3.347 \(\int \frac{(A+B x) (a+c x^2)^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=137 \[ -\frac{5}{2} a^{3/2} B c \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )-\frac{\left (a+c x^2\right )^{5/2} (A-B x)}{3 x^3}-\frac{5 \left (a+c x^2\right )^{3/2} (a B-A c x)}{6 x^2}-\frac{5 a c \sqrt{a+c x^2} (A-B x)}{2 x}+\frac{5}{2} a A c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right ) \]

[Out]

(-5*a*c*(A - B*x)*Sqrt[a + c*x^2])/(2*x) - (5*(a*B - A*c*x)*(a + c*x^2)^(3/2))/(6*x^2) - ((A - B*x)*(a + c*x^2
)^(5/2))/(3*x^3) + (5*a*A*c^(3/2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/2 - (5*a^(3/2)*B*c*ArcTanh[Sqrt[a + c*
x^2]/Sqrt[a]])/2

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Rubi [A]  time = 0.11025, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {813, 844, 217, 206, 266, 63, 208} \[ -\frac{5}{2} a^{3/2} B c \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )-\frac{\left (a+c x^2\right )^{5/2} (A-B x)}{3 x^3}-\frac{5 \left (a+c x^2\right )^{3/2} (a B-A c x)}{6 x^2}-\frac{5 a c \sqrt{a+c x^2} (A-B x)}{2 x}+\frac{5}{2} a A c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(5/2))/x^4,x]

[Out]

(-5*a*c*(A - B*x)*Sqrt[a + c*x^2])/(2*x) - (5*(a*B - A*c*x)*(a + c*x^2)^(3/2))/(6*x^2) - ((A - B*x)*(a + c*x^2
)^(5/2))/(3*x^3) + (5*a*A*c^(3/2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/2 - (5*a^(3/2)*B*c*ArcTanh[Sqrt[a + c*
x^2]/Sqrt[a]])/2

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 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 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a+c x^2\right )^{5/2}}{x^4} \, dx &=-\frac{(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}-\frac{5}{18} \int \frac{(-6 a B-6 A c x) \left (a+c x^2\right )^{3/2}}{x^3} \, dx\\ &=-\frac{5 (a B-A c x) \left (a+c x^2\right )^{3/2}}{6 x^2}-\frac{(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}+\frac{5}{48} \int \frac{(24 a A c+24 a B c x) \sqrt{a+c x^2}}{x^2} \, dx\\ &=-\frac{5 a c (A-B x) \sqrt{a+c x^2}}{2 x}-\frac{5 (a B-A c x) \left (a+c x^2\right )^{3/2}}{6 x^2}-\frac{(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}-\frac{5}{96} \int \frac{-48 a^2 B c-48 a A c^2 x}{x \sqrt{a+c x^2}} \, dx\\ &=-\frac{5 a c (A-B x) \sqrt{a+c x^2}}{2 x}-\frac{5 (a B-A c x) \left (a+c x^2\right )^{3/2}}{6 x^2}-\frac{(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}+\frac{1}{2} \left (5 a^2 B c\right ) \int \frac{1}{x \sqrt{a+c x^2}} \, dx+\frac{1}{2} \left (5 a A c^2\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx\\ &=-\frac{5 a c (A-B x) \sqrt{a+c x^2}}{2 x}-\frac{5 (a B-A c x) \left (a+c x^2\right )^{3/2}}{6 x^2}-\frac{(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}+\frac{1}{4} \left (5 a^2 B c\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+c x}} \, dx,x,x^2\right )+\frac{1}{2} \left (5 a A c^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )\\ &=-\frac{5 a c (A-B x) \sqrt{a+c x^2}}{2 x}-\frac{5 (a B-A c x) \left (a+c x^2\right )^{3/2}}{6 x^2}-\frac{(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}+\frac{5}{2} a A c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )+\frac{1}{2} \left (5 a^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x^2}\right )\\ &=-\frac{5 a c (A-B x) \sqrt{a+c x^2}}{2 x}-\frac{5 (a B-A c x) \left (a+c x^2\right )^{3/2}}{6 x^2}-\frac{(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}+\frac{5}{2} a A c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )-\frac{5}{2} a^{3/2} B c \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0272371, size = 94, normalized size = 0.69 \[ \frac{B c \left (a+c x^2\right )^{7/2} \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{c x^2}{a}+1\right )}{7 a^2}-\frac{a^2 A \sqrt{a+c x^2} \, _2F_1\left (-\frac{5}{2},-\frac{3}{2};-\frac{1}{2};-\frac{c x^2}{a}\right )}{3 x^3 \sqrt{\frac{c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(5/2))/x^4,x]

