-\frac {b e^2 x (f x)^{3+m} \sqrt {-1+c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}--\frac {\left (b \left (-c^4 d^2 (2+m) (3+m) (4+m) (5+m)-e (1+m)^2 \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right )\right ) x \sqrt {1-c^2 x^2}\right ) \int \frac {(f x)^m}{\sqrt {1-c^2 x^2}} \, dx}{c^3 (2+m) (4+m) \left (15+23 m+9 m^2+m^3\right ) \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2}} \\ & = -\frac {b e \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) x (f x)^{1+m} \sqrt {-1+c^2 x^2}}{c^3 f (2+m) (4+m) \left (15+8 m+m^2\right ) \sqrt {c^2 x^2}}-\frac {b e^2 x (f x)^{3+m} \sqrt {-1+c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {b \left (c^4 d^2 (2+m) (3+m) (4+m) (5+m)+e (1+m)^2 \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right )\right ) x (f x)^{1+m} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{c^3 f (1+m) (2+m) (4+m) \left (15+23 m+9 m^2+m^3\right ) \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 291, normalized size of antiderivative = 0.78 \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=x (f x)^m \left (\frac {a d^2}{1+m}+\frac {2 a d e x^2}{3+m}+\frac {a e^2 x^4}{5+m}+\frac {b d^2 \sec ^{-1}(c x)}{1+m}+\frac {2 b d e x^2 \sec ^{-1}(c x)}{3+m}+\frac {b e^2 x^4 \sec ^{-1}(c x)}{5+m}+\frac {b c d^2 \sqrt {1-\frac {1}{c^2 x^2}} x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{(1+m)^2 \sqrt {1-c^2 x^2}}+\frac {2 b c d e \sqrt {1-\frac {1}{c^2 x^2}} x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{(3+m)^2 \sqrt {1-c^2 x^2}}+\frac {b c e^2 \sqrt {1-\frac {1}{c^2 x^2}} x^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5+m}{2},\frac {7+m}{2},c^2 x^2\right )}{(5+m)^2 \sqrt {1-c^2 x^2}}\right ) \]

[In]

Integrate[(f*x)^m*(d + e*x^2)^2*(a + b*ArcSec[c*x]),x]

[Out]

x*(f*x)^m*((a*d^2)/(1 + m) + (2*a*d*e*x^2)/(3 + m) + (a*e^2*x^4)/(5 + m) + (b*d^2*ArcSec[c*x])/(1 + m) + (2*b*
d*e*x^2*ArcSec[c*x])/(3 + m) + (b*e^2*x^4*ArcSec[c*x])/(5 + m) + (b*c*d^2*Sqrt[1 - 1/(c^2*x^2)]*x*Hypergeometr
ic2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/((1 + m)^2*Sqrt[1 - c^2*x^2]) + (2*b*c*d*e*Sqrt[1 - 1/(c^2*x^2)]*x^
3*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, c^2*x^2])/((3 + m)^2*Sqrt[1 - c^2*x^2]) + (b*c*e^2*Sqrt[1 - 1/(
c^2*x^2)]*x^5*Hypergeometric2F1[1/2, (5 + m)/2, (7 + m)/2, c^2*x^2])/((5 + m)^2*Sqrt[1 - c^2*x^2]))

Maple [F]

\[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right )^{2} \left (a +b \,\operatorname {arcsec}\left (c x \right )\right )d x\]

[In]

int((f*x)^m*(e*x^2+d)^2*(a+b*arcsec(c*x)),x)

[Out]

int((f*x)^m*(e*x^2+d)^2*(a+b*arcsec(c*x)),x)

Fricas [F]

\[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]

[In]

integrate((f*x)^m*(e*x^2+d)^2*(a+b*arcsec(c*x)),x, algorithm="fricas")

[Out]

integral((a*e^2*x^4 + 2*a*d*e*x^2 + a*d^2 + (b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*arcsec(c*x))*(f*x)^m, x)

Sympy [F]

\[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int \left (f x\right )^{m} \left (a + b \operatorname {asec}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}\, dx \]

[In]

integrate((f*x)**m*(e*x**2+d)**2*(a+b*asec(c*x)),x)

[Out]

Integral((f*x)**m*(a + b*asec(c*x))*(d + e*x**2)**2, x)

Maxima [F]

\[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]

[In]

integrate((f*x)^m*(e*x^2+d)^2*(a+b*arcsec(c*x)),x, algorithm="maxima")

[Out]

a*e^2*f^m*x^5*x^m/(m + 5) + 2*a*d*e*f^m*x^3*x^m/(m + 3) + (f*x)^(m + 1)*a*d^2/(f*(m + 1)) + (((b*e^2*f^m*m^2 +
4*b*e^2*f^m*m + 3*b*e^2*f^m)*x^5 + 2*(b*d*e*f^m*m^2 + 6*b*d*e*f^m*m + 5*b*d*e*f^m)*x^3 + (b*d^2*f^m*m^2 + 8*b
*d^2*f^m*m + 15*b*d^2*f^m)*x)*x^m*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) - (m^3 + 9*m^2 + 23*m + 15)*integrate(-(
b*d^2*f^m*m^2 + 8*b*d^2*f^m*m + (b*e^2*f^m*m^2 + 4*b*e^2*f^m*m + 3*b*e^2*f^m)*x^4 + 15*b*d^2*f^m + 2*(b*d*e*f^
m*m^2 + 6*b*d*e*f^m*m + 5*b*d*e*f^m)*x^2)*sqrt(c*x + 1)*sqrt(c*x - 1)*x^m/(m^3 - (c^2*m^3 + 9*c^2*m^2 + 23*c^2
*m + 15*c^2)*x^2 + 9*m^2 + 23*m + 15), x))/(m^3 + 9*m^2 + 23*m + 15)

Giac [F]

\[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]

[In]

integrate((f*x)^m*(e*x^2+d)^2*(a+b*arcsec(c*x)),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^2*(b*arcsec(c*x) + a)*(f*x)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int {\left (f\,x\right )}^m\,{\left (e\,x^2+d\right )}^2\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

[In]

int((f*x)^m*(d + e*x^2)^2*(a + b*acos(1/(c*x))),x)

[Out]

int((f*x)^m*(d + e*x^2)^2*(a + b*acos(1/(c*x))), x)