\(\int \frac {-4+x^3}{x^4 \sqrt [4]{-1+x^3}} \, dx\) [69]

Optimal result

Integrand size = 18, antiderivative size = 16 \[ \int \frac {-4+x^3}{x^4 \sqrt [4]{-1+x^3}} \, dx=-\frac {4 \left (-1+x^3\right )^{3/4}}{3 x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {457, 75} \[ \int \frac {-4+x^3}{x^4 \sqrt [4]{-1+x^3}} \, dx=-\frac {4 \left (x^3-1\right )^{3/4}}{3 x^3} \]

[In]

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

[Out]

(-4*(-1 + x^3)^(3/4))/(3*x^3)

Rule 75

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {-4+x}{\sqrt [4]{-1+x} x^2} \, dx,x,x^3\right ) \\ & = -\frac {4 \left (-1+x^3\right )^{3/4}}{3 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {-4+x^3}{x^4 \sqrt [4]{-1+x^3}} \, dx=-\frac {4 \left (-1+x^3\right )^{3/4}}{3 x^3} \]

[In]

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

[Out]

(-4*(-1 + x^3)^(3/4))/(3*x^3)

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
trager	\(-\frac {4 \left (x^{3}-1\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(13\)
risch	\(-\frac {4 \left (x^{3}-1\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(13\)
pseudoelliptic	\(-\frac {4 \left (x^{3}-1\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(13\)
gosper	\(-\frac {4 \left (x^{2}+x +1\right ) \left (x -1\right )}{3 x^{3} \left (x^{3}-1\right )^{\frac {1}{4}}}\)	\(22\)
meijerg	\(\frac {2 \sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{4}} \left (-\frac {5 \pi \sqrt {2}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {9}{4}\right ], \left [2, 3\right ], x^{3}\right )}{32 \Gamma \left (\frac {3}{4}\right )}-\frac {\left (3-3 \ln \left (2\right )-\frac {\pi }{2}+3 \ln \left (x \right )+i \pi \right ) \pi \sqrt {2}}{4 \Gamma \left (\frac {3}{4}\right )}+\frac {\pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right ) x^{3}}\right )}{3 \pi \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{4}} \left (\frac {\pi \sqrt {2}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{4}\right ], \left [2, 2\right ], x^{3}\right )}{4 \Gamma \left (\frac {3}{4}\right )}+\frac {\left (-3 \ln \left (2\right )-\frac {\pi }{2}+3 \ln \left (x \right )+i \pi \right ) \pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right )}\right )}{6 \pi \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{4}}}\)	\(172\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {-4+x^3}{x^4 \sqrt [4]{-1+x^3}} \, dx=-\frac {4 \, {\left (x^{3} - 1\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]

[In]

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

[Out]

-4/3*(x^3 - 1)^(3/4)/x^3

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.92 (sec) , antiderivative size = 68, normalized size of antiderivative = 4.25 \[ \int \frac {-4+x^3}{x^4 \sqrt [4]{-1+x^3}} \, dx=- \frac {\Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{3}}} \right )}}{3 x^{\frac {3}{4}} \Gamma \left (\frac {5}{4}\right )} + \frac {4 \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{3}}} \right )}}{3 x^{\frac {15}{4}} \Gamma \left (\frac {9}{4}\right )} \]

[In]

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

[Out]

-gamma(1/4)*hyper((1/4, 1/4), (5/4,), exp_polar(2*I*pi)/x**3)/(3*x**(3/4)*gamma(5/4)) + 4*gamma(5/4)*hyper((1/
4, 5/4), (9/4,), exp_polar(2*I*pi)/x**3)/(3*x**(15/4)*gamma(9/4))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {-4+x^3}{x^4 \sqrt [4]{-1+x^3}} \, dx=-\frac {4 \, {\left (x^{3} - 1\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]

[In]

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

[Out]

-4/3*(x^3 - 1)^(3/4)/x^3

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {-4+x^3}{x^4 \sqrt [4]{-1+x^3}} \, dx=-\frac {4 \, {\left (x^{3} - 1\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]

[In]

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

[Out]

-4/3*(x^3 - 1)^(3/4)/x^3

Mupad [B] (verification not implemented)

Time = 4.98 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {-4+x^3}{x^4 \sqrt [4]{-1+x^3}} \, dx=-\frac {4\,{\left (x^3-1\right )}^{3/4}}{3\,x^3} \]

[In]

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

[Out]

-(4*(x^3 - 1)^(3/4))/(3*x^3)