\(\int \frac {\sqrt [4]{-x+x^4}}{x^8} \, dx\) [321]

Optimal result

Integrand size = 15, antiderivative size = 28 \[ \int \frac {\sqrt [4]{-x+x^4}}{x^8} \, dx=\frac {4 \sqrt [4]{-x+x^4} \left (-5+x^3+4 x^6\right )}{135 x^7} \]

[Out]

4/135*(x^4-x)^(1/4)*(4*x^6+x^3-5)/x^7

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2041, 2039} \[ \int \frac {\sqrt [4]{-x+x^4}}{x^8} \, dx=\frac {4 \left (x^4-x\right )^{5/4}}{27 x^8}+\frac {16 \left (x^4-x\right )^{5/4}}{135 x^5} \]

[In]

Int[(-x + x^4)^(1/4)/x^8,x]

[Out]

(4*(-x + x^4)^(5/4))/(27*x^8) + (16*(-x + x^4)^(5/4))/(135*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])

Rule 2041

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, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = \frac {4 \left (-x+x^4\right )^{5/4}}{27 x^8}+\frac {4}{9} \int \frac {\sqrt [4]{-x+x^4}}{x^5} \, dx \\ & = \frac {4 \left (-x+x^4\right )^{5/4}}{27 x^8}+\frac {16 \left (-x+x^4\right )^{5/4}}{135 x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt [4]{-x+x^4}}{x^8} \, dx=\frac {4 \left (x \left (-1+x^3\right )\right )^{5/4} \left (5+4 x^3\right )}{135 x^8} \]

[In]

Integrate[(-x + x^4)^(1/4)/x^8,x]

[Out]

(4*(x*(-1 + x^3))^(5/4)*(5 + 4*x^3))/(135*x^8)

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method	result	size
trager	\(\frac {4 \left (x^{4}-x \right )^{\frac {1}{4}} \left (4 x^{6}+x^{3}-5\right )}{135 x^{7}}\)	\(25\)
pseudoelliptic	\(\frac {4 \left (x^{4}-x \right )^{\frac {1}{4}} \left (4 x^{6}+x^{3}-5\right )}{135 x^{7}}\)	\(25\)
gosper	\(\frac {4 \left (x -1\right ) \left (x^{2}+x +1\right ) \left (4 x^{3}+5\right ) \left (x^{4}-x \right )^{\frac {1}{4}}}{135 x^{7}}\)	\(31\)
risch	\(\frac {4 {\left (x \left (x^{3}-1\right )\right )}^{\frac {1}{4}} \left (4 x^{9}-3 x^{6}-6 x^{3}+5\right )}{135 x^{7} \left (x^{3}-1\right )}\)	\(39\)
meijerg	\(-\frac {4 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{4}} \left (-\frac {4}{5} x^{6}-\frac {1}{5} x^{3}+1\right ) \left (-x^{3}+1\right )^{\frac {1}{4}}}{27 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{4}} x^{\frac {27}{4}}}\)	\(45\)

[In]

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

[Out]

4/135*(x^4-x)^(1/4)*(4*x^6+x^3-5)/x^7

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt [4]{-x+x^4}}{x^8} \, dx=\frac {4 \, {\left (4 \, x^{6} + x^{3} - 5\right )} {\left (x^{4} - x\right )}^{\frac {1}{4}}}{135 \, x^{7}} \]

[In]

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

[Out]

4/135*(4*x^6 + x^3 - 5)*(x^4 - x)^(1/4)/x^7

Sympy [F]

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

[In]

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

[Out]

Integral((x*(x - 1)*(x**2 + x + 1))**(1/4)/x**8, x)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt [4]{-x+x^4}}{x^8} \, dx=\frac {4 \, {\left (4 \, x^{7} + x^{4} - 5 \, x\right )} {\left (x^{2} + x + 1\right )}^{\frac {1}{4}} {\left (x - 1\right )}^{\frac {1}{4}}}{135 \, x^{\frac {31}{4}}} \]

[In]

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

[Out]

4/135*(4*x^7 + x^4 - 5*x)*(x^2 + x + 1)^(1/4)*(x - 1)^(1/4)/x^(31/4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt [4]{-x+x^4}}{x^8} \, dx=-\frac {4}{27} \, {\left (\frac {1}{x^{3}} - 1\right )}^{2} {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}} + \frac {4}{15} \, {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {5}{4}} \]

[In]

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

[Out]

-4/27*(1/x^3 - 1)^2*(-1/x^3 + 1)^(1/4) + 4/15*(-1/x^3 + 1)^(5/4)

Mupad [B] (verification not implemented)

Time = 5.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt [4]{-x+x^4}}{x^8} \, dx=\frac {4\,{\left (x^4-x\right )}^{1/4}\,\left (4\,x^6+x^3-5\right )}{135\,x^7} \]

[In]

int((x^4 - x)^(1/4)/x^8,x)

[Out]

(4*(x^4 - x)^(1/4)*(x^3 + 4*x^6 - 5))/(135*x^7)