3.7.30 \(\int \frac {x^4 (a+b x^2)^2}{(c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=197 \[ -\frac {c \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{9/2}}+\frac {x \sqrt {c+d x^2} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right )}{16 d^4}-\frac {x^3 \sqrt {c+d x^2} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right )}{24 c d^3}+\frac {x^5 (b c-a d)^2}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {463, 459, 321, 217, 206} \begin {gather*} -\frac {x^3 \sqrt {c+d x^2} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right )}{24 c d^3}+\frac {x \sqrt {c+d x^2} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right )}{16 d^4}-\frac {c \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{9/2}}+\frac {x^5 (b c-a d)^2}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x]

[Out]

((b*c - a*d)^2*x^5)/(c*d^2*Sqrt[c + d*x^2]) + ((35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*d
^4) - ((35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*x^3*Sqrt[c + d*x^2])/(24*c*d^3) + (b^2*x^5*Sqrt[c + d*x^2])/(6*d
^2) - (c*(35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(16*d^(9/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 321

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

Rule 459

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 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> -Simp[((b*c - a*
d)^2*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b^2*e*n*(p + 1)), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a + b
*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a,
b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx &=\frac {(b c-a d)^2 x^5}{c d^2 \sqrt {c+d x^2}}-\frac {\int \frac {x^4 \left (-a^2 d^2+5 (b c-a d)^2-b^2 c d x^2\right )}{\sqrt {c+d x^2}} \, dx}{c d^2}\\ &=\frac {(b c-a d)^2 x^5}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2}-\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) \int \frac {x^4}{\sqrt {c+d x^2}} \, dx}{6 c d^2}\\ &=\frac {(b c-a d)^2 x^5}{c d^2 \sqrt {c+d x^2}}-\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x^3 \sqrt {c+d x^2}}{24 c d^3}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2}+\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) \int \frac {x^2}{\sqrt {c+d x^2}} \, dx}{8 d^3}\\ &=\frac {(b c-a d)^2 x^5}{c d^2 \sqrt {c+d x^2}}+\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^4}-\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x^3 \sqrt {c+d x^2}}{24 c d^3}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2}-\frac {\left (c \left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{16 d^4}\\ &=\frac {(b c-a d)^2 x^5}{c d^2 \sqrt {c+d x^2}}+\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^4}-\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x^3 \sqrt {c+d x^2}}{24 c d^3}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2}-\frac {\left (c \left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{16 d^4}\\ &=\frac {(b c-a d)^2 x^5}{c d^2 \sqrt {c+d x^2}}+\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^4}-\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x^3 \sqrt {c+d x^2}}{24 c d^3}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2}-\frac {c \left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{9/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.16, size = 158, normalized size = 0.80 \begin {gather*} \sqrt {c+d x^2} \left (\frac {x \left (8 a^2 d^2-28 a b c d+19 b^2 c^2\right )}{16 d^4}+\frac {c x (b c-a d)^2}{d^4 \left (c+d x^2\right )}-\frac {b x^3 (11 b c-12 a d)}{24 d^3}+\frac {b^2 x^5}{6 d^2}\right )-\frac {c \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{16 d^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x]

[Out]

Sqrt[c + d*x^2]*(((19*b^2*c^2 - 28*a*b*c*d + 8*a^2*d^2)*x)/(16*d^4) - (b*(11*b*c - 12*a*d)*x^3)/(24*d^3) + (b^
2*x^5)/(6*d^2) + (c*(b*c - a*d)^2*x)/(d^4*(c + d*x^2))) - (c*(35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*Log[d*x +
Sqrt[d]*Sqrt[c + d*x^2]])/(16*d^(9/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.31, size = 171, normalized size = 0.87 \begin {gather*} \frac {\left (24 a^2 c d^2-60 a b c^2 d+35 b^2 c^3\right ) \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{16 d^{9/2}}+\frac {72 a^2 c d^2 x+24 a^2 d^3 x^3-180 a b c^2 d x-60 a b c d^2 x^3+24 a b d^3 x^5+105 b^2 c^3 x+35 b^2 c^2 d x^3-14 b^2 c d^2 x^5+8 b^2 d^3 x^7}{48 d^4 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^4*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x]

[Out]

(105*b^2*c^3*x - 180*a*b*c^2*d*x + 72*a^2*c*d^2*x + 35*b^2*c^2*d*x^3 - 60*a*b*c*d^2*x^3 + 24*a^2*d^3*x^3 - 14*
b^2*c*d^2*x^5 + 24*a*b*d^3*x^5 + 8*b^2*d^3*x^7)/(48*d^4*Sqrt[c + d*x^2]) + ((35*b^2*c^3 - 60*a*b*c^2*d + 24*a^
2*c*d^2)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(16*d^(9/2))

