3.17 \(\int \frac{g+h x}{(a+b x+c x^2) (a d+b d x+c d x^2)^2} \, dx\)

Optimal. Leaf size=140 \[ \frac{3 (b+2 c x) (2 c g-b h)}{2 d^2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{-2 a h+x (2 c g-b h)+b g}{2 d^2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{6 c (2 c g-b h) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{d^2 \left (b^2-4 a c\right )^{5/2}} \]

[Out]

-(b*g - 2*a*h + (2*c*g - b*h)*x)/(2*(b^2 - 4*a*c)*d^2*(a + b*x + c*x^2)^2) + (3*(2*c*g - b*h)*(b + 2*c*x))/(2*
(b^2 - 4*a*c)^2*d^2*(a + b*x + c*x^2)) - (6*c*(2*c*g - b*h)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*
a*c)^(5/2)*d^2)

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Rubi [A]  time = 0.13148, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {998, 638, 614, 618, 206} \[ \frac{3 (b+2 c x) (2 c g-b h)}{2 d^2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{-2 a h+x (2 c g-b h)+b g}{2 d^2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{6 c (2 c g-b h) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{d^2 \left (b^2-4 a c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*g - 2*a*h + (2*c*g - b*h)*x)/(2*(b^2 - 4*a*c)*d^2*(a + b*x + c*x^2)^2) + (3*(2*c*g - b*h)*(b + 2*c*x))/(2*
(b^2 - 4*a*c)^2*d^2*(a + b*x + c*x^2)) - (6*c*(2*c*g - b*h)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*
a*c)^(5/2)*d^2)

Rule 998

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_)
, x_Symbol] :> Dist[(c/f)^p, Int[(g + h*x)^m*(d + e*x + f*x^2)^(p + q), x], x] /; FreeQ[{a, b, c, d, e, f, g,
h, p, q}, x] && EqQ[c*d - a*f, 0] && EqQ[b*d - a*e, 0] && (IntegerQ[p] || GtQ[c/f, 0]) && ( !IntegerQ[q] || Le
afCount[d + e*x + f*x^2] <= LeafCount[a + b*x + c*x^2])

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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])

Rubi steps

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

Mathematica [A]  time = 0.153385, size = 131, normalized size = 0.94 \[ \frac{\frac{\left (b^2-4 a c\right ) (2 a h-b g+b h x-2 c g x)}{(a+x (b+c x))^2}-\frac{12 c (b h-2 c g) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+\frac{3 (b+2 c x) (2 c g-b h)}{a+x (b+c x)}}{2 d^2 \left (b^2-4 a c\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((b^2 - 4*a*c)*(-(b*g) + 2*a*h - 2*c*g*x + b*h*x))/(a + x*(b + c*x))^2 + (3*(2*c*g - b*h)*(b + 2*c*x))/(a + x
*(b + c*x)) - (12*c*(-2*c*g + b*h)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c])/(2*(b^2 - 4*a*c
)^2*d^2)

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Maple [B]  time = 0.161, size = 340, normalized size = 2.4 \begin{align*} -{\frac{bxh}{2\,{d}^{2} \left ( 4\,ac-{b}^{2} \right ) \left ( c{x}^{2}+bx+a \right ) ^{2}}}+{\frac{cxg}{{d}^{2} \left ( 4\,ac-{b}^{2} \right ) \left ( c{x}^{2}+bx+a \right ) ^{2}}}-{\frac{ah}{{d}^{2} \left ( 4\,ac-{b}^{2} \right ) \left ( c{x}^{2}+bx+a \right ) ^{2}}}+{\frac{bg}{2\,{d}^{2} \left ( 4\,ac-{b}^{2} \right ) \left ( c{x}^{2}+bx+a \right ) ^{2}}}-3\,{\frac{bcxh}{{d}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}+6\,{\frac{x{c}^{2}g}{{d}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}-{\frac{3\,{b}^{2}h}{2\,{d}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}+3\,{\frac{bcg}{{d}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}-6\,{\frac{bch}{{d}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{5/2}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+12\,{\frac{{c}^{2}g}{{d}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{5/2}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((h*x+g)/(c*x^2+b*x+a)/(c*d*x^2+b*d*x+a*d)^2,x)

[Out]

