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

Optimal. Leaf size=196 \[ \frac{3 \left (2+\sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right ),\frac{1}{2}\right )}{20\ 5^{3/4} \sqrt{x^4+5}}+\frac{3 \sqrt{x^4+5} x}{25 \left (x^2+\sqrt{5}\right )}-\frac{3 \sqrt{x^4+5}}{25 x}+\frac{3 x^2+2}{10 \sqrt{x^4+5} x}-\frac{3 \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{5\ 5^{3/4} \sqrt{x^4+5}} \]

[Out]

(2 + 3*x^2)/(10*x*Sqrt[5 + x^4]) - (3*Sqrt[5 + x^4])/(25*x) + (3*x*Sqrt[5 + x^4])/(25*(Sqrt[5] + x^2)) - (3*(S
qrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticE[2*ArcTan[x/5^(1/4)], 1/2])/(5*5^(3/4)*Sqrt[5 + x^4])
+ (3*(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])/(20
*5^(3/4)*Sqrt[5 + x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.081502, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1278, 1282, 1198, 220, 1196} \[ \frac{3 \sqrt{x^4+5} x}{25 \left (x^2+\sqrt{5}\right )}-\frac{3 \sqrt{x^4+5}}{25 x}+\frac{3 x^2+2}{10 \sqrt{x^4+5} x}+\frac{3 \left (2+\sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{20\ 5^{3/4} \sqrt{x^4+5}}-\frac{3 \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{5\ 5^{3/4} \sqrt{x^4+5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2 + 3*x^2)/(10*x*Sqrt[5 + x^4]) - (3*Sqrt[5 + x^4])/(25*x) + (3*x*Sqrt[5 + x^4])/(25*(Sqrt[5] + x^2)) - (3*(S
qrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticE[2*ArcTan[x/5^(1/4)], 1/2])/(5*5^(3/4)*Sqrt[5 + x^4])
+ (3*(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])/(20
*5^(3/4)*Sqrt[5 + x^4])

Rule 1278

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

Rule 1282

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(f*x)^(m + 1)*(a
+ c*x^4)^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^2*(m + 1)), Int[(f*x)^(m + 2)*(a + c*x^4)^p*(a*e*(m + 1) -
c*d*(m + 4*p + 5)*x^2), x], x] /; FreeQ[{a, c, d, e, f, p}, x] && LtQ[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{2+3 x^2}{x^2 \left (5+x^4\right )^{3/2}} \, dx &=\frac{2+3 x^2}{10 x \sqrt{5+x^4}}-\frac{1}{10} \int \frac{-6-3 x^2}{x^2 \sqrt{5+x^4}} \, dx\\ &=\frac{2+3 x^2}{10 x \sqrt{5+x^4}}-\frac{3 \sqrt{5+x^4}}{25 x}+\frac{1}{50} \int \frac{15+6 x^2}{\sqrt{5+x^4}} \, dx\\ &=\frac{2+3 x^2}{10 x \sqrt{5+x^4}}-\frac{3 \sqrt{5+x^4}}{25 x}-\frac{3 \int \frac{1-\frac{x^2}{\sqrt{5}}}{\sqrt{5+x^4}} \, dx}{5 \sqrt{5}}+\frac{1}{50} \left (3 \left (5+2 \sqrt{5}\right )\right ) \int \frac{1}{\sqrt{5+x^4}} \, dx\\ &=\frac{2+3 x^2}{10 x \sqrt{5+x^4}}-\frac{3 \sqrt{5+x^4}}{25 x}+\frac{3 x \sqrt{5+x^4}}{25 \left (\sqrt{5}+x^2\right )}-\frac{3 \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\ 5^{3/4} \sqrt{5+x^4}}+\frac{3 \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 )}{20\ 5^{3/4} \sqrt{5+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0652715, size = 71, normalized size = 0.36 \[ \frac{3 x \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{x^4}{5}\right )}{10 \sqrt{5}}-\frac{2 \, _2F_1\left (-\frac{1}{4},\frac{3}{2};\frac{3}{4};-\frac{x^4}{5}\right )}{5 \sqrt{5} x}+\frac{3 x}{10 \sqrt{x^4+5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x)/(10*Sqrt[5 + x^4]) - (2*Hypergeometric2F1[-1/4, 3/2, 3/4, -x^4/5])/(5*Sqrt[5]*x) + (3*x*Hypergeometric2F
1[1/4, 1/2, 5/4, -x^4/5])/(10*Sqrt[5])

________________________________________________________________________________________

Maple [C]  time = 0.018, size = 180, normalized size = 0.9 \begin{align*}{\frac{3\,x}{10}{\frac{1}{\sqrt{{x}^{4}+5}}}}+{\frac{3\,\sqrt{5}}{250\,\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{2}{25\,x}\sqrt{{x}^{4}+5}}-{\frac{{x}^{3}}{25}{\frac{1}{\sqrt{{x}^{4}+5}}}}+{\frac{{\frac{3\,i}{125}}}{\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((3*x^2+2)/x^2/(x^4+5)^(3/2),x)

[Out]

3/10*x/(x^4+5)^(1/2)+3/250*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)-2/25*(x^4+5)^(1/2)/x-1/25*x^3/(x^4+5)^(1/2)+3/125*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))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{4} + 5}{\left (3 \, x^{2} + 2\right )}}{x^{10} + 10 \, x^{6} + 25 \, x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + 5)*(3*x^2 + 2)/(x^10 + 10*x^6 + 25*x^2), x)

________________________________________________________________________________________

Sympy [C]  time = 6.07129, size = 75, normalized size = 0.38 \begin{align*} \frac{3 \sqrt{5} x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{3}{2} \\ \frac{5}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{100 \Gamma \left (\frac{5}{4}\right )} + \frac{\sqrt{5} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{3}{2} \\ \frac{3}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{50 x \Gamma \left (\frac{3}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(5)*x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), x**4*exp_polar(I*pi)/5)/(100*gamma(5/4)) + sqrt(5)*gamma(-1/4
)*hyper((-1/4, 3/2), (3/4,), x**4*exp_polar(I*pi)/5)/(50*x*gamma(3/4))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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