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

Optimal. Leaf size=180 \[ -\frac {3 \sqrt {x^4+5} x}{10 \left (x^2+\sqrt {5}\right )}+\frac {\left (3 x^2+2\right ) x}{10 \sqrt {x^4+5}}+\frac {\left (2-3 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{20 \sqrt [4]{5} \sqrt {x^4+5}}+\frac {3 \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{2\ 5^{3/4} \sqrt {x^4+5}} \]

[Out]

1/10*x*(3*x^2+2)/(x^4+5)^(1/2)-3/10*x*(x^4+5)^(1/2)/(x^2+5^(1/2))+3/10*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)+1/100*(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))*(2-3*5^(1/2))*(x^2+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 = 0.06, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1179, 1198, 220, 1196} \[ -\frac {3 \sqrt {x^4+5} x}{10 \left (x^2+\sqrt {5}\right )}+\frac {\left (3 x^2+2\right ) x}{10 \sqrt {x^4+5}}+\frac {\left (2-3 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{20 \sqrt [4]{5} \sqrt {x^4+5}}+\frac {3 \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{2\ 5^{3/4} \sqrt {x^4+5}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x^2)/(5 + x^4)^(3/2),x]

[Out]

(x*(2 + 3*x^2))/(10*Sqrt[5 + x^4]) - (3*x*Sqrt[5 + x^4])/(10*(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])/(2*5^(3/4)*Sqrt[5 + x^4]) + ((2 - 3*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^(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 1179

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)*(a + c*x^4)^(p + 1))/(
4*a*(p + 1)), x] + Dist[1/(4*a*(p + 1)), Int[Simp[d*(4*p + 5) + e*(4*p + 7)*x^2, x]*(a + c*x^4)^(p + 1), x], x
] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

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}{\left (5+x^4\right )^{3/2}} \, dx &=\frac {x \left (2+3 x^2\right )}{10 \sqrt {5+x^4}}-\frac {1}{10} \int \frac {-2+3 x^2}{\sqrt {5+x^4}} \, dx\\ &=\frac {x \left (2+3 x^2\right )}{10 \sqrt {5+x^4}}+\frac {3 \int \frac {1-\frac {x^2}{\sqrt {5}}}{\sqrt {5+x^4}} \, dx}{2 \sqrt {5}}-\frac {1}{10} \left (-2+3 \sqrt {5}\right ) \int \frac {1}{\sqrt {5+x^4}} \, dx\\ &=\frac {x \left (2+3 x^2\right )}{10 \sqrt {5+x^4}}-\frac {3 x \sqrt {5+x^4}}{10 \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 )}{2\ 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 )}{20 \sqrt [4]{5} \sqrt {5+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 66, normalized size = 0.37 \[ \frac {1}{25} x \left (\sqrt {5} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {x^4}{5}\right )+\sqrt {5} x^2 \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {x^4}{5}\right )+\frac {5}{\sqrt {x^4+5}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(5/Sqrt[5 + x^4] + Sqrt[5]*Hypergeometric2F1[1/4, 1/2, 5/4, -1/5*x^4] + Sqrt[5]*x^2*Hypergeometric2F1[3/4,
3/2, 7/4, -1/5*x^4]))/25

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 5} {\left (3 \, x^{2} + 2\right )}}{x^{8} + 10 \, x^{4} + 25}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + 5)*(3*x^2 + 2)/(x^8 + 10*x^4 + 25), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.01, size = 168, normalized size = 0.93 \[ \frac {3 x^{3}}{10 \sqrt {x^{4}+5}}+\frac {x}{5 \sqrt {x^{4}+5}}+\frac {\sqrt {5}\, \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )}{125 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {3 i \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \left (-\EllipticE \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )+\EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )\right )}{50 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10/(x^4+5)^(1/2)*x^3-3/50*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))+1/5/(x^4+5)
^(1/2)*x+1/125*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 \[ \int \frac {3 \, x^{2} + 2}{{\left (x^{4} + 5\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2 + 2)/(x^4 + 5)^(3/2),x)

[Out]

int((3*x^2 + 2)/(x^4 + 5)^(3/2), x)

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sympy [C]  time = 5.07, size = 73, normalized size = 0.41 \[ \frac {3 \sqrt {5} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{100 \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {5} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{50 \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(5)*x**3*gamma(3/4)*hyper((3/4, 3/2), (7/4,), x**4*exp_polar(I*pi)/5)/(100*gamma(7/4)) + sqrt(5)*x*gamma
(1/4)*hyper((1/4, 3/2), (5/4,), x**4*exp_polar(I*pi)/5)/(50*gamma(5/4))

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