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

Optimal. Leaf size=177 \[ -\frac{\sqrt{x^4+5} x}{5 \left (x^2+\sqrt{5}\right )}-\frac{\left (15-2 x^2\right ) x}{10 \sqrt{x^4+5}}-\frac{\left (2-3 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{4\ 5^{3/4} \sqrt{x^4+5}}+\frac{\left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{5^{3/4} \sqrt{x^4+5}} \]

[Out]

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

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Rubi [A]  time = 0.166431, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2 \[ -\frac{\sqrt{x^4+5} x}{5 \left (x^2+\sqrt{5}\right )}-\frac{\left (15-2 x^2\right ) x}{10 \sqrt{x^4+5}}-\frac{\left (2-3 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{4\ 5^{3/4} \sqrt{x^4+5}}+\frac{\left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{5^{3/4} \sqrt{x^4+5}} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 14.1709, size = 177, normalized size = 1. \[ - \frac{x \left (- 2 x^{2} + 15\right )}{10 \sqrt{x^{4} + 5}} - \frac{x \sqrt{x^{4} + 5}}{5 \left (x^{2} + \sqrt{5}\right )} + \frac{\sqrt [4]{5} \sqrt{\frac{x^{4} + 5}{\left (\frac{\sqrt{5} x^{2}}{5} + 1\right )^{2}}} \left (\frac{\sqrt{5} x^{2}}{5} + 1\right ) E\left (2 \operatorname{atan}{\left (\frac{5^{\frac{3}{4}} x}{5} \right )}\middle | \frac{1}{2}\right )}{5 \sqrt{x^{4} + 5}} - \frac{\sqrt [4]{5} \sqrt{\frac{x^{4} + 5}{\left (\frac{\sqrt{5} x^{2}}{5} + 1\right )^{2}}} \left (- 3 \sqrt{5} + 2\right ) \left (\frac{\sqrt{5} x^{2}}{5} + 1\right ) F\left (2 \operatorname{atan}{\left (\frac{5^{\frac{3}{4}} x}{5} \right )}\middle | \frac{1}{2}\right )}{20 \sqrt{x^{4} + 5}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-x*(-2*x**2 + 15)/(10*sqrt(x**4 + 5)) - x*sqrt(x**4 + 5)/(5*(x**2 + sqrt(5))) +
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)/(5*sqrt(x**4 + 5)) - 5**(1/4)*sqrt((x**4 + 5)/(sqrt
(5)*x**2/5 + 1)**2)*(-3*sqrt(5) + 2)*(sqrt(5)*x**2/5 + 1)*elliptic_f(2*atan(5**(
3/4)*x/5), 1/2)/(20*sqrt(x**4 + 5))

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Mathematica [C]  time = 0.18264, size = 85, normalized size = 0.48 \[ \frac{1}{10} \left (\frac{x \left (2 x^2-15\right )}{\sqrt{x^4+5}}-\sqrt [4]{-5} \left (3 \sqrt{5}+2 i\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac{1}{5}} x\right )\right |-1\right )+2 (-1)^{3/4} \sqrt [4]{5} E\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac{1}{5}} x\right )\right |-1\right )\right ) \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [C]  time = 0.019, size = 168, normalized size = 1. \[{\frac{{x}^{3}}{5}{\frac{1}{\sqrt{{x}^{4}+5}}}}-{\frac{{\frac{i}{25}}}{\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}} \left ({\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) -{\it EllipticE} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) \right ){\frac{1}{\sqrt{{x}^{4}+5}}}}-{\frac{3\,x}{2}{\frac{1}{\sqrt{{x}^{4}+5}}}}+{\frac{3\,\sqrt{5}}{50\,\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}}{\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ){\frac{1}{\sqrt{{x}^{4}+5}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/5*x^3/(x^4+5)^(1/2)-1/25*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))-3/2*x/(x^4+5)^(1/2)+3/50*5^(1/2)/
(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)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 11.1758, size = 75, normalized size = 0.42 \[ \frac{3 \sqrt{5} x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{3}{2} \\ \frac{9}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{100 \Gamma \left (\frac{9}{4}\right )} + \frac{\sqrt{5} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{3}{2} \\ \frac{7}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{50 \Gamma \left (\frac{7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

3*sqrt(5)*x**5*gamma(5/4)*hyper((5/4, 3/2), (9/4,), x**4*exp_polar(I*pi)/5)/(100
*gamma(9/4)) + sqrt(5)*x**3*gamma(3/4)*hyper((3/4, 3/2), (7/4,), x**4*exp_polar(
I*pi)/5)/(50*gamma(7/4))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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