3.18.47 \(\int (a+b x) (d+e x)^3 (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=172 \[ \frac {3 e^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6 (b d-a e)}{7 b^4}+\frac {e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5 (b d-a e)^2}{2 b^4}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e)^3}{5 b^4}+\frac {e^3 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^4} \]

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Rubi [A]  time = 0.18, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {770, 21, 43} \begin {gather*} \frac {3 e^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6 (b d-a e)}{7 b^4}+\frac {e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5 (b d-a e)^2}{2 b^4}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e)^3}{5 b^4}+\frac {e^3 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - a*e)^3*(a + b*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*b^4) + (e*(b*d - a*e)^2*(a + b*x)^5*Sqrt[a^2 + 2*
a*b*x + b^2*x^2])/(2*b^4) + (3*e^2*(b*d - a*e)*(a + b*x)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*b^4) + (e^3*(a +
b*x)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*b^4)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int (a+b x) (d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x) \left (a b+b^2 x\right )^3 (d+e x)^3 \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int (a+b x)^4 (d+e x)^3 \, dx}{a b+b^2 x}\\ &=\frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac {(b d-a e)^3 (a+b x)^4}{b^3}+\frac {3 e (b d-a e)^2 (a+b x)^5}{b^3}+\frac {3 e^2 (b d-a e) (a+b x)^6}{b^3}+\frac {e^3 (a+b x)^7}{b^3}\right ) \, dx}{a b+b^2 x}\\ &=\frac {(b d-a e)^3 (a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{5 b^4}+\frac {e (b d-a e)^2 (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{2 b^4}+\frac {3 e^2 (b d-a e) (a+b x)^6 \sqrt {a^2+2 a b x+b^2 x^2}}{7 b^4}+\frac {e^3 (a+b x)^7 \sqrt {a^2+2 a b x+b^2 x^2}}{8 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 212, normalized size = 1.23 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (70 a^4 \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )+56 a^3 b x \left (10 d^3+20 d^2 e x+15 d e^2 x^2+4 e^3 x^3\right )+28 a^2 b^2 x^2 \left (20 d^3+45 d^2 e x+36 d e^2 x^2+10 e^3 x^3\right )+8 a b^3 x^3 \left (35 d^3+84 d^2 e x+70 d e^2 x^2+20 e^3 x^3\right )+b^4 x^4 \left (56 d^3+140 d^2 e x+120 d e^2 x^2+35 e^3 x^3\right )\right )}{280 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(70*a^4*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + 56*a^3*b*x*(10*d^3 + 20*d^2*e*x + 1
5*d*e^2*x^2 + 4*e^3*x^3) + 28*a^2*b^2*x^2*(20*d^3 + 45*d^2*e*x + 36*d*e^2*x^2 + 10*e^3*x^3) + 8*a*b^3*x^3*(35*
d^3 + 84*d^2*e*x + 70*d*e^2*x^2 + 20*e^3*x^3) + b^4*x^4*(56*d^3 + 140*d^2*e*x + 120*d*e^2*x^2 + 35*e^3*x^3)))/
(280*(a + b*x))

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IntegrateAlgebraic [F]  time = 1.95, size = 0, normalized size = 0.00 \begin {gather*} \int (a+b x) (d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 225, normalized size = 1.31 \begin {gather*} \frac {1}{8} \, b^{4} e^{3} x^{8} + a^{4} d^{3} x + \frac {1}{7} \, {\left (3 \, b^{4} d e^{2} + 4 \, a b^{3} e^{3}\right )} x^{7} + \frac {1}{2} \, {\left (b^{4} d^{2} e + 4 \, a b^{3} d e^{2} + 2 \, a^{2} b^{2} e^{3}\right )} x^{6} + \frac {1}{5} \, {\left (b^{4} d^{3} + 12 \, a b^{3} d^{2} e + 18 \, a^{2} b^{2} d e^{2} + 4 \, a^{3} b e^{3}\right )} x^{5} + \frac {1}{4} \, {\left (4 \, a b^{3} d^{3} + 18 \, a^{2} b^{2} d^{2} e + 12 \, a^{3} b d e^{2} + a^{4} e^{3}\right )} x^{4} + {\left (2 \, a^{2} b^{2} d^{3} + 4 \, a^{3} b d^{2} e + a^{4} d e^{2}\right )} x^{3} + \frac {1}{2} \, {\left (4 \, a^{3} b d^{3} + 3 \, a^{4} d^{2} e\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^4*e^3*x^8 + a^4*d^3*x + 1/7*(3*b^4*d*e^2 + 4*a*b^3*e^3)*x^7 + 1/2*(b^4*d^2*e + 4*a*b^3*d*e^2 + 2*a^2*b^2
