3.132 \(\int x^2 \sin ^{-1}(a+b x)^2 \, dx\)

Optimal. Leaf size=220 \[ -\frac{2 a^2 x}{b^2}+\frac{a^3 \sin ^{-1}(a+b x)^2}{3 b^3}+\frac{2 a^2 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^3}+\frac{a (a+b x)^2}{2 b^3}-\frac{2 (a+b x)^3}{27 b^3}+\frac{a \sin ^{-1}(a+b x)^2}{2 b^3}-\frac{a (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^3}+\frac{2 (a+b x)^2 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{9 b^3}+\frac{4 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{9 b^3}+\frac{1}{3} x^3 \sin ^{-1}(a+b x)^2-\frac{4 x}{9 b^2} \]

[Out]

(-4*x)/(9*b^2) - (2*a^2*x)/b^2 + (a*(a + b*x)^2)/(2*b^3) - (2*(a + b*x)^3)/(27*b^3) + (4*Sqrt[1 - (a + b*x)^2]
*ArcSin[a + b*x])/(9*b^3) + (2*a^2*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/b^3 - (a*(a + b*x)*Sqrt[1 - (a + b*x
)^2]*ArcSin[a + b*x])/b^3 + (2*(a + b*x)^2*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(9*b^3) + (a*ArcSin[a + b*x]
^2)/(2*b^3) + (a^3*ArcSin[a + b*x]^2)/(3*b^3) + (x^3*ArcSin[a + b*x]^2)/3

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Rubi [A]  time = 0.39453, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {4805, 4743, 4763, 4641, 4677, 8, 4707, 30} \[ -\frac{2 a^2 x}{b^2}+\frac{a^3 \sin ^{-1}(a+b x)^2}{3 b^3}+\frac{2 a^2 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^3}+\frac{a (a+b x)^2}{2 b^3}-\frac{2 (a+b x)^3}{27 b^3}+\frac{a \sin ^{-1}(a+b x)^2}{2 b^3}-\frac{a (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^3}+\frac{2 (a+b x)^2 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{9 b^3}+\frac{4 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{9 b^3}+\frac{1}{3} x^3 \sin ^{-1}(a+b x)^2-\frac{4 x}{9 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcSin[a + b*x]^2,x]

[Out]

(-4*x)/(9*b^2) - (2*a^2*x)/b^2 + (a*(a + b*x)^2)/(2*b^3) - (2*(a + b*x)^3)/(27*b^3) + (4*Sqrt[1 - (a + b*x)^2]
*ArcSin[a + b*x])/(9*b^3) + (2*a^2*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/b^3 - (a*(a + b*x)*Sqrt[1 - (a + b*x
)^2]*ArcSin[a + b*x])/b^3 + (2*(a + b*x)^2*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(9*b^3) + (a*ArcSin[a + b*x]
^2)/(2*b^3) + (a^3*ArcSin[a + b*x]^2)/(3*b^3) + (x^3*ArcSin[a + b*x]^2)/3

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rule 4743

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

Rule 4763

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g},
x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && (m == 1 || p > 0 ||
(n == 1 && p > -1) || (m == 2 && p < -2))

Rule 4641

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSin[c*x])^
(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[n,
-1]

