3.1.66 \(\int \frac {x^{3/2}}{(a+b \sec (c+d \sqrt {x}))^2} \, dx\) [66]

3.1.66.1 Optimal result
3.1.66.2 Mathematica [A] (verified)
3.1.66.3 Rubi [A] (verified)
3.1.66.4 Maple [F]
3.1.66.5 Fricas [F]
3.1.66.6 Sympy [F]
3.1.66.7 Maxima [F(-2)]
3.1.66.8 Giac [F]
3.1.66.9 Mupad [F(-1)]

3.1.66.1 Optimal result

Integrand size = 22, antiderivative size = 1925 \[ \int \frac {x^{3/2}}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx =\text {Too large to display} \]

output
2*b^2*x^2*sin(c+d*x^(1/2))/a/(a^2-b^2)/d/(b+a*cos(c+d*x^(1/2)))-2*I*b^3*x^ 
2*ln(1+a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d 
-24*I*b^2*x*polylog(2,-a*exp(I*(c+d*x^(1/2)))/(b-I*(a^2-b^2)^(1/2)))/a^2/( 
a^2-b^2)/d^3-24*I*b^2*x*polylog(2,-a*exp(I*(c+d*x^(1/2)))/(b+I*(a^2-b^2)^( 
1/2)))/a^2/(a^2-b^2)/d^3-24*I*b^3*x*polylog(3,-a*exp(I*(c+d*x^(1/2)))/(b-( 
-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^3-4*I*b*x^2*ln(1+a*exp(I*(c+d*x^( 
1/2)))/(b+(-a^2+b^2)^(1/2)))/a^2/d/(-a^2+b^2)^(1/2)-48*I*b*x*polylog(3,-a* 
exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a^2/d^3/(-a^2+b^2)^(1/2)+8*b^2* 
x^(3/2)*ln(1+a*exp(I*(c+d*x^(1/2)))/(b-I*(a^2-b^2)^(1/2)))/a^2/(a^2-b^2)/d 
^2+8*b^2*x^(3/2)*ln(1+a*exp(I*(c+d*x^(1/2)))/(b+I*(a^2-b^2)^(1/2)))/a^2/(a 
^2-b^2)/d^2-8*b^3*x^(3/2)*polylog(2,-a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^ 
(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^2+8*b^3*x^(3/2)*polylog(2,-a*exp(I*(c+d*x^( 
1/2)))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^2+16*b*x^(3/2)*polylog 
(2,-a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a^2/d^2/(-a^2+b^2)^(1/2)- 
16*b*x^(3/2)*polylog(2,-a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a^2/d 
^2/(-a^2+b^2)^(1/2)+48*b^2*polylog(3,-a*exp(I*(c+d*x^(1/2)))/(b-I*(a^2-b^2 
)^(1/2)))*x^(1/2)/a^2/(a^2-b^2)/d^4+48*b^2*polylog(3,-a*exp(I*(c+d*x^(1/2) 
))/(b+I*(a^2-b^2)^(1/2)))*x^(1/2)/a^2/(a^2-b^2)/d^4+48*b^3*polylog(4,-a*ex 
p(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))*x^(1/2)/a^2/(-a^2+b^2)^(3/2)/d^4- 
48*b^3*polylog(4,-a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))*x^(1/2)/...
 
3.1.66.2 Mathematica [A] (verified)

Time = 14.04 (sec) , antiderivative size = 2254, normalized size of antiderivative = 1.17 \[ \int \frac {x^{3/2}}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx=\text {Result too large to show} \]

input
Integrate[x^(3/2)/(a + b*Sec[c + d*Sqrt[x]])^2,x]
 
output
(2*x^(5/2)*(b + a*Cos[c + d*Sqrt[x]])^2*Sec[c + d*Sqrt[x]]^2)/(5*a^2*(a + 
b*Sec[c + d*Sqrt[x]])^2) + (2*b*E^(I*c)*(b + a*Cos[c + d*Sqrt[x]])^2*((-2* 
I)*b*E^(I*c)*x^2 + ((1 + E^((2*I)*c))*(4*b*d^3*Sqrt[(-a^2 + b^2)*E^((2*I)* 
c)]*x^(3/2)*Log[1 + (a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + 
b^2)*E^((2*I)*c)])] + (2*I)*a^2*d^4*E^(I*c)*x^2*Log[1 + (a*E^(I*(2*c + d*S 
qrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - I*b^2*d^4*E^(I*c 
)*x^2*Log[1 + (a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E 
^((2*I)*c)])] + 4*b*d^3*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]*x^(3/2)*Log[1 + (a* 
E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - ( 
2*I)*a^2*d^4*E^(I*c)*x^2*Log[1 + (a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) + 
Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + I*b^2*d^4*E^(I*c)*x^2*Log[1 + (a*E^(I*( 
2*c + d*Sqrt[x])))/(b*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 4*d^2*( 
(3*I)*b*Sqrt[(-a^2 + b^2)*E^((2*I)*c)] - 2*a^2*d*E^(I*c)*Sqrt[x] + b^2*d*E 
^(I*c)*Sqrt[x])*x*PolyLog[2, -((a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sq 
rt[(-a^2 + b^2)*E^((2*I)*c)]))] + 4*d^2*((-3*I)*b*Sqrt[(-a^2 + b^2)*E^((2* 
I)*c)] - 2*a^2*d*E^(I*c)*Sqrt[x] + b^2*d*E^(I*c)*Sqrt[x])*x*PolyLog[2, -(( 
a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] 
+ 24*b*d*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]*Sqrt[x]*PolyLog[3, -((a*E^(I*(2*c 
+ d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + (24*I)*a^2 
*d^2*E^(I*c)*x*PolyLog[3, -((a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sq...
 
