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

Optimal. Leaf size=307 \[ -\frac {11}{13} \sqrt {x^4+5 x^2+3} x+\frac {43 \left (2 x^2+\sqrt {13}+5\right ) x}{13 \sqrt {x^4+5 x^2+3}}+\frac {11 \sqrt {\frac {3}{2 \left (5+\sqrt {13}\right )}} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) F\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{13 \sqrt {x^4+5 x^2+3}}-\frac {43 \sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{13 \sqrt {x^4+5 x^2+3}}+\frac {\left (11 x^2+8\right ) x^3}{13 \sqrt {x^4+5 x^2+3}} \]

[Out]

1/13*x^3*(11*x^2+8)/(x^4+5*x^2+3)^(1/2)+43/13*x*(5+2*x^2+13^(1/2))/(x^4+5*x^2+3)^(1/2)-11/13*x*(x^4+5*x^2+3)^(
1/2)+11/26*(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)-43/78*(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 = 0.16, antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1275, 1279, 1189, 1099, 1135} \[ \frac {\left (11 x^2+8\right ) x^3}{13 \sqrt {x^4+5 x^2+3}}-\frac {11}{13} \sqrt {x^4+5 x^2+3} x+\frac {43 \left (2 x^2+\sqrt {13}+5\right ) x}{13 \sqrt {x^4+5 x^2+3}}+\frac {11 \sqrt {\frac {3}{2 \left (5+\sqrt {13}\right )}} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) F\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{13 \sqrt {x^4+5 x^2+3}}-\frac {43 \sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{13 \sqrt {x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(43*x*(5 + Sqrt[13] + 2*x^2))/(13*Sqrt[3 + 5*x^2 + x^4]) + (x^3*(8 + 11*x^2))/(13*Sqrt[3 + 5*x^2 + x^4]) - (11
*x*Sqrt[3 + 5*x^2 + x^4])/13 - (43*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])/(13*Sqrt[3 + 5
*x^2 + x^4]) + (11*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])/(13*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 1275

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

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^4 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx &=\frac {x^3 \left (8+11 x^2\right )}{13 \sqrt {3+5 x^2+x^4}}+\frac {1}{13} \int \frac {x^2 \left (-24-33 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx\\ &=\frac {x^3 \left (8+11 x^2\right )}{13 \sqrt {3+5 x^2+x^4}}-\frac {11}{13} x \sqrt {3+5 x^2+x^4}-\frac {1}{39} \int \frac {-99-258 x^2}{\sqrt {3+5 x^2+x^4}} \, dx\\ &=\frac {x^3 \left (8+11 x^2\right )}{13 \sqrt {3+5 x^2+x^4}}-\frac {11}{13} x \sqrt {3+5 x^2+x^4}+\frac {33}{13} \int \frac {1}{\sqrt {3+5 x^2+x^4}} \, dx+\frac {86}{13} \int \frac {x^2}{\sqrt {3+5 x^2+x^4}} \, dx\\ &=\frac {43 x \left (5+\sqrt {13}+2 x^2\right )}{13 \sqrt {3+5 x^2+x^4}}+\frac {x^3 \left (8+11 x^2\right )}{13 \sqrt {3+5 x^2+x^4}}-\frac {11}{13} x \sqrt {3+5 x^2+x^4}-\frac {43 \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 )}{13 \sqrt {3+5 x^2+x^4}}+\frac {11 \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 )}{13 \sqrt {3+5 x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 219, normalized size = 0.71 \[ \frac {-2 x \left (47 x^2+33\right )-i \sqrt {2} \left (43 \sqrt {13}-182\right ) \sqrt {\frac {-2 x^2+\sqrt {13}-5}{\sqrt {13}-5}} \sqrt {2 x^2+\sqrt {13}+5} F\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right )|\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )+43 i \sqrt {2} \left (\sqrt {13}-5\right ) \sqrt {\frac {-2 x^2+\sqrt {13}-5}{\sqrt {13}-5}} \sqrt {2 x^2+\sqrt {13}+5} E\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right )|\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )}{26 \sqrt {x^4+5 x^2+3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*x*(33 + 47*x^2) + (43*I)*Sqrt[2]*(-5 + Sqrt[13])*Sqrt[(-5 + Sqrt[13] - 2*x^2)/(-5 + Sqrt[13])]*Sqrt[5 + Sq
rt[13] + 2*x^2]*EllipticE[I*ArcSinh[Sqrt[2/(5 + Sqrt[13])]*x], 19/6 + (5*Sqrt[13])/6] - I*Sqrt[2]*(-182 + 43*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])/(26*Sqrt[3 + 5*x^2 + x^4])

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3 \, x^{6} + 2 \, x^{4}\right )} \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{8} + 10 \, x^{6} + 31 \, x^{4} + 30 \, x^{2} + 9}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 240, normalized size = 0.78 \[ \frac {198 \sqrt {-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {-30+6 \sqrt {13}}\, x}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )}{13 \sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}}-\frac {6 \left (\frac {19}{26} x^{3}+\frac {15}{26} x \right )}{\sqrt {x^{4}+5 x^{2}+3}}-\frac {3096 \sqrt {-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {-30+6 \sqrt {13}}\, x}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )+\EllipticF \left (\frac {\sqrt {-30+6 \sqrt {13}}\, x}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )\right )}{13 \sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}\, \left (\sqrt {13}+5\right )}-\frac {4 \left (-\frac {5}{26} x^{3}-\frac {3}{13} x \right )}{\sqrt {x^{4}+5 x^{2}+3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*(19/26*x^3+15/26*x)/(x^4+5*x^2+3)^(1/2)+198/13/(-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))-3096/13/(-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)))-4*(-5/26*x^3-3/13*x)/(x^4+5*x^2+3)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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