Rule 4677

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4707

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSin[c*x])^n)/(e*m), x] + (Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)^(m
- 2)*(a + b*ArcSin[c*x])^n)/Sqrt[d + e*x^2], x], x] + Dist[(b*f*n*Sqrt[1 - c^2*x^2])/(c*m*Sqrt[d + e*x^2]), I
nt[(f*x)^(m - 1)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] &&
GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x^2 \sin ^{-1}(a+b x)^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right )^2 \sin ^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{3} x^3 \sin ^{-1}(a+b x)^2-\frac{2}{3} \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^3 \sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx,x,a+b x\right )\\ &=\frac{1}{3} x^3 \sin ^{-1}(a+b x)^2-\frac{2}{3} \operatorname{Subst}\left (\int \left (-\frac{a^3 \sin ^{-1}(x)}{b^3 \sqrt{1-x^2}}+\frac{3 a^2 x \sin ^{-1}(x)}{b^3 \sqrt{1-x^2}}-\frac{3 a x^2 \sin ^{-1}(x)}{b^3 \sqrt{1-x^2}}+\frac{x^3 \sin ^{-1}(x)}{b^3 \sqrt{1-x^2}}\right ) \, dx,x,a+b x\right )\\ &=\frac{1}{3} x^3 \sin ^{-1}(a+b x)^2-\frac{2 \operatorname{Subst}\left (\int \frac{x^3 \sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx,x,a+b x\right )}{3 b^3}+\frac{(2 a) \operatorname{Subst}\left (\int \frac{x^2 \sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx,x,a+b x\right )}{b^3}-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{x \sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx,x,a+b x\right )}{b^3}+\frac{\left (2 a^3\right ) \operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx,x,a+b x\right )}{3 b^3}\\ &=\frac{2 a^2 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^3}-\frac{a (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^3}+\frac{2 (a+b x)^2 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{9 b^3}+\frac{a^3 \sin ^{-1}(a+b x)^2}{3 b^3}+\frac{1}{3} x^3 \sin ^{-1}(a+b x)^2-\frac{2 \operatorname{Subst}\left (\int x^2 \, dx,x,a+b x\right )}{9 b^3}-\frac{4 \operatorname{Subst}\left (\int \frac{x \sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx,x,a+b x\right )}{9 b^3}+\frac{a \operatorname{Subst}(\int x \, dx,x,a+b x)}{b^3}+\frac{a \operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx,x,a+b x\right )}{b^3}-\frac{\left (2 a^2\right ) \operatorname{Subst}(\int 1 \, dx,x,a+b x)}{b^3}\\ &=-\frac{2 a^2 x}{b^2}+\frac{a (a+b x)^2}{2 b^3}-\frac{2 (a+b x)^3}{27 b^3}+\frac{4 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{9 b^3}+\frac{2 a^2 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^3}-\frac{a (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^3}+\frac{2 (a+b x)^2 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{9 b^3}+\frac{a \sin ^{-1}(a+b x)^2}{2 b^3}+\frac{a^3 \sin ^{-1}(a+b x)^2}{3 b^3}+\frac{1}{3} x^3 \sin ^{-1}(a+b x)^2-\frac{4 \operatorname{Subst}(\int 1 \, dx,x,a+b x)}{9 b^3}\\ &=-\frac{4 x}{9 b^2}-\frac{2 a^2 x}{b^2}+\frac{a (a+b x)^2}{2 b^3}-\frac{2 (a+b x)^3}{27 b^3}+\frac{4 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{9 b^3}+\frac{2 a^2 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^3}-\frac{a (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^3}+\frac{2 (a+b x)^2 \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{9 b^3}+\frac{a \sin ^{-1}(a+b x)^2}{2 b^3}+\frac{a^3 \sin ^{-1}(a+b x)^2}{3 b^3}+\frac{1}{3} x^3 \sin ^{-1}(a+b x)^2\\ \end{align*}

Mathematica [A]  time = 0.154691, size = 111, normalized size = 0.5 \[ \frac{-b x \left (66 a^2-15 a b x+4 b^2 x^2+24\right )+9 \left (2 a^3+3 a+2 b^3 x^3\right ) \sin ^{-1}(a+b x)^2+6 \sqrt{-a^2-2 a b x-b^2 x^2+1} \left (11 a^2-5 a b x+2 b^2 x^2+4\right ) \sin ^{-1}(a+b x)}{54 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcSin[a + b*x]^2,x]

[Out]

(-(b*x*(24 + 66*a^2 - 15*a*b*x + 4*b^2*x^2)) + 6*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*(4 + 11*a^2 - 5*a*b*x + 2*b
^2*x^2)*ArcSin[a + b*x] + 9*(3*a + 2*a^3 + 2*b^3*x^3)*ArcSin[a + b*x]^2)/(54*b^3)

