3.1.42 \(\int \frac {2+3 x^2}{x^2 \sqrt {5+x^4}} \, dx\) [42]

Optimal. Leaf size=173 \[ -\frac {2 \sqrt {5+x^4}}{5 x}+\frac {2 x \sqrt {5+x^4}}{5 \left (\sqrt {5}+x^2\right )}-\frac {2 \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{5^{3/4} \sqrt {5+x^4}}+\frac {\left (2+3 \sqrt {5}\right ) \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{2\ 5^{3/4} \sqrt {5+x^4}} \]

[Out]

-2/5*(x^4+5)^(1/2)/x+2/5*x*(x^4+5)^(1/2)/(x^2+5^(1/2))-2/5*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)+1/10*(cos(2*arctan(1/5*x*5^(3/4)))^2)^(1/2)/cos(2*arctan(1/5*x*5^(3/4)))*EllipticF(
sin(2*arctan(1/5*x*5^(3/4))),1/2*2^(1/2))*(x^2+5^(1/2))*(2+3*5^(1/2))*((x^4+5)/(x^2+5^(1/2))^2)^(1/2)*5^(1/4)/
(x^4+5)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 173, 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 = {1296, 1212, 226, 1210} \begin {gather*} \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 \text {ArcTan}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{2\ 5^{3/4} \sqrt {x^4+5}}-\frac {2 \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \text {ArcTan}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{5^{3/4} \sqrt {x^4+5}}-\frac {2 \sqrt {x^4+5}}{5 x}+\frac {2 \sqrt {x^4+5} x}{5 \left (x^2+\sqrt {5}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[5 + x^4])/(5*x) + (2*x*Sqrt[5 + x^4])/(5*(Sqrt[5] + x^2)) - (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)*Sq
rt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x/5^(1/4)], 1/2])/(2*5^(3/4)*Sqrt[5 + x^4])

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1212

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 1296

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])

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 10.12, size = 81, normalized size = 0.47 \begin {gather*} \frac {1}{5} \left (-\frac {2 \sqrt {5+x^4}}{x}-2 (-1)^{3/4} \sqrt [4]{5} E\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac {1}{5}} x\right )\right |-1\right )-\sqrt [4]{-5} \left (-2 i+3 \sqrt {5}\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac {1}{5}} x\right )\right |-1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*Sqrt[5 + x^4])/x - 2*(-1)^(3/4)*5^(1/4)*EllipticE[I*ArcSinh[(-1/5)^(1/4)*x], -1] - (-5)^(1/4)*(-2*I + 3*S
qrt[5])*EllipticF[I*ArcSinh[(-1/5)^(1/4)*x], -1])/5

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Maple [C] Result contains complex when optimal does not.
time = 0.14, size = 158, normalized size = 0.91

method	result	size
meijerg	\(-\frac {2 \sqrt {5}\, \hypergeom \left (\left [-\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{4}\right ], -\frac {x^{4}}{5}\right )}{5 x}+\frac {3 \sqrt {5}\, x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {5}{4}\right ], -\frac {x^{4}}{5}\right )}{5}\)	\(38\)
default	\(\frac {3 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {2 \sqrt {x^{4}+5}}{5 x}+\frac {2 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(158\)
risch	\(\frac {3 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {2 \sqrt {x^{4}+5}}{5 x}+\frac {2 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(158\)
elliptic	\(\frac {3 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {2 \sqrt {x^{4}+5}}{5 x}+\frac {2 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(158\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/25*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*(x^4+5)^(1/2)/x+2/25*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))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [C] Result contains complex when optimal does not.
time = 0.85, size = 75, normalized size = 0.43 \begin {gather*} \frac {3 \sqrt {5} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{20 \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 {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{10 x \Gamma \left (\frac {3}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.50, size = 48, normalized size = 0.28 \begin {gather*} \frac {3\,\sqrt {5}\,x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {5}{4};\ -\frac {x^4}{5}\right )}{5}-\frac {2\,\sqrt {\frac {5}{x^4}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{4};\ \frac {7}{4};\ -\frac {5}{x^4}\right )}{3\,x\,\sqrt {x^4+5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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