\(\int \frac {1}{x^2 \sqrt {-x+x^4}} \, dx\) [131]

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int \frac {1}{x^2 \sqrt {-x+x^4}} \, dx=\frac {2 \sqrt {-x+x^4}}{3 x^2} \]

[Out]

2/3*(x^4-x)^(1/2)/x^2

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

[In]

Int[1/(x^2*Sqrt[-x + x^4]),x]

[Out]

(2*Sqrt[-x + x^4])/(3*x^2)

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 {2 \sqrt {-x+x^4}}{3 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \sqrt {-x+x^4}} \, dx=\frac {2 \sqrt {x \left (-1+x^3\right )}}{3 x^2} \]

[In]

Integrate[1/(x^2*Sqrt[-x + x^4]),x]

[Out]

(2*Sqrt[x*(-1 + x^3)])/(3*x^2)

Maple [A] (verified)

Time = 2.94 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

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

[In]

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

[Out]

2/3*(x^4-x)^(1/2)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^2 \sqrt {-x+x^4}} \, dx=\frac {2 \, \sqrt {x^{4} - x}}{3 \, x^{2}} \]

[In]

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

[Out]

2/3*sqrt(x^4 - x)/x^2

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {1}{x^2 \sqrt {-x+x^4}} \, dx=\frac {2 \, {\left (x^{4} - x\right )}}{3 \, \sqrt {x^{2} + x + 1} \sqrt {x - 1} x^{\frac {5}{2}}} \]

[In]

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

[Out]

2/3*(x^4 - x)/(sqrt(x^2 + x + 1)*sqrt(x - 1)*x^(5/2))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x^2 \sqrt {-x+x^4}} \, dx=\frac {2}{3} \, \sqrt {-\frac {1}{x^{3}} + 1} \]

[In]

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

[Out]

2/3*sqrt(-1/x^3 + 1)

Mupad [B] (verification not implemented)

Time = 4.96 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^2 \sqrt {-x+x^4}} \, dx=\frac {2\,\sqrt {x^4-x}}{3\,x^2} \]

[In]

int(1/(x^2*(x^4 - x)^(1/2)),x)

[Out]

(2*(x^4 - x)^(1/2))/(3*x^2)