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

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*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*Sqrt[5])*(Sqrt[5] + x
^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x/5^(1/4)], 1/2])/(6*Sq
rt[5 + x^4])

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Rubi [A]  time = 0.213611, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2 \[ \frac{2}{3} \sqrt{x^4+5} x+\frac{3}{5} \sqrt{x^4+5} x^3-\frac{9 \sqrt{x^4+5} x}{x^2+\sqrt{5}}-\frac{\sqrt [4]{5} \left (27+2 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{6 \sqrt{x^4+5}}+\frac{9 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}} \]

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*Sqrt[5])*(Sqrt[5] + x
^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x/5^(1/4)], 1/2])/(6*Sq
rt[5 + x^4])

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Rubi in Sympy [A]  time = 18.9082, size = 185, normalized size = 1. \[ \frac{3 x^{3} \sqrt{x^{4} + 5}}{5} + \frac{2 x \sqrt{x^{4} + 5}}{3} - \frac{9 x \sqrt{x^{4} + 5}}{x^{2} + \sqrt{5}} + \frac{9 \sqrt [4]{5} \sqrt{\frac{x^{4} + 5}{\left (\frac{\sqrt{5} x^{2}}{5} + 1\right )^{2}}} \left (\frac{\sqrt{5} x^{2}}{5} + 1\right ) E\left (2 \operatorname{atan}{\left (\frac{5^{\frac{3}{4}} x}{5} \right )}\middle | \frac{1}{2}\right )}{\sqrt{x^{4} + 5}} - \frac{\sqrt [4]{5} \sqrt{\frac{x^{4} + 5}{\left (\frac{\sqrt{5} x^{2}}{5} + 1\right )^{2}}} \left (10 \sqrt{5} + 135\right ) \left (\frac{\sqrt{5} x^{2}}{5} + 1\right ) F\left (2 \operatorname{atan}{\left (\frac{5^{\frac{3}{4}} x}{5} \right )}\middle | \frac{1}{2}\right )}{30 \sqrt{x^{4} + 5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**4*(3*x**2+2)/(x**4+5)**(1/2),x)

[Out]

3*x**3*sqrt(x**4 + 5)/5 + 2*x*sqrt(x**4 + 5)/3 - 9*x*sqrt(x**4 + 5)/(x**2 + sqrt
(5)) + 9*5**(1/4)*sqrt((x**4 + 5)/(sqrt(5)*x**2/5 + 1)**2)*(sqrt(5)*x**2/5 + 1)*
elliptic_e(2*atan(5**(3/4)*x/5), 1/2)/sqrt(x**4 + 5) - 5**(1/4)*sqrt((x**4 + 5)/
(sqrt(5)*x**2/5 + 1)**2)*(10*sqrt(5) + 135)*(sqrt(5)*x**2/5 + 1)*elliptic_f(2*at
an(5**(3/4)*x/5), 1/2)/(30*sqrt(x**4 + 5))

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Mathematica [C]  time = 0.254498, size = 96, normalized size = 0.52 \[ \frac{1}{15} \left (\frac{x \left (9 x^6+10 x^4+45 x^2+50\right )}{\sqrt{x^4+5}}+5 \sqrt [4]{-5} \left (2 \sqrt{5}-27 i\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac{1}{5}} x\right )\right |-1\right )\right )+9 (-1)^{3/4} \sqrt [4]{5} E\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac{1}{5}} x\right )\right |-1\right ) \]

Antiderivative was successfully verified.

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

[Out]

9*(-1)^(3/4)*5^(1/4)*EllipticE[I*ArcSinh[(-1/5)^(1/4)*x], -1] + ((x*(50 + 45*x^2
+ 10*x^4 + 9*x^6))/Sqrt[5 + x^4] + 5*(-5)^(1/4)*(-27*I + 2*Sqrt[5])*EllipticF[I
*ArcSinh[(-1/5)^(1/4)*x], -1])/15

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Maple [C]  time = 0.025, size = 168, normalized size = 0.9 \[{\frac{2\,x}{3}\sqrt{{x}^{4}+5}}-{\frac{2\,\sqrt{5}}{15\,\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}}{\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ){\frac{1}{\sqrt{{x}^{4}+5}}}}+{\frac{3\,{x}^{3}}{5}\sqrt{{x}^{4}+5}}-{\frac{{\frac{9\,i}{5}}}{\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}} \left ({\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) -{\it EllipticE} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) \right ){\frac{1}{\sqrt{{x}^{4}+5}}}} \]

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]

2/3*x*(x^4+5)^(1/2)-2/15*5^(1/2)/(I*5^(1/2))^(1/2)*(25-5*I*5^(1/2)*x^2)^(1/2)*(2
5+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)+3/5*x^3*(x^4+5)^(1/2)-9/5*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., size = 0, normalized size = 0. \[ \int \frac{{\left (3 \, x^{2} + 2\right )} x^{4}}{\sqrt{x^{4} + 5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 4.37928, 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*gamma(5/4)*hyper((1/2, 5/4), (9/4,), x**4*exp_polar
(I*pi)/5)/(10*gamma(9/4))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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