3.954 \(\int \frac{x^{10}}{(1+x^4)^{3/2}} \, dx\)

Optimal. Leaf size=140 \[ -\frac{21 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}(x),\frac{1}{2}\right )}{20 \sqrt{x^4+1}}-\frac{x^7}{2 \sqrt{x^4+1}}+\frac{7}{10} \sqrt{x^4+1} x^3-\frac{21 \sqrt{x^4+1} x}{10 \left (x^2+1\right )}+\frac{21 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{10 \sqrt{x^4+1}} \]

[Out]

-x^7/(2*Sqrt[1 + x^4]) + (7*x^3*Sqrt[1 + x^4])/10 - (21*x*Sqrt[1 + x^4])/(10*(1 + x^2)) + (21*(1 + x^2)*Sqrt[(
1 + x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/2])/(10*Sqrt[1 + x^4]) - (21*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)
^2]*EllipticF[2*ArcTan[x], 1/2])/(20*Sqrt[1 + x^4])

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Rubi [A]  time = 0.031527, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {288, 321, 305, 220, 1196} \[ -\frac{x^7}{2 \sqrt{x^4+1}}+\frac{7}{10} \sqrt{x^4+1} x^3-\frac{21 \sqrt{x^4+1} x}{10 \left (x^2+1\right )}-\frac{21 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{20 \sqrt{x^4+1}}+\frac{21 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{10 \sqrt{x^4+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^10/(1 + x^4)^(3/2),x]

[Out]

-x^7/(2*Sqrt[1 + x^4]) + (7*x^3*Sqrt[1 + x^4])/10 - (21*x*Sqrt[1 + x^4])/(10*(1 + x^2)) + (21*(1 + x^2)*Sqrt[(
1 + x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/2])/(10*Sqrt[1 + x^4]) - (21*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)
^2]*EllipticF[2*ArcTan[x], 1/2])/(20*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 305

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

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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{x^{10}}{\left (1+x^4\right )^{3/2}} \, dx &=-\frac{x^7}{2 \sqrt{1+x^4}}+\frac{7}{2} \int \frac{x^6}{\sqrt{1+x^4}} \, dx\\ &=-\frac{x^7}{2 \sqrt{1+x^4}}+\frac{7}{10} x^3 \sqrt{1+x^4}-\frac{21}{10} \int \frac{x^2}{\sqrt{1+x^4}} \, dx\\ &=-\frac{x^7}{2 \sqrt{1+x^4}}+\frac{7}{10} x^3 \sqrt{1+x^4}-\frac{21}{10} \int \frac{1}{\sqrt{1+x^4}} \, dx+\frac{21}{10} \int \frac{1-x^2}{\sqrt{1+x^4}} \, dx\\ &=-\frac{x^7}{2 \sqrt{1+x^4}}+\frac{7}{10} x^3 \sqrt{1+x^4}-\frac{21 x \sqrt{1+x^4}}{10 \left (1+x^2\right )}+\frac{21 \left (1+x^2\right ) \sqrt{\frac{1+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{10 \sqrt{1+x^4}}-\frac{21 \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 )}{20 \sqrt{1+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0093168, size = 47, normalized size = 0.34 \[ \frac{x^3 \left (7 \sqrt{x^4+1} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-x^4\right )+x^4-7\right )}{5 \sqrt{x^4+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/(1 + x^4)^(3/2),x]

[Out]

(x^3*(-7 + x^4 + 7*Sqrt[1 + x^4]*Hypergeometric2F1[3/4, 3/2, 7/4, -x^4]))/(5*Sqrt[1 + x^4])

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Maple [C]  time = 0.009, size = 107, normalized size = 0.8 \begin{align*}{\frac{{x}^{3}}{2}{\frac{1}{\sqrt{{x}^{4}+1}}}}+{\frac{{x}^{3}}{5}\sqrt{{x}^{4}+1}}-{\frac{{\frac{21\,i}{10}} \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}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(x^4+1)^(3/2),x)

[Out]

1/2*x^3/(x^4+1)^(1/2)+1/5*x^3*(x^4+1)^(1/2)-21/10*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)*(EllipticF(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. \begin{align*} \int \frac{x^{10}}{{\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^10/(x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^10/(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^{10}}{x^{8} + 2 \, x^{4} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(x^4+1)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^4 + 1)*x^10/(x^8 + 2*x^4 + 1), x)

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Sympy [C]  time = 1.36664, size = 29, normalized size = 0.21 \begin{align*} \frac{x^{11} \Gamma \left (\frac{11}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{11}{4} \\ \frac{15}{4} \end{matrix}\middle |{x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac{15}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10/(x**4+1)**(3/2),x)

[Out]

x**11*gamma(11/4)*hyper((3/2, 11/4), (15/4,), x**4*exp_polar(I*pi))/(4*gamma(15/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{10}}{{\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^10/(x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(x^10/(x^4 + 1)^(3/2), x)