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

Optimal. Leaf size=189 \[ -\frac{\left (2-9 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right ),\frac{1}{2}\right )}{30 \sqrt [4]{5} \sqrt{x^4+5}}+\frac{3 \sqrt{x^4+5} x}{5 \left (x^2+\sqrt{5}\right )}-\frac{3 \sqrt{x^4+5}}{5 x}-\frac{2 \sqrt{x^4+5}}{15 x^3}-\frac{3 \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{5^{3/4} \sqrt{x^4+5}} \]

[Out]

(-2*Sqrt[5 + x^4])/(15*x^3) - (3*Sqrt[5 + x^4])/(5*x) + (3*x*Sqrt[5 + x^4])/(5*(Sqrt[5] + x^2)) - (3*(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 -
9*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x/5^(1/4)], 1/2])/(30*5^(1/4)*
Sqrt[5 + x^4])

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Rubi [A]  time = 0.0860867, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1282, 1198, 220, 1196} \[ \frac{3 \sqrt{x^4+5} x}{5 \left (x^2+\sqrt{5}\right )}-\frac{3 \sqrt{x^4+5}}{5 x}-\frac{2 \sqrt{x^4+5}}{15 x^3}-\frac{\left (2-9 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{30 \sqrt [4]{5} \sqrt{x^4+5}}-\frac{3 \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{5^{3/4} \sqrt{x^4+5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[5 + x^4])/(15*x^3) - (3*Sqrt[5 + x^4])/(5*x) + (3*x*Sqrt[5 + x^4])/(5*(Sqrt[5] + x^2)) - (3*(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 -
9*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x/5^(1/4)], 1/2])/(30*5^(1/4)*
Sqrt[5 + x^4])

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^4 \sqrt{5+x^4}} \, dx &=-\frac{2 \sqrt{5+x^4}}{15 x^3}-\frac{1}{15} \int \frac{-45+2 x^2}{x^2 \sqrt{5+x^4}} \, dx\\ &=-\frac{2 \sqrt{5+x^4}}{15 x^3}-\frac{3 \sqrt{5+x^4}}{5 x}+\frac{1}{75} \int \frac{-10+45 x^2}{\sqrt{5+x^4}} \, dx\\ &=-\frac{2 \sqrt{5+x^4}}{15 x^3}-\frac{3 \sqrt{5+x^4}}{5 x}-\frac{3 \int \frac{1-\frac{x^2}{\sqrt{5}}}{\sqrt{5+x^4}} \, dx}{\sqrt{5}}+\frac{1}{15} \left (-2+9 \sqrt{5}\right ) \int \frac{1}{\sqrt{5+x^4}} \, dx\\ &=-\frac{2 \sqrt{5+x^4}}{15 x^3}-\frac{3 \sqrt{5+x^4}}{5 x}+\frac{3 x \sqrt{5+x^4}}{5 \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^{3/4} \sqrt{5+x^4}}-\frac{\left (2-9 \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 )}{30 \sqrt [4]{5} \sqrt{5+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0255959, size = 54, normalized size = 0.29 \[ -\frac{9 x^2 \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{x^4}{5}\right )+2 \, _2F_1\left (-\frac{3}{4},\frac{1}{2};\frac{1}{4};-\frac{x^4}{5}\right )}{3 \sqrt{5} x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*Hypergeometric2F1[-3/4, 1/2, 1/4, -x^4/5] + 9*x^2*Hypergeometric2F1[-1/4, 1/2, 3/4, -x^4/5])/(3*Sqrt[5]*x^
3)

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Maple [C]  time = 0.017, size = 170, normalized size = 0.9 \begin{align*} -{\frac{3}{5\,x}\sqrt{{x}^{4}+5}}+{\frac{{\frac{3\,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}}}}-{\frac{2}{15\,{x}^{3}}\sqrt{{x}^{4}+5}}-{\frac{2\,\sqrt{5}}{375\,\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}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/5*(x^4+5)^(1/2)/x+3/25*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)-EllipticE(1/5*x*5^(1/2)*(I*5^(1/2))^(1/2),I))-2/15*(x^4+5)^(
1/2)/x^3-2/375*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)*E
llipticF(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{3 \, x^{2} + 2}{\sqrt{x^{4} + 5} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{8} + 5 \, x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 1.89739, size = 80, normalized size = 0.42 \begin{align*} \frac{3 \sqrt{5} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{1}{2} \\ \frac{3}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{20 x \Gamma \left (\frac{3}{4}\right )} + \frac{\sqrt{5} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{10 x^{3} \Gamma \left (\frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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