[Out]

-(a^2*A*Sqrt[a + c*x^2]*Hypergeometric2F1[-5/2, -3/2, -1/2, -((c*x^2)/a)])/(3*x^3*Sqrt[1 + (c*x^2)/a]) + (B*c*
(a + c*x^2)^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, 1 + (c*x^2)/a])/(7*a^2)

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Maple [A]  time = 0.01, size = 207, normalized size = 1.5 \begin{align*} -{\frac{A}{3\,a{x}^{3}} \left ( c{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{4\,Ac}{3\,{a}^{2}x} \left ( c{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{4\,A{c}^{2}x}{3\,{a}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,A{c}^{2}x}{3\,a} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,A{c}^{2}x}{2}\sqrt{c{x}^{2}+a}}+{\frac{5\,aA}{2}{c}^{{\frac{3}{2}}}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ) }-{\frac{B}{2\,a{x}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{Bc}{2\,a} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,Bc}{6} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,Bc}{2}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ) }+{\frac{5\,aBc}{2}\sqrt{c{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(5/2)/x^4,x)

[Out]

-1/3*A/a/x^3*(c*x^2+a)^(7/2)-4/3*A/a^2*c/x*(c*x^2+a)^(7/2)+4/3*A/a^2*c^2*x*(c*x^2+a)^(5/2)+5/3*A/a*c^2*x*(c*x^
2+a)^(3/2)+5/2*A*c^2*x*(c*x^2+a)^(1/2)+5/2*A*a*c^(3/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))-1/2*B/a/x^2*(c*x^2+a)^(7/
2)+1/2*B/a*c*(c*x^2+a)^(5/2)+5/6*B*c*(c*x^2+a)^(3/2)-5/2*B*a^(3/2)*c*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2))/x)+5/2
*B*a*c*(c*x^2+a)^(1/2)

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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((B*x+A)*(c*x^2+a)^(5/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.87295, size = 1300, normalized size = 9.49 \begin{align*} \left [\frac{15 \, A a c^{\frac{3}{2}} x^{3} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 15 \, B a^{\frac{3}{2}} c x^{3} \log \left (-\frac{c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (2 \, B c^{2} x^{5} + 3 \, A c^{2} x^{4} + 14 \, B a c x^{3} - 14 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt{c x^{2} + a}}{12 \, x^{3}}, -\frac{30 \, A a \sqrt{-c} c x^{3} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) - 15 \, B a^{\frac{3}{2}} c x^{3} \log \left (-\frac{c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (2 \, B c^{2} x^{5} + 3 \, A c^{2} x^{4} + 14 \, B a c x^{3} - 14 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt{c x^{2} + a}}{12 \, x^{3}}, \frac{30 \, B \sqrt{-a} a c x^{3} \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) + 15 \, A a c^{\frac{3}{2}} x^{3} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 2 \,{\left (2 \, B c^{2} x^{5} + 3 \, A c^{2} x^{4} + 14 \, B a c x^{3} - 14 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt{c x^{2} + a}}{12 \, x^{3}}, -\frac{15 \, A a \sqrt{-c} c x^{3} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) - 15 \, B \sqrt{-a} a c x^{3} \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) -{\left (2 \, B c^{2} x^{5} + 3 \, A c^{2} x^{4} + 14 \, B a c x^{3} - 14 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt{c x^{2} + a}}{6 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x^4,x, algorithm="fricas")

[Out]

[1/12*(15*A*a*c^(3/2)*x^3*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 15*B*a^(3/2)*c*x^3*log(-(c*x^2 - 2
*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*B*c^2*x^5 + 3*A*c^2*x^4 + 14*B*a*c*x^3 - 14*A*a*c*x^2 - 3*B*a^2*x
- 2*A*a^2)*sqrt(c*x^2 + a))/x^3, -1/12*(30*A*a*sqrt(-c)*c*x^3*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 15*B*a^(3/2
)*c*x^3*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(2*B*c^2*x^5 + 3*A*c^2*x^4 + 14*B*a*c*x^3 - 14
*A*a*c*x^2 - 3*B*a^2*x - 2*A*a^2)*sqrt(c*x^2 + a))/x^3, 1/12*(30*B*sqrt(-a)*a*c*x^3*arctan(sqrt(-a)/sqrt(c*x^2
+ a)) + 15*A*a*c^(3/2)*x^3*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(2*B*c^2*x^5 + 3*A*c^2*x^4 + 1
4*B*a*c*x^3 - 14*A*a*c*x^2 - 3*B*a^2*x - 2*A*a^2)*sqrt(c*x^2 + a))/x^3, -1/6*(15*A*a*sqrt(-c)*c*x^3*arctan(sqr
t(-c)*x/sqrt(c*x^2 + a)) - 15*B*sqrt(-a)*a*c*x^3*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (2*B*c^2*x^5 + 3*A*c^2*x^4
+ 14*B*a*c*x^3 - 14*A*a*c*x^2 - 3*B*a^2*x - 2*A*a^2)*sqrt(c*x^2 + a))/x^3]

