3.947 \(\int \frac{x^{12}}{(1+x^4)^{3/2}} \, dx\)

Optimal. Leaf size=90 \[ \frac{15 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}(x),\frac{1}{2}\right )}{28 \sqrt{x^4+1}}-\frac{x^9}{2 \sqrt{x^4+1}}+\frac{9}{14} \sqrt{x^4+1} x^5-\frac{15}{14} \sqrt{x^4+1} x \]

[Out]

-x^9/(2*Sqrt[1 + x^4]) - (15*x*Sqrt[1 + x^4])/14 + (9*x^5*Sqrt[1 + x^4])/14 + (15*(1 + x^2)*Sqrt[(1 + x^4)/(1
+ x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(28*Sqrt[1 + x^4])

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Rubi [A]  time = 0.0199829, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {288, 321, 220} \[ -\frac{x^9}{2 \sqrt{x^4+1}}+\frac{9}{14} \sqrt{x^4+1} x^5-\frac{15}{14} \sqrt{x^4+1} x+\frac{15 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{28 \sqrt{x^4+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^12/(1 + x^4)^(3/2),x]

[Out]

-x^9/(2*Sqrt[1 + x^4]) - (15*x*Sqrt[1 + x^4])/14 + (9*x^5*Sqrt[1 + x^4])/14 + (15*(1 + x^2)*Sqrt[(1 + x^4)/(1
+ x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(28*Sqrt[1 + x^4])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{x^{12}}{\left (1+x^4\right )^{3/2}} \, dx &=-\frac{x^9}{2 \sqrt{1+x^4}}+\frac{9}{2} \int \frac{x^8}{\sqrt{1+x^4}} \, dx\\ &=-\frac{x^9}{2 \sqrt{1+x^4}}+\frac{9}{14} x^5 \sqrt{1+x^4}-\frac{45}{14} \int \frac{x^4}{\sqrt{1+x^4}} \, dx\\ &=-\frac{x^9}{2 \sqrt{1+x^4}}-\frac{15}{14} x \sqrt{1+x^4}+\frac{9}{14} x^5 \sqrt{1+x^4}+\frac{15}{14} \int \frac{1}{\sqrt{1+x^4}} \, dx\\ &=-\frac{x^9}{2 \sqrt{1+x^4}}-\frac{15}{14} x \sqrt{1+x^4}+\frac{9}{14} x^5 \sqrt{1+x^4}+\frac{15 \left (1+x^2\right ) \sqrt{\frac{1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{28 \sqrt{1+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0104249, size = 52, normalized size = 0.58 \[ \frac{x \left (15 \sqrt{x^4+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-x^4\right )+2 x^8-6 x^4-15\right )}{14 \sqrt{x^4+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^12/(1 + x^4)^(3/2),x]

[Out]

(x*(-15 - 6*x^4 + 2*x^8 + 15*Sqrt[1 + x^4]*Hypergeometric2F1[1/4, 1/2, 5/4, -x^4]))/(14*Sqrt[1 + x^4])

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Maple [C]  time = 0.02, size = 94, normalized size = 1. \begin{align*} -{\frac{x}{2}{\frac{1}{\sqrt{{x}^{4}+1}}}}+{\frac{{x}^{5}}{7}\sqrt{{x}^{4}+1}}-{\frac{4\,x}{7}\sqrt{{x}^{4}+1}}+{\frac{15\,{\it EllipticF} \left ( x \left ( 1/2\,\sqrt{2}+i/2\sqrt{2} \right ) ,i \right ) }{7\,\sqrt{2}+7\,i\sqrt{2}}\sqrt{1-i{x}^{2}}\sqrt{1+i{x}^{2}}{\frac{1}{\sqrt{{x}^{4}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12/(x^4+1)^(3/2),x)

[Out]

-1/2*x/(x^4+1)^(1/2)+1/7*x^5*(x^4+1)^(1/2)-4/7*x*(x^4+1)^(1/2)+15/14/(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)*EllipticF(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. \begin{align*} \int \frac{x^{12}}{{\left (x^{4} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12/(x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^12/(x^4 + 1)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{4} + 1} x^{12}}{x^{8} + 2 \, x^{4} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12/(x^4+1)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^4 + 1)*x^12/(x^8 + 2*x^4 + 1), x)

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Sympy [C]  time = 1.36499, size = 29, normalized size = 0.32 \begin{align*} \frac{x^{13} \Gamma \left (\frac{13}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{13}{4} \\ \frac{17}{4} \end{matrix}\middle |{x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac{17}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**12/(x**4+1)**(3/2),x)

[Out]

x**13*gamma(13/4)*hyper((3/2, 13/4), (17/4,), x**4*exp_polar(I*pi))/(4*gamma(17/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{12}}{{\left (x^{4} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12/(x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(x^12/(x^4 + 1)^(3/2), x)