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

Optimal. Leaf size=270 \[ \sqrt {x^4+5 x^2+3} x-\frac {4 \left (2 x^2+\sqrt {13}+5\right ) x}{\sqrt {x^4+5 x^2+3}}-\frac {\sqrt {\frac {3}{2 \left (5+\sqrt {13}\right )}} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) F\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{\sqrt {x^4+5 x^2+3}}+\frac {2 \sqrt {\frac {2}{3} \left (5+\sqrt {13}\right )} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{\sqrt {x^4+5 x^2+3}} \]

[Out]

-4*x*(5+2*x^2+13^(1/2))/(x^4+5*x^2+3)^(1/2)+x*(x^4+5*x^2+3)^(1/2)-1/2*(1/(36+x^2*(30+6*13^(1/2))))^(1/2)*(36+x
^2*(30+6*13^(1/2)))^(1/2)*EllipticF(x*(30+6*13^(1/2))^(1/2)/(36+x^2*(30+6*13^(1/2)))^(1/2),1/6*(-78+30*13^(1/2
))^(1/2))*(6+x^2*(5+13^(1/2)))*6^(1/2)/(5+13^(1/2))^(1/2)*((6+x^2*(5-13^(1/2)))/(6+x^2*(5+13^(1/2))))^(1/2)/(x
^4+5*x^2+3)^(1/2)+2/3*(1/(36+x^2*(30+6*13^(1/2))))^(1/2)*(36+x^2*(30+6*13^(1/2)))^(1/2)*EllipticE(x*(30+6*13^(
1/2))^(1/2)/(36+x^2*(30+6*13^(1/2)))^(1/2),1/6*(-78+30*13^(1/2))^(1/2))*(6+x^2*(5+13^(1/2)))*(30+6*13^(1/2))^(
1/2)*((6+x^2*(5-13^(1/2)))/(6+x^2*(5+13^(1/2))))^(1/2)/(x^4+5*x^2+3)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1279, 1189, 1099, 1135} \[ \sqrt {x^4+5 x^2+3} x-\frac {4 \left (2 x^2+\sqrt {13}+5\right ) x}{\sqrt {x^4+5 x^2+3}}-\frac {\sqrt {\frac {3}{2 \left (5+\sqrt {13}\right )}} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) F\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{\sqrt {x^4+5 x^2+3}}+\frac {2 \sqrt {\frac {2}{3} \left (5+\sqrt {13}\right )} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{\sqrt {x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*x*(5 + Sqrt[13] + 2*x^2))/Sqrt[3 + 5*x^2 + x^4] + x*Sqrt[3 + 5*x^2 + x^4] + (2*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 + Sq
rt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/Sqrt[3 + 5*x^2 + x^4] - (Sqrt[3/(2*(5 + Sqrt[13]))]*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[3 + 5*x^2 + x^4]

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]

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 1279

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

Rubi steps

\begin {align*} \int \frac {x^2 \left (2+3 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx &=x \sqrt {3+5 x^2+x^4}-\frac {1}{3} \int \frac {9+24 x^2}{\sqrt {3+5 x^2+x^4}} \, dx\\ &=x \sqrt {3+5 x^2+x^4}-3 \int \frac {1}{\sqrt {3+5 x^2+x^4}} \, dx-8 \int \frac {x^2}{\sqrt {3+5 x^2+x^4}} \, dx\\ &=-\frac {4 x \left (5+\sqrt {13}+2 x^2\right )}{\sqrt {3+5 x^2+x^4}}+x \sqrt {3+5 x^2+x^4}+\frac {2 \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 )}{\sqrt {3+5 x^2+x^4}}-\frac {\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*}

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Mathematica [C]  time = 0.24, size = 222, normalized size = 0.82 \[ \frac {i \sqrt {2} \left (4 \sqrt {13}-17\right ) \sqrt {\frac {-2 x^2+\sqrt {13}-5}{\sqrt {13}-5}} \sqrt {2 x^2+\sqrt {13}+5} F\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right )|\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )-4 i \sqrt {2} \left (\sqrt {13}-5\right ) \sqrt {\frac {-2 x^2+\sqrt {13}-5}{\sqrt {13}-5}} \sqrt {2 x^2+\sqrt {13}+5} E\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right )|\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )+2 x \left (x^4+5 x^2+3\right )}{2 \sqrt {x^4+5 x^2+3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*x*(3 + 5*x^2 + x^4) - (4*I)*Sqrt[2]*(-5 + Sqrt[13])*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]*(-17 + 4*S
qrt[13])*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])/(2*Sqrt[3 + 5*x^2 + x^4])

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {3 \, x^{4} + 2 \, x^{2}}{\sqrt {x^{4} + 5 \, x^{2} + 3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(3*x^2+2)/(x^4+5*x^2+3)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (3 \, x^{2} + 2\right )} x^{2}}{\sqrt {x^{4} + 5 \, x^{2} + 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(3*x^2+2)/(x^4+5*x^2+3)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 208, normalized size = 0.77 \[ \sqrt {x^{4}+5 x^{2}+3}\, x -\frac {18 \sqrt {-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {-30+6 \sqrt {13}}\, x}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )}{\sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}}+\frac {288 \sqrt {-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {-30+6 \sqrt {13}}\, x}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )+\EllipticF \left (\frac {\sqrt {-30+6 \sqrt {13}}\, x}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )\right )}{\sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}\, \left (\sqrt {13}+5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(3*x^2+2)/(x^4+5*x^2+3)^(1/2),x)

[Out]

(x^4+5*x^2+3)^(1/2)*x-18/(-30+6*13^(1/2))^(1/2)*(-(-5/6+1/6*13^(1/2))*x^2+1)^(1/2)*(-(-5/6-1/6*13^(1/2))*x^2+1
)^(1/2)/(x^4+5*x^2+3)^(1/2)*EllipticF(1/6*(-30+6*13^(1/2))^(1/2)*x,5/6*3^(1/2)+1/6*39^(1/2))+288/(-30+6*13^(1/
2))^(1/2)*(-(-5/6+1/6*13^(1/2))*x^2+1)^(1/2)*(-(-5/6-1/6*13^(1/2))*x^2+1)^(1/2)/(x^4+5*x^2+3)^(1/2)/(13^(1/2)+
5)*(EllipticF(1/6*(-30+6*13^(1/2))^(1/2)*x,5/6*3^(1/2)+1/6*39^(1/2))-EllipticE(1/6*(-30+6*13^(1/2))^(1/2)*x,5/
6*3^(1/2)+1/6*39^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (3 \, x^{2} + 2\right )} x^{2}}{\sqrt {x^{4} + 5 \, x^{2} + 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(3*x^2+2)/(x^4+5*x^2+3)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,\left (3\,x^2+2\right )}{\sqrt {x^4+5\,x^2+3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(3*x^2 + 2))/(5*x^2 + x^4 + 3)^(1/2),x)

[Out]

int((x^2*(3*x^2 + 2))/(5*x^2 + x^4 + 3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (3 x^{2} + 2\right )}{\sqrt {x^{4} + 5 x^{2} + 3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(3*x**2+2)/(x**4+5*x**2+3)**(1/2),x)

[Out]

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

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