*e^3)*x^6 + 1/5*(b^4*d^3 + 12*a*b^3*d^2*e + 18*a^2*b^2*d*e^2 + 4*a^3*b*e^3)*x^5 + 1/4*(4*a*b^3*d^3 + 18*a^2*b^
2*d^2*e + 12*a^3*b*d*e^2 + a^4*e^3)*x^4 + (2*a^2*b^2*d^3 + 4*a^3*b*d^2*e + a^4*d*e^2)*x^3 + 1/2*(4*a^3*b*d^3 +
3*a^4*d^2*e)*x^2

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giac [B]  time = 0.21, size = 360, normalized size = 2.09 \begin {gather*} \frac {1}{8} \, b^{4} x^{8} e^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{7} \, b^{4} d x^{7} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, b^{4} d^{2} x^{6} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, b^{4} d^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {4}{7} \, a b^{3} x^{7} e^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, a b^{3} d x^{6} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {12}{5} \, a b^{3} d^{2} x^{5} e \mathrm {sgn}\left (b x + a\right ) + a b^{3} d^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + a^{2} b^{2} x^{6} e^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {18}{5} \, a^{2} b^{2} d x^{5} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {9}{2} \, a^{2} b^{2} d^{2} x^{4} e \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{2} b^{2} d^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {4}{5} \, a^{3} b x^{5} e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{3} b d x^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + 4 \, a^{3} b d^{2} x^{3} e \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{3} b d^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, a^{4} x^{4} e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} d x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, a^{4} d^{2} x^{2} e \mathrm {sgn}\left (b x + a\right ) + a^{4} d^{3} x \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^4*x^8*e^3*sgn(b*x + a) + 3/7*b^4*d*x^7*e^2*sgn(b*x + a) + 1/2*b^4*d^2*x^6*e*sgn(b*x + a) + 1/5*b^4*d^3*x
^5*sgn(b*x + a) + 4/7*a*b^3*x^7*e^3*sgn(b*x + a) + 2*a*b^3*d*x^6*e^2*sgn(b*x + a) + 12/5*a*b^3*d^2*x^5*e*sgn(b
*x + a) + a*b^3*d^3*x^4*sgn(b*x + a) + a^2*b^2*x^6*e^3*sgn(b*x + a) + 18/5*a^2*b^2*d*x^5*e^2*sgn(b*x + a) + 9/
2*a^2*b^2*d^2*x^4*e*sgn(b*x + a) + 2*a^2*b^2*d^3*x^3*sgn(b*x + a) + 4/5*a^3*b*x^5*e^3*sgn(b*x + a) + 3*a^3*b*d
*x^4*e^2*sgn(b*x + a) + 4*a^3*b*d^2*x^3*e*sgn(b*x + a) + 2*a^3*b*d^3*x^2*sgn(b*x + a) + 1/4*a^4*x^4*e^3*sgn(b*
x + a) + a^4*d*x^3*e^2*sgn(b*x + a) + 3/2*a^4*d^2*x^2*e*sgn(b*x + a) + a^4*d^3*x*sgn(b*x + a)

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maple [B]  time = 0.05, size = 264, normalized size = 1.53 \begin {gather*} \frac {\left (35 e^{3} b^{4} x^{7}+160 x^{6} e^{3} a \,b^{3}+120 x^{6} d \,e^{2} b^{4}+280 x^{5} e^{3} a^{2} b^{2}+560 x^{5} d \,e^{2} a \,b^{3}+140 x^{5} d^{2} e \,b^{4}+224 x^{4} e^{3} a^{3} b +1008 x^{4} d \,e^{2} a^{2} b^{2}+672 x^{4} d^{2} e a \,b^{3}+56 x^{4} d^{3} b^{4}+70 x^{3} e^{3} a^{4}+840 x^{3} d \,e^{2} a^{3} b +1260 x^{3} d^{2} e \,a^{2} b^{2}+280 x^{3} d^{3} a \,b^{3}+280 a^{4} d \,e^{2} x^{2}+1120 a^{3} b \,d^{2} e \,x^{2}+560 a^{2} b^{2} d^{3} x^{2}+420 x \,d^{2} e \,a^{4}+560 x \,d^{3} a^{3} b +280 d^{3} a^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x}{280 \left (b x +a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/280*x*(35*b^4*e^3*x^7+160*a*b^3*e^3*x^6+120*b^4*d*e^2*x^6+280*a^2*b^2*e^3*x^5+560*a*b^3*d*e^2*x^5+140*b^4*d^
2*e*x^5+224*a^3*b*e^3*x^4+1008*a^2*b^2*d*e^2*x^4+672*a*b^3*d^2*e*x^4+56*b^4*d^3*x^4+70*a^4*e^3*x^3+840*a^3*b*d
