\(\int \frac {1}{x^8 \sqrt {1+x^4}} \, dx\) [931]

Optimal result

Integrand size = 13, antiderivative size = 76 \[ \int \frac {1}{x^8 \sqrt {1+x^4}} \, dx=-\frac {\sqrt {1+x^4}}{7 x^7}+\frac {5 \sqrt {1+x^4}}{21 x^3}+\frac {5 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{42 \sqrt {1+x^4}} \]

[Out]

-1/7*(x^4+1)^(1/2)/x^7+5/21*(x^4+1)^(1/2)/x^3+5/42*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*Ellipti
cF(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 = 0.01 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {331, 226} \[ \int \frac {1}{x^8 \sqrt {1+x^4}} \, dx=\frac {5 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{42 \sqrt {x^4+1}}-\frac {\sqrt {x^4+1}}{7 x^7}+\frac {5 \sqrt {x^4+1}}{21 x^3} \]

[In]

Int[1/(x^8*Sqrt[1 + x^4]),x]

[Out]

-1/7*Sqrt[1 + x^4]/x^7 + (5*Sqrt[1 + x^4])/(21*x^3) + (5*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*Arc
Tan[x], 1/2])/(42*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 331

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {1+x^4}}{7 x^7}-\frac {5}{7} \int \frac {1}{x^4 \sqrt {1+x^4}} \, dx \\ & = -\frac {\sqrt {1+x^4}}{7 x^7}+\frac {5 \sqrt {1+x^4}}{21 x^3}+\frac {5}{21} \int \frac {1}{\sqrt {1+x^4}} \, dx \\ & = -\frac {\sqrt {1+x^4}}{7 x^7}+\frac {5 \sqrt {1+x^4}}{21 x^3}+\frac {5 \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 )}{42 \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 = 10.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.29 \[ \int \frac {1}{x^8 \sqrt {1+x^4}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {1}{2},-\frac {3}{4},-x^4\right )}{7 x^7} \]

[In]

Integrate[1/(x^8*Sqrt[1 + x^4]),x]

[Out]

-1/7*Hypergeometric2F1[-7/4, 1/2, -3/4, -x^4]/x^7

Maple [C] (verified)

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

Time = 4.32 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.22

method	result	size
meijerg	\(-\frac {{}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {7}{4},\frac {1}{2};-\frac {3}{4};-x^{4}\right )}{7 x^{7}}\)	\(17\)
default	\(-\frac {\sqrt {x^{4}+1}}{7 x^{7}}+\frac {5 \sqrt {x^{4}+1}}{21 x^{3}}+\frac {5 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{21 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(86\)
risch	\(\frac {5 x^{8}+2 x^{4}-3}{21 x^{7} \sqrt {x^{4}+1}}+\frac {5 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{21 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(86\)
elliptic	\(-\frac {\sqrt {x^{4}+1}}{7 x^{7}}+\frac {5 \sqrt {x^{4}+1}}{21 x^{3}}+\frac {5 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{21 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(86\)

[In]

int(1/x^8/(x^4+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/7/x^7*hypergeom([-7/4,1/2],[-3/4],-x^4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x^8 \sqrt {1+x^4}} \, dx=\frac {-5 i \, \sqrt {i} x^{7} F(\arcsin \left (\sqrt {i} x\right )\,|\,-1) + {\left (5 \, x^{4} - 3\right )} \sqrt {x^{4} + 1}}{21 \, x^{7}} \]

[In]

integrate(1/x^8/(x^4+1)^(1/2),x, algorithm="fricas")

[Out]

1/21*(-5*I*sqrt(I)*x^7*elliptic_f(arcsin(sqrt(I)*x), -1) + (5*x^4 - 3)*sqrt(x^4 + 1))/x^7

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.50 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^8 \sqrt {1+x^4}} \, dx=\frac {\Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, \frac {1}{2} \\ - \frac {3}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x^{7} \Gamma \left (- \frac {3}{4}\right )} \]

[In]

integrate(1/x**8/(x**4+1)**(1/2),x)

[Out]

gamma(-7/4)*hyper((-7/4, 1/2), (-3/4,), x**4*exp_polar(I*pi))/(4*x**7*gamma(-3/4))

Maxima [F]

\[ \int \frac {1}{x^8 \sqrt {1+x^4}} \, dx=\int { \frac {1}{\sqrt {x^{4} + 1} x^{8}} \,d x } \]

[In]

integrate(1/x^8/(x^4+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(x^4 + 1)*x^8), x)

Giac [F]

\[ \int \frac {1}{x^8 \sqrt {1+x^4}} \, dx=\int { \frac {1}{\sqrt {x^{4} + 1} x^{8}} \,d x } \]

[In]

integrate(1/x^8/(x^4+1)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(x^4 + 1)*x^8), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^8 \sqrt {1+x^4}} \, dx=\int \frac {1}{x^8\,\sqrt {x^4+1}} \,d x \]

[In]

int(1/(x^8*(x^4 + 1)^(1/2)),x)

[Out]

int(1/(x^8*(x^4 + 1)^(1/2)), x)