\(\int \frac {x^6}{\sqrt {1+x^4}} \, dx\) [933]

Optimal result

Integrand size = 13, antiderivative size = 124 \[ \int \frac {x^6}{\sqrt {1+x^4}} \, dx=\frac {1}{5} x^3 \sqrt {1+x^4}-\frac {3 x \sqrt {1+x^4}}{5 \left (1+x^2\right )}+\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{5 \sqrt {1+x^4}}-\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{10 \sqrt {1+x^4}} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {327, 311, 226, 1210} \[ \int \frac {x^6}{\sqrt {1+x^4}} \, dx=-\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{10 \sqrt {x^4+1}}+\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{5 \sqrt {x^4+1}}+\frac {1}{5} \sqrt {x^4+1} x^3-\frac {3 \sqrt {x^4+1} x}{5 \left (x^2+1\right )} \]

[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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

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 327

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 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps \begin{align*} \text {integral}& = \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*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.27 \[ \int \frac {x^6}{\sqrt {1+x^4}} \, dx=\frac {1}{5} x^3 \left (\sqrt {1+x^4}-\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-x^4\right )\right ) \]

[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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 4.44 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.14

method	result	size
meijerg	\(\frac {x^{7} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {7}{4};\frac {11}{4};-x^{4}\right )}{7}\)	\(17\)
default	\(\frac {x^{3} \sqrt {x^{4}+1}}{5}-\frac {3 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-E\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{5 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(95\)
risch	\(\frac {x^{3} \sqrt {x^{4}+1}}{5}-\frac {3 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-E\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{5 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(95\)
elliptic	\(\frac {x^{3} \sqrt {x^{4}+1}}{5}-\frac {3 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-E\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{5 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(95\)

[In]

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

[Out]

1/7*x^7*hypergeom([1/2,7/4],[11/4],-x^4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.41 \[ \int \frac {x^6}{\sqrt {1+x^4}} \, dx=\frac {-3 i \, \sqrt {i} x E(\arcsin \left (\frac {\sqrt {i}}{x}\right )\,|\,-1) + 3 i \, \sqrt {i} x F(\arcsin \left (\frac {\sqrt {i}}{x}\right )\,|\,-1) + \sqrt {x^{4} + 1} {\left (x^{4} - 3\right )}}{5 \, x} \]

[In]

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

[Out]

1/5*(-3*I*sqrt(I)*x*elliptic_e(arcsin(sqrt(I)/x), -1) + 3*I*sqrt(I)*x*elliptic_f(arcsin(sqrt(I)/x), -1) + sqrt
(x^4 + 1)*(x^4 - 3))/x

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.40 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.23 \[ \int \frac {x^6}{\sqrt {1+x^4}} \, dx=\frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} \]

[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))

Maxima [F]

\[ \int \frac {x^6}{\sqrt {1+x^4}} \, dx=\int { \frac {x^{6}}{\sqrt {x^{4} + 1}} \,d x } \]

[In]

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

[Out]

integrate(x^6/sqrt(x^4 + 1), x)

Giac [F]

\[ \int \frac {x^6}{\sqrt {1+x^4}} \, dx=\int { \frac {x^{6}}{\sqrt {x^{4} + 1}} \,d x } \]

[In]

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

[Out]

integrate(x^6/sqrt(x^4 + 1), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\sqrt {1+x^4}} \, dx=\int \frac {x^6}{\sqrt {x^4+1}} \,d x \]

[In]

int(x^6/(x^4 + 1)^(1/2),x)

[Out]

int(x^6/(x^4 + 1)^(1/2), x)