\(\int x^2 (a^2+2 a b x^3+b^2 x^6)^{3/2} \, dx\) [30]

Optimal result

Integrand size = 26, antiderivative size = 36 \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {\left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{12 b} \]

[Out]

1/12*(b*x^3+a)*(b^2*x^6+2*a*b*x^3+a^2)^(3/2)/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1366, 623} \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {\left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{12 b} \]

[In]

Int[x^2*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]

[Out]

((a + b*x^3)*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2))/(12*b)

Rule 623

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

Rule 1366

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx,x,x^3\right ) \\ & = \frac {\left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{12 b} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(112\) vs. \(2(36)=72\).

Time = 0.71 (sec) , antiderivative size = 112, normalized size of antiderivative = 3.11 \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {x^3 \left (4 a^3+6 a^2 b x^3+4 a b^2 x^6+b^3 x^9\right ) \left (\sqrt {a^2} b x^3+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )\right )}{12 \left (-a^2-a b x^3+\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )} \]

[In]

Integrate[x^2*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]

[Out]

(x^3*(4*a^3 + 6*a^2*b*x^3 + 4*a*b^2*x^6 + b^3*x^9)*(Sqrt[a^2]*b*x^3 + a*(Sqrt[a^2] - Sqrt[(a + b*x^3)^2])))/(1
2*(-a^2 - a*b*x^3 + Sqrt[a^2]*Sqrt[(a + b*x^3)^2]))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.64

method	result	size
pseudoelliptic	\(\frac {\left (b \,x^{3}+a \right )^{4} \operatorname {csgn}\left (b \,x^{3}+a \right )}{12 b}\)	\(23\)
default	\(\frac {\left (b \,x^{3}+a \right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{12 b}\)	\(24\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (b \,x^{3}+a \right )^{3}}{12 b}\)	\(26\)
gosper	\(\frac {x^{3} \left (b^{3} x^{9}+4 b^{2} x^{6} a +6 a^{2} b \,x^{3}+4 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{12 \left (b \,x^{3}+a \right )^{3}}\)	\(57\)

[In]

int(x^2*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/12*(b*x^3+a)^4*csgn(b*x^3+a)/b

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97 \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {1}{12} \, b^{3} x^{12} + \frac {1}{3} \, a b^{2} x^{9} + \frac {1}{2} \, a^{2} b x^{6} + \frac {1}{3} \, a^{3} x^{3} \]

[In]

integrate(x^2*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/12*b^3*x^12 + 1/3*a*b^2*x^9 + 1/2*a^2*b*x^6 + 1/3*a^3*x^3

Sympy [F]

\[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\int x^{2} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}\, dx \]

[In]

integrate(x**2*(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.44 \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {1}{12} \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} x^{3} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a}{12 \, b} \]

[In]

integrate(x^2*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*x^3 + 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*a/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.22 \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {1}{12} \, {\left (2 \, {\left (b x^{6} + 2 \, a x^{3}\right )} a^{2} + {\left (b x^{6} + 2 \, a x^{3}\right )}^{2} b\right )} \mathrm {sgn}\left (b x^{3} + a\right ) \]

[In]

integrate(x^2*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")

[Out]

1/12*(2*(b*x^6 + 2*a*x^3)*a^2 + (b*x^6 + 2*a*x^3)^2*b)*sgn(b*x^3 + a)

Mupad [B] (verification not implemented)

Time = 8.48 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {\left (b^2\,x^3+a\,b\right )\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}}{12\,b^2} \]

[In]

int(x^2*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2),x)

[Out]

((a*b + b^2*x^3)*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2))/(12*b^2)