\(\int \frac {x^2}{(a^2+2 a b x^3+b^2 x^6)^{5/2}} \, dx\) [111]

Optimal result

Integrand size = 26, antiderivative size = 38 \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {1}{12 b \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \]

[Out]

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

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} \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {1}{12 b \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \]

[In]

Int[x^2/(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

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

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 )^{5/2}} \, dx,x,x^3\right ) \\ & = -\frac {1}{12 b \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(267\) vs. \(2(38)=76\).

Time = 0.59 (sec) , antiderivative size = 267, normalized size of antiderivative = 7.03 \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {x^3 \left (-\sqrt {a^2} b^7 x^{21}-a^3 b^4 x^{12} \sqrt {\left (a+b x^3\right )^2}+a^2 b^5 x^{15} \sqrt {\left (a+b x^3\right )^2}-a b^6 x^{18} \sqrt {\left (a+b x^3\right )^2}+4 a^7 \left (\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )+2 a^5 b^2 x^6 \left (2 \sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )+2 a^6 b x^3 \left (3 \sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )+a^4 b^3 x^9 \left (\sqrt {a^2}+\sqrt {\left (a+b x^3\right )^2}\right )\right )}{12 a^8 \left (a+b x^3\right )^3 \left (a^2+a b x^3-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )} \]

[In]

Integrate[x^2/(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

-1/12*(x^3*(-(Sqrt[a^2]*b^7*x^21) - a^3*b^4*x^12*Sqrt[(a + b*x^3)^2] + a^2*b^5*x^15*Sqrt[(a + b*x^3)^2] - a*b^
6*x^18*Sqrt[(a + b*x^3)^2] + 4*a^7*(Sqrt[a^2] - Sqrt[(a + b*x^3)^2]) + 2*a^5*b^2*x^6*(2*Sqrt[a^2] - Sqrt[(a +
b*x^3)^2]) + 2*a^6*b*x^3*(3*Sqrt[a^2] - Sqrt[(a + b*x^3)^2]) + a^4*b^3*x^9*(Sqrt[a^2] + Sqrt[(a + b*x^3)^2])))
/(a^8*(a + b*x^3)^3*(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 = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.61

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right )}{12 \left (b \,x^{3}+a \right )^{4} b}\)	\(23\)
gosper	\(-\frac {b \,x^{3}+a}{12 b {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(24\)
default	\(-\frac {b \,x^{3}+a}{12 b {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(24\)
risch	\(-\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}}{12 \left (b \,x^{3}+a \right )^{5} b}\)	\(26\)

[In]

int(x^2/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {1}{12 \, {\left (b^{5} x^{12} + 4 \, a b^{4} x^{9} + 6 \, a^{2} b^{3} x^{6} + 4 \, a^{3} b^{2} x^{3} + a^{4} b\right )}} \]

[In]

integrate(x^2/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/12/(b^5*x^12 + 4*a*b^4*x^9 + 6*a^2*b^3*x^6 + 4*a^3*b^2*x^3 + a^4*b)

Sympy [F]

\[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\int \frac {x^{2}}{\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(x**2/(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)

[Out]

Integral(x**2/((a + b*x**3)**2)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.42 \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {1}{12 \, {\left (x^{3} + \frac {a}{b}\right )}^{4} b^{5}} \]

[In]

integrate(x^2/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")

[Out]

-1/12/((x^3 + a/b)^4*b^5)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.63 \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {1}{12 \, {\left (b x^{3} + a\right )}^{4} b \mathrm {sgn}\left (b x^{3} + a\right )} \]

[In]

integrate(x^2/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.37 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {x^2}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{12\,b\,{\left (b\,x^3+a\right )}^5} \]

[In]

int(x^2/(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2),x)

[Out]

-(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2)/(12*b*(a + b*x^3)^5)