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

Optimal. Leaf size=171 \[ \frac {4 \sqrt {x^4+5} x}{x^2+\sqrt {5}}-\frac {\left (2-x^2\right ) \sqrt {x^4+5}}{x}+\frac {\sqrt [4]{5} \left (2+\sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}}-\frac {4 \sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}} \]

[Out]

-(-x^2+2)*(x^4+5)^(1/2)/x+4*x*(x^4+5)^(1/2)/(x^2+5^(1/2))-4*5^(1/4)*(cos(2*arctan(1/5*x*5^(3/4)))^2)^(1/2)/cos
(2*arctan(1/5*x*5^(3/4)))*EllipticE(sin(2*arctan(1/5*x*5^(3/4))),1/2*2^(1/2))*(x^2+5^(1/2))*((x^4+5)/(x^2+5^(1
/2))^2)^(1/2)/(x^4+5)^(1/2)+5^(1/4)*(cos(2*arctan(1/5*x*5^(3/4)))^2)^(1/2)/cos(2*arctan(1/5*x*5^(3/4)))*Ellipt
icF(sin(2*arctan(1/5*x*5^(3/4))),1/2*2^(1/2))*(2+5^(1/2))*(x^2+5^(1/2))*((x^4+5)/(x^2+5^(1/2))^2)^(1/2)/(x^4+5
)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1272, 1198, 220, 1196} \[ \frac {4 \sqrt {x^4+5} x}{x^2+\sqrt {5}}-\frac {\left (2-x^2\right ) \sqrt {x^4+5}}{x}+\frac {\sqrt [4]{5} \left (2+\sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}}-\frac {4 \sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2 - x^2)*Sqrt[5 + x^4])/x) + (4*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) - (4*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])/Sqrt[5 + x^4]

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]

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 1272

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*(d*(m + 4*p + 3) + e*(m + 1)*x^2))/(f*(m + 1)*(m + 4*p + 3)), x] + Dist[(4*p)/(f^2*(m + 1)*(m + 4*p
+ 3)), Int[(f*x)^(m + 2)*(a + c*x^4)^(p - 1)*(a*e*(m + 1) - c*d*(m + 4*p + 3)*x^2), x], x] /; FreeQ[{a, c, d,
e, f}, x] && GtQ[p, 0] && LtQ[m, -1] && m + 4*p + 3 != 0 && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 53, normalized size = 0.31 \[ 3 \sqrt {5} x \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {x^4}{5}\right )-\frac {2 \sqrt {5} \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};-\frac {x^4}{5}\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 5} {\left (3 \, x^{2} + 2\right )}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{4} + 5} {\left (3 \, x^{2} + 2\right )}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 167, normalized size = 0.98 \[ \sqrt {x^{4}+5}\, x +\frac {2 \sqrt {5}\, \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {2 \sqrt {x^{4}+5}}{x}+\frac {4 i \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \left (-\EllipticE \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )+\EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{4} + 5} {\left (3 \, x^{2} + 2\right )}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.41, size = 61, normalized size = 0.36 \[ \frac {3\,x\,\sqrt {x^4+5}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {1}{4};\ \frac {5}{4};\ -\frac {x^4}{5}\right )}{\sqrt {\frac {x^4}{5}+1}}+\frac {2\,\sqrt {x^4+5}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},-\frac {1}{4};\ \frac {3}{4};\ -\frac {5}{x^4}\right )}{x\,\sqrt {\frac {5}{x^4}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 2.30, size = 78, normalized size = 0.46 \[ \frac {3 \sqrt {5} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt {5} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{2 x \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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