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

Optimal. Leaf size=356 \[ -\frac {4210}{429} \sqrt {x^4+5 x^2+3} x+\frac {176723 \left (2 x^2+\sqrt {13}+5\right ) x}{4290 \sqrt {x^4+5 x^2+3}}+\frac {2105 \sqrt {\frac {2}{3 \left (5+\sqrt {13}\right )}} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) F\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{143 \sqrt {x^4+5 x^2+3}}-\frac {176723 \sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{4290 \sqrt {x^4+5 x^2+3}}+\frac {1}{143} \left (33 x^2+71\right ) \left (x^4+5 x^2+3\right )^{3/2} x^5-\frac {1}{429} \left (272 x^2+283\right ) \sqrt {x^4+5 x^2+3} x^5+\frac {1251}{715} \sqrt {x^4+5 x^2+3} x^3 \]

[Out]

1/143*x^5*(33*x^2+71)*(x^4+5*x^2+3)^(3/2)+176723/4290*x*(5+2*x^2+13^(1/2))/(x^4+5*x^2+3)^(1/2)-4210/429*x*(x^4
+5*x^2+3)^(1/2)+1251/715*x^3*(x^4+5*x^2+3)^(1/2)-1/429*x^5*(272*x^2+283)*(x^4+5*x^2+3)^(1/2)+2105/429*(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)-176723/25740*(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 = 0.26, antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1273, 1279, 1189, 1099, 1135} \[ \frac {1}{143} \left (33 x^2+71\right ) \left (x^4+5 x^2+3\right )^{3/2} x^5-\frac {1}{429} \left (272 x^2+283\right ) \sqrt {x^4+5 x^2+3} x^5+\frac {1251}{715} \sqrt {x^4+5 x^2+3} x^3-\frac {4210}{429} \sqrt {x^4+5 x^2+3} x+\frac {176723 \left (2 x^2+\sqrt {13}+5\right ) x}{4290 \sqrt {x^4+5 x^2+3}}+\frac {2105 \sqrt {\frac {2}{3 \left (5+\sqrt {13}\right )}} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) F\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{143 \sqrt {x^4+5 x^2+3}}-\frac {176723 \sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{4290 \sqrt {x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(176723*x*(5 + Sqrt[13] + 2*x^2))/(4290*Sqrt[3 + 5*x^2 + x^4]) - (4210*x*Sqrt[3 + 5*x^2 + x^4])/429 + (1251*x^
3*Sqrt[3 + 5*x^2 + x^4])/715 - (x^5*(283 + 272*x^2)*Sqrt[3 + 5*x^2 + x^4])/429 + (x^5*(71 + 33*x^2)*(3 + 5*x^2
+ x^4)^(3/2))/143 - (176723*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])/(4290*Sqrt[3 + 5*x^2
+ x^4]) + (2105*Sqrt[2/(3*(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])/(143*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 1273

