\(\int \frac {-1+x^8}{x^4 \sqrt [4]{-x^2+x^6}} \, dx\) [187]

Optimal result

Integrand size = 22, antiderivative size = 20 \[ \int \frac {-1+x^8}{x^4 \sqrt [4]{-x^2+x^6}} \, dx=\frac {2 \left (-x^2+x^6\right )^{7/4}}{7 x^7} \]

[Out]

2/7*(x^6-x^2)^(7/4)/x^7

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(41\) vs. \(2(20)=40\).

Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.05, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {2077, 2050, 2036, 335, 252, 251, 2049} \[ \int \frac {-1+x^8}{x^4 \sqrt [4]{-x^2+x^6}} \, dx=\frac {2 \left (x^6-x^2\right )^{3/4}}{7 x}-\frac {2 \left (x^6-x^2\right )^{3/4}}{7 x^5} \]

[In]

Int[(-1 + x^8)/(x^4*(-x^2 + x^6)^(1/4)),x]

[Out]

(-2*(-x^2 + x^6)^(3/4))/(7*x^5) + (2*(-x^2 + x^6)^(3/4))/(7*x)

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2049

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

Rule 2050

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, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2077

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{x^4 \sqrt [4]{-x^2+x^6}}+\frac {x^4}{\sqrt [4]{-x^2+x^6}}\right ) \, dx \\ & = -\int \frac {1}{x^4 \sqrt [4]{-x^2+x^6}} \, dx+\int \frac {x^4}{\sqrt [4]{-x^2+x^6}} \, dx \\ & = -\frac {2 \left (-x^2+x^6\right )^{3/4}}{7 x^5}+\frac {2 \left (-x^2+x^6\right )^{3/4}}{7 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x^8}{x^4 \sqrt [4]{-x^2+x^6}} \, dx=\frac {2 \left (x^2 \left (-1+x^4\right )\right )^{7/4}}{7 x^7} \]

[In]

Integrate[(-1 + x^8)/(x^4*(-x^2 + x^6)^(1/4)),x]

[Out]

(2*(x^2*(-1 + x^4))^(7/4))/(7*x^7)

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10

method	result	size
trager	\(\frac {2 \left (x^{4}-1\right ) \left (x^{6}-x^{2}\right )^{\frac {3}{4}}}{7 x^{5}}\)	\(22\)
pseudoelliptic	\(\frac {2 \left (x^{4}-1\right ) \left (x^{6}-x^{2}\right )^{\frac {3}{4}}}{7 x^{5}}\)	\(22\)
risch	\(\frac {\frac {2}{7} x^{8}-\frac {4}{7} x^{4}+\frac {2}{7}}{x^{3} \left (x^{2} \left (x^{4}-1\right )\right )^{\frac {1}{4}}}\)	\(27\)
gosper	\(\frac {2 \left (x^{4}-1\right ) \left (x -1\right ) \left (1+x \right ) \left (x^{2}+1\right )}{7 \left (x^{6}-x^{2}\right )^{\frac {1}{4}} x^{3}}\)	\(33\)
meijerg	\(\frac {2 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {7}{8}, \frac {1}{4}\right ], \left [\frac {1}{8}\right ], x^{4}\right )}{7 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{\frac {7}{2}}}+\frac {2 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} x^{\frac {9}{2}} \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {9}{8}\right ], \left [\frac {17}{8}\right ], x^{4}\right )}{9 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}}}\)	\(66\)

[In]

int((x^8-1)/x^4/(x^6-x^2)^(1/4),x,method=_RETURNVERBOSE)

[Out]

2/7*(x^4-1)/x^5*(x^6-x^2)^(3/4)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-1+x^8}{x^4 \sqrt [4]{-x^2+x^6}} \, dx=\frac {2 \, {\left (x^{6} - x^{2}\right )}^{\frac {3}{4}} {\left (x^{4} - 1\right )}}{7 \, x^{5}} \]

[In]

integrate((x^8-1)/x^4/(x^6-x^2)^(1/4),x, algorithm="fricas")

[Out]

2/7*(x^6 - x^2)^(3/4)*(x^4 - 1)/x^5

Sympy [F]

\[ \int \frac {-1+x^8}{x^4 \sqrt [4]{-x^2+x^6}} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )}{x^{4} \sqrt [4]{x^{2} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}}\, dx \]

[In]

integrate((x**8-1)/x**4/(x**6-x**2)**(1/4),x)

[Out]

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

Maxima [F]

\[ \int \frac {-1+x^8}{x^4 \sqrt [4]{-x^2+x^6}} \, dx=\int { \frac {x^{8} - 1}{{\left (x^{6} - x^{2}\right )}^{\frac {1}{4}} x^{4}} \,d x } \]

[In]

integrate((x^8-1)/x^4/(x^6-x^2)^(1/4),x, algorithm="maxima")

[Out]

integrate((x^8 - 1)/((x^6 - x^2)^(1/4)*x^4), x)

Giac [F]

\[ \int \frac {-1+x^8}{x^4 \sqrt [4]{-x^2+x^6}} \, dx=\int { \frac {x^{8} - 1}{{\left (x^{6} - x^{2}\right )}^{\frac {1}{4}} x^{4}} \,d x } \]

[In]

integrate((x^8-1)/x^4/(x^6-x^2)^(1/4),x, algorithm="giac")

[Out]

integrate((x^8 - 1)/((x^6 - x^2)^(1/4)*x^4), x)

Mupad [B] (verification not implemented)

Time = 5.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int \frac {-1+x^8}{x^4 \sqrt [4]{-x^2+x^6}} \, dx=\frac {2\,{\left (x^6-x^2\right )}^{3/4}}{7\,x}-\frac {2\,{\left (x^6-x^2\right )}^{3/4}}{7\,x^5} \]

[In]

int((x^8 - 1)/(x^4*(x^6 - x^2)^(1/4)),x)

[Out]

(2*(x^6 - x^2)^(3/4))/(7*x) - (2*(x^6 - x^2)^(3/4))/(7*x^5)