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

Optimal. Leaf size=214 \[ -\frac {9 \sqrt {x^4+5}}{50 x}-\frac {\sqrt {x^4+5}}{15 x^3}+\frac {9 \sqrt {x^4+5} x}{50 \left (x^2+\sqrt {5}\right )}+\frac {\left (27-2 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{60\ 5^{3/4} \sqrt {x^4+5}}-\frac {9 \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{10\ 5^{3/4} \sqrt {x^4+5}}+\frac {3 x^2+2}{10 \sqrt {x^4+5} x^3} \]

[Out]

1/10*(3*x^2+2)/x^3/(x^4+5)^(1/2)-1/15*(x^4+5)^(1/2)/x^3-9/50*(x^4+5)^(1/2)/x+9/50*x*(x^4+5)^(1/2)/(x^2+5^(1/2)
)-9/50*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)+1/300*(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))*(27-2*5^(
1/2))*(x^2+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 = 0.11, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1278, 1282, 1198, 220, 1196} \[ \frac {9 \sqrt {x^4+5} x}{50 \left (x^2+\sqrt {5}\right )}-\frac {9 \sqrt {x^4+5}}{50 x}-\frac {\sqrt {x^4+5}}{15 x^3}+\frac {3 x^2+2}{10 \sqrt {x^4+5} x^3}+\frac {\left (27-2 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{60\ 5^{3/4} \sqrt {x^4+5}}-\frac {9 \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{10\ 5^{3/4} \sqrt {x^4+5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2 + 3*x^2)/(10*x^3*Sqrt[5 + x^4]) - Sqrt[5 + x^4]/(15*x^3) - (9*Sqrt[5 + x^4])/(50*x) + (9*x*Sqrt[5 + x^4])/(
50*(Sqrt[5] + x^2)) - (9*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticE[2*ArcTan[x/5^(1/4)], 1/2]
)/(10*5^(3/4)*Sqrt[5 + x^4]) + ((27 - 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])/(60*5^(3/4)*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]

Rule 1278

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)*Simp
[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 1282

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^4 \left (5+x^4\right )^{3/2}} \, dx &=\frac {2+3 x^2}{10 x^3 \sqrt {5+x^4}}-\frac {1}{10} \int \frac {-10-9 x^2}{x^4 \sqrt {5+x^4}} \, dx\\ &=\frac {2+3 x^2}{10 x^3 \sqrt {5+x^4}}-\frac {\sqrt {5+x^4}}{15 x^3}+\frac {1}{150} \int \frac {135-10 x^2}{x^2 \sqrt {5+x^4}} \, dx\\ &=\frac {2+3 x^2}{10 x^3 \sqrt {5+x^4}}-\frac {\sqrt {5+x^4}}{15 x^3}-\frac {9 \sqrt {5+x^4}}{50 x}-\frac {1}{750} \int \frac {50-135 x^2}{\sqrt {5+x^4}} \, dx\\ &=\frac {2+3 x^2}{10 x^3 \sqrt {5+x^4}}-\frac {\sqrt {5+x^4}}{15 x^3}-\frac {9 \sqrt {5+x^4}}{50 x}-\frac {9 \int \frac {1-\frac {x^2}{\sqrt {5}}}{\sqrt {5+x^4}} \, dx}{10 \sqrt {5}}-\frac {1}{150} \left (10-27 \sqrt {5}\right ) \int \frac {1}{\sqrt {5+x^4}} \, dx\\ &=\frac {2+3 x^2}{10 x^3 \sqrt {5+x^4}}-\frac {\sqrt {5+x^4}}{15 x^3}-\frac {9 \sqrt {5+x^4}}{50 x}+\frac {9 x \sqrt {5+x^4}}{50 \left (\sqrt {5}+x^2\right )}-\frac {9 \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 )}{10\ 5^{3/4} \sqrt {5+x^4}}+\frac {\left (27-2 \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 )}{60\ 5^{3/4} \sqrt {5+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 54, normalized size = 0.25 \[ -\frac {2 \, _2F_1\left (-\frac {3}{4},\frac {3}{2};\frac {1}{4};-\frac {x^4}{5}\right )+9 x^2 \, _2F_1\left (-\frac {1}{4},\frac {3}{2};\frac {3}{4};-\frac {x^4}{5}\right )}{15 \sqrt {5} x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(2*Hypergeometric2F1[-3/4, 3/2, 1/4, -1/5*x^4] + 9*x^2*Hypergeometric2F1[-1/4, 3/2, 3/4, -1/5*x^4])/(Sqr
t[5]*x^3)

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 5} {\left (3 \, x^{2} + 2\right )}}{x^{12} + 10 \, x^{8} + 25 \, x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + 5)*(3*x^2 + 2)/(x^12 + 10*x^8 + 25*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 192, normalized size = 0.90 \[ -\frac {3 x^{3}}{50 \sqrt {x^{4}+5}}-\frac {x}{25 \sqrt {x^{4}+5}}-\frac {\sqrt {5}\, \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )}{375 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {3 \sqrt {x^{4}+5}}{25 x}-\frac {2 \sqrt {x^{4}+5}}{75 x^{3}}+\frac {9 i \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \left (-\EllipticE \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )+\EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )\right )}{250 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/25*(x^4+5)^(1/2)/x-3/50/(x^4+5)^(1/2)*x^3+9/250*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))-2/75*(x^4+5)^(1/2)/x^3-1/25/(x^4+5)^(1/2)*x-1/375*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 \[ \int \frac {3 \, x^{2} + 2}{{\left (x^{4} + 5\right )}^{\frac {3}{2}} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 8.15, size = 80, normalized size = 0.37 \[ \frac {3 \sqrt {5} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{100 x \Gamma \left (\frac {3}{4}\right )} + \frac {\sqrt {5} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{50 x^{3} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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