*e^2*x^3+1260*a^2*b^2*d^2*e*x^3+280*a*b^3*d^3*x^3+280*a^4*d*e^2*x^2+1120*a^3*b*d^2*e*x^2+560*a^2*b^2*d^3*x^2+4
20*a^4*d^2*e*x+560*a^3*b*d^3*x+280*a^4*d^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [B]  time = 0.65, size = 693, normalized size = 4.03 \begin {gather*} \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} e^{3} x^{3}}{8 \, b} + \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d^{3} x + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{4} e^{3} x}{4 \, b^{3}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a e^{3} x^{2}}{56 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} d^{3}}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{5} e^{3}}{4 \, b^{4}} + \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2} e^{3} x}{56 \, b^{3}} - \frac {69 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{3} e^{3}}{280 \, b^{4}} - \frac {{\left (3 \, b d e^{2} + a e^{3}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3} x}{4 \, b^{3}} + \frac {3 \, {\left (b d^{2} e + a d e^{2}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} x}{4 \, b^{2}} - \frac {{\left (b d^{3} + 3 \, a d^{2} e\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a x}{4 \, b} + \frac {{\left (3 \, b d e^{2} + a e^{3}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} x^{2}}{7 \, b^{2}} - \frac {{\left (3 \, b d e^{2} + a e^{3}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{4}}{4 \, b^{4}} + \frac {3 \, {\left (b d^{2} e + a d e^{2}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3}}{4 \, b^{3}} - \frac {{\left (b d^{3} + 3 \, a d^{2} e\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2}}{4 \, b^{2}} - \frac {3 \, {\left (3 \, b d e^{2} + a e^{3}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a x}{14 \, b^{3}} + \frac {{\left (b d^{2} e + a d e^{2}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} x}{2 \, b^{2}} + \frac {17 \, {\left (3 \, b d e^{2} + a e^{3}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2}}{70 \, b^{4}} - \frac {7 \, {\left (b d^{2} e + a d e^{2}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a}{10 \, b^{3}} + \frac {{\left (b d^{3} + 3 \, a d^{2} e\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}{5 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*e^3*x^3/b + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*d^3*x + 1/4*(b^2*x^2 + 2
*a*b*x + a^2)^(3/2)*a^4*e^3*x/b^3 - 11/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*e^3*x^2/b^2 + 1/4*(b^2*x^2 + 2*a*b
*x + a^2)^(3/2)*a^2*d^3/b + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^5*e^3/b^4 + 13/56*(b^2*x^2 + 2*a*b*x + a^2)^
(5/2)*a^2*e^3*x/b^3 - 69/280*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^3*e^3/b^4 - 1/4*(3*b*d*e^2 + a*e^3)*(b^2*x^2 +
2*a*b*x + a^2)^(3/2)*a^3*x/b^3 + 3/4*(b*d^2*e + a*d*e^2)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2*x/b^2 - 1/4*(b*d^
3 + 3*a*d^2*e)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*x/b + 1/7*(3*b*d*e^2 + a*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)
*x^2/b^2 - 1/4*(3*b*d*e^2 + a*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^4/b^4 + 3/4*(b*d^2*e + a*d*e^2)*(b^2*x^2
+ 2*a*b*x + a^2)^(3/2)*a^3/b^3 - 1/4*(b*d^3 + 3*a*d^2*e)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2/b^2 - 3/14*(3*b*d
*e^2 + a*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*x/b^3 + 1/2*(b*d^2*e + a*d*e^2)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2
)*x/b^2 + 17/70*(3*b*d*e^2 + a*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^2/b^4 - 7/10*(b*d^2*e + a*d*e^2)*(b^2*x^
2 + 2*a*b*x + a^2)^(5/2)*a/b^3 + 1/5*(b*d^3 + 3*a*d^2*e)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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