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

Optimal. Leaf size=199 \[ -\frac{\left (14-3 x^2\right ) \left (x^4+5\right )^{3/2}}{7 x}+\frac{6}{35} x \left (14 x^2+25\right ) \sqrt{x^4+5}+\frac{24 x \sqrt{x^4+5}}{x^2+\sqrt{5}}+\frac{6 \sqrt [4]{5} \left (14+5 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{7 \sqrt{x^4+5}}-\frac{24 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}} \]

[Out]

(24*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) + (6*x*(25 + 14*x^2)*Sqrt[5 + x^4])/35 - ((
14 - 3*x^2)*(5 + x^4)^(3/2))/(7*x) - (24*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] + (6*5^(1/
4)*(14 + 5*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[
2*ArcTan[x/5^(1/4)], 1/2])/(7*Sqrt[5 + x^4])

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Rubi [A]  time = 0.188189, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25 \[ -\frac{\left (14-3 x^2\right ) \left (x^4+5\right )^{3/2}}{7 x}+\frac{6}{35} x \left (14 x^2+25\right ) \sqrt{x^4+5}+\frac{24 x \sqrt{x^4+5}}{x^2+\sqrt{5}}+\frac{6 \sqrt [4]{5} \left (14+5 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{7 \sqrt{x^4+5}}-\frac{24 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}} \]

Antiderivative was successfully verified.

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

[Out]

(24*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) + (6*x*(25 + 14*x^2)*Sqrt[5 + x^4])/35 - ((
14 - 3*x^2)*(5 + x^4)^(3/2))/(7*x) - (24*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] + (6*5^(1/
4)*(14 + 5*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[
2*ArcTan[x/5^(1/4)], 1/2])/(7*Sqrt[5 + x^4])

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Rubi in Sympy [A]  time = 17.5969, size = 197, normalized size = 0.99 \[ \frac{2 x \left (42 x^{2} + 75\right ) \sqrt{x^{4} + 5}}{35} + \frac{24 x \sqrt{x^{4} + 5}}{x^{2} + \sqrt{5}} - \frac{24 \sqrt [4]{5} \sqrt{\frac{x^{4} + 5}{\left (\frac{\sqrt{5} x^{2}}{5} + 1\right )^{2}}} \left (\frac{\sqrt{5} x^{2}}{5} + 1\right ) E\left (2 \operatorname{atan}{\left (\frac{5^{\frac{3}{4}} x}{5} \right )}\middle | \frac{1}{2}\right )}{\sqrt{x^{4} + 5}} + \frac{2 \sqrt [4]{5} \sqrt{\frac{x^{4} + 5}{\left (\frac{\sqrt{5} x^{2}}{5} + 1\right )^{2}}} \left (15 \sqrt{5} + 42\right ) \left (\frac{\sqrt{5} x^{2}}{5} + 1\right ) F\left (2 \operatorname{atan}{\left (\frac{5^{\frac{3}{4}} x}{5} \right )}\middle | \frac{1}{2}\right )}{7 \sqrt{x^{4} + 5}} - \frac{\left (- 3 x^{2} + 14\right ) \left (x^{4} + 5\right )^{\frac{3}{2}}}{7 x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

2*x*(42*x**2 + 75)*sqrt(x**4 + 5)/35 + 24*x*sqrt(x**4 + 5)/(x**2 + sqrt(5)) - 24
*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) + 2*5**(1/4)*sqrt((x**4 + 5)/(sqrt(
5)*x**2/5 + 1)**2)*(15*sqrt(5) + 42)*(sqrt(5)*x**2/5 + 1)*elliptic_f(2*atan(5**(
3/4)*x/5), 1/2)/(7*sqrt(x**4 + 5)) - (-3*x**2 + 14)*(x**4 + 5)**(3/2)/(7*x)

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Mathematica [C]  time = 0.172841, size = 125, normalized size = 0.63 \[ \frac{15 x^{10}+14 x^8+300 x^6-280 x^4+60 \sqrt [4]{-5} \left (-5 \sqrt{5}+14 i\right ) \sqrt{x^4+5} x F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac{1}{5}} x\right )\right |-1\right )-840 (-1)^{3/4} \sqrt [4]{5} \sqrt{x^4+5} x E\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac{1}{5}} x\right )\right |-1\right )+1125 x^2-1750}{35 x \sqrt{x^4+5}} \]

Antiderivative was successfully verified.

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

[Out]

(-1750 + 1125*x^2 - 280*x^4 + 300*x^6 + 14*x^8 + 15*x^10 - 840*(-1)^(3/4)*5^(1/4
)*x*Sqrt[5 + x^4]*EllipticE[I*ArcSinh[(-1/5)^(1/4)*x], -1] + 60*(-5)^(1/4)*(14*I
- 5*Sqrt[5])*x*Sqrt[5 + x^4]*EllipticF[I*ArcSinh[(-1/5)^(1/4)*x], -1])/(35*x*Sq
rt[5 + x^4])

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Maple [C]  time = 0.022, size = 192, normalized size = 1. \[{\frac{3\,{x}^{5}}{7}\sqrt{{x}^{4}+5}}+{\frac{45\,x}{7}\sqrt{{x}^{4}+5}}+{\frac{12\,\sqrt{5}}{7\,\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}}{\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ){\frac{1}{\sqrt{{x}^{4}+5}}}}-10\,{\frac{\sqrt{{x}^{4}+5}}{x}}+{\frac{2\,{x}^{3}}{5}\sqrt{{x}^{4}+5}}+{\frac{{\frac{24\,i}{5}}}{\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}} \left ({\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) -{\it EllipticE} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) \right ){\frac{1}{\sqrt{{x}^{4}+5}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

3/7*x^5*(x^4+5)^(1/2)+45/7*x*(x^4+5)^(1/2)+12/7*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)-10*(x^4+5)^(1/2)/x+2/5*x^3*(x^4+5)^(1/2)+24/5*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. \[ \int \frac{{\left (x^{4} + 5\right )}^{\frac{3}{2}}{\left (3 \, x^{2} + 2\right )}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (3 \, x^{6} + 2 \, x^{4} + 15 \, x^{2} + 10\right )} \sqrt{x^{4} + 5}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((3*x^6 + 2*x^4 + 15*x^2 + 10)*sqrt(x^4 + 5)/x^2, x)

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Sympy [A]  time = 6.98045, size = 160, normalized size = 0.8 \[ \frac{3 \sqrt{5} x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac{9}{4}\right )} + \frac{\sqrt{5} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{2 \Gamma \left (\frac{7}{4}\right )} + \frac{15 \sqrt{5} x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{5 \sqrt{5} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{2 x \Gamma \left (\frac{3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

3*sqrt(5)*x**5*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), x**4*exp_polar(I*pi)/5)/(4*
gamma(9/4)) + sqrt(5)*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), x**4*exp_polar(
I*pi)/5)/(2*gamma(7/4)) + 15*sqrt(5)*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), x**
4*exp_polar(I*pi)/5)/(4*gamma(5/4)) + 5*sqrt(5)*gamma(-1/4)*hyper((-1/2, -1/4),
(3/4,), x**4*exp_polar(I*pi)/5)/(2*x*gamma(3/4))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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