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Sympy [A]  time = 10.819, size = 277, normalized size = 2.02 \begin{align*} - \frac{2 A a^{\frac{3}{2}} c}{x \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{A \sqrt{a} c^{2} x \sqrt{1 + \frac{c x^{2}}{a}}}{2} - \frac{2 A \sqrt{a} c^{2} x}{\sqrt{1 + \frac{c x^{2}}{a}}} - \frac{A a^{2} \sqrt{c} \sqrt{\frac{a}{c x^{2}} + 1}}{3 x^{2}} - \frac{A a c^{\frac{3}{2}} \sqrt{\frac{a}{c x^{2}} + 1}}{3} + \frac{5 A a c^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{2} - \frac{5 B a^{\frac{3}{2}} c \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{c} x} \right )}}{2} - \frac{B a^{2} \sqrt{c} \sqrt{\frac{a}{c x^{2}} + 1}}{2 x} + \frac{2 B a^{2} \sqrt{c}}{x \sqrt{\frac{a}{c x^{2}} + 1}} + \frac{2 B a c^{\frac{3}{2}} x}{\sqrt{\frac{a}{c x^{2}} + 1}} + B c^{2} \left (\begin{cases} \frac{\sqrt{a} x^{2}}{2} & \text{for}\: c = 0 \\\frac{\left (a + c x^{2}\right )^{\frac{3}{2}}}{3 c} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(5/2)/x**4,x)

[Out]

-2*A*a**(3/2)*c/(x*sqrt(1 + c*x**2/a)) + A*sqrt(a)*c**2*x*sqrt(1 + c*x**2/a)/2 - 2*A*sqrt(a)*c**2*x/sqrt(1 + c
*x**2/a) - A*a**2*sqrt(c)*sqrt(a/(c*x**2) + 1)/(3*x**2) - A*a*c**(3/2)*sqrt(a/(c*x**2) + 1)/3 + 5*A*a*c**(3/2)
*asinh(sqrt(c)*x/sqrt(a))/2 - 5*B*a**(3/2)*c*asinh(sqrt(a)/(sqrt(c)*x))/2 - B*a**2*sqrt(c)*sqrt(a/(c*x**2) + 1
)/(2*x) + 2*B*a**2*sqrt(c)/(x*sqrt(a/(c*x**2) + 1)) + 2*B*a*c**(3/2)*x/sqrt(a/(c*x**2) + 1) + B*c**2*Piecewise
((sqrt(a)*x**2/2, Eq(c, 0)), ((a + c*x**2)**(3/2)/(3*c), True))

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Giac [B]  time = 1.20972, size = 323, normalized size = 2.36 \begin{align*} \frac{5 \, B a^{2} c \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{5}{2} \, A a c^{\frac{3}{2}} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right ) + \frac{1}{6} \,{\left (14 \, B a c +{\left (2 \, B c^{2} x + 3 \, A c^{2}\right )} x\right )} \sqrt{c x^{2} + a} + \frac{3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{5} B a^{2} c + 18 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{4} A a^{2} c^{\frac{3}{2}} - 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} A a^{3} c^{\frac{3}{2}} - 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} B a^{4} c + 14 \, A a^{4} c^{\frac{3}{2}}}{3 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} - a\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x^4,x, algorithm="giac")

[Out]

5*B*a^2*c*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) - 5/2*A*a*c^(3/2)*log(abs(-sqrt(c)*x + sqrt
(c*x^2 + a))) + 1/6*(14*B*a*c + (2*B*c^2*x + 3*A*c^2)*x)*sqrt(c*x^2 + a) + 1/3*(3*(sqrt(c)*x - sqrt(c*x^2 + a)
)^5*B*a^2*c + 18*(sqrt(c)*x - sqrt(c*x^2 + a))^4*A*a^2*c^(3/2) - 24*(sqrt(c)*x - sqrt(c*x^2 + a))^2*A*a^3*c^(3
/2) - 3*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^4*c + 14*A*a^4*c^(3/2))/((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^3