\(\int \frac {\sqrt [4]{-x^3+x^4}}{x^3} \, dx\) [178]

Optimal result

Integrand size = 17, antiderivative size = 20 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^3} \, dx=\frac {4 \left (-x^3+x^4\right )^{5/4}}{5 x^5} \]

[Out]

4/5*(x^4-x^3)^(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} \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^3} \, dx=\frac {4 \left (x^4-x^3\right )^{5/4}}{5 x^5} \]

[In]

Int[(-x^3 + x^4)^(1/4)/x^3,x]

[Out]

(4*(-x^3 + 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 {4 \left (-x^3+x^4\right )^{5/4}}{5 x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^3} \, dx=\frac {4 \left ((-1+x) x^3\right )^{5/4}}{5 x^5} \]

[In]

Integrate[(-x^3 + x^4)^(1/4)/x^3,x]

[Out]

(4*((-1 + x)*x^3)^(5/4))/(5*x^5)

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90

method	result	size
pseudoelliptic	\(\frac {4 \left (x -1\right ) \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}{5 x^{2}}\)	\(18\)
gosper	\(\frac {4 \left (x -1\right ) \left (x^{4}-x^{3}\right )^{\frac {1}{4}}}{5 x^{2}}\)	\(20\)
trager	\(\frac {4 \left (x -1\right ) \left (x^{4}-x^{3}\right )^{\frac {1}{4}}}{5 x^{2}}\)	\(20\)
meijerg	\(-\frac {4 \operatorname {signum}\left (x -1\right )^{\frac {1}{4}} \left (1-x \right )^{\frac {5}{4}}}{5 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}} x^{\frac {5}{4}}}\)	\(27\)
risch	\(\frac {4 \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}} \left (x^{2}-2 x +1\right )}{5 x^{2} \left (x -1\right )}\)	\(28\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^3} \, dx=\frac {4 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} {\left (x - 1\right )}}{5 \, x^{2}} \]

[In]

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

[Out]

4/5*(x^4 - x^3)^(1/4)*(x - 1)/x^2

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate((x^4 - x^3)^(1/4)/x^3, x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^3} \, dx=\frac {4}{5} \, {\left (-\frac {1}{x} + 1\right )}^{\frac {5}{4}} \]

[In]

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

[Out]

4/5*(-1/x + 1)^(5/4)

Mupad [B] (verification not implemented)

Time = 5.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{x^3} \, dx=\frac {4\,x\,{\left (x^4-x^3\right )}^{1/4}-4\,{\left (x^4-x^3\right )}^{1/4}}{5\,x^2} \]

[In]

int((x^4 - x^3)^(1/4)/x^3,x)

[Out]

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