3.168 \(\int \frac{(2+3 x^2) (3+5 x^2+x^4)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=331 \[ \frac{103 \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right ),\frac{1}{6} \left (5 \sqrt{13}-13\right )\right )}{\sqrt{6 \left (5+\sqrt{13}\right )} \sqrt{x^4+5 x^2+3}}-\frac{\left (2-5 x^2\right ) \left (x^4+5 x^2+3\right )^{3/2}}{5 x^5}-\frac{\left (40-87 x^2\right ) \sqrt{x^4+5 x^2+3}}{5 x^3}-\frac{722 \sqrt{x^4+5 x^2+3}}{15 x}+\frac{361 x \left (2 x^2+\sqrt{13}+5\right )}{15 \sqrt{x^4+5 x^2+3}}-\frac{361 \sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{15 \sqrt{x^4+5 x^2+3}} \]

[Out]

(361*x*(5 + Sqrt[13] + 2*x^2))/(15*Sqrt[3 + 5*x^2 + x^4]) - (722*Sqrt[3 + 5*x^2 + x^4])/(15*x) - ((40 - 87*x^2
)*Sqrt[3 + 5*x^2 + x^4])/(5*x^3) - ((2 - 5*x^2)*(3 + 5*x^2 + x^4)^(3/2))/(5*x^5) - (361*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 + S
qrt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/(15*Sqrt[3 + 5*x^2 + x^4]) + (103*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])
/(Sqrt[6*(5 + Sqrt[13])]*Sqrt[3 + 5*x^2 + x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.202751, antiderivative size = 331, 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 = {1271, 1281, 1189, 1099, 1135} \[ -\frac{\left (2-5 x^2\right ) \left (x^4+5 x^2+3\right )^{3/2}}{5 x^5}-\frac{\left (40-87 x^2\right ) \sqrt{x^4+5 x^2+3}}{5 x^3}-\frac{722 \sqrt{x^4+5 x^2+3}}{15 x}+\frac{361 x \left (2 x^2+\sqrt{13}+5\right )}{15 \sqrt{x^4+5 x^2+3}}+\frac{103 \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) F\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{\sqrt{6 \left (5+\sqrt{13}\right )} \sqrt{x^4+5 x^2+3}}-\frac{361 \sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{15 \sqrt{x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

Int[((2 + 3*x^2)*(3 + 5*x^2 + x^4)^(3/2))/x^6,x]

[Out]

