\(\int \frac {x}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\) [650]

Optimal result

Integrand size = 24, antiderivative size = 38 \[ \int \frac {x}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {1}{8 b \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \]

[Out]

-1/8/b/(b*x^2+a)/(b^2*x^4+2*a*b*x^2+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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1121, 621} \[ \int \frac {x}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {1}{8 b \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \]

[In]

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

[Out]

-1/8*1/(b*(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 1121

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \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 ) \\ & = -\frac {1}{8 b \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {x}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {a+b x^2}{8 b \left (\left (a+b x^2\right )^2\right )^{5/2}} \]

[In]

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

[Out]

-1/8*(a + b*x^2)/(b*((a + b*x^2)^2)^(5/2))

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	\(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right )}{8 \left (b \,x^{2}+a \right )^{4} b}\)	\(23\)
gosper	\(-\frac {b \,x^{2}+a}{8 b {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(24\)
default	\(-\frac {b \,x^{2}+a}{8 b {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(24\)
risch	\(-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}}{8 \left (b \,x^{2}+a \right )^{5} b}\)	\(26\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int \frac {x}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {1}{8 \, {\left (b^{5} x^{8} + 4 \, a b^{4} x^{6} + 6 \, a^{2} b^{3} x^{4} + 4 \, a^{3} b^{2} x^{2} + a^{4} b\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int \frac {x}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {1}{8 \, {\left (b^{5} x^{8} + 4 \, a b^{4} x^{6} + 6 \, a^{2} b^{3} x^{4} + 4 \, a^{3} b^{2} x^{2} + a^{4} b\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.63 \[ \int \frac {x}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {1}{8 \, {\left (b x^{2} + a\right )}^{4} b \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

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

[Out]

-1/8/((b*x^2 + a)^4*b*sgn(b*x^2 + a))

Mupad [B] (verification not implemented)

Time = 13.51 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {x}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,b\,{\left (b\,x^2+a\right )}^5} \]

[In]

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

[Out]

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