\(\int \frac {\sqrt [4]{-x^2+x^4}}{x^4} \, dx\) [174]

Optimal result

Integrand size = 17, antiderivative size = 20 \[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4} \, dx=\frac {2 \left (-x^2+x^4\right )^{5/4}}{5 x^5} \]

[Out]

2/5*(x^4-x^2)^(5/4)/x^5

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2039} \[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4} \, dx=\frac {2 \left (x^4-x^2\right )^{5/4}}{5 x^5} \]

[In]

Int[(-x^2 + x^4)^(1/4)/x^4,x]

[Out]

(2*(-x^2 + x^4)^(5/4))/(5*x^5)

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 \left (-x^2+x^4\right )^{5/4}}{5 x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4} \, dx=\frac {2 \left (x^2 \left (-1+x^2\right )\right )^{5/4}}{5 x^5} \]

[In]

Integrate[(-x^2 + x^4)^(1/4)/x^4,x]

[Out]

(2*(x^2*(-1 + x^2))^(5/4))/(5*x^5)

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10

method	result	size
trager	\(\frac {2 \left (x^{2}-1\right ) \left (x^{4}-x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}\)	\(22\)
pseudoelliptic	\(\frac {2 \left (x^{2}-1\right ) \left (x^{4}-x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}\)	\(22\)
gosper	\(\frac {2 \left (1+x \right ) \left (x -1\right ) \left (x^{4}-x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}\)	\(23\)
meijerg	\(-\frac {2 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{4}} \left (-x^{2}+1\right )^{\frac {5}{4}}}{5 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{4}} x^{\frac {5}{2}}}\)	\(33\)
risch	\(\frac {2 \left (x^{2} \left (x^{2}-1\right )\right )^{\frac {1}{4}} \left (x^{4}-2 x^{2}+1\right )}{5 x^{3} \left (x^{2}-1\right )}\)	\(34\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4} \, dx=\frac {2 \, {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} - 1\right )}}{5 \, x^{3}} \]

[In]

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

[Out]

2/5*(x^4 - x^2)^(1/4)*(x^2 - 1)/x^3

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4} \, dx=\frac {2 \, {\left (x^{3} - x\right )} {\left (x + 1\right )}^{\frac {1}{4}} {\left (x - 1\right )}^{\frac {1}{4}}}{5 \, x^{\frac {7}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4} \, dx=\frac {2}{5} \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {5}{4}} \]

[In]

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

[Out]

2/5*(-1/x^2 + 1)^(5/4)

Mupad [B] (verification not implemented)

Time = 5.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int \frac {\sqrt [4]{-x^2+x^4}}{x^4} \, dx=\frac {2\,{\left (x^4-x^2\right )}^{1/4}}{5\,x}-\frac {2\,{\left (x^4-x^2\right )}^{1/4}}{5\,x^3} \]

[In]

int((x^4 - x^2)^(1/4)/x^4,x)

[Out]

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