\(\int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x^4} \, dx\) [177]

Optimal result

Integrand size = 20, antiderivative size = 20 \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x^4} \, dx=\frac {4 \left (-x^3+x^4\right )^{9/4}}{9 x^9} \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(41\) vs. \(2(20)=40\).

Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.05, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2077, 2041, 2039} \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x^4} \, dx=\frac {4 \left (x^4-x^3\right )^{5/4}}{9 x^5}-\frac {4 \left (x^4-x^3\right )^{5/4}}{9 x^6} \]

[In]

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

[Out]

(-4*(-x^3 + x^4)^(5/4))/(9*x^6) + (4*(-x^3 + x^4)^(5/4))/(9*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])

Rule 2077

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {\sqrt [4]{-x^3+x^4}}{x^4}+\frac {\sqrt [4]{-x^3+x^4}}{x^3}\right ) \, dx \\ & = -\int \frac {\sqrt [4]{-x^3+x^4}}{x^4} \, dx+\int \frac {\sqrt [4]{-x^3+x^4}}{x^3} \, dx \\ & = -\frac {4 \left (-x^3+x^4\right )^{5/4}}{9 x^6}+\frac {4 \left (-x^3+x^4\right )^{5/4}}{5 x^5}-\frac {4}{9} \int \frac {\sqrt [4]{-x^3+x^4}}{x^3} \, dx \\ & = -\frac {4 \left (-x^3+x^4\right )^{5/4}}{9 x^6}+\frac {4 \left (-x^3+x^4\right )^{5/4}}{9 x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x^4} \, dx=\frac {4 \left ((-1+x) x^3\right )^{9/4}}{9 x^9} \]

[In]

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

[Out]

(4*((-1 + x)*x^3)^(9/4))/(9*x^9)

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
pseudoelliptic	\(\frac {4 \left (x -1\right )^{2} \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}{9 x^{3}}\)	\(20\)
gosper	\(\frac {4 \left (x -1\right )^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}}{9 x^{3}}\)	\(22\)
trager	\(\frac {4 \left (x^{2}-2 x +1\right ) \left (x^{4}-x^{3}\right )^{\frac {1}{4}}}{9 x^{3}}\)	\(25\)
risch	\(\frac {4 \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}} \left (x^{3}-3 x^{2}+3 x -1\right )}{9 \left (x -1\right ) x^{3}}\)	\(33\)
meijerg	\(\frac {4 \operatorname {signum}\left (x -1\right )^{\frac {1}{4}} \left (-\frac {4}{5} x^{2}-\frac {1}{5} x +1\right ) \left (1-x \right )^{\frac {1}{4}}}{9 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}} x^{\frac {9}{4}}}-\frac {4 \operatorname {signum}\left (x -1\right )^{\frac {1}{4}} \left (1-x \right )^{\frac {5}{4}}}{5 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}} x^{\frac {5}{4}}}\)	\(64\)

[In]

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

[Out]

4/9/x^3*(x-1)^2*(x^3*(x-1))^(1/4)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x^4} \, dx=\frac {4 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} {\left (x^{2} - 2 \, x + 1\right )}}{9 \, x^{3}} \]

[In]

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

[Out]

4/9*(x^4 - x^3)^(1/4)*(x^2 - 2*x + 1)/x^3

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x^4} \, dx=\frac {4}{9} \, {\left (\frac {1}{x} - 1\right )}^{2} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} \]

[In]

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

[Out]

4/9*(1/x - 1)^2*(-1/x + 1)^(1/4)

Mupad [B] (verification not implemented)

Time = 5.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.45 \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x^4} \, dx=\frac {4\,x^2\,{\left (x^4-x^3\right )}^{1/4}-8\,x\,{\left (x^4-x^3\right )}^{1/4}+4\,{\left (x^4-x^3\right )}^{1/4}}{9\,x^3} \]

[In]

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

[Out]

(4*x^2*(x^4 - x^3)^(1/4) - 8*x*(x^4 - x^3)^(1/4) + 4*(x^4 - x^3)^(1/4))/(9*x^3)