\(\int \frac {\sqrt [3]{x+x^4} (-2-x^3+x^6)}{x^6} \, dx\) [112]

Optimal result

Integrand size = 23, antiderivative size = 16 \[ \int \frac {\sqrt [3]{x+x^4} \left (-2-x^3+x^6\right )}{x^6} \, dx=\frac {3 \left (x+x^4\right )^{7/3}}{7 x^7} \]

[Out]

3/7*(x^4+x)^(7/3)/x^7

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(47\) vs. \(2(16)=32\).

Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.94, number of steps used = 12, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2077, 2029, 2057, 371, 2045, 2050} \[ \int \frac {\sqrt [3]{x+x^4} \left (-2-x^3+x^6\right )}{x^6} \, dx=\frac {3}{7} \sqrt [3]{x^4+x} x+\frac {3 \sqrt [3]{x^4+x}}{7 x^5}+\frac {6 \sqrt [3]{x^4+x}}{7 x^2} \]

[In]

Int[((x + x^4)^(1/3)*(-2 - x^3 + x^6))/x^6,x]

[Out]

(3*(x + x^4)^(1/3))/(7*x^5) + (6*(x + x^4)^(1/3))/(7*x^2) + (3*x*(x + x^4)^(1/3))/7

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2029

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j + b*x^n)^p/(n*p + 1)), x] + Dist[a
*(n - j)*(p/(n*p + 1)), Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0,
j, n] && GtQ[p, 0] && NeQ[n*p + 1, 0]

Rule 2045

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b*
x^n)^p/(c*(m + j*p + 1))), x] - Dist[b*p*((n - j)/(c^n*(m + j*p + 1))), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2050

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, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 2077

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = \int \left (\sqrt [3]{x+x^4}-\frac {2 \sqrt [3]{x+x^4}}{x^6}-\frac {\sqrt [3]{x+x^4}}{x^3}\right ) \, dx \\ & = -\left (2 \int \frac {\sqrt [3]{x+x^4}}{x^6} \, dx\right )+\int \sqrt [3]{x+x^4} \, dx-\int \frac {\sqrt [3]{x+x^4}}{x^3} \, dx \\ & = \frac {3 \sqrt [3]{x+x^4}}{7 x^5}+\frac {3 \sqrt [3]{x+x^4}}{5 x^2}+\frac {3}{7} x \sqrt [3]{x+x^4}-\frac {3}{7} \int \frac {1}{x^2 \left (x+x^4\right )^{2/3}} \, dx+\frac {3}{7} \int \frac {x}{\left (x+x^4\right )^{2/3}} \, dx-\frac {3}{5} \int \frac {x}{\left (x+x^4\right )^{2/3}} \, dx \\ & = \frac {3 \sqrt [3]{x+x^4}}{7 x^5}+\frac {6 \sqrt [3]{x+x^4}}{7 x^2}+\frac {3}{7} x \sqrt [3]{x+x^4}+\frac {6}{35} \int \frac {x}{\left (x+x^4\right )^{2/3}} \, dx+\frac {\left (3 x^{2/3} \left (1+x^3\right )^{2/3}\right ) \int \frac {\sqrt [3]{x}}{\left (1+x^3\right )^{2/3}} \, dx}{7 \left (x+x^4\right )^{2/3}}-\frac {\left (3 x^{2/3} \left (1+x^3\right )^{2/3}\right ) \int \frac {\sqrt [3]{x}}{\left (1+x^3\right )^{2/3}} \, dx}{5 \left (x+x^4\right )^{2/3}} \\ & = \frac {3 \sqrt [3]{x+x^4}}{7 x^5}+\frac {6 \sqrt [3]{x+x^4}}{7 x^2}+\frac {3}{7} x \sqrt [3]{x+x^4}-\frac {9 x^2 \left (1+x^3\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {4}{9},\frac {2}{3},\frac {13}{9},-x^3\right )}{70 \left (x+x^4\right )^{2/3}}+\frac {\left (6 x^{2/3} \left (1+x^3\right )^{2/3}\right ) \int \frac {\sqrt [3]{x}}{\left (1+x^3\right )^{2/3}} \, dx}{35 \left (x+x^4\right )^{2/3}} \\ & = \frac {3 \sqrt [3]{x+x^4}}{7 x^5}+\frac {6 \sqrt [3]{x+x^4}}{7 x^2}+\frac {3}{7} x \sqrt [3]{x+x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.91 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int \frac {\sqrt [3]{x+x^4} \left (-2-x^3+x^6\right )}{x^6} \, dx=\frac {3 \left (1+x^3\right )^2 \sqrt [3]{x+x^4}}{7 x^5} \]

[In]

Integrate[((x + x^4)^(1/3)*(-2 - x^3 + x^6))/x^6,x]

