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

Optimal. Leaf size=185 \[ \frac {2}{3} \sqrt {x^4+5} x+\frac {3}{5} \sqrt {x^4+5} x^3-\frac {9 \sqrt {x^4+5} x}{x^2+\sqrt {5}}-\frac {\sqrt [4]{5} \left (27+2 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{6 \sqrt {x^4+5}}+\frac {9 \sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}} \]

[Out]

2/3*x*(x^4+5)^(1/2)+3/5*x^3*(x^4+5)^(1/2)-9*x*(x^4+5)^(1/2)/(x^2+5^(1/2))+9*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)-1/6*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))*(x^2+5^(1/2))*(27+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 = 0.09, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1280, 1198, 220, 1196} \[ \frac {3}{5} \sqrt {x^4+5} x^3-\frac {9 \sqrt {x^4+5} x}{x^2+\sqrt {5}}+\frac {2}{3} \sqrt {x^4+5} x-\frac {\sqrt [4]{5} \left (27+2 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{6 \sqrt {x^4+5}}+\frac {9 \sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[5 + x^4])/3 + (3*x^3*Sqrt[5 + x^4])/5 - (9*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) + (9*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] - (5^(1/4)*(27 + 2*S
qrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x/5^(1/4)], 1/2])/(6*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 1280

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*f*(f*x)^(m - 1)*
(a + c*x^4)^(p + 1))/(c*(m + 4*p + 3)), x] - Dist[f^2/(c*(m + 4*p + 3)), Int[(f*x)^(m - 2)*(a + c*x^4)^p*(a*e*
(m - 1) - c*d*(m + 4*p + 3)*x^2), x], x] /; FreeQ[{a, c, d, e, f, p}, x] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] &
& IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {x^4 \left (2+3 x^2\right )}{\sqrt {5+x^4}} \, dx &=\frac {3}{5} x^3 \sqrt {5+x^4}-\frac {1}{5} \int \frac {x^2 \left (45-10 x^2\right )}{\sqrt {5+x^4}} \, dx\\ &=\frac {2}{3} x \sqrt {5+x^4}+\frac {3}{5} x^3 \sqrt {5+x^4}+\frac {1}{15} \int \frac {-50-135 x^2}{\sqrt {5+x^4}} \, dx\\ &=\frac {2}{3} x \sqrt {5+x^4}+\frac {3}{5} x^3 \sqrt {5+x^4}+\left (9 \sqrt {5}\right ) \int \frac {1-\frac {x^2}{\sqrt {5}}}{\sqrt {5+x^4}} \, dx-\frac {1}{3} \left (10+27 \sqrt {5}\right ) \int \frac {1}{\sqrt {5+x^4}} \, dx\\ &=\frac {2}{3} x \sqrt {5+x^4}+\frac {3}{5} x^3 \sqrt {5+x^4}-\frac {9 x \sqrt {5+x^4}}{\sqrt {5}+x^2}+\frac {9 \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 {\sqrt [4]{5} \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 )}{6 \sqrt {5+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 74, normalized size = 0.40 \[ \frac {1}{15} x \left (-10 \sqrt {5} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {x^4}{5}\right )-9 \sqrt {5} x^2 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {x^4}{5}\right )+\left (9 x^2+10\right ) \sqrt {x^4+5}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*((10 + 9*x^2)*Sqrt[5 + x^4] - 10*Sqrt[5]*Hypergeometric2F1[1/4, 1/2, 5/4, -1/5*x^4] - 9*Sqrt[5]*x^2*Hyperge
ometric2F1[1/2, 3/4, 7/4, -1/5*x^4]))/15

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {3 \, x^{6} + 2 \, x^{4}}{\sqrt {x^{4} + 5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((3*x^6 + 2*x^4)/sqrt(x^4 + 5), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 168, normalized size = 0.91 \[ \frac {3 \sqrt {x^{4}+5}\, x^{3}}{5}+\frac {2 \sqrt {x^{4}+5}\, x}{3}-\frac {2 \sqrt {5}\, \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )}{15 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {9 i \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \left (-\EllipticE \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )+\EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/5*(x^4+5)^(1/2)*x^3-9/5*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))+2/3*(x^4+5)^(
1/2)*x-2/15*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)*Ell
ipticF(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 \[ \int \frac {{\left (3 \, x^{2} + 2\right )} x^{4}}{\sqrt {x^{4} + 5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 2.55, size = 75, normalized size = 0.41 \[ \frac {3 \sqrt {5} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{20 \Gamma \left (\frac {11}{4}\right )} + \frac {\sqrt {5} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{10 \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(5)*x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), x**4*exp_polar(I*pi)/5)/(20*gamma(11/4)) + sqrt(5)*x**5*g
amma(5/4)*hyper((1/2, 5/4), (9/4,), x**4*exp_polar(I*pi)/5)/(10*gamma(9/4))

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