\(\int \frac {x^3}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\) [649]

Optimal result

Integrand size = 26, antiderivative size = 69 \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {1}{6 b^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac {a}{8 b^2 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \]

[Out]

-1/6/b^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/2)+1/8*a/b^2/(b*x^2+a)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, 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} \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {a}{8 b^2 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}-\frac {1}{6 b^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \]

[In]

Int[x^3/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

-1/6*1/(b^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)) + a/(8*b^2*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(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 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 )^{5/2}} \, dx,x,x^2\right ) \\ & = -\frac {1}{6 b^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}-\frac {a \text {Subst}\left (\int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx,x,x^2\right )}{2 b} \\ & = -\frac {1}{6 b^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac {a}{8 b^2 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(232\) vs. \(2(69)=138\).

Time = 0.46 (sec) , antiderivative size = 232, normalized size of antiderivative = 3.36 \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {x^4 \left (3 \sqrt {a^2} b^6 x^{12}+3 a^3 b^3 x^6 \sqrt {\left (a+b x^2\right )^2}-3 a^2 b^4 x^8 \sqrt {\left (a+b x^2\right )^2}+3 a b^5 x^{10} \sqrt {\left (a+b x^2\right )^2}+a^4 b^2 x^4 \left (\sqrt {a^2}-3 \sqrt {\left (a+b x^2\right )^2}\right )+6 a^6 \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )+2 a^5 b x^2 \left (2 \sqrt {a^2}+\sqrt {\left (a+b x^2\right )^2}\right )\right )}{24 a^7 \left (a+b x^2\right )^3 \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]

[In]

Integrate[x^3/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

-1/24*(x^4*(3*Sqrt[a^2]*b^6*x^12 + 3*a^3*b^3*x^6*Sqrt[(a + b*x^2)^2] - 3*a^2*b^4*x^8*Sqrt[(a + b*x^2)^2] + 3*a
*b^5*x^10*Sqrt[(a + b*x^2)^2] + a^4*b^2*x^4*(Sqrt[a^2] - 3*Sqrt[(a + b*x^2)^2]) + 6*a^6*(Sqrt[a^2] - Sqrt[(a +
b*x^2)^2]) + 2*a^5*b*x^2*(2*Sqrt[a^2] + Sqrt[(a + b*x^2)^2])))/(a^7*(a + b*x^2)^3*(a^2 + a*b*x^2 - 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 = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.45

method	result	size
pseudoelliptic	\(-\frac {\left (4 b \,x^{2}+a \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{24 \left (b \,x^{2}+a \right )^{4} b^{2}}\)	\(31\)
gosper	\(-\frac {\left (b \,x^{2}+a \right ) \left (4 b \,x^{2}+a \right )}{24 b^{2} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(32\)
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (4 b \,x^{2}+a \right )}{24 b^{2} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(32\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {x^{2}}{6 b}-\frac {a}{24 b^{2}}\right )}{\left (b \,x^{2}+a \right )^{5}}\)	\(37\)

[In]

int(x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/24/(b*x^2+a)^4*(4*b*x^2+a)/b^2*csgn(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {4 \, b x^{2} + a}{24 \, {\left (b^{6} x^{8} + 4 \, a b^{5} x^{6} + 6 \, a^{2} b^{4} x^{4} + 4 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}} \]

[In]

integrate(x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/24*(4*b*x^2 + a)/(b^6*x^8 + 4*a*b^5*x^6 + 6*a^2*b^4*x^4 + 4*a^3*b^3*x^2 + a^4*b^2)

Sympy [F]

\[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^{3}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(x**3/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {4 \, b x^{2} + a}{24 \, {\left (b^{6} x^{8} + 4 \, a b^{5} x^{6} + 6 \, a^{2} b^{4} x^{4} + 4 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}} \]

[In]

integrate(x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

-1/24*(4*b*x^2 + a)/(b^6*x^8 + 4*a*b^5*x^6 + 6*a^2*b^4*x^4 + 4*a^3*b^3*x^2 + a^4*b^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.46 \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {4 \, b x^{2} + a}{24 \, {\left (b x^{2} + a\right )}^{4} b^{2} \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

integrate(x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

-1/24*(4*b*x^2 + a)/((b*x^2 + a)^4*b^2*sgn(b*x^2 + a))

Mupad [B] (verification not implemented)

Time = 13.69 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.61 \[ \int \frac {x^3}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {\left (4\,b\,x^2+a\right )\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{24\,b^2\,{\left (b\,x^2+a\right )}^5} \]

[In]

int(x^3/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

-((a + 4*b*x^2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(24*b^2*(a + b*x^2)^5)