\(\int \frac {1}{x \sqrt [4]{-x^3+x^4}} \, dx\) [176]

Optimal result

Integrand size = 17, antiderivative size = 20 \[ \int \frac {1}{x \sqrt [4]{-x^3+x^4}} \, dx=\frac {4 \left (-x^3+x^4\right )^{3/4}}{3 x^3} \]

[Out]

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

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

[In]

Int[1/(x*(-x^3 + x^4)^(1/4)),x]

[Out]

(4*(-x^3 + x^4)^(3/4))/(3*x^3)

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 )^{3/4}}{3 x^3} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

Integrate[1/(x*(-x^3 + x^4)^(1/4)),x]

[Out]

(4*(-1 + x))/(3*((-1 + x)*x^3)^(1/4))

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {\frac {4 x}{3}-\frac {4}{3}}{\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}\)	\(15\)
pseudoelliptic	\(\frac {4 \left (x^{3} \left (x -1\right )\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(15\)
gosper	\(\frac {\frac {4 x}{3}-\frac {4}{3}}{\left (x^{4}-x^{3}\right )^{\frac {1}{4}}}\)	\(17\)
trager	\(\frac {4 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(17\)
meijerg	\(-\frac {4 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}} \left (1-x \right )^{\frac {3}{4}}}{3 \operatorname {signum}\left (x -1\right )^{\frac {1}{4}} x^{\frac {3}{4}}}\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x \sqrt [4]{-x^3+x^4}} \, dx=\frac {4 \, {\left (x^{4} - x^{3}\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]

[In]

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

[Out]

4/3*(x^4 - x^3)^(3/4)/x^3

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate(1/((x^4 - x^3)^(1/4)*x), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55 \[ \int \frac {1}{x \sqrt [4]{-x^3+x^4}} \, dx=\frac {4}{3} \, {\left (-\frac {1}{x} + 1\right )}^{\frac {3}{4}} \]

[In]

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

[Out]

4/3*(-1/x + 1)^(3/4)

Mupad [B] (verification not implemented)

Time = 5.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x \sqrt [4]{-x^3+x^4}} \, dx=\frac {4\,{\left (x^4-x^3\right )}^{3/4}}{3\,x^3} \]

[In]

int(1/(x*(x^4 - x^3)^(1/4)),x)

[Out]

(4*(x^4 - x^3)^(3/4))/(3*x^3)