-1/2/d^2/(4*a*c-b^2)/(c*x^2+b*x+a)^2*x*b*h+1/d^2/(4*a*c-b^2)/(c*x^2+b*x+a)^2*x*c*g-1/d^2/(4*a*c-b^2)/(c*x^2+b*
x+a)^2*a*h+1/2/d^2/(4*a*c-b^2)/(c*x^2+b*x+a)^2*b*g-3/d^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)*x*c*b*h+6/d^2/(4*a*c-b^2)
^2/(c*x^2+b*x+a)*x*c^2*g-3/2/d^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)*b^2*h+3/d^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)*b*c*g-6/d
^2/(4*a*c-b^2)^(5/2)*c*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b*h+12/d^2/(4*a*c-b^2)^(5/2)*c^2*arctan((2*c*x+b)/(
4*a*c-b^2)^(1/2))*g

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.93127, size = 2399, normalized size = 17.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(6*(2*(b^2*c^3 - 4*a*c^4)*g - (b^3*c^2 - 4*a*b*c^3)*h)*x^3 + 9*(2*(b^3*c^2 - 4*a*b*c^3)*g - (b^4*c - 4*a*
b^2*c^2)*h)*x^2 - 6*(2*a^2*c^2*g - a^2*b*c*h + (2*c^4*g - b*c^3*h)*x^4 + 2*(2*b*c^3*g - b^2*c^2*h)*x^3 + (2*(b
^2*c^2 + 2*a*c^3)*g - (b^3*c + 2*a*b*c^2)*h)*x^2 + 2*(2*a*b*c^2*g - a*b^2*c*h)*x)*sqrt(b^2 - 4*a*c)*log((2*c^2
*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) - (b^5 - 14*a*b^3*c + 40*a^2*
b*c^2)*g - (a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2)*h + 2*(2*(b^4*c + a*b^2*c^2 - 20*a^2*c^3)*g - (b^5 + a*b^3*c - 2
0*a^2*b*c^2)*h)*x)/((b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*d^2*x^4 + 2*(b^7*c - 12*a*b^5*c^2 +
48*a^2*b^3*c^3 - 64*a^3*b*c^4)*d^2*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*d
^2*x^2 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*d^2*x + (a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2
*c^2 - 64*a^5*c^3)*d^2), 1/2*(6*(2*(b^2*c^3 - 4*a*c^4)*g - (b^3*c^2 - 4*a*b*c^3)*h)*x^3 + 9*(2*(b^3*c^2 - 4*a*
b*c^3)*g - (b^4*c - 4*a*b^2*c^2)*h)*x^2 - 12*(2*a^2*c^2*g - a^2*b*c*h + (2*c^4*g - b*c^3*h)*x^4 + 2*(2*b*c^3*g
- b^2*c^2*h)*x^3 + (2*(b^2*c^2 + 2*a*c^3)*g - (b^3*c + 2*a*b*c^2)*h)*x^2 + 2*(2*a*b*c^2*g - a*b^2*c*h)*x)*sqr
t(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - (b^5 - 14*a*b^3*c + 40*a^2*b*c^2)*g -
(a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2)*h + 2*(2*(b^4*c + a*b^2*c^2 - 20*a^2*c^3)*g - (b^5 + a*b^3*c - 20*a^2*b*c^2
)*h)*x)/((b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*d^2*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3
*c^3 - 64*a^3*b*c^4)*d^2*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*d^2*x^2 + 2*
(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*d^2*x + (a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a
^5*c^3)*d^2)]

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Sympy [B]  time = 3.77541, size = 709, normalized size = 5.06 \begin{align*} \frac{3 c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) \log{\left (x + \frac{- 192 a^{3} c^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 144 a^{2} b^{2} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) - 36 a b^{4} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 3 b^{6} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 3 b^{2} c h - 6 b c^{2} g}{6 b c^{2} h - 12 c^{3} g} \right )}}{d^{2}} - \frac{3 c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) \log{\left (x + \frac{192 a^{3} c^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) - 144 a^{2} b^{2} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 36 a b^{4} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) - 3 b^{6} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 3 b^{2} c h - 6 b c^{2} g}{6 b c^{2} h - 12 c^{3} g} \right )}}{d^{2}} - \frac{8 a^{2} c h + a b^{2} h - 10 a b c g + b^{3} g + x^{3} \left (6 b c^{2} h - 12 c^{3} g\right ) + x^{2} \left (9 b^{2} c h - 18 b c^{2} g\right ) + x \left (10 a b c h - 20 a c^{2} g + 2 b^{3} h - 4 b^{2} c g\right )}{32 a^{4} c^{2} d^{2} - 16 a^{3} b^{2} c d^{2} + 2 a^{2} b^{4} d^{2} + x^{4} \left (32 a^{2} c^{4} d^{2} - 16 a b^{2} c^{3} d^{2} + 2 b^{4} c^{2} d^{2}\right ) + x^{3} \left (64 a^{2} b c^{3} d^{2} - 32 a b^{3} c^{2} d^{2} + 4 b^{5} c d^{2}\right ) + x^{2} \left (64 a^{3} c^{3} d^{2} - 12 a b^{4} c d^{2} + 2 b^{6} d^{2}\right ) + x \left (64 a^{3} b c^{2} d^{2} - 32 a^{2} b^{3} c d^{2} + 4 a b^{5} d^{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)/(c*x**2+b*x+a)/(c*d*x**2+b*d*x+a*d)**2,x)

