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

Optimal. Leaf size=312 \[ \frac{19 \sqrt{\frac{3}{2 \left (5+\sqrt{13}\right )}} \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right ),\frac{1}{6} \left (5 \sqrt{13}-13\right )\right )}{\sqrt{x^4+5 x^2+3}}-\frac{\left (14-3 x^2\right ) \left (x^4+5 x^2+3\right )^{3/2}}{7 x}+\frac{1}{35} x \left (129 x^2+655\right ) \sqrt{x^4+5 x^2+3}+\frac{412 x \left (2 x^2+\sqrt{13}+5\right )}{35 \sqrt{x^4+5 x^2+3}}-\frac{206 \sqrt{\frac{2}{3} \left (5+\sqrt{13}\right )} \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{35 \sqrt{x^4+5 x^2+3}} \]

[Out]

(412*x*(5 + Sqrt[13] + 2*x^2))/(35*Sqrt[3 + 5*x^2 + x^4]) + (x*(655 + 129*x^2)*Sqrt[3 + 5*x^2 + x^4])/35 - ((1
4 - 3*x^2)*(3 + 5*x^2 + x^4)^(3/2))/(7*x) - (206*Sqrt[(2*(5 + Sqrt[13]))/3]*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])/(35*Sqrt[3 + 5*x^2 + x^4]) + (19*Sqrt[3/(2*(5 + Sqrt[13]))]*Sqrt[(6 + (5 - Sqrt[13])*x^2)/(6 + (5 + Sqrt[1
3])*x^2)]*(6 + (5 + Sqrt[13])*x^2)*EllipticF[ArcTan[Sqrt[(5 + Sqrt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/Sqrt[3 +
5*x^2 + x^4]

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Rubi [A]  time = 0.150656, antiderivative size = 312, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1271, 1176, 1189, 1099, 1135} \[ -\frac{\left (14-3 x^2\right ) \left (x^4+5 x^2+3\right )^{3/2}}{7 x}+\frac{1}{35} x \left (129 x^2+655\right ) \sqrt{x^4+5 x^2+3}+\frac{412 x \left (2 x^2+\sqrt{13}+5\right )}{35 \sqrt{x^4+5 x^2+3}}+\frac{19 \sqrt{\frac{3}{2 \left (5+\sqrt{13}\right )}} \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) F\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{\sqrt{x^4+5 x^2+3}}-\frac{206 \sqrt{\frac{2}{3} \left (5+\sqrt{13}\right )} \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{35 \sqrt{x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(412*x*(5 + Sqrt[13] + 2*x^2))/(35*Sqrt[3 + 5*x^2 + x^4]) + (x*(655 + 129*x^2)*Sqrt[3 + 5*x^2 + x^4])/35 - ((1
4 - 3*x^2)*(3 + 5*x^2 + x^4)^(3/2))/(7*x) - (206*Sqrt[(2*(5 + Sqrt[13]))/3]*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])/(35*Sqrt[3 + 5*x^2 + x^4]) + (19*Sqrt[3/(2*(5 + Sqrt[13]))]*Sqrt[(6 + (5 - Sqrt[13])*x^2)/(6 + (5 + Sqrt[1
3])*x^2)]*(6 + (5 + Sqrt[13])*x^2)*EllipticF[ArcTan[Sqrt[(5 + Sqrt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/Sqrt[3 +
5*x^2 + x^4]

Rule 1271

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*(d*(m + 4*p + 3) + e*(m + 1)*x^2))/(f*(m + 1)*(m + 4*p + 3)), x] + Dist[(2*p
)/(f^2*(m + 1)*(m + 4*p + 3)), Int[(f*x)^(m + 2)*(a + b*x^2 + c*x^4)^(p - 1)*Simp[2*a*e*(m + 1) - b*d*(m + 4*p
+ 3) + (b*e*(m + 1) - 2*c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c
, 0] && GtQ[p, 0] && LtQ[m, -1] && m + 4*p + 3 != 0 && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1176

