3.203 \(\int \frac{2+3 x^2}{x^4 (3+5 x^2+x^4)^{3/2}} \, dx\)

Optimal. Leaf size=326 \[ -\frac{5 \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right ),\frac{1}{6} \left (5 \sqrt{13}-13\right )\right )}{351 \sqrt{6 \left (5+\sqrt{13}\right )} \sqrt{x^4+5 x^2+3}}-\frac{133 x \left (2 x^2+\sqrt{13}+5\right )}{1053 \sqrt{x^4+5 x^2+3}}+\frac{266 \sqrt{x^4+5 x^2+3}}{1053 x}-\frac{5 \sqrt{x^4+5 x^2+3}}{351 x^3}-\frac{8 x^2+7}{39 x^3 \sqrt{x^4+5 x^2+3}}+\frac{133 \sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{1053 \sqrt{x^4+5 x^2+3}} \]

[Out]

(-133*x*(5 + Sqrt[13] + 2*x^2))/(1053*Sqrt[3 + 5*x^2 + x^4]) - (7 + 8*x^2)/(39*x^3*Sqrt[3 + 5*x^2 + x^4]) - (5
*Sqrt[3 + 5*x^2 + x^4])/(351*x^3) + (266*Sqrt[3 + 5*x^2 + x^4])/(1053*x) + (133*Sqrt[(5 + Sqrt[13])/6]*Sqrt[(6
+ (5 - Sqrt[13])*x^2)/(6 + (5 + Sqrt[13])*x^2)]*(6 + (5 + Sqrt[13])*x^2)*EllipticE[ArcTan[Sqrt[(5 + Sqrt[13])
/6]*x], (-13 + 5*Sqrt[13])/6])/(1053*Sqrt[3 + 5*x^2 + x^4]) - (5*Sqrt[(6 + (5 - Sqrt[13])*x^2)/(6 + (5 + Sqrt[
13])*x^2)]*(6 + (5 + Sqrt[13])*x^2)*EllipticF[ArcTan[Sqrt[(5 + Sqrt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/(351*Sq
rt[6*(5 + Sqrt[13])]*Sqrt[3 + 5*x^2 + x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.205836, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1277, 1281, 1189, 1099, 1135} \[ -\frac{133 x \left (2 x^2+\sqrt{13}+5\right )}{1053 \sqrt{x^4+5 x^2+3}}+\frac{266 \sqrt{x^4+5 x^2+3}}{1053 x}-\frac{5 \sqrt{x^4+5 x^2+3}}{351 x^3}-\frac{8 x^2+7}{39 x^3 \sqrt{x^4+5 x^2+3}}-\frac{5 \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) F\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{351 \sqrt{6 \left (5+\sqrt{13}\right )} \sqrt{x^4+5 x^2+3}}+\frac{133 \sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{1053 \sqrt{x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-133*x*(5 + Sqrt[13] + 2*x^2))/(1053*Sqrt[3 + 5*x^2 + x^4]) - (7 + 8*x^2)/(39*x^3*Sqrt[3 + 5*x^2 + x^4]) - (5
*Sqrt[3 + 5*x^2 + x^4])/(351*x^3) + (266*Sqrt[3 + 5*x^2 + x^4])/(1053*x) + (133*Sqrt[(5 + Sqrt[13])/6]*Sqrt[(6
+ (5 - Sqrt[13])*x^2)/(6 + (5 + Sqrt[13])*x^2)]*(6 + (5 + Sqrt[13])*x^2)*EllipticE[ArcTan[Sqrt[(5 + Sqrt[13])
/6]*x], (-13 + 5*Sqrt[13])/6])/(1053*Sqrt[3 + 5*x^2 + x^4]) - (5*Sqrt[(6 + (5 - Sqrt[13])*x^2)/(6 + (5 + Sqrt[
13])*x^2)]*(6 + (5 + Sqrt[13])*x^2)*EllipticF[ArcTan[Sqrt[(5 + Sqrt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/(351*Sq
rt[6*(5 + Sqrt[13])]*Sqrt[3 + 5*x^2 + x^4])

Rule 1277

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[((f
*x)^(m + 1)*(a + b*x^2 + c*x^4)^(p + 1)*(d*(b^2 - 2*a*c) - a*b*e + (b*d - 2*a*e)*c*x^2))/(2*a*f*(p + 1)*(b^2 -
4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(f*x)^m*(a + b*x^2 + c*x^4)^(p + 1)*Simp[d*(b^2*(m + 2*
(p + 1) + 1) - 2*a*c*(m + 4*(p + 1) + 1)) - a*b*e*(m + 1) + c*(m + 2*(2*p + 3) + 1)*(b*d - 2*a*e)*x^2, x], x],
x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p] && (IntegerQ[p] |
| IntegerQ[m])