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*(b*e*2*p + c*d*(m + 4*p + 3) + c*e*(4*p + m + 1)*x^2))/(c*f*(4*p + m + 1)*(m
+ 4*p + 3)), x] + Dist[(2*p)/(c*(4*p + m + 1)*(m + 4*p + 3)), Int[(f*x)^m*(a + b*x^2 + c*x^4)^(p - 1)*Simp[2*
a*c*d*(m + 4*p + 3) - a*b*e*(m + 1) + (2*a*c*e*(4*p + m + 1) + b*c*d*(m + 4*p + 3) - b^2*e*(m + 2*p + 1))*x^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && NeQ[4*p + m + 1, 0] && N
eQ[m + 4*p + 3, 0] && 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 x^4 \left (2+3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2} \, dx &=\frac {1}{143} x^5 \left (71+33 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac {3}{143} \int x^4 \left (-69-272 x^2\right ) \sqrt {3+5 x^2+x^4} \, dx\\ &=-\frac {1}{429} x^5 \left (283+272 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{143} x^5 \left (71+33 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac {\int \frac {x^4 \left (16674+26271 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx}{3003}\\ &=\frac {1251}{715} x^3 \sqrt {3+5 x^2+x^4}-\frac {1}{429} x^5 \left (283+272 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{143} x^5 \left (71+33 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}-\frac {\int \frac {x^2 \left (236439+442050 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx}{15015}\\ &=-\frac {4210}{429} x \sqrt {3+5 x^2+x^4}+\frac {1251}{715} x^3 \sqrt {3+5 x^2+x^4}-\frac {1}{429} x^5 \left (283+272 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{143} x^5 \left (71+33 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac {\int \frac {1326150+3711183 x^2}{\sqrt {3+5 x^2+x^4}} \, dx}{45045}\\ &=-\frac {4210}{429} x \sqrt {3+5 x^2+x^4}+\frac {1251}{715} x^3 \sqrt {3+5 x^2+x^4}-\frac {1}{429} x^5 \left (283+272 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{143} x^5 \left (71+33 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac {4210}{143} \int \frac {1}{\sqrt {3+5 x^2+x^4}} \, dx+\frac {176723 \int \frac {x^2}{\sqrt {3+5 x^2+x^4}} \, dx}{2145}\\ &=\frac {176723 x \left (5+\sqrt {13}+2 x^2\right )}{4290 \sqrt {3+5 x^2+x^4}}-\frac {4210}{429} x \sqrt {3+5 x^2+x^4}+\frac {1251}{715} x^3 \sqrt {3+5 x^2+x^4}-\frac {1}{429} x^5 \left (283+272 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{143} x^5 \left (71+33 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}-\frac {176723 \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 )}{4290 \sqrt {3+5 x^2+x^4}}+\frac {2105 \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 ) 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 )}{143 \sqrt {3+5 x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.27, size = 249, normalized size = 0.70 \[ \frac {-i \sqrt {2} \left (176723 \sqrt {13}-757315\right ) \sqrt {\frac {-2 x^2+\sqrt {13}-5}{\sqrt {13}-5}} \sqrt {2 x^2+\sqrt {13}+5} F\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right )|\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )+176723 i \sqrt {2} \left (\sqrt {13}-5\right ) \sqrt {\frac {-2 x^2+\sqrt {13}-5}{\sqrt {13}-5}} \sqrt {2 x^2+\sqrt {13}+5} E\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right )|\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )+4 x \left (495 x^{14}+6015 x^{12}+24635 x^{10}+39650 x^8+29003 x^6+3055 x^4-93991 x^2-63150\right )}{8580 \sqrt {x^4+5 x^2+3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*x*(-63150 - 93991*x^2 + 3055*x^4 + 29003*x^6 + 39650*x^8 + 24635*x^10 + 6015*x^12 + 495*x^14) + (176723*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*A
rcSinh[Sqrt[2/(5 + Sqrt[13])]*x], 19/6 + (5*Sqrt[13])/6] - I*Sqrt[2]*(-757315 + 176723*Sqrt[13])*Sqrt[(-5 + Sq
rt[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])/(8580*Sqrt[3 + 5*x^2 + x^4])

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (3 \, x^{10} + 17 \, x^{8} + 19 \, x^{6} + 6 \, x^{4}\right )} \sqrt {x^{4} + 5 \, x^{2} + 3}, x\right ) \]

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^10 + 17*x^8 + 19*x^6 + 6*x^4)*sqrt(x^4 + 5*x^2 + 3), x)

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

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

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maple [A]  time = 0.02, size = 294, normalized size = 0.83 \[ \frac {3 \sqrt {x^{4}+5 x^{2}+3}\, x^{11}}{13}+\frac {236 \sqrt {x^{4}+5 x^{2}+3}\, x^{9}}{143}+\frac {1090 \sqrt {x^{4}+5 x^{2}+3}\, x^{7}}{429}+\frac {356 \sqrt {x^{4}+5 x^{2}+3}\, x^{5}}{429}+\frac {1251 \sqrt {x^{4}+5 x^{2}+3}\, x^{3}}{715}-\frac {4210 \sqrt {x^{4}+5 x^{2}+3}\, x}{429}+\frac {25260 \sqrt {-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {-30+6 \sqrt {13}}\, x}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )}{143 \sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}}-\frac {2120676 \sqrt {-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {-30+6 \sqrt {13}}\, x}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )+\EllipticF \left (\frac {\sqrt {-30+6 \sqrt {13}}\, x}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )\right )}{715 \sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}\, \left (\sqrt {13}+5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/13*x^11*(x^4+5*x^2+3)^(1/2)+236/143*x^9*(x^4+5*x^2+3)^(1/2)+1090/429*(x^4+5*x^2+3)^(1/2)*x^7+356/429*(x^4+5*
x^2+3)^(1/2)*x^5+1251/715*(x^4+5*x^2+3)^(1/2)*x^3-4210/429*(x^4+5*x^2+3)^(1/2)*x+25260/143/(-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))-2120676/715/(-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 \[ \int {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}} {\left (3 \, x^{2} + 2\right )} x^{4}\,{d x} \]

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

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

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 \[ \int x^{4} \left (3 x^{2} + 2\right ) \left (x^{4} + 5 x^{2} + 3\right )^{\frac {3}{2}}\, dx \]

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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