\(\int \frac {1}{x^2 \sqrt [3]{x^2+x^3}} \, dx\) [212]

Optimal result

Integrand size = 15, antiderivative size = 23 \[ \int \frac {1}{x^2 \sqrt [3]{x^2+x^3}} \, dx=\frac {3 (-2+3 x) \left (x^2+x^3\right )^{2/3}}{10 x^3} \]

[Out]

3/10*(-2+3*x)*(x^3+x^2)^(2/3)/x^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2041, 2039} \[ \int \frac {1}{x^2 \sqrt [3]{x^2+x^3}} \, dx=\frac {9 \left (x^3+x^2\right )^{2/3}}{10 x^2}-\frac {3 \left (x^3+x^2\right )^{2/3}}{5 x^3} \]

[In]

Int[1/(x^2*(x^2 + x^3)^(1/3)),x]

[Out]

(-3*(x^2 + x^3)^(2/3))/(5*x^3) + (9*(x^2 + x^3)^(2/3))/(10*x^2)

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])

Rule 2041

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*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = -\frac {3 \left (x^2+x^3\right )^{2/3}}{5 x^3}-\frac {3}{5} \int \frac {1}{x \sqrt [3]{x^2+x^3}} \, dx \\ & = -\frac {3 \left (x^2+x^3\right )^{2/3}}{5 x^3}+\frac {9 \left (x^2+x^3\right )^{2/3}}{10 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \sqrt [3]{x^2+x^3}} \, dx=\frac {3 \left (x^2 (1+x)\right )^{2/3} (-2+3 x)}{10 x^3} \]

[In]

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

[Out]

(3*(x^2*(1 + x))^(2/3)*(-2 + 3*x))/(10*x^3)

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70

method	result	size
meijerg	\(-\frac {3 \left (1-\frac {3 x}{2}\right ) \left (1+x \right )^{\frac {2}{3}}}{5 x^{\frac {5}{3}}}\)	\(16\)
trager	\(\frac {3 \left (3 x -2\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}}{10 x^{3}}\)	\(20\)
pseudoelliptic	\(\frac {3 \left (3 x -2\right ) \left (x^{2} \left (1+x \right )\right )^{\frac {2}{3}}}{10 x^{3}}\)	\(20\)
gosper	\(\frac {3 \left (1+x \right ) \left (3 x -2\right )}{10 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}}\)	\(23\)
risch	\(\frac {-\frac {3}{5}+\frac {3}{10} x +\frac {9}{10} x^{2}}{x \left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}}}\)	\(23\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^2 \sqrt [3]{x^2+x^3}} \, dx=\frac {3 \, {\left (x^{3} + x^{2}\right )}^{\frac {2}{3}} {\left (3 \, x - 2\right )}}{10 \, x^{3}} \]

[In]

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

[Out]

3/10*(x^3 + x^2)^(2/3)*(3*x - 2)/x^3

Sympy [F]

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

[In]

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

[Out]

Integral(1/(x**2*(x**2*(x + 1))**(1/3)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^2 \sqrt [3]{x^2+x^3}} \, dx=-\frac {3}{5} \, {\left (\frac {1}{x} + 1\right )}^{\frac {5}{3}} + \frac {3}{2} \, {\left (\frac {1}{x} + 1\right )}^{\frac {2}{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.59 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {1}{x^2 \sqrt [3]{x^2+x^3}} \, dx=\frac {9\,x\,{\left (x^3+x^2\right )}^{2/3}-6\,{\left (x^3+x^2\right )}^{2/3}}{10\,x^3} \]

[In]

int(1/(x^2*(x^2 + x^3)^(1/3)),x)

[Out]

(9*x*(x^2 + x^3)^(2/3) - 6*(x^2 + x^3)^(2/3))/(10*x^3)