3.932 \(\int \frac{x^{10}}{\sqrt{1+x^4}} \, dx\)

Optimal. Leaf size=140 \[ \frac{7 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}(x),\frac{1}{2}\right )}{30 \sqrt{x^4+1}}+\frac{1}{9} \sqrt{x^4+1} x^7-\frac{7}{45} \sqrt{x^4+1} x^3+\frac{7 \sqrt{x^4+1} x}{15 \left (x^2+1\right )}-\frac{7 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{15 \sqrt{x^4+1}} \]

[Out]

(-7*x^3*Sqrt[1 + x^4])/45 + (x^7*Sqrt[1 + x^4])/9 + (7*x*Sqrt[1 + x^4])/(15*(1 + x^2)) - (7*(1 + x^2)*Sqrt[(1
+ x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/2])/(15*Sqrt[1 + x^4]) + (7*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]
*EllipticF[2*ArcTan[x], 1/2])/(30*Sqrt[1 + x^4])

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Rubi [A]  time = 0.0309806, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {321, 305, 220, 1196} \[ \frac{1}{9} \sqrt{x^4+1} x^7-\frac{7}{45} \sqrt{x^4+1} x^3+\frac{7 \sqrt{x^4+1} x}{15 \left (x^2+1\right )}+\frac{7 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{30 \sqrt{x^4+1}}-\frac{7 \left (x^2+1\right ) \sqrt{\frac{x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{2}\right )}{15 \sqrt{x^4+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^10/Sqrt[1 + x^4],x]

[Out]

(-7*x^3*Sqrt[1 + x^4])/45 + (x^7*Sqrt[1 + x^4])/9 + (7*x*Sqrt[1 + x^4])/(15*(1 + x^2)) - (7*(1 + x^2)*Sqrt[(1
+ x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/2])/(15*Sqrt[1 + x^4]) + (7*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]
*EllipticF[2*ArcTan[x], 1/2])/(30*Sqrt[1 + x^4])

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}}{\sqrt{1+x^4}} \, dx &=\frac{1}{9} x^7 \sqrt{1+x^4}-\frac{7}{9} \int \frac{x^6}{\sqrt{1+x^4}} \, dx\\ &=-\frac{7}{45} x^3 \sqrt{1+x^4}+\frac{1}{9} x^7 \sqrt{1+x^4}+\frac{7}{15} \int \frac{x^2}{\sqrt{1+x^4}} \, dx\\ &=-\frac{7}{45} x^3 \sqrt{1+x^4}+\frac{1}{9} x^7 \sqrt{1+x^4}+\frac{7}{15} \int \frac{1}{\sqrt{1+x^4}} \, dx-\frac{7}{15} \int \frac{1-x^2}{\sqrt{1+x^4}} \, dx\\ &=-\frac{7}{45} x^3 \sqrt{1+x^4}+\frac{1}{9} x^7 \sqrt{1+x^4}+\frac{7 x \sqrt{1+x^4}}{15 \left (1+x^2\right )}-\frac{7 \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 )}{15 \sqrt{1+x^4}}+\frac{7 \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 )}{30 \sqrt{1+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0122631, size = 42, normalized size = 0.3 \[ \frac{1}{45} x^3 \left (7 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-x^4\right )+\sqrt{x^4+1} \left (5 x^4-7\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/Sqrt[1 + x^4],x]

[Out]

(x^3*(Sqrt[1 + x^4]*(-7 + 5*x^4) + 7*Hypergeometric2F1[1/2, 3/4, 7/4, -x^4]))/45

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Maple [C]  time = 0.024, size = 107, normalized size = 0.8 \begin{align*}{\frac{{x}^{7}}{9}\sqrt{{x}^{4}+1}}-{\frac{7\,{x}^{3}}{45}\sqrt{{x}^{4}+1}}+{\frac{{\frac{7\,i}{15}} \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)^(1/2),x)

[Out]

1/9*x^7*(x^4+1)^(1/2)-7/45*x^3*(x^4+1)^(1/2)+7/15*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}}{\sqrt{x^{4} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^10/sqrt(x^4 + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{10}}{\sqrt{x^{4} + 1}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^10/sqrt(x^4 + 1), x)

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Sympy [C]  time = 1.25854, size = 29, normalized size = 0.21 \begin{align*} \frac{x^{11} \Gamma \left (\frac{11}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{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)**(1/2),x)

[Out]

x**11*gamma(11/4)*hyper((1/2, 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}}{\sqrt{x^{4} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^10/sqrt(x^4 + 1), x)