\(\int \frac {x^2}{(a^2+2 a b x^3+b^2 x^6)^{3/2}} \, dx\) [100]

Optimal result

Integrand size = 26, antiderivative size = 38 \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {1}{6 b \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, 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, 621} \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {1}{6 b \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[In]

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

[Out]

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

Rule 621

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[2*((a + b*x + c*x^2)^(p + 1)/((2*p + 1)*(b + 2*
c*x))), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && LtQ[p, -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 \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx,x,x^3\right ) \\ & = -\frac {1}{6 b \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(143\) vs. \(2(38)=76\).

Time = 0.35 (sec) , antiderivative size = 143, normalized size of antiderivative = 3.76 \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {x^3 \left (2 a^4+a^3 b x^3-a b^3 x^9+a \sqrt {a^2} b x^3 \sqrt {\left (a+b x^3\right )^2}-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2} \left (2 a^2+b^2 x^6\right )\right )}{6 a^4 \left (a+b x^3\right ) \left (\sqrt {a^2} b x^3+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )\right )} \]

[In]

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

[Out]

-1/6*(x^3*(2*a^4 + a^3*b*x^3 - a*b^3*x^9 + a*Sqrt[a^2]*b*x^3*Sqrt[(a + b*x^3)^2] - Sqrt[a^2]*Sqrt[(a + b*x^3)^
2]*(2*a^2 + b^2*x^6)))/(a^4*(a + b*x^3)*(Sqrt[a^2]*b*x^3 + a*(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 = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.61

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.68 \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {1}{6 \, {\left (b^{3} x^{6} + 2 \, a b^{2} x^{3} + a^{2} b\right )}} \]

[In]

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

[Out]

-1/6/(b^3*x^6 + 2*a*b^2*x^3 + a^2*b)

Sympy [F]

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

[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 = 16, normalized size of antiderivative = 0.42 \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {1}{6 \, {\left (x^{3} + \frac {a}{b}\right )}^{2} b^{3}} \]

[In]

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

[Out]

-1/6/((x^3 + a/b)^2*b^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.63 \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {1}{6 \, {\left (b x^{3} + a\right )}^{2} b \mathrm {sgn}\left (b x^{3} + a\right )} \]

[In]

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

[Out]

-1/6/((b*x^3 + a)^2*b*sgn(b*x^3 + a))

Mupad [B] (verification not implemented)

Time = 8.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{6\,b\,{\left (b\,x^3+a\right )}^3} \]

[In]

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

[Out]

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