3.933 \(\int \frac {x^6}{\sqrt {1+x^4}} \, dx\)

Optimal. Leaf size=124 \[ \frac {1}{5} \sqrt {x^4+1} x^3-\frac {3 \sqrt {x^4+1} x}{5 \left (x^2+1\right )}-\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{10 \sqrt {x^4+1}}+\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{5 \sqrt {x^4+1}} \]

[Out]

1/5*x^3*(x^4+1)^(1/2)-3/5*x*(x^4+1)^(1/2)/(x^2+1)+3/5*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*Elli
pticE(sin(2*arctan(x)),1/2*2^(1/2))*((x^4+1)/(x^2+1)^2)^(1/2)/(x^4+1)^(1/2)-3/10*(x^2+1)*(cos(2*arctan(x))^2)^
(1/2)/cos(2*arctan(x))*EllipticF(sin(2*arctan(x)),1/2*2^(1/2))*((x^4+1)/(x^2+1)^2)^(1/2)/(x^4+1)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {321, 305, 220, 1196} \[ \frac {1}{5} \sqrt {x^4+1} x^3-\frac {3 \sqrt {x^4+1} x}{5 \left (x^2+1\right )}-\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{10 \sqrt {x^4+1}}+\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{5 \sqrt {x^4+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/Sqrt[1 + x^4],x]

[Out]

(x^3*Sqrt[1 + x^4])/5 - (3*x*Sqrt[1 + x^4])/(5*(1 + x^2)) + (3*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticE
[2*ArcTan[x], 1/2])/(5*Sqrt[1 + x^4]) - (3*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/
(10*Sqrt[1 + x^4])

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 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 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 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^6}{\sqrt {1+x^4}} \, dx &=\frac {1}{5} x^3 \sqrt {1+x^4}-\frac {3}{5} \int \frac {x^2}{\sqrt {1+x^4}} \, dx\\ &=\frac {1}{5} x^3 \sqrt {1+x^4}-\frac {3}{5} \int \frac {1}{\sqrt {1+x^4}} \, dx+\frac {3}{5} \int \frac {1-x^2}{\sqrt {1+x^4}} \, dx\\ &=\frac {1}{5} x^3 \sqrt {1+x^4}-\frac {3 x \sqrt {1+x^4}}{5 \left (1+x^2\right )}+\frac {3 \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 )}{5 \sqrt {1+x^4}}-\frac {3 \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 )}{10 \sqrt {1+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 34, normalized size = 0.27 \[ \frac {1}{5} x^3 \left (\sqrt {x^4+1}-\, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-x^4\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/Sqrt[1 + x^4],x]

[Out]

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

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{6}}{\sqrt {x^{4} + 1}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(x^4+1)^(1/2),x, algorithm="fricas")

[Out]

integral(x^6/sqrt(x^4 + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{\sqrt {x^{4} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(x^4+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^6/sqrt(x^4 + 1), x)

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maple [C]  time = 0.01, size = 95, normalized size = 0.77 \[ \frac {\sqrt {x^{4}+1}\, x^{3}}{5}-\frac {3 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (-\EllipticE \left (\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) x , i\right )+\EllipticF \left (\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) x , i\right )\right )}{5 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(x^4+1)^(1/2),x)

[Out]

1/5*(x^4+1)^(1/2)*x^3-3/5*I/(1/2*2^(1/2)+1/2*I*2^(1/2))*(-I*x^2+1)^(1/2)*(I*x^2+1)^(1/2)/(x^4+1)^(1/2)*(Ellipt
icF((1/2*2^(1/2)+1/2*I*2^(1/2))*x,I)-EllipticE((1/2*2^(1/2)+1/2*I*2^(1/2))*x,I))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{\sqrt {x^{4} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(x^4+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^6/sqrt(x^4 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^6}{\sqrt {x^4+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(x^4 + 1)^(1/2),x)

[Out]

int(x^6/(x^4 + 1)^(1/2), x)

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sympy [C]  time = 0.91, size = 29, normalized size = 0.23 \[ \frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(x**4+1)**(1/2),x)

[Out]

x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), x**4*exp_polar(I*pi))/(4*gamma(11/4))

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