\(\int \frac {4+x^3}{x^2 (1+x^3)^{3/4}} \, dx\) [49]

Optimal result

Integrand size = 18, antiderivative size = 14 \[ \int \frac {4+x^3}{x^2 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4 \sqrt [4]{1+x^3}}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {460} \[ \int \frac {4+x^3}{x^2 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4 \sqrt [4]{x^3+1}}{x} \]

[In]

Int[(4 + x^3)/(x^2*(1 + x^3)^(3/4)),x]

[Out]

(-4*(1 + x^3)^(1/4))/x

Rule 460

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

Rubi steps \begin{align*} \text {integral}& = -\frac {4 \sqrt [4]{1+x^3}}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.90 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {4+x^3}{x^2 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4 \sqrt [4]{1+x^3}}{x} \]

[In]

Integrate[(4 + x^3)/(x^2*(1 + x^3)^(3/4)),x]

[Out]

(-4*(1 + x^3)^(1/4))/x

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
trager	\(-\frac {4 \left (x^{3}+1\right )^{\frac {1}{4}}}{x}\)	\(13\)
risch	\(-\frac {4 \left (x^{3}+1\right )^{\frac {1}{4}}}{x}\)	\(13\)
gosper	\(-\frac {4 \left (1+x \right ) \left (x^{2}-x +1\right )}{x \left (x^{3}+1\right )^{\frac {3}{4}}}\)	\(24\)
meijerg	\(\frac {x^{2} \operatorname {hypergeom}\left (\left [\frac {2}{3}, \frac {3}{4}\right ], \left [\frac {5}{3}\right ], -x^{3}\right )}{2}-\frac {4 \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {3}{4}\right ], \left [\frac {2}{3}\right ], -x^{3}\right )}{x}\)	\(34\)

[In]

int((x^3+4)/x^2/(x^3+1)^(3/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {4+x^3}{x^2 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4 \, {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{x} \]

[In]

integrate((x^3+4)/x^2/(x^3+1)^(3/4),x, algorithm="fricas")

[Out]

-4*(x^3 + 1)^(1/4)/x

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.80 (sec) , antiderivative size = 63, normalized size of antiderivative = 4.50 \[ \int \frac {4+x^3}{x^2 \left (1+x^3\right )^{3/4}} \, dx=\frac {x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {3}{4} \\ \frac {5}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} + \frac {4 \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{4} \\ \frac {2}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} \]

[In]

integrate((x**3+4)/x**2/(x**3+1)**(3/4),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43 \[ \int \frac {4+x^3}{x^2 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4 \, {\left (x^{2} - x + 1\right )}^{\frac {1}{4}} {\left (x + 1\right )}^{\frac {1}{4}}}{x} \]

[In]

integrate((x^3+4)/x^2/(x^3+1)^(3/4),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {4+x^3}{x^2 \left (1+x^3\right )^{3/4}} \, dx=\int { \frac {x^{3} + 4}{{\left (x^{3} + 1\right )}^{\frac {3}{4}} x^{2}} \,d x } \]

[In]

integrate((x^3+4)/x^2/(x^3+1)^(3/4),x, algorithm="giac")

[Out]

integrate((x^3 + 4)/((x^3 + 1)^(3/4)*x^2), x)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {4+x^3}{x^2 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4\,{\left (x^3+1\right )}^{1/4}}{x} \]

[In]

int((x^3 + 4)/(x^2*(x^3 + 1)^(3/4)),x)

[Out]

-(4*(x^3 + 1)^(1/4))/x