________________________________________________________________________________________

fricas [A]  time = 1.44, size = 431, normalized size = 2.19 \begin {gather*} \left [\frac {3 \, {\left (35 \, b^{2} c^{4} - 60 \, a b c^{3} d + 24 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (8 \, b^{2} d^{4} x^{7} - 2 \, {\left (7 \, b^{2} c d^{3} - 12 \, a b d^{4}\right )} x^{5} + {\left (35 \, b^{2} c^{2} d^{2} - 60 \, a b c d^{3} + 24 \, a^{2} d^{4}\right )} x^{3} + 3 \, {\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{96 \, {\left (d^{6} x^{2} + c d^{5}\right )}}, \frac {3 \, {\left (35 \, b^{2} c^{4} - 60 \, a b c^{3} d + 24 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (8 \, b^{2} d^{4} x^{7} - 2 \, {\left (7 \, b^{2} c d^{3} - 12 \, a b d^{4}\right )} x^{5} + {\left (35 \, b^{2} c^{2} d^{2} - 60 \, a b c d^{3} + 24 \, a^{2} d^{4}\right )} x^{3} + 3 \, {\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{48 \, {\left (d^{6} x^{2} + c d^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/96*(3*(35*b^2*c^4 - 60*a*b*c^3*d + 24*a^2*c^2*d^2 + (35*b^2*c^3*d - 60*a*b*c^2*d^2 + 24*a^2*c*d^3)*x^2)*sqr
t(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(8*b^2*d^4*x^7 - 2*(7*b^2*c*d^3 - 12*a*b*d^4)*x^5 + (
35*b^2*c^2*d^2 - 60*a*b*c*d^3 + 24*a^2*d^4)*x^3 + 3*(35*b^2*c^3*d - 60*a*b*c^2*d^2 + 24*a^2*c*d^3)*x)*sqrt(d*x
^2 + c))/(d^6*x^2 + c*d^5), 1/48*(3*(35*b^2*c^4 - 60*a*b*c^3*d + 24*a^2*c^2*d^2 + (35*b^2*c^3*d - 60*a*b*c^2*d
^2 + 24*a^2*c*d^3)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (8*b^2*d^4*x^7 - 2*(7*b^2*c*d^3 - 12*a*b
*d^4)*x^5 + (35*b^2*c^2*d^2 - 60*a*b*c*d^3 + 24*a^2*d^4)*x^3 + 3*(35*b^2*c^3*d - 60*a*b*c^2*d^2 + 24*a^2*c*d^3
)*x)*sqrt(d*x^2 + c))/(d^6*x^2 + c*d^5)]

________________________________________________________________________________________

giac [A]  time = 0.48, size = 175, normalized size = 0.89 \begin {gather*} \frac {{\left ({\left (2 \, {\left (\frac {4 \, b^{2} x^{2}}{d} - \frac {7 \, b^{2} c d^{5} - 12 \, a b d^{6}}{d^{7}}\right )} x^{2} + \frac {35 \, b^{2} c^{2} d^{4} - 60 \, a b c d^{5} + 24 \, a^{2} d^{6}}{d^{7}}\right )} x^{2} + \frac {3 \, {\left (35 \, b^{2} c^{3} d^{3} - 60 \, a b c^{2} d^{4} + 24 \, a^{2} c d^{5}\right )}}{d^{7}}\right )} x}{48 \, \sqrt {d x^{2} + c}} + \frac {{\left (35 \, b^{2} c^{3} - 60 \, a b c^{2} d + 24 \, a^{2} c d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{16 \, d^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

1/48*((2*(4*b^2*x^2/d - (7*b^2*c*d^5 - 12*a*b*d^6)/d^7)*x^2 + (35*b^2*c^2*d^4 - 60*a*b*c*d^5 + 24*a^2*d^6)/d^7
)*x^2 + 3*(35*b^2*c^3*d^3 - 60*a*b*c^2*d^4 + 24*a^2*c*d^5)/d^7)*x/sqrt(d*x^2 + c) + 1/16*(35*b^2*c^3 - 60*a*b*
c^2*d + 24*a^2*c*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(9/2)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 263, normalized size = 1.34 \begin {gather*} \frac {b^{2} x^{7}}{6 \sqrt {d \,x^{2}+c}\, d}+\frac {a b \,x^{5}}{2 \sqrt {d \,x^{2}+c}\, d}-\frac {7 b^{2} c \,x^{5}}{24 \sqrt {d \,x^{2}+c}\, d^{2}}+\frac {a^{2} x^{3}}{2 \sqrt {d \,x^{2}+c}\, d}-\frac {5 a b c \,x^{3}}{4 \sqrt {d \,x^{2}+c}\, d^{2}}+\frac {35 b^{2} c^{2} x^{3}}{48 \sqrt {d \,x^{2}+c}\, d^{3}}+\frac {3 a^{2} c x}{2 \sqrt {d \,x^{2}+c}\, d^{2}}-\frac {15 a b \,c^{2} x}{4 \sqrt {d \,x^{2}+c}\, d^{3}}+\frac {35 b^{2} c^{3} x}{16 \sqrt {d \,x^{2}+c}\, d^{4}}-\frac {3 a^{2} c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {5}{2}}}+\frac {15 a b \,c^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{4 d^{\frac {7}{2}}}-\frac {35 b^{2} c^{3} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{16 d^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^2+a)^2/(d*x^2+c)^(3/2),x)

