\(\int \frac {1}{x^3 \sqrt [4]{-x+x^4}} \, dx\) [132]

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int \frac {1}{x^3 \sqrt [4]{-x+x^4}} \, dx=\frac {4 \left (-x+x^4\right )^{3/4}}{9 x^3} \]

[Out]

4/9*(x^4-x)^(3/4)/x^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2039} \[ \int \frac {1}{x^3 \sqrt [4]{-x+x^4}} \, dx=\frac {4 \left (x^4-x\right )^{3/4}}{9 x^3} \]

[In]

Int[1/(x^3*(-x + x^4)^(1/4)),x]

[Out]

(4*(-x + x^4)^(3/4))/(9*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])

Rubi steps \begin{align*} \text {integral}& = \frac {4 \left (-x+x^4\right )^{3/4}}{9 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 9.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 \sqrt [4]{-x+x^4}} \, dx=\frac {4 \left (x \left (-1+x^3\right )\right )^{3/4}}{9 x^3} \]

[In]

Integrate[1/(x^3*(-x + x^4)^(1/4)),x]

[Out]

(4*(x*(-1 + x^3))^(3/4))/(9*x^3)

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
trager	\(\frac {4 \left (x^{4}-x \right )^{\frac {3}{4}}}{9 x^{3}}\)	\(15\)
pseudoelliptic	\(\frac {4 \left (x^{4}-x \right )^{\frac {3}{4}}}{9 x^{3}}\)	\(15\)
risch	\(\frac {\frac {4 x^{3}}{9}-\frac {4}{9}}{x^{2} {\left (x \left (x^{3}-1\right )\right )}^{\frac {1}{4}}}\)	\(20\)
gosper	\(\frac {4 \left (x -1\right ) \left (x^{2}+x +1\right )}{9 x^{2} \left (x^{4}-x \right )^{\frac {1}{4}}}\)	\(24\)
meijerg	\(-\frac {4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{4}} \left (-x^{3}+1\right )^{\frac {3}{4}}}{9 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{4}} x^{\frac {9}{4}}}\)	\(33\)

[In]

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

[Out]

4/9*(x^4-x)^(3/4)/x^3

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^3 \sqrt [4]{-x+x^4}} \, dx=\frac {4 \, {\left (x^{4} - x\right )}^{\frac {3}{4}}}{9 \, x^{3}} \]

[In]

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

[Out]

4/9*(x^4 - x)^(3/4)/x^3

Sympy [F]

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

[In]

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

[Out]

Integral(1/(x**3*(x*(x - 1)*(x**2 + x + 1))**(1/4)), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {1}{x^3 \sqrt [4]{-x+x^4}} \, dx=\frac {4 \, {\left (x^{4} - x\right )}}{9 \, {\left (x^{2} + x + 1\right )}^{\frac {1}{4}} {\left (x - 1\right )}^{\frac {1}{4}} x^{\frac {13}{4}}} \]

[In]

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

[Out]

4/9*(x^4 - x)/((x^2 + x + 1)^(1/4)*(x - 1)^(1/4)*x^(13/4))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x^3 \sqrt [4]{-x+x^4}} \, dx=\frac {4}{9} \, {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {3}{4}} \]

[In]

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

[Out]

4/9*(-1/x^3 + 1)^(3/4)

Mupad [B] (verification not implemented)

Time = 4.98 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^3 \sqrt [4]{-x+x^4}} \, dx=\frac {4\,{\left (x^4-x\right )}^{3/4}}{9\,x^3} \]

[In]

int(1/(x^3*(x^4 - x)^(1/4)),x)

[Out]

(4*(x^4 - x)^(3/4))/(9*x^3)