3.1.53 \(\int \frac {2+3 x^2}{x^2 (5+x^4)^{3/2}} \, dx\) [53]

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

[Out]

1/10*(3*x^2+2)/x/(x^4+5)^(1/2)-3/25*(x^4+5)^(1/2)/x+3/25*x*(x^4+5)^(1/2)/(x^2+5^(1/2))-3/25*5^(1/4)*(cos(2*arc
tan(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)+3/100*5^(1/4)*(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))*(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 = 0.06, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1292, 1296, 1212, 226, 1210} \begin {gather*} \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 \text {ArcTan}\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 \text {ArcTan}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{5\ 5^{3/4} \sqrt {x^4+5}}-\frac {3 \sqrt {x^4+5}}{25 x}+\frac {3 \sqrt {x^4+5} x}{25 \left (x^2+\sqrt {5}\right )}+\frac {3 x^2+2}{10 \sqrt {x^4+5} x} \end {gather*}

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 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 1292

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)*Si
mp[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 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 \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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.09, size = 108, normalized size = 0.55 \begin {gather*} -\frac {20-15 x^2+6 x^4+6 (-1)^{3/4} \sqrt [4]{5} x \sqrt {5+x^4} E\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac {1}{5}} x\right )\right |-1\right )+3 \sqrt [4]{-5} \left (-2 i+\sqrt {5}\right ) x \sqrt {5+x^4} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac {1}{5}} x\right )\right |-1\right )}{50 x \sqrt {5+x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/50*(20 - 15*x^2 + 6*x^4 + 6*(-1)^(3/4)*5^(1/4)*x*Sqrt[5 + x^4]*EllipticE[I*ArcSinh[(-1/5)^(1/4)*x], -1] + 3
*(-5)^(1/4)*(-2*I + Sqrt[5])*x*Sqrt[5 + x^4]*EllipticF[I*ArcSinh[(-1/5)^(1/4)*x], -1])/(x*Sqrt[5 + x^4])

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Maple [C] Result contains complex when optimal does not.
time = 0.18, size = 180, normalized size = 0.92

method	result	size
meijerg	\(-\frac {2 \sqrt {5}\, \hypergeom \left (\left [-\frac {1}{4}, \frac {3}{2}\right ], \left [\frac {3}{4}\right ], -\frac {x^{4}}{5}\right )}{25 x}+\frac {3 \sqrt {5}\, x \hypergeom \left (\left [\frac {1}{4}, \frac {3}{2}\right ], \left [\frac {5}{4}\right ], -\frac {x^{4}}{5}\right )}{25}\)	\(38\)
risch	\(-\frac {6 x^{4}-15 x^{2}+20}{50 x \sqrt {x^{4}+5}}+\frac {3 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{125 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {3 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{250 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(170\)
elliptic	\(-\frac {2 \left (\frac {1}{50} x^{3}-\frac {3}{20} x \right )}{\sqrt {x^{4}+5}}-\frac {2 \sqrt {x^{4}+5}}{25 x}+\frac {3 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{125 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {3 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{250 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(176\)
default	\(\frac {3 x}{10 \sqrt {x^{4}+5}}+\frac {3 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{250 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {x^{3}}{25 \sqrt {x^{4}+5}}-\frac {2 \sqrt {x^{4}+5}}{25 x}+\frac {3 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{125 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(180\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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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)^(3/2),x, algorithm="maxima")

[Out]

integrate((3*x^2 + 2)/((x^4 + 5)^(3/2)*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)^(3/2),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 = 3.01, size = 75, normalized size = 0.38 \begin {gather*} \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 {gather*}

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))

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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)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.46, size = 48, normalized size = 0.24 \begin {gather*} \frac {3\,\sqrt {5}\,x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {3}{2};\ \frac {5}{4};\ -\frac {x^4}{5}\right )}{25}-\frac {2\,{\left (\frac {5}{x^4}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {7}{4};\ \frac {11}{4};\ -\frac {5}{x^4}\right )}{7\,x\,{\left (x^4+5\right )}^{3/2}} \end {gather*}

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*5^(1/2)*x*hypergeom([1/4, 3/2], 5/4, -x^4/5))/25 - (2*(5/x^4 + 1)^(3/2)*hypergeom([3/2, 7/4], 11/4, -5/x^4)
)/(7*x*(x^4 + 5)^(3/2))

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