(361*x*(5 + Sqrt[13] + 2*x^2))/(15*Sqrt[3 + 5*x^2 + x^4]) - (722*Sqrt[3 + 5*x^2 + x^4])/(15*x) - ((40 - 87*x^2
)*Sqrt[3 + 5*x^2 + x^4])/(5*x^3) - ((2 - 5*x^2)*(3 + 5*x^2 + x^4)^(3/2))/(5*x^5) - (361*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 + S
qrt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/(15*Sqrt[3 + 5*x^2 + x^4]) + (103*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])
/(Sqrt[6*(5 + Sqrt[13])]*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 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{\left (2+3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{x^6} \, dx &=-\frac{\left (2-5 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{5 x^5}-\frac{1}{5} \int \frac{\left (-120-87 x^2\right ) \sqrt{3+5 x^2+x^4}}{x^4} \, dx\\ &=-\frac{\left (40-87 x^2\right ) \sqrt{3+5 x^2+x^4}}{5 x^3}-\frac{\left (2-5 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{5 x^5}+\frac{1}{15} \int \frac{2166+1545 x^2}{x^2 \sqrt{3+5 x^2+x^4}} \, dx\\ &=-\frac{722 \sqrt{3+5 x^2+x^4}}{15 x}-\frac{\left (40-87 x^2\right ) \sqrt{3+5 x^2+x^4}}{5 x^3}-\frac{\left (2-5 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{5 x^5}-\frac{1}{45} \int \frac{-4635-2166 x^2}{\sqrt{3+5 x^2+x^4}} \, dx\\ &=-\frac{722 \sqrt{3+5 x^2+x^4}}{15 x}-\frac{\left (40-87 x^2\right ) \sqrt{3+5 x^2+x^4}}{5 x^3}-\frac{\left (2-5 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{5 x^5}+\frac{722}{15} \int \frac{x^2}{\sqrt{3+5 x^2+x^4}} \, dx+103 \int \frac{1}{\sqrt{3+5 x^2+x^4}} \, dx\\ &=\frac{361 x \left (5+\sqrt{13}+2 x^2\right )}{15 \sqrt{3+5 x^2+x^4}}-\frac{722 \sqrt{3+5 x^2+x^4}}{15 x}-\frac{\left (40-87 x^2\right ) \sqrt{3+5 x^2+x^4}}{5 x^3}-\frac{\left (2-5 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}}{5 x^5}-\frac{361 \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 )}{15 \sqrt{3+5 x^2+x^4}}+\frac{103 \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{6 \left (5+\sqrt{13}\right )} \sqrt{3+5 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.316542, size = 244, normalized size = 0.74 \[ \frac{-i \sqrt{2} \left (361 \sqrt{13}-260\right ) \sqrt{\frac{-2 x^2+\sqrt{13}-5}{\sqrt{13}-5}} \sqrt{2 x^2+\sqrt{13}+5} x^5 \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{2}{5+\sqrt{13}}} x\right ),\frac{19}{6}+\frac{5 \sqrt{13}}{6}\right )+30 x^{10}-634 x^8-4040 x^6-3438 x^4-810 x^2+361 i \sqrt{2} \left (\sqrt{13}-5\right ) \sqrt{\frac{-2 x^2+\sqrt{13}-5}{\sqrt{13}-5}} \sqrt{2 x^2+\sqrt{13}+5} x^5 E\left (i \sinh ^{-1}\left (\sqrt{\frac{2}{5+\sqrt{13}}} x\right )|\frac{19}{6}+\frac{5 \sqrt{13}}{6}\right )-108}{30 x^5 \sqrt{x^4+5 x^2+3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((2 + 3*x^2)*(3 + 5*x^2 + x^4)^(3/2))/x^6,x]

[Out]

(-108 - 810*x^2 - 3438*x^4 - 4040*x^6 - 634*x^8 + 30*x^10 + (361*I)*Sqrt[2]*(-5 + Sqrt[13])*x^5*Sqrt[(-5 + Sqr
t[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]*(-260 + 361*Sqrt[13])*x^5*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])/(30*x^5*Sqrt[3 + 5*
x^2 + x^4])

________________________________________________________________________________________

Maple [A]  time = 0.02, size = 259, normalized size = 0.8 \begin{align*} -{\frac{6}{5\,{x}^{5}}\sqrt{{x}^{4}+5\,{x}^{2}+3}}-7\,{\frac{\sqrt{{x}^{4}+5\,{x}^{2}+3}}{{x}^{3}}}-{\frac{392}{15\,x}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+618\,{\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{8664}{5\,\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}}}}+x\sqrt{{x}^{4}+5\,{x}^{2}+3} \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^6,x)

[Out]

-6/5/x^5*(x^4+5*x^2+3)^(1/2)-7*(x^4+5*x^2+3)^(1/2)/x^3-392/15*(x^4+5*x^2+3)^(1/2)/x+618/(-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))-8664/5/(-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)))+x*(x^4+5*x^2+3)^(1/2)

________________________________________________________________________________________

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^{6}}\,{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^6,x, algorithm="maxima")

[Out]

integrate((x^4 + 5*x^2 + 3)^(3/2)*(3*x^2 + 2)/x^6, x)

________________________________________________________________________________________

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^{6}}, 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^6,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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^{6}}\, 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**6,x)

[Out]

Integral((3*x**2 + 2)*(x**4 + 5*x**2 + 3)**(3/2)/x**6, x)

________________________________________________________________________________________

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^{6}}\,{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^6,x, algorithm="giac")

[Out]

integrate((x^4 + 5*x^2 + 3)^(3/2)*(3*x^2 + 2)/x^6, x)