\(\int \frac {2+x}{x^2 \sqrt [4]{x^2+x^3}} \, dx\) [127]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, 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 = {1604} \[ \int \frac {2+x}{x^2 \sqrt [4]{x^2+x^3}} \, dx=-\frac {4 \left (x^3+x^2\right )^{3/4}}{3 x^3} \]

[In]

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

[Out]

(-4*(x^2 + x^3)^(3/4))/(3*x^3)

Rule 1604

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S
imp[Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*(Rr^(n + 1)/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x,
r])), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp
, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n
}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {4 \left (x^2+x^3\right )^{3/4}}{3 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {2+x}{x^2 \sqrt [4]{x^2+x^3}} \, dx=-\frac {4 \left (x^2 (1+x)\right )^{3/4}}{3 x^3} \]

[In]

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

[Out]

(-4*(x^2*(1 + x))^(3/4))/(3*x^3)

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
trager	\(-\frac {4 \left (x^{3}+x^{2}\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(15\)
gosper	\(-\frac {4 \left (1+x \right )}{3 x \left (x^{3}+x^{2}\right )^{\frac {1}{4}}}\)	\(18\)
risch	\(-\frac {4 \left (1+x \right )}{3 x \left (x^{2} \left (1+x \right )\right )^{\frac {1}{4}}}\)	\(18\)
meijerg	\(-\frac {4 \operatorname {hypergeom}\left (\left [-\frac {3}{2}, \frac {1}{4}\right ], \left [-\frac {1}{2}\right ], -x \right )}{3 x^{\frac {3}{2}}}-\frac {2 \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {1}{4}\right ], \left [\frac {1}{2}\right ], -x \right )}{\sqrt {x}}\)	\(30\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {2+x}{x^2 \sqrt [4]{x^2+x^3}} \, dx=-\frac {4 \, {\left (x^{3} + x^{2}\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]

[In]

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

[Out]

-4/3*(x^3 + x^2)^(3/4)/x^3

Sympy [F]

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

[In]

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

[Out]

Integral((x + 2)/(x**2*(x**2*(x + 1))**(1/4)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.61 \[ \int \frac {2+x}{x^2 \sqrt [4]{x^2+x^3}} \, dx=-\frac {4}{3} \, {\left (\frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {3}{4}} \]

[In]

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

[Out]

-4/3*(1/x + 1/x^2)^(3/4)

Mupad [B] (verification not implemented)

Time = 4.94 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {2+x}{x^2 \sqrt [4]{x^2+x^3}} \, dx=-\frac {4\,{\left (x^3+x^2\right )}^{3/4}}{3\,x^3} \]

[In]

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

[Out]

-(4*(x^2 + x^3)^(3/4))/(3*x^3)