[Out]

1/6*b^2*x^7/d/(d*x^2+c)^(1/2)-7/24*b^2*c/d^2*x^5/(d*x^2+c)^(1/2)+35/48*b^2*c^2/d^3*x^3/(d*x^2+c)^(1/2)+35/16*b
^2*c^3/d^4*x/(d*x^2+c)^(1/2)-35/16*b^2*c^3/d^(9/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))+1/2*a*b*x^5/d/(d*x^2+c)^(1/2)
-5/4*a*b*c/d^2*x^3/(d*x^2+c)^(1/2)-15/4*a*b*c^2/d^3*x/(d*x^2+c)^(1/2)+15/4*a*b*c^2/d^(7/2)*ln(d^(1/2)*x+(d*x^2
+c)^(1/2))+1/2*a^2*x^3/d/(d*x^2+c)^(1/2)+3/2*a^2*c/d^2*x/(d*x^2+c)^(1/2)-3/2*a^2*c/d^(5/2)*ln(d^(1/2)*x+(d*x^2
+c)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 0.96, size = 241, normalized size = 1.22 \begin {gather*} \frac {b^{2} x^{7}}{6 \, \sqrt {d x^{2} + c} d} - \frac {7 \, b^{2} c x^{5}}{24 \, \sqrt {d x^{2} + c} d^{2}} + \frac {a b x^{5}}{2 \, \sqrt {d x^{2} + c} d} + \frac {35 \, b^{2} c^{2} x^{3}}{48 \, \sqrt {d x^{2} + c} d^{3}} - \frac {5 \, a b c x^{3}}{4 \, \sqrt {d x^{2} + c} d^{2}} + \frac {a^{2} x^{3}}{2 \, \sqrt {d x^{2} + c} d} + \frac {35 \, b^{2} c^{3} x}{16 \, \sqrt {d x^{2} + c} d^{4}} - \frac {15 \, a b c^{2} x}{4 \, \sqrt {d x^{2} + c} d^{3}} + \frac {3 \, a^{2} c x}{2 \, \sqrt {d x^{2} + c} d^{2}} - \frac {35 \, b^{2} c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{16 \, d^{\frac {9}{2}}} + \frac {15 \, a b c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{4 \, d^{\frac {7}{2}}} - \frac {3 \, a^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, d^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

1/6*b^2*x^7/(sqrt(d*x^2 + c)*d) - 7/24*b^2*c*x^5/(sqrt(d*x^2 + c)*d^2) + 1/2*a*b*x^5/(sqrt(d*x^2 + c)*d) + 35/
48*b^2*c^2*x^3/(sqrt(d*x^2 + c)*d^3) - 5/4*a*b*c*x^3/(sqrt(d*x^2 + c)*d^2) + 1/2*a^2*x^3/(sqrt(d*x^2 + c)*d) +
35/16*b^2*c^3*x/(sqrt(d*x^2 + c)*d^4) - 15/4*a*b*c^2*x/(sqrt(d*x^2 + c)*d^3) + 3/2*a^2*c*x/(sqrt(d*x^2 + c)*d
^2) - 35/16*b^2*c^3*arcsinh(d*x/sqrt(c*d))/d^(9/2) + 15/4*a*b*c^2*arcsinh(d*x/sqrt(c*d))/d^(7/2) - 3/2*a^2*c*a
rcsinh(d*x/sqrt(c*d))/d^(5/2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4\,{\left (b\,x^2+a\right )}^2}{{\left (d\,x^2+c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x)

[Out]

int((x^4*(a + b*x^2)^2)/(c + d*x^2)^(3/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} \left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b*x**2+a)**2/(d*x**2+c)**(3/2),x)

[Out]

Integral(x**4*(a + b*x**2)**2/(c + d*x**2)**(3/2), x)

________________________________________________________________________________________