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Maple [A]  time = 0.061, size = 231, normalized size = 1.1 \begin{align*}{\frac{1}{{b}^{3}} \left ( -{\frac{a}{2} \left ( 2\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) ^{2}+2\,\arcsin \left ( bx+a \right ) \sqrt{1- \left ( bx+a \right ) ^{2}} \left ( bx+a \right ) - \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}- \left ( bx+a \right ) ^{2} \right ) }+{\frac{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{2} \left ( \left ( bx+a \right ) ^{2}-3 \right ) \left ( bx+a \right ) }{3}}-{\frac{2\,bx}{3}}-{\frac{2\,a}{3}}+{\frac{2\,\arcsin \left ( bx+a \right ) }{3}\sqrt{1- \left ( bx+a \right ) ^{2}}}+{\frac{2\,\arcsin \left ( bx+a \right ) \left ( -1+ \left ( bx+a \right ) ^{2} \right ) }{9}\sqrt{1- \left ( bx+a \right ) ^{2}}}-{\frac{ \left ( 2\, \left ( bx+a \right ) ^{2}-6 \right ) \left ( bx+a \right ) }{27}}+{a}^{2} \left ( \left ( \arcsin \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) -2\,bx-2\,a+2\,\sqrt{1- \left ( bx+a \right ) ^{2}}\arcsin \left ( bx+a \right ) \right ) + \left ( \arcsin \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arcsin(b*x+a)^2,x)

[Out]

1/b^3*(-1/2*a*(2*arcsin(b*x+a)^2*(b*x+a)^2+2*arcsin(b*x+a)*(1-(b*x+a)^2)^(1/2)*(b*x+a)-arcsin(b*x+a)^2-(b*x+a)
^2)+1/3*arcsin(b*x+a)^2*((b*x+a)^2-3)*(b*x+a)-2/3*b*x-2/3*a+2/3*(1-(b*x+a)^2)^(1/2)*arcsin(b*x+a)+2/9*arcsin(b
*x+a)*(-1+(b*x+a)^2)*(1-(b*x+a)^2)^(1/2)-2/27*((b*x+a)^2-3)*(b*x+a)+a^2*(arcsin(b*x+a)^2*(b*x+a)-2*b*x-2*a+2*(
1-(b*x+a)^2)^(1/2)*arcsin(b*x+a))+arcsin(b*x+a)^2*(b*x+a))

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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(x^2*arcsin(b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.77983, size = 266, normalized size = 1.21 \begin{align*} -\frac{4 \, b^{3} x^{3} - 15 \, a b^{2} x^{2} + 6 \,{\left (11 \, a^{2} + 4\right )} b x - 9 \,{\left (2 \, b^{3} x^{3} + 2 \, a^{3} + 3 \, a\right )} \arcsin \left (b x + a\right )^{2} - 6 \,{\left (2 \, b^{2} x^{2} - 5 \, a b x + 11 \, a^{2} + 4\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} \arcsin \left (b x + a\right )}{54 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/54*(4*b^3*x^3 - 15*a*b^2*x^2 + 6*(11*a^2 + 4)*b*x - 9*(2*b^3*x^3 + 2*a^3 + 3*a)*arcsin(b*x + a)^2 - 6*(2*b^
2*x^2 - 5*a*b*x + 11*a^2 + 4)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*arcsin(b*x + a))/b^3

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Sympy [A]  time = 1.81073, size = 243, normalized size = 1.1 \begin{align*} \begin{cases} \frac{a^{3} \operatorname{asin}^{2}{\left (a + b x \right )}}{3 b^{3}} - \frac{11 a^{2} x}{9 b^{2}} + \frac{11 a^{2} \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}{\left (a + b x \right )}}{9 b^{3}} + \frac{5 a x^{2}}{18 b} - \frac{5 a x \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}{\left (a + b x \right )}}{9 b^{2}} + \frac{a \operatorname{asin}^{2}{\left (a + b x \right )}}{2 b^{3}} + \frac{x^{3} \operatorname{asin}^{2}{\left (a + b x \right )}}{3} - \frac{2 x^{3}}{27} + \frac{2 x^{2} \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}{\left (a + b x \right )}}{9 b} - \frac{4 x}{9 b^{2}} + \frac{4 \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}{\left (a + b x \right )}}{9 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \operatorname{asin}^{2}{\left (a \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*asin(b*x+a)**2,x)

