\(\int \frac {x^3}{(a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\) [636]

Optimal result

Integrand size = 26, antiderivative size = 41 \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {x^4}{4 a \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

1/4*x^4/a/(b*x^2+a)/((b*x^2+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.68, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 654, 621} \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {a}{4 b^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {1}{2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

Int[x^3/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

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

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 654

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

Rule 1125

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx,x,x^2\right ) \\ & = -\frac {1}{2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a \text {Subst}\left (\int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx,x,x^2\right )}{2 b} \\ & = -\frac {1}{2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a}{4 b^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(119\) vs. \(2(41)=82\).

Time = 0.30 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.90 \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {x^4 \left (a^3+a b^2 x^4-a \sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}+\sqrt {a^2} b x^2 \sqrt {\left (a+b x^2\right )^2}\right )}{4 a^3 \left (a+b x^2\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )} \]

[In]

Integrate[x^3/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(-\frac {\left (2 b \,x^{2}+a \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{4 \left (b \,x^{2}+a \right )^{2} b^{2}}\)	\(31\)
gosper	\(-\frac {\left (b \,x^{2}+a \right ) \left (2 b \,x^{2}+a \right )}{4 b^{2} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\)	\(32\)
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (2 b \,x^{2}+a \right )}{4 b^{2} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\)	\(32\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {x^{2}}{2 b}-\frac {a}{4 b^{2}}\right )}{\left (b \,x^{2}+a \right )^{3}}\)	\(37\)

[In]

int(x^3/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/(b*x^2+a)^2*(2*b*x^2+a)/b^2*csgn(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {2 \, b x^{2} + a}{4 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} \]

[In]

integrate(x^3/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

-1/4*(2*b*x^2 + a)/(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2)

Sympy [F]

\[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {x^{3}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**3/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {2 \, b x^{2} + a}{4 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} \]

[In]

integrate(x^3/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

-1/4*(2*b*x^2 + a)/(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {2 \, b x^{2} + a}{4 \, {\left (b x^{2} + a\right )}^{2} b^{2} \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

integrate(x^3/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

-1/4*(2*b*x^2 + a)/((b*x^2 + a)^2*b^2*sgn(b*x^2 + a))

Mupad [B] (verification not implemented)

Time = 13.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {\left (2\,b\,x^2+a\right )\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,b^2\,{\left (b\,x^2+a\right )}^3} \]

[In]

int(x^3/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

-((a + 2*b*x^2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(4*b^2*(a + b*x^2)^3)