\(\int \frac {x (d+e x^2)}{(a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\) [82]

Optimal result

Integrand size = 31, antiderivative size = 77 \[ \int \frac {x \left (d+e x^2\right )}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {e}{2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b d-a e}{4 b^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {1261, 654, 621} \[ \int \frac {x \left (d+e x^2\right )}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {b d-a e}{4 b^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {e}{2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

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

[Out]

-1/2*e/(b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (b*d - a*e)/(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 1261

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {d+e x}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx,x,x^2\right ) \\ & = -\frac {e}{2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {(b d-a e) \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 {e}{2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b d-a e}{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. \(181\) vs. \(2(77)=154\).

Time = 0.65 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.35 \[ \int \frac {x \left (d+e x^2\right )}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {x^2 \left (a^3 b d x^2+a^2 b^2 e x^6+a^4 \left (2 d+e x^2\right )+a \left (-b^3 d x^6+\sqrt {a^2} b x^2 \sqrt {\left (a+b x^2\right )^2} \left (d+e x^2\right )\right )-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2} \left (b^2 d x^4+a^2 \left (2 d+e x^2\right )\right )\right )}{4 a^4 \left (a+b x^2\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )} \]

[In]

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

[Out]

-1/4*(x^2*(a^3*b*d*x^2 + a^2*b^2*e*x^6 + a^4*(2*d + e*x^2) + a*(-(b^3*d*x^6) + Sqrt[a^2]*b*x^2*Sqrt[(a + b*x^2
)^2]*(d + e*x^2)) - Sqrt[a^2]*Sqrt[(a + b*x^2)^2]*(b^2*d*x^4 + a^2*(2*d + e*x^2))))/(a^4*(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 = 37, normalized size of antiderivative = 0.48

method	result	size
pseudoelliptic	\(-\frac {\left (\left (2 e \,x^{2}+d \right ) b +a e \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{4 \left (b \,x^{2}+a \right )^{2} b^{2}}\)	\(37\)
gosper	\(-\frac {\left (b \,x^{2}+a \right ) \left (2 e \,x^{2} b +a e +b d \right )}{4 b^{2} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\)	\(38\)
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (2 e \,x^{2} b +a e +b d \right )}{4 b^{2} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\)	\(38\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {e \,x^{2}}{2 b}-\frac {a e +b d}{4 b^{2}}\right )}{\left (b \,x^{2}+a \right )^{3}}\)	\(44\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.55 \[ \int \frac {x \left (d+e x^2\right )}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {2 \, b e x^{2} + b d + a e}{4 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int \frac {x \left (d+e x^2\right )}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {{\left (2 \, b x^{2} + a\right )} e}{4 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} - \frac {d}{4 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{2} + a^{2} b\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.49 \[ \int \frac {x \left (d+e x^2\right )}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {2 \, b e x^{2} + b d + a e}{4 \, {\left (b x^{2} + a\right )}^{2} b^{2} \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.73 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.62 \[ \int \frac {x \left (d+e x^2\right )}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (2\,b\,e\,x^2+a\,e+b\,d\right )}{4\,b^2\,{\left (b\,x^2+a\right )}^3} \]

[In]

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

[Out]

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