3.1.66.3 Rubi [A] (verified)

Time = 3.19 (sec) , antiderivative size = 1926, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4692, 3042, 4679, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^{3/2}}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx\)

\(\Big \downarrow \) 4692

\(\displaystyle 2 \int \frac {x^2}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2}d\sqrt {x}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int \frac {x^2}{\left (a+b \csc \left (c+d \sqrt {x}+\frac {\pi }{2}\right )\right )^2}d\sqrt {x}\)

\(\Big \downarrow \) 4679

\(\displaystyle 2 \int \left (-\frac {2 b x^2}{a^2 \left (b+a \cos \left (c+d \sqrt {x}\right )\right )}+\frac {x^2}{a^2}+\frac {b^2 x^2}{a^2 \left (b+a \cos \left (c+d \sqrt {x}\right )\right )^2}\right )d\sqrt {x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (-\frac {i x^2 \log \left (\frac {e^{i \left (c+d \sqrt {x}\right )} a}{b-\sqrt {b^2-a^2}}+1\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d}+\frac {i x^2 \log \left (\frac {e^{i \left (c+d \sqrt {x}\right )} a}{b+\sqrt {b^2-a^2}}+1\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d}-\frac {4 x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^2}+\frac {4 x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^2}-\frac {12 i x \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^3}+\frac {12 i x \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^3}+\frac {24 \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^4}-\frac {24 \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^4}+\frac {24 i \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^5}-\frac {24 i \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^5}-\frac {i x^2 b^2}{a^2 \left (a^2-b^2\right ) d}+\frac {4 x^{3/2} \log \left (\frac {e^{i \left (c+d \sqrt {x}\right )} a}{b-i \sqrt {a^2-b^2}}+1\right ) b^2}{a^2 \left (a^2-b^2\right ) d^2}+\frac {4 x^{3/2} \log \left (\frac {e^{i \left (c+d \sqrt {x}\right )} a}{b+i \sqrt {a^2-b^2}}+1\right ) b^2}{a^2 \left (a^2-b^2\right ) d^2}-\frac {12 i x \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-i \sqrt {a^2-b^2}}\right ) b^2}{a^2 \left (a^2-b^2\right ) d^3}-\frac {12 i x \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+i \sqrt {a^2-b^2}}\right ) b^2}{a^2 \left (a^2-b^2\right ) d^3}+\frac {24 \sqrt {x} \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-i \sqrt {a^2-b^2}}\right ) b^2}{a^2 \left (a^2-b^2\right ) d^4}+\frac {24 \sqrt {x} \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+i \sqrt {a^2-b^2}}\right ) b^2}{a^2 \left (a^2-b^2\right ) d^4}+\frac {24 i \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-i \sqrt {a^2-b^2}}\right ) b^2}{a^2 \left (a^2-b^2\right ) d^5}+\frac {24 i \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+i \sqrt {a^2-b^2}}\right ) b^2}{a^2 \left (a^2-b^2\right ) d^5}+\frac {x^2 \sin \left (c+d \sqrt {x}\right ) b^2}{a \left (a^2-b^2\right ) d \left (b+a \cos \left (c+d \sqrt {x}\right )\right )}+\frac {2 i x^2 \log \left (\frac {e^{i \left (c+d \sqrt {x}\right )} a}{b-\sqrt {b^2-a^2}}+1\right ) b}{a^2 \sqrt {b^2-a^2} d}-\frac {2 i x^2 \log \left (\frac {e^{i \left (c+d \sqrt {x}\right )} a}{b+\sqrt {b^2-a^2}}+1\right ) b}{a^2 \sqrt {b^2-a^2} d}+\frac {8 x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^2}-\frac {8 x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^2}+\frac {24 i x \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^3}-\frac {24 i x \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^3}-\frac {48 \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^4}+\frac {48 \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^4}-\frac {48 i \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^5}+\frac {48 i \operatorname {PolyLog}\left (5,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^5}+\frac {x^{5/2}}{5 a^2}\right )\)

input
Int[x^(3/2)/(a + b*Sec[c + d*Sqrt[x]])^2,x]
 