[Out]

3*c*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g)*log(x + (-192*a**3*c**4*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g)
+ 144*a**2*b**2*c**3*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) - 36*a*b**4*c**2*sqrt(-1/(4*a*c - b**2)**5)*(b*h
- 2*c*g) + 3*b**6*c*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) + 3*b**2*c*h - 6*b*c**2*g)/(6*b*c**2*h - 12*c**3
*g))/d**2 - 3*c*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g)*log(x + (192*a**3*c**4*sqrt(-1/(4*a*c - b**2)**5)*(b*
h - 2*c*g) - 144*a**2*b**2*c**3*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) + 36*a*b**4*c**2*sqrt(-1/(4*a*c - b**
2)**5)*(b*h - 2*c*g) - 3*b**6*c*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) + 3*b**2*c*h - 6*b*c**2*g)/(6*b*c**2*
h - 12*c**3*g))/d**2 - (8*a**2*c*h + a*b**2*h - 10*a*b*c*g + b**3*g + x**3*(6*b*c**2*h - 12*c**3*g) + x**2*(9*
b**2*c*h - 18*b*c**2*g) + x*(10*a*b*c*h - 20*a*c**2*g + 2*b**3*h - 4*b**2*c*g))/(32*a**4*c**2*d**2 - 16*a**3*b
**2*c*d**2 + 2*a**2*b**4*d**2 + x**4*(32*a**2*c**4*d**2 - 16*a*b**2*c**3*d**2 + 2*b**4*c**2*d**2) + x**3*(64*a
**2*b*c**3*d**2 - 32*a*b**3*c**2*d**2 + 4*b**5*c*d**2) + x**2*(64*a**3*c**3*d**2 - 12*a*b**4*c*d**2 + 2*b**6*d
**2) + x*(64*a**3*b*c**2*d**2 - 32*a**2*b**3*c*d**2 + 4*a*b**5*d**2))

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Giac [A]  time = 1.208, size = 296, normalized size = 2.11 \begin{align*} \frac{6 \,{\left (2 \, c^{2} g - b c h\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} d^{2} - 8 \, a b^{2} c d^{2} + 16 \, a^{2} c^{2} d^{2}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{12 \, c^{3} g x^{3} - 6 \, b c^{2} h x^{3} + 18 \, b c^{2} g x^{2} - 9 \, b^{2} c h x^{2} + 4 \, b^{2} c g x + 20 \, a c^{2} g x - 2 \, b^{3} h x - 10 \, a b c h x - b^{3} g + 10 \, a b c g - a b^{2} h - 8 \, a^{2} c h}{2 \,{\left (b^{4} d^{2} - 8 \, a b^{2} c d^{2} + 16 \, a^{2} c^{2} d^{2}\right )}{\left (c x^{2} + b x + a\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*(2*c^2*g - b*c*h)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^4*d^2 - 8*a*b^2*c*d^2 + 16*a^2*c^2*d^2)*sqrt(-b
^2 + 4*a*c)) + 1/2*(12*c^3*g*x^3 - 6*b*c^2*h*x^3 + 18*b*c^2*g*x^2 - 9*b^2*c*h*x^2 + 4*b^2*c*g*x + 20*a*c^2*g*x
- 2*b^3*h*x - 10*a*b*c*h*x - b^3*g + 10*a*b*c*g - a*b^2*h - 8*a^2*c*h)/((b^4*d^2 - 8*a*b^2*c*d^2 + 16*a^2*c^2
*d^2)*(c*x^2 + b*x + a)^2)