[Out]

Piecewise((a**3*asin(a + b*x)**2/(3*b**3) - 11*a**2*x/(9*b**2) + 11*a**2*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)
*asin(a + b*x)/(9*b**3) + 5*a*x**2/(18*b) - 5*a*x*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)/(9*b**2)
+ a*asin(a + b*x)**2/(2*b**3) + x**3*asin(a + b*x)**2/3 - 2*x**3/27 + 2*x**2*sqrt(-a**2 - 2*a*b*x - b**2*x**2
+ 1)*asin(a + b*x)/(9*b) - 4*x/(9*b**2) + 4*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)/(9*b**3), Ne(
b, 0)), (x**3*asin(a)**2/3, True))

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Giac [A]  time = 1.2404, size = 366, normalized size = 1.66 \begin{align*} \frac{{\left (b x + a\right )} a^{2} \arcsin \left (b x + a\right )^{2}}{b^{3}} + \frac{{\left ({\left (b x + a\right )}^{2} - 1\right )}{\left (b x + a\right )} \arcsin \left (b x + a\right )^{2}}{3 \, b^{3}} - \frac{{\left ({\left (b x + a\right )}^{2} - 1\right )} a \arcsin \left (b x + a\right )^{2}}{b^{3}} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )} a \arcsin \left (b x + a\right )}{b^{3}} + \frac{2 \, \sqrt{-{\left (b x + a\right )}^{2} + 1} a^{2} \arcsin \left (b x + a\right )}{b^{3}} - \frac{2 \,{\left (b x + a\right )} a^{2}}{b^{3}} + \frac{{\left (b x + a\right )} \arcsin \left (b x + a\right )^{2}}{3 \, b^{3}} - \frac{a \arcsin \left (b x + a\right )^{2}}{2 \, b^{3}} - \frac{2 \,{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac{3}{2}} \arcsin \left (b x + a\right )}{9 \, b^{3}} - \frac{2 \,{\left ({\left (b x + a\right )}^{2} - 1\right )}{\left (b x + a\right )}}{27 \, b^{3}} + \frac{{\left ({\left (b x + a\right )}^{2} - 1\right )} a}{2 \, b^{3}} + \frac{2 \, \sqrt{-{\left (b x + a\right )}^{2} + 1} \arcsin \left (b x + a\right )}{3 \, b^{3}} - \frac{14 \,{\left (b x + a\right )}}{27 \, b^{3}} + \frac{a}{4 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsin(b*x+a)^2,x, algorithm="giac")

[Out]

(b*x + a)*a^2*arcsin(b*x + a)^2/b^3 + 1/3*((b*x + a)^2 - 1)*(b*x + a)*arcsin(b*x + a)^2/b^3 - ((b*x + a)^2 - 1
)*a*arcsin(b*x + a)^2/b^3 - sqrt(-(b*x + a)^2 + 1)*(b*x + a)*a*arcsin(b*x + a)/b^3 + 2*sqrt(-(b*x + a)^2 + 1)*
a^2*arcsin(b*x + a)/b^3 - 2*(b*x + a)*a^2/b^3 + 1/3*(b*x + a)*arcsin(b*x + a)^2/b^3 - 1/2*a*arcsin(b*x + a)^2/
b^3 - 2/9*(-(b*x + a)^2 + 1)^(3/2)*arcsin(b*x + a)/b^3 - 2/27*((b*x + a)^2 - 1)*(b*x + a)/b^3 + 1/2*((b*x + a)
^2 - 1)*a/b^3 + 2/3*sqrt(-(b*x + a)^2 + 1)*arcsin(b*x + a)/b^3 - 14/27*(b*x + a)/b^3 + 1/4*a/b^3