Rule 1281

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(
f*x)^(m + 1)*(a + b*x^2 + c*x^4)^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^2*(m + 1)), Int[(f*x)^(m + 2)*(a + b
*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1189

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}
, Dist[d, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[e, Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b +
q)/a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1099

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[((2*a + (b +
q)*x^2)*Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticF[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)]
)/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSq
rtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1135

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(x*(b +
q + 2*c*x^2))/(2*c*Sqrt[a + b*x^2 + c*x^4]), x] - Simp[(Rt[(b + q)/(2*a), 2]*(2*a + (b + q)*x^2)*Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticE[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)])/(2*c*Sqrt[a + b*x^2
+ c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; Fre
eQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{2+3 x^2}{x^4 \left (3+5 x^2+x^4\right )^{3/2}} \, dx &=-\frac{7+8 x^2}{39 x^3 \sqrt{3+5 x^2+x^4}}-\frac{1}{39} \int \frac{-5+24 x^2}{x^4 \sqrt{3+5 x^2+x^4}} \, dx\\ &=-\frac{7+8 x^2}{39 x^3 \sqrt{3+5 x^2+x^4}}-\frac{5 \sqrt{3+5 x^2+x^4}}{351 x^3}+\frac{1}{351} \int \frac{-266-5 x^2}{x^2 \sqrt{3+5 x^2+x^4}} \, dx\\ &=-\frac{7+8 x^2}{39 x^3 \sqrt{3+5 x^2+x^4}}-\frac{5 \sqrt{3+5 x^2+x^4}}{351 x^3}+\frac{266 \sqrt{3+5 x^2+x^4}}{1053 x}-\frac{\int \frac{15+266 x^2}{\sqrt{3+5 x^2+x^4}} \, dx}{1053}\\ &=-\frac{7+8 x^2}{39 x^3 \sqrt{3+5 x^2+x^4}}-\frac{5 \sqrt{3+5 x^2+x^4}}{351 x^3}+\frac{266 \sqrt{3+5 x^2+x^4}}{1053 x}-\frac{5}{351} \int \frac{1}{\sqrt{3+5 x^2+x^4}} \, dx-\frac{266 \int \frac{x^2}{\sqrt{3+5 x^2+x^4}} \, dx}{1053}\\ &=-\frac{133 x \left (5+\sqrt{13}+2 x^2\right )}{1053 \sqrt{3+5 x^2+x^4}}-\frac{7+8 x^2}{39 x^3 \sqrt{3+5 x^2+x^4}}-\frac{5 \sqrt{3+5 x^2+x^4}}{351 x^3}+\frac{266 \sqrt{3+5 x^2+x^4}}{1053 x}+\frac{133 \sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} \sqrt{\frac{6+\left (5-\sqrt{13}\right ) x^2}{6+\left (5+\sqrt{13}\right ) x^2}} \left (6+\left (5+\sqrt{13}\right ) x^2\right ) E\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{1053 \sqrt{3+5 x^2+x^4}}-\frac{5 \sqrt{\frac{6+\left (5-\sqrt{13}\right ) x^2}{6+\left (5+\sqrt{13}\right ) x^2}} \left (6+\left (5+\sqrt{13}\right ) x^2\right ) F\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{351 \sqrt{6 \left (5+\sqrt{13}\right )} \sqrt{3+5 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.321328, size = 234, normalized size = 0.72 \[ \frac{i \sqrt{2} \left (133 \sqrt{13}-650\right ) \sqrt{\frac{-2 x^2+\sqrt{13}-5}{\sqrt{13}-5}} \sqrt{2 x^2+\sqrt{13}+5} x^3 \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{2}{5+\sqrt{13}}} x\right ),\frac{19}{6}+\frac{5 \sqrt{13}}{6}\right )+532 x^6+2630 x^4+1014 x^2-133 i \sqrt{2} \left (\sqrt{13}-5\right ) \sqrt{\frac{-2 x^2+\sqrt{13}-5}{\sqrt{13}-5}} \sqrt{2 x^2+\sqrt{13}+5} x^3 E\left (i \sinh ^{-1}\left (\sqrt{\frac{2}{5+\sqrt{13}}} x\right )|\frac{19}{6}+\frac{5 \sqrt{13}}{6}\right )-468}{2106 x^3 \sqrt{x^4+5 x^2+3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-468 + 1014*x^2 + 2630*x^4 + 532*x^6 - (133*I)*Sqrt[2]*(-5 + Sqrt[13])*x^3*Sqrt[(-5 + Sqrt[13] - 2*x^2)/(-5 +
Sqrt[13])]*Sqrt[5 + Sqrt[13] + 2*x^2]*EllipticE[I*ArcSinh[Sqrt[2/(5 + Sqrt[13])]*x], 19/6 + (5*Sqrt[13])/6] +
I*Sqrt[2]*(-650 + 133*Sqrt[13])*x^3*Sqrt[(-5 + Sqrt[13] - 2*x^2)/(-5 + Sqrt[13])]*Sqrt[5 + Sqrt[13] + 2*x^2]*
EllipticF[I*ArcSinh[Sqrt[2/(5 + Sqrt[13])]*x], 19/6 + (5*Sqrt[13])/6])/(2106*x^3*Sqrt[3 + 5*x^2 + x^4])

