\(\int \frac {1}{x^8 (1+x^4)^{3/2}} \, dx\) [952]

Optimal result

Integrand size = 13, antiderivative size = 92 \[ \int \frac {1}{x^8 \left (1+x^4\right )^{3/2}} \, dx=\frac {1}{2 x^7 \sqrt {1+x^4}}-\frac {9 \sqrt {1+x^4}}{14 x^7}+\frac {15 \sqrt {1+x^4}}{14 x^3}+\frac {15 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{28 \sqrt {1+x^4}} \]

[Out]

1/2/x^7/(x^4+1)^(1/2)-9/14*(x^4+1)^(1/2)/x^7+15/14*(x^4+1)^(1/2)/x^3+15/28*(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 = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {296, 331, 226} \[ \int \frac {1}{x^8 \left (1+x^4\right )^{3/2}} \, dx=\frac {15 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{28 \sqrt {x^4+1}}-\frac {9 \sqrt {x^4+1}}{14 x^7}+\frac {1}{2 x^7 \sqrt {x^4+1}}+\frac {15 \sqrt {x^4+1}}{14 x^3} \]

[In]

Int[1/(x^8*(1 + x^4)^(3/2)),x]

[Out]

1/(2*x^7*Sqrt[1 + x^4]) - (9*Sqrt[1 + x^4])/(14*x^7) + (15*Sqrt[1 + x^4])/(14*x^3) + (15*(1 + x^2)*Sqrt[(1 + x
^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(28*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 296

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

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 {1}{2 x^7 \sqrt {1+x^4}}+\frac {9}{2} \int \frac {1}{x^8 \sqrt {1+x^4}} \, dx \\ & = \frac {1}{2 x^7 \sqrt {1+x^4}}-\frac {9 \sqrt {1+x^4}}{14 x^7}-\frac {45}{14} \int \frac {1}{x^4 \sqrt {1+x^4}} \, dx \\ & = \frac {1}{2 x^7 \sqrt {1+x^4}}-\frac {9 \sqrt {1+x^4}}{14 x^7}+\frac {15 \sqrt {1+x^4}}{14 x^3}+\frac {15}{14} \int \frac {1}{\sqrt {1+x^4}} \, dx \\ & = \frac {1}{2 x^7 \sqrt {1+x^4}}-\frac {9 \sqrt {1+x^4}}{14 x^7}+\frac {15 \sqrt {1+x^4}}{14 x^3}+\frac {15 \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 )}{28 \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 = 0.24 \[ \int \frac {1}{x^8 \left (1+x^4\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {3}{2},-\frac {3}{4},-x^4\right )}{7 x^7} \]

[In]

Integrate[1/(x^8*(1 + x^4)^(3/2)),x]

[Out]

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

Maple [C] (verified)

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

Time = 4.33 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.18

method	result	size
meijerg	\(-\frac {{}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {7}{4},\frac {3}{2};-\frac {3}{4};-x^{4}\right )}{7 x^{7}}\)	\(17\)
risch	\(\frac {15 x^{8}+6 x^{4}-2}{14 x^{7} \sqrt {x^{4}+1}}+\frac {15 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{14 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(86\)
default	\(-\frac {\sqrt {x^{4}+1}}{7 x^{7}}+\frac {4 \sqrt {x^{4}+1}}{7 x^{3}}+\frac {x}{2 \sqrt {x^{4}+1}}+\frac {15 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{14 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(96\)
elliptic	\(-\frac {\sqrt {x^{4}+1}}{7 x^{7}}+\frac {4 \sqrt {x^{4}+1}}{7 x^{3}}+\frac {x}{2 \sqrt {x^{4}+1}}+\frac {15 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{14 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(96\)

[In]

int(1/x^8/(x^4+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^8 \left (1+x^4\right )^{3/2}} \, dx=-\frac {15 \, \sqrt {i} {\left (i \, x^{11} + i \, x^{7}\right )} F(\arcsin \left (\sqrt {i} x\right )\,|\,-1) - {\left (15 \, x^{8} + 6 \, x^{4} - 2\right )} \sqrt {x^{4} + 1}}{14 \, {\left (x^{11} + x^{7}\right )}} \]

[In]

integrate(1/x^8/(x^4+1)^(3/2),x, algorithm="fricas")

[Out]

-1/14*(15*sqrt(I)*(I*x^11 + I*x^7)*elliptic_f(arcsin(sqrt(I)*x), -1) - (15*x^8 + 6*x^4 - 2)*sqrt(x^4 + 1))/(x^
11 + x^7)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.59 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.39 \[ \int \frac {1}{x^8 \left (1+x^4\right )^{3/2}} \, dx=\frac {\Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, \frac {3}{2} \\ - \frac {3}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x^{7} \Gamma \left (- \frac {3}{4}\right )} \]

[In]

integrate(1/x**8/(x**4+1)**(3/2),x)

[Out]

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

Maxima [F]

\[ \int \frac {1}{x^8 \left (1+x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{4} + 1\right )}^{\frac {3}{2}} x^{8}} \,d x } \]

[In]

integrate(1/x^8/(x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((x^4 + 1)^(3/2)*x^8), x)

Giac [F]

\[ \int \frac {1}{x^8 \left (1+x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{4} + 1\right )}^{\frac {3}{2}} x^{8}} \,d x } \]

[In]

integrate(1/x^8/(x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((x^4 + 1)^(3/2)*x^8), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^8 \left (1+x^4\right )^{3/2}} \, dx=\int \frac {1}{x^8\,{\left (x^4+1\right )}^{3/2}} \,d x \]

[In]

int(1/(x^8*(x^4 + 1)^(3/2)),x)

[Out]

int(1/(x^8*(x^4 + 1)^(3/2)), x)