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

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{\left (2+3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{x^2} \, dx &=-\frac{\left (14-3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{7 x}-\frac{3}{7} \int \left (-88-43 x^2\right ) \sqrt{3+5 x^2+x^4} \, dx\\ &=\frac{1}{35} x \left (655+129 x^2\right ) \sqrt{3+5 x^2+x^4}-\frac{\left (14-3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{7 x}-\frac{1}{35} \int \frac{-1995-824 x^2}{\sqrt{3+5 x^2+x^4}} \, dx\\ &=\frac{1}{35} x \left (655+129 x^2\right ) \sqrt{3+5 x^2+x^4}-\frac{\left (14-3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{7 x}+\frac{824}{35} \int \frac{x^2}{\sqrt{3+5 x^2+x^4}} \, dx+57 \int \frac{1}{\sqrt{3+5 x^2+x^4}} \, dx\\ &=\frac{412 x \left (5+\sqrt{13}+2 x^2\right )}{35 \sqrt{3+5 x^2+x^4}}+\frac{1}{35} x \left (655+129 x^2\right ) \sqrt{3+5 x^2+x^4}-\frac{\left (14-3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{7 x}-\frac{206 \sqrt{\frac{2}{3} \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 )}{35 \sqrt{3+5 x^2+x^4}}+\frac{19 \sqrt{\frac{3}{2 \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 ) 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 )}{\sqrt{3+5 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.307957, size = 235, normalized size = 0.75 \[ \frac{-i \sqrt{2} \left (412 \sqrt{13}-65\right ) \sqrt{\frac{-2 x^2+\sqrt{13}-5}{\sqrt{13}-5}} \sqrt{2 x^2+\sqrt{13}+5} x \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{2}{5+\sqrt{13}}} x\right ),\frac{19}{6}+\frac{5 \sqrt{13}}{6}\right )+30 x^{10}+418 x^8+2130 x^6+3884 x^4+412 i \sqrt{2} \left (\sqrt{13}-5\right ) \sqrt{\frac{-2 x^2+\sqrt{13}-5}{\sqrt{13}-5}} \sqrt{2 x^2+\sqrt{13}+5} x E\left (i \sinh ^{-1}\left (\sqrt{\frac{2}{5+\sqrt{13}}} x\right )|\frac{19}{6}+\frac{5 \sqrt{13}}{6}\right )-1260}{70 x \sqrt{x^4+5 x^2+3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-1260 + 3884*x^4 + 2130*x^6 + 418*x^8 + 30*x^10 + (412*I)*Sqrt[2]*(-5 + Sqrt[13])*x*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]*(-65 + 412*Sqrt[13])*x*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])/(70*x*Sqrt[3 + 5*x^2 + x^4])

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Maple [A]  time = 0.016, size = 260, normalized size = 0.8 \begin{align*}{\frac{3\,{x}^{5}}{7}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{134\,{x}^{3}}{35}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+10\,x\sqrt{{x}^{4}+5\,{x}^{2}+3}+342\,{\frac{\sqrt{1- \left ( -5/6+1/6\,\sqrt{13} \right ){x}^{2}}\sqrt{1- \left ( -5/6-1/6\,\sqrt{13} \right ){x}^{2}}{\it EllipticF} \left ( 1/6\,x\sqrt{-30+6\,\sqrt{13}},5/6\,\sqrt{3}+1/6\,\sqrt{39} \right ) }{\sqrt{-30+6\,\sqrt{13}}\sqrt{{x}^{4}+5\,{x}^{2}+3}}}-{\frac{29664}{35\,\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}}}}-6\,{\frac{\sqrt{{x}^{4}+5\,{x}^{2}+3}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/7*x^5*(x^4+5*x^2+3)^(1/2)+134/35*x^3*(x^4+5*x^2+3)^(1/2)+10*x*(x^4+5*x^2+3)^(1/2)+342/(-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))-29664/35/(-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)))-6*(x^4+5*x^2+3)^(1/2)
/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (3 \, x^{6} + 17 \, x^{4} + 19 \, x^{2} + 6\right )} \sqrt{x^{4} + 5 \, x^{2} + 3}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((3*x^6 + 17*x^4 + 19*x^2 + 6)*sqrt(x^4 + 5*x^2 + 3)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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