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

Optimal. Leaf size=201 \[ -\frac{2 \left (27-2 x^2\right ) \sqrt{x^4+5}}{3 x}+\frac{36 x \sqrt{x^4+5}}{x^2+\sqrt{5}}+\frac{2 \sqrt [4]{5} \left (27+2 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{3 \sqrt{x^4+5}}-\frac{36 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}}-\frac{\left (10-9 x^2\right ) \left (x^4+5\right )^{3/2}}{15 x^3} \]

[Out]

(-2*(27 - 2*x^2)*Sqrt[5 + x^4])/(3*x) + (36*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) - (
(10 - 9*x^2)*(5 + x^4)^(3/2))/(15*x^3) - (36*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] + (2*5
^(1/4)*(27 + 2*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*Ellipt
icF[2*ArcTan[x/5^(1/4)], 1/2])/(3*Sqrt[5 + x^4])

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Rubi [A]  time = 0.207704, antiderivative size = 201, 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 \[ -\frac{2 \left (27-2 x^2\right ) \sqrt{x^4+5}}{3 x}+\frac{36 x \sqrt{x^4+5}}{x^2+\sqrt{5}}+\frac{2 \sqrt [4]{5} \left (27+2 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{3 \sqrt{x^4+5}}-\frac{36 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}}-\frac{\left (10-9 x^2\right ) \left (x^4+5\right )^{3/2}}{15 x^3} \]

Antiderivative was successfully verified.

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

[Out]

(-2*(27 - 2*x^2)*Sqrt[5 + x^4])/(3*x) + (36*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) - (
(10 - 9*x^2)*(5 + x^4)^(3/2))/(15*x^3) - (36*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] + (2*5
^(1/4)*(27 + 2*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*Ellipt
icF[2*ArcTan[x/5^(1/4)], 1/2])/(3*Sqrt[5 + x^4])

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Rubi in Sympy [A]  time = 18.466, size = 199, normalized size = 0.99 \[ \frac{36 x \sqrt{x^{4} + 5}}{x^{2} + \sqrt{5}} - \frac{36 \sqrt [4]{5} \sqrt{\frac{x^{4} + 5}{\left (\frac{\sqrt{5} x^{2}}{5} + 1\right )^{2}}} \left (\frac{\sqrt{5} x^{2}}{5} + 1\right ) E\left (2 \operatorname{atan}{\left (\frac{5^{\frac{3}{4}} x}{5} \right )}\middle | \frac{1}{2}\right )}{\sqrt{x^{4} + 5}} + \frac{2 \sqrt [4]{5} \sqrt{\frac{x^{4} + 5}{\left (\frac{\sqrt{5} x^{2}}{5} + 1\right )^{2}}} \left (10 \sqrt{5} + 135\right ) \left (\frac{\sqrt{5} x^{2}}{5} + 1\right ) F\left (2 \operatorname{atan}{\left (\frac{5^{\frac{3}{4}} x}{5} \right )}\middle | \frac{1}{2}\right )}{15 \sqrt{x^{4} + 5}} - \frac{2 \left (- 10 x^{2} + 135\right ) \sqrt{x^{4} + 5}}{15 x} - \frac{\left (- 9 x^{2} + 10\right ) \left (x^{4} + 5\right )^{\frac{3}{2}}}{15 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

36*x*sqrt(x**4 + 5)/(x**2 + sqrt(5)) - 36*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)*(10*sqrt(5) + 135)*(
sqrt(5)*x**2/5 + 1)*elliptic_f(2*atan(5**(3/4)*x/5), 1/2)/(15*sqrt(x**4 + 5)) -
2*(-10*x**2 + 135)*sqrt(x**4 + 5)/(15*x) - (-9*x**2 + 10)*(x**4 + 5)**(3/2)/(15*
x**3)

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Mathematica [C]  time = 0.169942, size = 124, normalized size = 0.62 \[ \frac{9 x^{10}+10 x^8-180 x^6-1125 x^2+20 \sqrt [4]{-5} \left (-2 \sqrt{5}+27 i\right ) \sqrt{x^4+5} x^3 F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac{1}{5}} x\right )\right |-1\right )-540 (-1)^{3/4} \sqrt [4]{5} \sqrt{x^4+5} x^3 E\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac{1}{5}} x\right )\right |-1\right )-250}{15 x^3 \sqrt{x^4+5}} \]

Antiderivative was successfully verified.

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

[Out]

(-250 - 1125*x^2 - 180*x^6 + 10*x^8 + 9*x^10 - 540*(-1)^(3/4)*5^(1/4)*x^3*Sqrt[5
+ x^4]*EllipticE[I*ArcSinh[(-1/5)^(1/4)*x], -1] + 20*(-5)^(1/4)*(27*I - 2*Sqrt[
5])*x^3*Sqrt[5 + x^4]*EllipticF[I*ArcSinh[(-1/5)^(1/4)*x], -1])/(15*x^3*Sqrt[5 +
x^4])

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Maple [C]  time = 0.025, size = 192, normalized size = 1. \[ -{\frac{10}{3\,{x}^{3}}\sqrt{{x}^{4}+5}}+{\frac{2\,x}{3}\sqrt{{x}^{4}+5}}+{\frac{8\,\sqrt{5}}{15\,\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}}{\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ){\frac{1}{\sqrt{{x}^{4}+5}}}}-15\,{\frac{\sqrt{{x}^{4}+5}}{x}}+{\frac{3\,{x}^{3}}{5}\sqrt{{x}^{4}+5}}+{\frac{{\frac{36\,i}{5}}}{\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}} \left ({\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) -{\it EllipticE} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) \right ){\frac{1}{\sqrt{{x}^{4}+5}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-10/3*(x^4+5)^(1/2)/x^3+2/3*x*(x^4+5)^(1/2)+8/15*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)-15*(x^4+5)^(1/2)/x+3/5*x^3*(x^4+5)^(1/2)+36/5*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. \[ \int \frac{{\left (x^{4} + 5\right )}^{\frac{3}{2}}{\left (3 \, x^{2} + 2\right )}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 7.08515, size = 163, normalized size = 0.81 \[ \frac{3 \sqrt{5} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} + \frac{\sqrt{5} x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{2 \Gamma \left (\frac{5}{4}\right )} + \frac{15 \sqrt{5} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{4 x \Gamma \left (\frac{3}{4}\right )} + \frac{5 \sqrt{5} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{2 x^{3} \Gamma \left (\frac{1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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