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

Optimal. Leaf size=177 \[ -\frac{\left (2-3 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right ),\frac{1}{2}\right )}{4\ 5^{3/4} \sqrt{x^4+5}}-\frac{\sqrt{x^4+5} x}{5 \left (x^2+\sqrt{5}\right )}-\frac{\left (15-2 x^2\right ) x}{10 \sqrt{x^4+5}}+\frac{\left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{5^{3/4} \sqrt{x^4+5}} \]

[Out]

-(x*(15 - 2*x^2))/(10*Sqrt[5 + x^4]) - (x*Sqrt[5 + x^4])/(5*(Sqrt[5] + x^2)) + ((Sqrt[5] + x^2)*Sqrt[(5 + x^4)
/(Sqrt[5] + x^2)^2]*EllipticE[2*ArcTan[x/5^(1/4)], 1/2])/(5^(3/4)*Sqrt[5 + x^4]) - ((2 - 3*Sqrt[5])*(Sqrt[5] +
x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x/5^(1/4)], 1/2])/(4*5^(3/4)*Sqrt[5 + x^4])

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Rubi [A]  time = 0.0691724, antiderivative size = 177, 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 = {1276, 1198, 220, 1196} \[ -\frac{\sqrt{x^4+5} x}{5 \left (x^2+\sqrt{5}\right )}-\frac{\left (15-2 x^2\right ) x}{10 \sqrt{x^4+5}}-\frac{\left (2-3 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{4\ 5^{3/4} \sqrt{x^4+5}}+\frac{\left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{5^{3/4} \sqrt{x^4+5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*(15 - 2*x^2))/(10*Sqrt[5 + x^4]) - (x*Sqrt[5 + x^4])/(5*(Sqrt[5] + x^2)) + ((Sqrt[5] + x^2)*Sqrt[(5 + x^4)
/(Sqrt[5] + x^2)^2]*EllipticE[2*ArcTan[x/5^(1/4)], 1/2])/(5^(3/4)*Sqrt[5 + x^4]) - ((2 - 3*Sqrt[5])*(Sqrt[5] +
x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x/5^(1/4)], 1/2])/(4*5^(3/4)*Sqrt[5 + x^4])

Rule 1276

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(f*(f*x)^(m - 1)*(
a + c*x^4)^(p + 1)*(a*e - c*d*x^2))/(4*a*c*(p + 1)), x] - Dist[f^2/(4*a*c*(p + 1)), Int[(f*x)^(m - 2)*(a + c*x
^4)^(p + 1)*(a*e*(m - 1) - c*d*(4*p + 4 + m + 1)*x^2), x], x] /; FreeQ[{a, c, d, e, f}, x] && LtQ[p, -1] && Gt
Q[m, 1] && 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 )}{\left (5+x^4\right )^{3/2}} \, dx &=-\frac{x \left (15-2 x^2\right )}{10 \sqrt{5+x^4}}+\frac{1}{10} \int \frac{15-2 x^2}{\sqrt{5+x^4}} \, dx\\ &=-\frac{x \left (15-2 x^2\right )}{10 \sqrt{5+x^4}}+\frac{\int \frac{1-\frac{x^2}{\sqrt{5}}}{\sqrt{5+x^4}} \, dx}{\sqrt{5}}+\frac{1}{10} \left (15-2 \sqrt{5}\right ) \int \frac{1}{\sqrt{5+x^4}} \, dx\\ &=-\frac{x \left (15-2 x^2\right )}{10 \sqrt{5+x^4}}-\frac{x \sqrt{5+x^4}}{5 \left (\sqrt{5}+x^2\right )}+\frac{\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 )}{5^{3/4} \sqrt{5+x^4}}-\frac{\left (2-3 \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 )}{4\ 5^{3/4} \sqrt{5+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0321569, size = 68, normalized size = 0.38 \[ \frac{1}{150} x \left (4 \sqrt{5} x^2 \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\frac{x^4}{5}\right )+45 \sqrt{5} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{x^4}{5}\right )-\frac{225}{\sqrt{x^4+5}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-225/Sqrt[5 + x^4] + 45*Sqrt[5]*Hypergeometric2F1[1/4, 1/2, 5/4, -x^4/5] + 4*Sqrt[5]*x^2*Hypergeometric2F1
[3/4, 3/2, 7/4, -x^4/5]))/150

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Maple [C]  time = 0.013, size = 168, normalized size = 1. \begin{align*} -{\frac{3\,x}{2}{\frac{1}{\sqrt{{x}^{4}+5}}}}+{\frac{3\,\sqrt{5}}{50\,\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{{x}^{3}}{5}{\frac{1}{\sqrt{{x}^{4}+5}}}}-{\frac{{\frac{i}{25}}}{\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)^(3/2),x)

[Out]

-3/2*x/(x^4+5)^(1/2)+3/50*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)+1/5*x^3/(x^4+5)^(1/2)-1/25*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)-Elli
pticE(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}}{{\left (x^{4} + 5\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (3 \, x^{4} + 2 \, x^{2}\right )} \sqrt{x^{4} + 5}}{x^{8} + 10 \, x^{4} + 25}, 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)^(3/2),x, algorithm="fricas")

[Out]

integral((3*x^4 + 2*x^2)*sqrt(x^4 + 5)/(x^8 + 10*x^4 + 25), x)

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Sympy [C]  time = 4.28565, size = 75, normalized size = 0.42 \begin{align*} \frac{3 \sqrt{5} x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{3}{2} \\ \frac{9}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{100 \Gamma \left (\frac{9}{4}\right )} + \frac{\sqrt{5} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{3}{2} \\ \frac{7}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{50 \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)**(3/2),x)

[Out]

3*sqrt(5)*x**5*gamma(5/4)*hyper((5/4, 3/2), (9/4,), x**4*exp_polar(I*pi)/5)/(100*gamma(9/4)) + sqrt(5)*x**3*ga
mma(3/4)*hyper((3/4, 3/2), (7/4,), x**4*exp_polar(I*pi)/5)/(50*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}}{{\left (x^{4} + 5\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="giac")

[Out]

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