\(\int \frac {1}{x^8 \sqrt [4]{-x^2+x^4}} \, dx\) [460]

Optimal result

Integrand size = 17, antiderivative size = 37 \[ \int \frac {1}{x^8 \sqrt [4]{-x^2+x^4}} \, dx=\frac {2 \left (-x^2+x^4\right )^{3/4} \left (77+84 x^2+96 x^4+128 x^6\right )}{1155 x^9} \]

[Out]

2/1155*(x^4-x^2)^(3/4)*(128*x^6+96*x^4+84*x^2+77)/x^9

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(81\) vs. \(2(37)=74\).

Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.19, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2041, 2039} \[ \int \frac {1}{x^8 \sqrt [4]{-x^2+x^4}} \, dx=\frac {2 \left (x^4-x^2\right )^{3/4}}{15 x^9}+\frac {8 \left (x^4-x^2\right )^{3/4}}{55 x^7}+\frac {64 \left (x^4-x^2\right )^{3/4}}{385 x^5}+\frac {256 \left (x^4-x^2\right )^{3/4}}{1155 x^3} \]

[In]

Int[1/(x^8*(-x^2 + x^4)^(1/4)),x]

[Out]

(2*(-x^2 + x^4)^(3/4))/(15*x^9) + (8*(-x^2 + x^4)^(3/4))/(55*x^7) + (64*(-x^2 + x^4)^(3/4))/(385*x^5) + (256*(
-x^2 + x^4)^(3/4))/(1155*x^3)

Rule 2039

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2041

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = \frac {2 \left (-x^2+x^4\right )^{3/4}}{15 x^9}+\frac {4}{5} \int \frac {1}{x^6 \sqrt [4]{-x^2+x^4}} \, dx \\ & = \frac {2 \left (-x^2+x^4\right )^{3/4}}{15 x^9}+\frac {8 \left (-x^2+x^4\right )^{3/4}}{55 x^7}+\frac {32}{55} \int \frac {1}{x^4 \sqrt [4]{-x^2+x^4}} \, dx \\ & = \frac {2 \left (-x^2+x^4\right )^{3/4}}{15 x^9}+\frac {8 \left (-x^2+x^4\right )^{3/4}}{55 x^7}+\frac {64 \left (-x^2+x^4\right )^{3/4}}{385 x^5}+\frac {128}{385} \int \frac {1}{x^2 \sqrt [4]{-x^2+x^4}} \, dx \\ & = \frac {2 \left (-x^2+x^4\right )^{3/4}}{15 x^9}+\frac {8 \left (-x^2+x^4\right )^{3/4}}{55 x^7}+\frac {64 \left (-x^2+x^4\right )^{3/4}}{385 x^5}+\frac {256 \left (-x^2+x^4\right )^{3/4}}{1155 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.34 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^8 \sqrt [4]{-x^2+x^4}} \, dx=\frac {2 \left (x^2 \left (-1+x^2\right )\right )^{3/4} \left (77+84 x^2+96 x^4+128 x^6\right )}{1155 x^9} \]

[In]

Integrate[1/(x^8*(-x^2 + x^4)^(1/4)),x]

[Out]

(2*(x^2*(-1 + x^2))^(3/4)*(77 + 84*x^2 + 96*x^4 + 128*x^6))/(1155*x^9)