output
2*(((-I)*b^2*x^2)/(a^2*(a^2 - b^2)*d) + x^(5/2)/(5*a^2) + (4*b^2*x^(3/2)*L 
og[1 + (a*E^(I*(c + d*Sqrt[x])))/(b - I*Sqrt[a^2 - b^2])])/(a^2*(a^2 - b^2 
)*d^2) + (4*b^2*x^(3/2)*Log[1 + (a*E^(I*(c + d*Sqrt[x])))/(b + I*Sqrt[a^2 
- b^2])])/(a^2*(a^2 - b^2)*d^2) - (I*b^3*x^2*Log[1 + (a*E^(I*(c + d*Sqrt[x 
])))/(b - Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d) + ((2*I)*b*x^2*Lo 
g[1 + (a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + 
b^2]*d) + (I*b^3*x^2*Log[1 + (a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^ 
2])])/(a^2*(-a^2 + b^2)^(3/2)*d) - ((2*I)*b*x^2*Log[1 + (a*E^(I*(c + d*Sqr 
t[x])))/(b + Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*d) - ((12*I)*b^2*x* 
PolyLog[2, -((a*E^(I*(c + d*Sqrt[x])))/(b - I*Sqrt[a^2 - b^2]))])/(a^2*(a^ 
2 - b^2)*d^3) - ((12*I)*b^2*x*PolyLog[2, -((a*E^(I*(c + d*Sqrt[x])))/(b + 
I*Sqrt[a^2 - b^2]))])/(a^2*(a^2 - b^2)*d^3) - (4*b^3*x^(3/2)*PolyLog[2, -( 
(a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b^2)^(3/2 
)*d^2) + (8*b*x^(3/2)*PolyLog[2, -((a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^ 
2 + b^2]))])/(a^2*Sqrt[-a^2 + b^2]*d^2) + (4*b^3*x^(3/2)*PolyLog[2, -((a*E 
^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b^2)^(3/2)*d^ 
2) - (8*b*x^(3/2)*PolyLog[2, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + 
b^2]))])/(a^2*Sqrt[-a^2 + b^2]*d^2) + (24*b^2*Sqrt[x]*PolyLog[3, -((a*E^(I 
*(c + d*Sqrt[x])))/(b - I*Sqrt[a^2 - b^2]))])/(a^2*(a^2 - b^2)*d^4) + (24* 
b^2*Sqrt[x]*PolyLog[3, -((a*E^(I*(c + d*Sqrt[x])))/(b + I*Sqrt[a^2 - b^...
 

3.1.66.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4679
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si 
n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt 
Q[m, 0]
 

rule 4692
Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 
 1)/n], 0] && IntegerQ[p]
 
3.1.66.4 Maple [F]

\[\int \frac {x^{\frac {3}{2}}}{\left (a +b \sec \left (c +d \sqrt {x}\right )\right )^{2}}d x\]

input
int(x^(3/2)/(a+b*sec(c+d*x^(1/2)))^2,x)
 
output
int(x^(3/2)/(a+b*sec(c+d*x^(1/2)))^2,x)
 
3.1.66.5 Fricas [F]

\[ \int \frac {x^{3/2}}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx=\int { \frac {x^{\frac {3}{2}}}{{\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )}^{2}} \,d x } \]

input
integrate(x^(3/2)/(a+b*sec(c+d*x^(1/2)))^2,x, algorithm="fricas")
 
output
integral(x^(3/2)/(b^2*sec(d*sqrt(x) + c)^2 + 2*a*b*sec(d*sqrt(x) + c) + a^ 
2), x)
 
3.1.66.6 Sympy [F]

\[ \int \frac {x^{3/2}}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx=\int \frac {x^{\frac {3}{2}}}{\left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )^{2}}\, dx \]

input
integrate(x**(3/2)/(a+b*sec(c+d*x**(1/2)))**2,x)
 
output
Integral(x**(3/2)/(a + b*sec(c + d*sqrt(x)))**2, x)
 
3.1.66.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {x^{3/2}}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx=\text {Exception raised: ValueError} \]

input
integrate(x^(3/2)/(a+b*sec(c+d*x^(1/2)))^2,x, algorithm="maxima")
 
output
Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 
3.1.66.8 Giac [F]

\[ \int \frac {x^{3/2}}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx=\int { \frac {x^{\frac {3}{2}}}{{\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )}^{2}} \,d x } \]

input
integrate(x^(3/2)/(a+b*sec(c+d*x^(1/2)))^2,x, algorithm="giac")
 
output
integrate(x^(3/2)/(b*sec(d*sqrt(x) + c) + a)^2, x)
 
3.1.66.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^{3/2}}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx=\int \frac {x^{3/2}}{{\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right )}^2} \,d x \]

input
int(x^(3/2)/(a + b/cos(c + d*x^(1/2)))^2,x)
 
output
int(x^(3/2)/(a + b/cos(c + d*x^(1/2)))^2, x)