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

Optimal. Leaf size=166 \[ \frac{\sqrt [4]{5} \left (2-\sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right ),\frac{1}{2}\right )}{2 \sqrt{x^4+5}}+\frac{2 \sqrt{x^4+5} x}{x^2+\sqrt{5}}+\sqrt{x^4+5} x-\frac{2 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}} \]

[Out]

x*Sqrt[5 + x^4] + (2*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) - (2*5^(1/4)*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x
^2)^2]*EllipticE[2*ArcTan[x/5^(1/4)], 1/2])/Sqrt[5 + x^4] + (5^(1/4)*(2 - Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x
^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x/5^(1/4)], 1/2])/(2*Sqrt[5 + x^4])

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Rubi [A]  time = 0.0663092, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1280, 1198, 220, 1196} \[ \frac{2 \sqrt{x^4+5} x}{x^2+\sqrt{5}}+\sqrt{x^4+5} x+\frac{\sqrt [4]{5} \left (2-\sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{2 \sqrt{x^4+5}}-\frac{2 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*Sqrt[5 + x^4] + (2*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) - (2*5^(1/4)*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x
^2)^2]*EllipticE[2*ArcTan[x/5^(1/4)], 1/2])/Sqrt[5 + x^4] + (5^(1/4)*(2 - Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x
^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x/5^(1/4)], 1/2])/(2*Sqrt[5 + x^4])

Rule 1280

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

Rule 1198

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{x^2 \left (2+3 x^2\right )}{\sqrt{5+x^4}} \, dx &=x \sqrt{5+x^4}-\frac{1}{3} \int \frac{15-6 x^2}{\sqrt{5+x^4}} \, dx\\ &=x \sqrt{5+x^4}-\left (2 \sqrt{5}\right ) \int \frac{1-\frac{x^2}{\sqrt{5}}}{\sqrt{5+x^4}} \, dx-\left (5-2 \sqrt{5}\right ) \int \frac{1}{\sqrt{5+x^4}} \, dx\\ &=x \sqrt{5+x^4}+\frac{2 x \sqrt{5+x^4}}{\sqrt{5}+x^2}-\frac{2 \sqrt [4]{5} \left (\sqrt{5}+x^2\right ) \sqrt{\frac{5+x^4}{\left (\sqrt{5}+x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{5+x^4}}+\frac{\sqrt [4]{5} \left (2-\sqrt{5}\right ) \left (\sqrt{5}+x^2\right ) \sqrt{\frac{5+x^4}{\left (\sqrt{5}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{2 \sqrt{5+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0245802, size = 66, normalized size = 0.4 \[ \frac{2 x^3 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{x^4}{5}\right )}{3 \sqrt{5}}-\sqrt{5} x \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{x^4}{5}\right )+\sqrt{x^4+5} x \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*Sqrt[5 + x^4] - Sqrt[5]*x*Hypergeometric2F1[1/4, 1/2, 5/4, -x^4/5] + (2*x^3*Hypergeometric2F1[1/2, 3/4, 7/4,
-x^4/5])/(3*Sqrt[5])

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Maple [C]  time = 0.013, size = 155, normalized size = 0.9 \begin{align*} x\sqrt{{x}^{4}+5}-{\frac{\sqrt{5}}{5\,\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}}{\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ){\frac{1}{\sqrt{{x}^{4}+5}}}}+{\frac{{\frac{2\,i}{5}}}{\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}} \left ({\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) -{\it EllipticE} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) \right ){\frac{1}{\sqrt{{x}^{4}+5}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(x^4+5)^(1/2)-1/5*5^(1/2)/(I*5^(1/2))^(1/2)*(25-5*I*5^(1/2)*x^2)^(1/2)*(25+5*I*5^(1/2)*x^2)^(1/2)/(x^4+5)^(1
/2)*EllipticF(1/5*x*5^(1/2)*(I*5^(1/2))^(1/2),I)+2/5*I/(I*5^(1/2))^(1/2)*(25-5*I*5^(1/2)*x^2)^(1/2)*(25+5*I*5^
(1/2)*x^2)^(1/2)/(x^4+5)^(1/2)*(EllipticF(1/5*x*5^(1/2)*(I*5^(1/2))^(1/2),I)-EllipticE(1/5*x*5^(1/2)*(I*5^(1/2
))^(1/2),I))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 1.91011, size = 75, normalized size = 0.45 \begin{align*} \frac{3 \sqrt{5} x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{20 \Gamma \left (\frac{9}{4}\right )} + \frac{\sqrt{5} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{10 \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(5)*x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), x**4*exp_polar(I*pi)/5)/(20*gamma(9/4)) + sqrt(5)*x**3*gam
ma(3/4)*hyper((1/2, 3/4), (7/4,), x**4*exp_polar(I*pi)/5)/(10*gamma(7/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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