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

Optimal. Leaf size=192 \[ \frac{10}{7} \sqrt{x^4+5} x+\frac{4 \sqrt{x^4+5} x}{x^2+\sqrt{5}}+\frac{\sqrt [4]{5} \left (14-5 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{7 \sqrt{x^4+5}}-\frac{4 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}}+\frac{1}{35} \left (15 x^2+14\right ) \sqrt{x^4+5} x^3 \]

[Out]

(10*x*Sqrt[5 + x^4])/7 + (4*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) + (x^3*(14 + 15*x^2
)*Sqrt[5 + x^4])/35 - (4*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)*(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 = 0.22468, antiderivative size = 192, 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 \[ \frac{10}{7} \sqrt{x^4+5} x+\frac{4 \sqrt{x^4+5} x}{x^2+\sqrt{5}}+\frac{\sqrt [4]{5} \left (14-5 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{7 \sqrt{x^4+5}}-\frac{4 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}}+\frac{1}{35} \left (15 x^2+14\right ) \sqrt{x^4+5} x^3 \]

Antiderivative was successfully verified.

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

[Out]

(10*x*Sqrt[5 + x^4])/7 + (4*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) + (x^3*(14 + 15*x^2
)*Sqrt[5 + x^4])/35 - (4*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)*(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 = 19.9341, size = 190, normalized size = 0.99 \[ \frac{x^{3} \left (15 x^{2} + 14\right ) \sqrt{x^{4} + 5}}{35} + \frac{10 x \sqrt{x^{4} + 5}}{7} + \frac{4 x \sqrt{x^{4} + 5}}{x^{2} + \sqrt{5}} - \frac{4 \sqrt [4]{5} \sqrt{\frac{x^{4} + 5}{\left (\frac{\sqrt{5} x^{2}}{5} + 1\right )^{2}}} \left (\frac{\sqrt{5} x^{2}}{5} + 1\right ) E\left (2 \operatorname{atan}{\left (\frac{5^{\frac{3}{4}} x}{5} \right )}\middle | \frac{1}{2}\right )}{\sqrt{x^{4} + 5}} + \frac{\sqrt [4]{5} \sqrt{\frac{x^{4} + 5}{\left (\frac{\sqrt{5} x^{2}}{5} + 1\right )^{2}}} \left (- 15 \sqrt{5} + 42\right ) \left (\frac{\sqrt{5} x^{2}}{5} + 1\right ) F\left (2 \operatorname{atan}{\left (\frac{5^{\frac{3}{4}} x}{5} \right )}\middle | \frac{1}{2}\right )}{21 \sqrt{x^{4} + 5}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

x**3*(15*x**2 + 14)*sqrt(x**4 + 5)/35 + 10*x*sqrt(x**4 + 5)/7 + 4*x*sqrt(x**4 +
5)/(x**2 + sqrt(5)) - 4*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)*s
qrt((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)/(21*sqrt(x**4 + 5))

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Mathematica [C]  time = 0.189907, size = 101, normalized size = 0.53 \[ \frac{x \left (15 x^8+14 x^6+125 x^4+70 x^2+250\right )}{35 \sqrt{x^4+5}}+\frac{2}{7} \sqrt [4]{-5} \left (5 \sqrt{5}+14 i\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac{1}{5}} x\right )\right |-1\right )-4 (-1)^{3/4} \sqrt [4]{5} E\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-\frac{1}{5}} x\right )\right |-1\right ) \]

Antiderivative was successfully verified.

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

[Out]

(x*(250 + 70*x^2 + 125*x^4 + 14*x^6 + 15*x^8))/(35*Sqrt[5 + x^4]) - 4*(-1)^(3/4)
*5^(1/4)*EllipticE[I*ArcSinh[(-1/5)^(1/4)*x], -1] + (2*(-5)^(1/4)*(14*I + 5*Sqrt
[5])*EllipticF[I*ArcSinh[(-1/5)^(1/4)*x], -1])/7

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Maple [C]  time = 0.017, size = 180, normalized size = 0.9 \[{\frac{2\,{x}^{3}}{5}\sqrt{{x}^{4}+5}}+{\frac{{\frac{4\,i}{5}}}{\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}} \left ({\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) -{\it EllipticE} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) \right ){\frac{1}{\sqrt{{x}^{4}+5}}}}+{\frac{3\,{x}^{5}}{7}\sqrt{{x}^{4}+5}}+{\frac{10\,x}{7}\sqrt{{x}^{4}+5}}-{\frac{2\,\sqrt{5}}{7\,\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}}{\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ){\frac{1}{\sqrt{{x}^{4}+5}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

2/5*x^3*(x^4+5)^(1/2)+4/5*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))+3/7*x^5*(x^4+5)^(1/2)+10/7*x*(x^4+
5)^(1/2)-2/7*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)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 3.67311, size = 78, normalized size = 0.41 \[ \frac{3 \sqrt{5} x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac{9}{4}\right )} + \frac{\sqrt{5} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{2 \Gamma \left (\frac{7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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