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

Optimal. Leaf size=308 \[ \frac{5 \sqrt{\frac{2}{3 \left (5+\sqrt{13}\right )}} \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right ),\frac{1}{6} \left (5 \sqrt{13}-13\right )\right )}{\sqrt{x^4+5 x^2+3}}+\frac{1}{3} x \left (x^2+3\right ) \left (x^4+5 x^2+3\right )^{3/2}-\frac{1}{15} x \left (12 x^2+5\right ) \sqrt{x^4+5 x^2+3}+\frac{203 x \left (2 x^2+\sqrt{13}+5\right )}{30 \sqrt{x^4+5 x^2+3}}-\frac{203 \sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{30 \sqrt{x^4+5 x^2+3}} \]

[Out]

(203*x*(5 + Sqrt[13] + 2*x^2))/(30*Sqrt[3 + 5*x^2 + x^4]) - (x*(5 + 12*x^2)*Sqrt[3 + 5*x^2 + x^4])/15 + (x*(3
+ x^2)*(3 + 5*x^2 + x^4)^(3/2))/3 - (203*Sqrt[(5 + Sqrt[13])/6]*Sqrt[(6 + (5 - Sqrt[13])*x^2)/(6 + (5 + Sqrt[1
3])*x^2)]*(6 + (5 + Sqrt[13])*x^2)*EllipticE[ArcTan[Sqrt[(5 + Sqrt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/(30*Sqrt
[3 + 5*x^2 + x^4]) + (5*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])/Sqrt[3 + 5*x^2 + x^4]

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Rubi [A]  time = 0.152772, antiderivative size = 308, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {1176, 1189, 1099, 1135} \[ \frac{1}{3} x \left (x^2+3\right ) \left (x^4+5 x^2+3\right )^{3/2}-\frac{1}{15} x \left (12 x^2+5\right ) \sqrt{x^4+5 x^2+3}+\frac{203 x \left (2 x^2+\sqrt{13}+5\right )}{30 \sqrt{x^4+5 x^2+3}}+\frac{5 \sqrt{\frac{2}{3 \left (5+\sqrt{13}\right )}} \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) F\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{\sqrt{x^4+5 x^2+3}}-\frac{203 \sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} \sqrt{\frac{\left (5-\sqrt{13}\right ) x^2+6}{\left (5+\sqrt{13}\right ) x^2+6}} \left (\left (5+\sqrt{13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt{\frac{1}{6} \left (5+\sqrt{13}\right )} x\right )|\frac{1}{6} \left (-13+5 \sqrt{13}\right )\right )}{30 \sqrt{x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(203*x*(5 + Sqrt[13] + 2*x^2))/(30*Sqrt[3 + 5*x^2 + x^4]) - (x*(5 + 12*x^2)*Sqrt[3 + 5*x^2 + x^4])/15 + (x*(3
+ x^2)*(3 + 5*x^2 + x^4)^(3/2))/3 - (203*Sqrt[(5 + Sqrt[13])/6]*Sqrt[(6 + (5 - Sqrt[13])*x^2)/(6 + (5 + Sqrt[1
3])*x^2)]*(6 + (5 + Sqrt[13])*x^2)*EllipticE[ArcTan[Sqrt[(5 + Sqrt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/(30*Sqrt
[3 + 5*x^2 + x^4]) + (5*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])/Sqrt[3 + 5*x^2 + x^4]

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 \left (2+3 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2} \, dx &=\frac{1}{3} x \left (3+x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac{1}{21} \int \left (63-84 x^2\right ) \sqrt{3+5 x^2+x^4} \, dx\\ &=-\frac{1}{15} x \left (5+12 x^2\right ) \sqrt{3+5 x^2+x^4}+\frac{1}{3} x \left (3+x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac{1}{315} \int \frac{3150+4263 x^2}{\sqrt{3+5 x^2+x^4}} \, dx\\ &=-\frac{1}{15} x \left (5+12 x^2\right ) \sqrt{3+5 x^2+x^4}+\frac{1}{3} x \left (3+x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+10 \int \frac{1}{\sqrt{3+5 x^2+x^4}} \, dx+\frac{203}{15} \int \frac{x^2}{\sqrt{3+5 x^2+x^4}} \, dx\\ &=\frac{203 x \left (5+\sqrt{13}+2 x^2\right )}{30 \sqrt{3+5 x^2+x^4}}-\frac{1}{15} x \left (5+12 x^2\right ) \sqrt{3+5 x^2+x^4}+\frac{1}{3} x \left (3+x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}-\frac{203 \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 )}{30 \sqrt{3+5 x^2+x^4}}+\frac{5 \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 )}{\sqrt{3+5 x^2+x^4}}\\ \end{align*}

Mathematica [F]  time = 0, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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Maple [A]  time = 0.012, size = 260, normalized size = 0.8 \begin{align*}{\frac{{x}^{7}}{3}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{8\,{x}^{5}}{3}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{26\,{x}^{3}}{5}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{8\,x}{3}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+60\,{\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{2436}{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}}}} \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)

[Out]

1/3*x^7*(x^4+5*x^2+3)^(1/2)+8/3*x^5*(x^4+5*x^2+3)^(1/2)+26/5*x^3*(x^4+5*x^2+3)^(1/2)+8/3*x*(x^4+5*x^2+3)^(1/2)
+60/(-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))-2436/5/(-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)
))

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

[Out]

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

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

[Out]

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

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

[Out]

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

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

[Out]

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