________________________________________________________________________________________

Maple [A]  time = 0.02, size = 274, normalized size = 0.8 \begin{align*}{\frac{23}{81\,x}\sqrt{{x}^{4}+5\,{x}^{2}+3}}-6\,{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}} \left ({\frac{19\,{x}^{3}}{234}}+{\frac{40\,x}{117}} \right ) }-{\frac{10}{117\,\sqrt{-30+6\,\sqrt{13}}}\sqrt{1- \left ( -{\frac{5}{6}}+{\frac{\sqrt{13}}{6}} \right ){x}^{2}}\sqrt{1- \left ( -{\frac{5}{6}}-{\frac{\sqrt{13}}{6}} \right ){x}^{2}}{\it EllipticF} \left ({\frac{x\sqrt{-30+6\,\sqrt{13}}}{6}},{\frac{5\,\sqrt{3}}{6}}+{\frac{\sqrt{39}}{6}} \right ){\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}}+{\frac{1064}{117\,\sqrt{-30+6\,\sqrt{13}} \left ( \sqrt{13}+5 \right ) }\sqrt{1- \left ( -{\frac{5}{6}}+{\frac{\sqrt{13}}{6}} \right ){x}^{2}}\sqrt{1- \left ( -{\frac{5}{6}}-{\frac{\sqrt{13}}{6}} \right ){x}^{2}} \left ({\it EllipticF} \left ({\frac{x\sqrt{-30+6\,\sqrt{13}}}{6}},{\frac{5\,\sqrt{3}}{6}}+{\frac{\sqrt{39}}{6}} \right ) -{\it EllipticE} \left ({\frac{x\sqrt{-30+6\,\sqrt{13}}}{6}},{\frac{5\,\sqrt{3}}{6}}+{\frac{\sqrt{39}}{6}} \right ) \right ){\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}}-{\frac{2}{27\,{x}^{3}}\sqrt{{x}^{4}+5\,{x}^{2}+3}}-4\,{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}} \left ( -{\frac{40\,{x}^{3}}{351}}-{\frac{343\,x}{702}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

23/81*(x^4+5*x^2+3)^(1/2)/x-6*(19/234*x^3+40/117*x)/(x^4+5*x^2+3)^(1/2)-10/117/(-30+6*13^(1/2))^(1/2)*(1-(-5/6
+1/6*13^(1/2))*x^2)^(1/2)*(1-(-5/6-1/6*13^(1/2))*x^2)^(1/2)/(x^4+5*x^2+3)^(1/2)*EllipticF(1/6*x*(-30+6*13^(1/2
))^(1/2),5/6*3^(1/2)+1/6*39^(1/2))+1064/117/(-30+6*13^(1/2))^(1/2)*(1-(-5/6+1/6*13^(1/2))*x^2)^(1/2)*(1-(-5/6-
1/6*13^(1/2))*x^2)^(1/2)/(x^4+5*x^2+3)^(1/2)/(13^(1/2)+5)*(EllipticF(1/6*x*(-30+6*13^(1/2))^(1/2),5/6*3^(1/2)+
1/6*39^(1/2))-EllipticE(1/6*x*(-30+6*13^(1/2))^(1/2),5/6*3^(1/2)+1/6*39^(1/2)))-2/27*(x^4+5*x^2+3)^(1/2)/x^3-4
*(-40/351*x^3-343/702*x)/(x^4+5*x^2+3)^(1/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{3 \, x^{2} + 2}{{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{3}{2}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{4} + 5 \, x^{2} + 3}{\left (3 \, x^{2} + 2\right )}}{x^{12} + 10 \, x^{10} + 31 \, x^{8} + 30 \, x^{6} + 9 \, x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + 5*x^2 + 3)*(3*x^2 + 2)/(x^12 + 10*x^10 + 31*x^8 + 30*x^6 + 9*x^4), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{3 x^{2} + 2}{x^{4} \left (x^{4} + 5 x^{2} + 3\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((3*x**2 + 2)/(x**4*(x**4 + 5*x**2 + 3)**(3/2)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{3 \, x^{2} + 2}{{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{3}{2}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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