3.951 \(\int \frac{x^{14}}{\left (1+x^4\right )^{3/2}} \, dx\)

Optimal. Leaf size=156 \[ -\frac{x^{11}}{2 \sqrt{x^4+1}}+\frac{11}{18} \sqrt{x^4+1} x^7-\frac{77}{90} \sqrt{x^4+1} x^3+\frac{77 \sqrt{x^4+1} x}{30 \left (x^2+1\right )}+\frac{77 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{60 \sqrt{x^4+1}}-\frac{77 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{30 \sqrt{x^4+1}} \]

[Out]

-x^11/(2*Sqrt[1 + x^4]) - (77*x^3*Sqrt[1 + x^4])/90 + (11*x^7*Sqrt[1 + x^4])/18
+ (77*x*Sqrt[1 + x^4])/(30*(1 + x^2)) - (77*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2
]*EllipticE[2*ArcTan[x], 1/2])/(30*Sqrt[1 + x^4]) + (77*(1 + x^2)*Sqrt[(1 + x^4)
/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(60*Sqrt[1 + x^4])

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Rubi [A]  time = 0.112814, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385 \[ -\frac{x^{11}}{2 \sqrt{x^4+1}}+\frac{11}{18} \sqrt{x^4+1} x^7-\frac{77}{90} \sqrt{x^4+1} x^3+\frac{77 \sqrt{x^4+1} x}{30 \left (x^2+1\right )}+\frac{77 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{60 \sqrt{x^4+1}}-\frac{77 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{30 \sqrt{x^4+1}} \]

Antiderivative was successfully verified.

[In]  Int[x^14/(1 + x^4)^(3/2),x]

[Out]

-x^11/(2*Sqrt[1 + x^4]) - (77*x^3*Sqrt[1 + x^4])/90 + (11*x^7*Sqrt[1 + x^4])/18
+ (77*x*Sqrt[1 + x^4])/(30*(1 + x^2)) - (77*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2
]*EllipticE[2*ArcTan[x], 1/2])/(30*Sqrt[1 + x^4]) + (77*(1 + x^2)*Sqrt[(1 + x^4)
/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(60*Sqrt[1 + x^4])

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Rubi in Sympy [A]  time = 10.2372, size = 143, normalized size = 0.92 \[ - \frac{x^{11}}{2 \sqrt{x^{4} + 1}} + \frac{11 x^{7} \sqrt{x^{4} + 1}}{18} - \frac{77 x^{3} \sqrt{x^{4} + 1}}{90} + \frac{77 x \sqrt{x^{4} + 1}}{30 \left (x^{2} + 1\right )} - \frac{77 \sqrt{\frac{x^{4} + 1}{\left (x^{2} + 1\right )^{2}}} \left (x^{2} + 1\right ) E\left (2 \operatorname{atan}{\left (x \right )}\middle | \frac{1}{2}\right )}{30 \sqrt{x^{4} + 1}} + \frac{77 \sqrt{\frac{x^{4} + 1}{\left (x^{2} + 1\right )^{2}}} \left (x^{2} + 1\right ) F\left (2 \operatorname{atan}{\left (x \right )}\middle | \frac{1}{2}\right )}{60 \sqrt{x^{4} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**14/(x**4+1)**(3/2),x)

[Out]

-x**11/(2*sqrt(x**4 + 1)) + 11*x**7*sqrt(x**4 + 1)/18 - 77*x**3*sqrt(x**4 + 1)/9
0 + 77*x*sqrt(x**4 + 1)/(30*(x**2 + 1)) - 77*sqrt((x**4 + 1)/(x**2 + 1)**2)*(x**
2 + 1)*elliptic_e(2*atan(x), 1/2)/(30*sqrt(x**4 + 1)) + 77*sqrt((x**4 + 1)/(x**2
+ 1)**2)*(x**2 + 1)*elliptic_f(2*atan(x), 1/2)/(60*sqrt(x**4 + 1))

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Mathematica [C]  time = 0.0957604, size = 72, normalized size = 0.46 \[ \frac{1}{90} \left (\frac{\left (10 x^8-22 x^4-77\right ) x^3}{\sqrt{x^4+1}}+231 (-1)^{3/4} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )-231 (-1)^{3/4} E\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )\right ) \]

Antiderivative was successfully verified.

[In]  Integrate[x^14/(1 + x^4)^(3/2),x]

[Out]

((x^3*(-77 - 22*x^4 + 10*x^8))/Sqrt[1 + x^4] - 231*(-1)^(3/4)*EllipticE[I*ArcSin
h[(-1)^(1/4)*x], -1] + 231*(-1)^(3/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*x], -1])/90

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Maple [C]  time = 0.012, size = 119, normalized size = 0.8 \[ -{\frac{{x}^{3}}{2}{\frac{1}{\sqrt{{x}^{4}+1}}}}+{\frac{{x}^{7}}{9}\sqrt{{x}^{4}+1}}-{\frac{16\,{x}^{3}}{45}\sqrt{{x}^{4}+1}}+{\frac{{\frac{77\,i}{30}} \left ({\it EllipticF} \left ( x \left ({\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) ,i \right ) -{\it EllipticE} \left ( x \left ({\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) ,i \right ) \right ) }{{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2}}\sqrt{1-i{x}^{2}}\sqrt{1+i{x}^{2}}{\frac{1}{\sqrt{{x}^{4}+1}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^14/(x^4+1)^(3/2),x)

[Out]

-1/2*x^3/(x^4+1)^(1/2)+1/9*x^7*(x^4+1)^(1/2)-16/45*x^3*(x^4+1)^(1/2)+77/30*I/(1/
2*2^(1/2)+1/2*I*2^(1/2))*(1-I*x^2)^(1/2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*(Elliptic
F(x*(1/2*2^(1/2)+1/2*I*2^(1/2)),I)-EllipticE(x*(1/2*2^(1/2)+1/2*I*2^(1/2)),I))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{x^{14}}{{\left (x^{4} + 1\right )}^{\frac{3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^14/(x^4 + 1)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^14/(x^4 + 1)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{x^{14}}{{\left (x^{4} + 1\right )}^{\frac{3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^14/(x^4 + 1)^(3/2),x, algorithm="fricas")

[Out]

integral(x^14/(x^4 + 1)^(3/2), x)

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Sympy [A]  time = 8.61967, size = 29, normalized size = 0.19 \[ \frac{x^{15} \Gamma \left (\frac{15}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{15}{4} \\ \frac{19}{4} \end{matrix}\middle |{x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac{19}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x**14/(x**4+1)**(3/2),x)

[Out]

x**15*gamma(15/4)*hyper((3/2, 15/4), (19/4,), x**4*exp_polar(I*pi))/(4*gamma(19/
4))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{x^{14}}{{\left (x^{4} + 1\right )}^{\frac{3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^14/(x^4 + 1)^(3/2),x, algorithm="giac")

[Out]

integrate(x^14/(x^4 + 1)^(3/2), x)