[Out]

(3*(1 + x^3)^2*(x + x^4)^(1/3))/(7*x^5)

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25

method	result	size
pseudoelliptic	\(\frac {3 \left (x^{3}+1\right )^{2} \left (x^{4}+x \right )^{\frac {1}{3}}}{7 x^{5}}\)	\(20\)
trager	\(\frac {3 \left (x^{6}+2 x^{3}+1\right ) \left (x^{4}+x \right )^{\frac {1}{3}}}{7 x^{5}}\)	\(23\)
gosper	\(\frac {3 \left (x^{3}+1\right ) \left (x^{4}+x \right )^{\frac {1}{3}} \left (1+x \right ) \left (x^{2}-x +1\right )}{7 x^{5}}\)	\(29\)
risch	\(\frac {3 {\left (x \left (x^{3}+1\right )\right )}^{\frac {1}{3}} \left (x^{9}+3 x^{6}+3 x^{3}+1\right )}{7 x^{5} \left (x^{3}+1\right )}\)	\(37\)
meijerg	\(\frac {3 x^{\frac {4}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {4}{9}\right ], \left [\frac {13}{9}\right ], -x^{3}\right )}{4}+\frac {3 \operatorname {hypergeom}\left (\left [-\frac {5}{9}, -\frac {1}{3}\right ], \left [\frac {4}{9}\right ], -x^{3}\right )}{5 x^{\frac {5}{3}}}+\frac {3 \operatorname {hypergeom}\left (\left [-\frac {14}{9}, -\frac {1}{3}\right ], \left [-\frac {5}{9}\right ], -x^{3}\right )}{7 x^{\frac {14}{3}}}\)	\(50\)

[In]

int((x^4+x)^(1/3)*(x^6-x^3-2)/x^6,x,method=_RETURNVERBOSE)

[Out]

3/7/x^5*(x^3+1)^2*(x^4+x)^(1/3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt [3]{x+x^4} \left (-2-x^3+x^6\right )}{x^6} \, dx=\frac {3 \, {\left (x^{6} + 2 \, x^{3} + 1\right )} {\left (x^{4} + x\right )}^{\frac {1}{3}}}{7 \, x^{5}} \]

[In]

integrate((x^4+x)^(1/3)*(x^6-x^3-2)/x^6,x, algorithm="fricas")

[Out]

3/7*(x^6 + 2*x^3 + 1)*(x^4 + x)^(1/3)/x^5

Sympy [F]

\[ \int \frac {\sqrt [3]{x+x^4} \left (-2-x^3+x^6\right )}{x^6} \, dx=\int \frac {\sqrt [3]{x \left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (x + 1\right ) \left (x^{3} - 2\right ) \left (x^{2} - x + 1\right )}{x^{6}}\, dx \]

[In]

integrate((x**4+x)**(1/3)*(x**6-x**3-2)/x**6,x)

[Out]

Integral((x*(x + 1)*(x**2 - x + 1))**(1/3)*(x + 1)*(x**3 - 2)*(x**2 - x + 1)/x**6, x)

Maxima [F]

\[ \int \frac {\sqrt [3]{x+x^4} \left (-2-x^3+x^6\right )}{x^6} \, dx=\int { \frac {{\left (x^{6} - x^{3} - 2\right )} {\left (x^{4} + x\right )}^{\frac {1}{3}}}{x^{6}} \,d x } \]

[In]

integrate((x^4+x)^(1/3)*(x^6-x^3-2)/x^6,x, algorithm="maxima")

[Out]

integrate((x^6 - x^3 - 2)*(x^4 + x)^(1/3)/x^6, x)

Giac [F]

\[ \int \frac {\sqrt [3]{x+x^4} \left (-2-x^3+x^6\right )}{x^6} \, dx=\int { \frac {{\left (x^{6} - x^{3} - 2\right )} {\left (x^{4} + x\right )}^{\frac {1}{3}}}{x^{6}} \,d x } \]

[In]

integrate((x^4+x)^(1/3)*(x^6-x^3-2)/x^6,x, algorithm="giac")

[Out]

integrate((x^6 - x^3 - 2)*(x^4 + x)^(1/3)/x^6, x)

Mupad [B] (verification not implemented)

Time = 5.45 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt [3]{x+x^4} \left (-2-x^3+x^6\right )}{x^6} \, dx=\frac {3\,{\left (x^3+1\right )}^2\,{\left (x^4+x\right )}^{1/3}}{7\,x^5} \]

[In]

int(-((x + x^4)^(1/3)*(x^3 - x^6 + 2))/x^6,x)

[Out]

(3*(x^3 + 1)^2*(x + x^4)^(1/3))/(7*x^5)