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

Optimal. Leaf size=155 \[ \frac {3 \sqrt {x^4+5} x}{x^2+\sqrt {5}}+\frac {\left (2+3 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}-\frac {3 \sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1198, 220, 1196} \[ \frac {3 \sqrt {x^4+5} x}{x^2+\sqrt {5}}+\frac {\left (2+3 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}-\frac {3 \sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x^2)/Sqrt[5 + x^4],x]

[Out]

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

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 48, normalized size = 0.31 \[ \frac {x \left (2 \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {x^4}{5}\right )+x^2 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {x^4}{5}\right )\right )}{\sqrt {5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(2*Hypergeometric2F1[1/4, 1/2, 5/4, -1/5*x^4] + x^2*Hypergeometric2F1[1/2, 3/4, 7/4, -1/5*x^4]))/Sqrt[5]

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {3 \, x^{2} + 2}{\sqrt {x^{4} + 5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.01, size = 146, normalized size = 0.94 \[ \frac {2 \sqrt {5}\, \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {3 i \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \left (-\EllipticE \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )+\EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/5*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))+2/25*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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {3\,x^2+2}{\sqrt {x^4+5}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2 + 2)/(x^4 + 5)^(1/2),x)

[Out]

int((3*x^2 + 2)/(x^4 + 5)^(1/2), x)

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sympy [C]  time = 1.71, size = 73, normalized size = 0.47 \[ \frac {3 \sqrt {5} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{20 \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {5} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{10 \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(5)*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), x**4*exp_polar(I*pi)/5)/(20*gamma(7/4)) + sqrt(5)*x*gamma(
1/4)*hyper((1/4, 1/2), (5/4,), x**4*exp_polar(I*pi)/5)/(10*gamma(5/4))

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