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92

method	result	size
trager	\(\frac {2 \left (x^{4}-x^{2}\right )^{\frac {3}{4}} \left (128 x^{6}+96 x^{4}+84 x^{2}+77\right )}{1155 x^{9}}\)	\(34\)
pseudoelliptic	\(\frac {2 \left (x^{4}-x^{2}\right )^{\frac {3}{4}} \left (128 x^{6}+96 x^{4}+84 x^{2}+77\right )}{1155 x^{9}}\)	\(34\)
risch	\(\frac {-\frac {2}{165} x^{2}-\frac {2}{15}-\frac {8}{385} x^{4}-\frac {64}{1155} x^{6}+\frac {256}{1155} x^{8}}{x^{7} \left (x^{2} \left (x^{2}-1\right )\right )^{\frac {1}{4}}}\)	\(39\)
gosper	\(\frac {2 \left (1+x \right ) \left (x -1\right ) \left (128 x^{6}+96 x^{4}+84 x^{2}+77\right )}{1155 x^{7} \left (x^{4}-x^{2}\right )^{\frac {1}{4}}}\)	\(40\)
meijerg	\(-\frac {2 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{4}} \left (\frac {128}{77} x^{6}+\frac {96}{77} x^{4}+\frac {12}{11} x^{2}+1\right ) \left (-x^{2}+1\right )^{\frac {3}{4}}}{15 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{4}} x^{\frac {15}{2}}}\)	\(50\)

[In]

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

[Out]

2/1155*(x^4-x^2)^(3/4)*(128*x^6+96*x^4+84*x^2+77)/x^9

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^8 \sqrt [4]{-x^2+x^4}} \, dx=\frac {2 \, {\left (128 \, x^{6} + 96 \, x^{4} + 84 \, x^{2} + 77\right )} {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{1155 \, x^{9}} \]

[In]

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

[Out]

2/1155*(128*x^6 + 96*x^4 + 84*x^2 + 77)*(x^4 - x^2)^(3/4)/x^9

Sympy [F]

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

[In]

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

[Out]

Integral(1/(x**8*(x**2*(x - 1)*(x + 1))**(1/4)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^8 \sqrt [4]{-x^2+x^4}} \, dx=\frac {2 \, {\left (128 \, x^{9} - 32 \, x^{7} - 12 \, x^{5} - 7 \, x^{3} - 77 \, x\right )}}{1155 \, {\left (x + 1\right )}^{\frac {1}{4}} {\left (x - 1\right )}^{\frac {1}{4}} x^{\frac {17}{2}}} \]

[In]

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

[Out]

2/1155*(128*x^9 - 32*x^7 - 12*x^5 - 7*x^3 - 77*x)/((x + 1)^(1/4)*(x - 1)^(1/4)*x^(17/2))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.59 \[ \int \frac {1}{x^8 \sqrt [4]{-x^2+x^4}} \, dx=\frac {2}{15} \, {\left (\frac {1}{x^{2}} - 1\right )}^{3} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {3}{4}} + \frac {6}{11} \, {\left (\frac {1}{x^{2}} - 1\right )}^{2} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {3}{4}} - \frac {6}{7} \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {7}{4}} + \frac {2}{3} \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {3}{4}} \]

[In]

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

[Out]

2/15*(1/x^2 - 1)^3*(-1/x^2 + 1)^(3/4) + 6/11*(1/x^2 - 1)^2*(-1/x^2 + 1)^(3/4) - 6/7*(-1/x^2 + 1)^(7/4) + 2/3*(
-1/x^2 + 1)^(3/4)

Mupad [B] (verification not implemented)

Time = 4.99 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.76 \[ \int \frac {1}{x^8 \sqrt [4]{-x^2+x^4}} \, dx=\frac {256\,{\left (x^4-x^2\right )}^{3/4}}{1155\,x^3}+\frac {64\,{\left (x^4-x^2\right )}^{3/4}}{385\,x^5}+\frac {8\,{\left (x^4-x^2\right )}^{3/4}}{55\,x^7}+\frac {2\,{\left (x^4-x^2\right )}^{3/4}}{15\,x^9} \]

[In]

int(1/(x^8*(x^4 - x^2)^(1/4)),x)

[Out]

(256*(x^4 - x^2)^(3/4))/(1155*x^3) + (64*(x^4 - x^2)^(3/4))/(385*x^5) + (8*(x^4 - x^2)^(3/4))/(55*x^7) + (2*(x
^4 - x^2)^(3/4))/(15*x^9)