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

Optimal. Leaf size=16 \[ \frac {3 \left (x^4+x\right )^{7/3}}{7 x^7} \]

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Rubi [B]  time = 0.19, antiderivative size = 47, normalized size of antiderivative = 2.94, number of steps used = 12, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2052, 2004, 2032, 364, 2020, 2025} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2004

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 2020

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 2025

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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[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 2052

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*} \int \frac {\sqrt [3]{x+x^4} \left (-2-x^3+x^6\right )}{x^6} \, dx &=\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} \, _2F_1\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*}

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Mathematica [C]  time = 0.02, size = 83, normalized size = 5.19 \begin {gather*} \frac {3 \sqrt [3]{x^4+x} \left (28 x^3 \, _2F_1\left (-\frac {5}{9},-\frac {1}{3};\frac {4}{9};-x^3\right )+20 \, _2F_1\left (-\frac {14}{9},-\frac {1}{3};-\frac {5}{9};-x^3\right )+35 x^6 \, _2F_1\left (-\frac {1}{3},\frac {4}{9};\frac {13}{9};-x^3\right )\right )}{140 x^5 \sqrt [3]{x^3+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(x + x^4)^(1/3)*(20*Hypergeometric2F1[-14/9, -1/3, -5/9, -x^3] + 28*x^3*Hypergeometric2F1[-5/9, -1/3, 4/9,
-x^3] + 35*x^6*Hypergeometric2F1[-1/3, 4/9, 13/9, -x^3]))/(140*x^5*(1 + x^3)^(1/3))

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IntegrateAlgebraic [A]  time = 0.20, size = 16, normalized size = 1.00 \begin {gather*} \frac {3 \left (x+x^4\right )^{7/3}}{7 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(x + x^4)^(7/3))/(7*x^7)

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fricas [A]  time = 0.46, size = 22, normalized size = 1.38 \begin {gather*} \frac {3 \, {\left (x^{6} + 2 \, x^{3} + 1\right )} {\left (x^{4} + x\right )}^{\frac {1}{3}}}{7 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{6} - x^{3} - 2\right )} {\left (x^{4} + x\right )}^{\frac {1}{3}}}{x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.10, size = 23, normalized size = 1.44

method	result	size
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	\(\frac {3 \hypergeom \left (\left [-\frac {1}{3}, \frac {4}{9}\right ], \left [\frac {13}{9}\right ], -x^{3}\right ) x^{\frac {4}{3}}}{4}+\frac {3 \hypergeom \left (\left [-\frac {5}{9}, -\frac {1}{3}\right ], \left [\frac {4}{9}\right ], -x^{3}\right )}{5 x^{\frac {5}{3}}}+\frac {3 \hypergeom \left (\left [-\frac {14}{9}, -\frac {1}{3}\right ], \left [-\frac {5}{9}\right ], -x^{3}\right )}{7 x^{\frac {14}{3}}}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{6} - x^{3} - 2\right )} {\left (x^{4} + x\right )}^{\frac {1}{3}}}{x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.33, size = 19, normalized size = 1.19 \begin {gather*} \frac {3\,{\left (x^3+1\right )}^2\,{\left (x^4+x\right )}^{1/3}}{7\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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