\(\int \frac {1}{(-1+x^3) \sqrt [4]{-x+x^4}} \, dx\) [205]

Optimal result

Integrand size = 19, antiderivative size = 22 \[ \int \frac {1}{\left (-1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=-\frac {4 \left (-x+x^4\right )^{3/4}}{3 \left (-1+x^3\right )} \]

[Out]

-4*(x^4-x)^(3/4)/(3*x^3-3)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2081, 270} \[ \int \frac {1}{\left (-1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=-\frac {4 x}{3 \sqrt [4]{x^4-x}} \]

[In]

Int[1/((-1 + x^3)*(-x + x^4)^(1/4)),x]

[Out]

(-4*x)/(3*(-x + x^4)^(1/4))

Rule 270

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] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

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

Mathematica [A] (verified)

Time = 10.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\left (-1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=-\frac {4 x}{3 \sqrt [4]{x \left (-1+x^3\right )}} \]

[In]

Integrate[1/((-1 + x^3)*(-x + x^4)^(1/4)),x]

[Out]

(-4*x)/(3*(x*(-1 + x^3))^(1/4))

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59

method	result	size
gosper	\(-\frac {4 x}{3 \left (x^{4}-x \right )^{\frac {1}{4}}}\)	\(13\)
risch	\(-\frac {4 x}{3 {\left (x \left (x^{3}-1\right )\right )}^{\frac {1}{4}}}\)	\(13\)
pseudoelliptic	\(-\frac {4 x}{3 \left (x^{4}-x \right )^{\frac {1}{4}}}\)	\(13\)
trager	\(-\frac {4 \left (x^{4}-x \right )^{\frac {3}{4}}}{3 \left (x^{3}-1\right )}\)	\(19\)
meijerg	\(-\frac {4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{4}} x^{\frac {3}{4}}}{3 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{4}} \left (-x^{3}+1\right )^{\frac {1}{4}}}\)	\(33\)

[In]

int(1/(x^3-1)/(x^4-x)^(1/4),x,method=_RETURNVERBOSE)

[Out]

-4/3*x/(x^4-x)^(1/4)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (-1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=-\frac {4 \, {\left (x^{4} - x\right )}^{\frac {3}{4}}}{3 \, {\left (x^{3} - 1\right )}} \]

[In]

integrate(1/(x^3-1)/(x^4-x)^(1/4),x, algorithm="fricas")

[Out]

-4/3*(x^4 - x)^(3/4)/(x^3 - 1)

Sympy [F]

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

[In]

integrate(1/(x**3-1)/(x**4-x)**(1/4),x)

[Out]

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

Maxima [F]

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

[In]

integrate(1/(x^3-1)/(x^4-x)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((x^4 - x)^(1/4)*(x^3 - 1)), x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\left (-1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=-\frac {4}{3 \, {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}}} \]

[In]

integrate(1/(x^3-1)/(x^4-x)^(1/4),x, algorithm="giac")

[Out]

-4/3/(-1/x^3 + 1)^(1/4)

Mupad [B] (verification not implemented)

Time = 5.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (-1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=-\frac {4\,{\left (x^4-x\right )}^{3/4}}{3\,\left (x^3-1\right )} \]

[In]

int(1/((x^4 - x)^(1/4)*(x^3 - 1)),x)

[Out]

-(4*(x^4 - x)^(3/4))/(3*(x^3 - 1))