3.5.12 \(\int \frac {(1+x^3)^{2/3} (1-x^3+2 x^6)}{x^{12}} \, dx\)

Optimal. Leaf size=33 \[ \frac {\left (x^3+1\right )^{2/3} \left (-227 x^9-142 x^6+45 x^3-40\right )}{440 x^{11}} \]

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Rubi [A]  time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.48, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1486, 453, 271, 264} \begin {gather*} -\frac {\left (x^3+1\right )^{5/3}}{11 x^{11}}+\frac {17 \left (x^3+1\right )^{5/3}}{88 x^8}-\frac {227 \left (x^3+1\right )^{5/3}}{440 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 + x^3)^(2/3)*(1 - x^3 + 2*x^6))/x^12,x]

[Out]

-1/11*(1 + x^3)^(5/3)/x^11 + (17*(1 + x^3)^(5/3))/(88*x^8) - (227*(1 + x^3)^(5/3))/(440*x^5)

Rule 264

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

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 453

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] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 1486

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Sy
mbol] :> Simp[(c^p*(f*x)^(m + 2*n*p - n + 1)*(d + e*x^n)^(q + 1))/(e*f^(2*n*p - n + 1)*(m + 2*n*p + n*q + 1)),
x] + Dist[1/(e*(m + 2*n*p + n*q + 1)), Int[(f*x)^m*(d + e*x^n)^q*ExpandToSum[e*(m + 2*n*p + n*q + 1)*((a + b*
x^n + c*x^(2*n))^p - c^p*x^(2*n*p)) - d*c^p*(m + 2*n*p - n + 1)*x^(2*n*p - n), x], x], x] /; FreeQ[{a, b, c, d
, e, f, m, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IGtQ[p, 0] && GtQ[2*n*p, n - 1] &&
!IntegerQ[q] && NeQ[m + 2*n*p + n*q + 1, 0]

Rubi steps

\begin {align*} \int \frac {\left (1+x^3\right )^{2/3} \left (1-x^3+2 x^6\right )}{x^{12}} \, dx &=-\frac {2 \left (1+x^3\right )^{5/3}}{3 x^8}-\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3} \left (-3+19 x^3\right )}{x^{12}} \, dx\\ &=-\frac {\left (1+x^3\right )^{5/3}}{11 x^{11}}-\frac {2 \left (1+x^3\right )^{5/3}}{3 x^8}-\frac {227}{33} \int \frac {\left (1+x^3\right )^{2/3}}{x^9} \, dx\\ &=-\frac {\left (1+x^3\right )^{5/3}}{11 x^{11}}+\frac {17 \left (1+x^3\right )^{5/3}}{88 x^8}+\frac {227}{88} \int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx\\ &=-\frac {\left (1+x^3\right )^{5/3}}{11 x^{11}}+\frac {17 \left (1+x^3\right )^{5/3}}{88 x^8}-\frac {227 \left (1+x^3\right )^{5/3}}{440 x^5}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 33, normalized size = 1.00 \begin {gather*} -\frac {\left (x^3+1\right )^{2/3} \left (227 x^9+142 x^6-45 x^3+40\right )}{440 x^{11}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + x^3)^(2/3)*(1 - x^3 + 2*x^6))/x^12,x]

[Out]

-1/440*((1 + x^3)^(2/3)*(40 - 45*x^3 + 142*x^6 + 227*x^9))/x^11

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IntegrateAlgebraic [A]  time = 0.11, size = 33, normalized size = 1.00 \begin {gather*} \frac {\left (1+x^3\right )^{2/3} \left (-40+45 x^3-142 x^6-227 x^9\right )}{440 x^{11}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((1 + x^3)^(2/3)*(1 - x^3 + 2*x^6))/x^12,x]

[Out]

((1 + x^3)^(2/3)*(-40 + 45*x^3 - 142*x^6 - 227*x^9))/(440*x^11)

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fricas [A]  time = 0.64, size = 29, normalized size = 0.88 \begin {gather*} -\frac {{\left (227 \, x^{9} + 142 \, x^{6} - 45 \, x^{3} + 40\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{440 \, x^{11}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)^(2/3)*(2*x^6-x^3+1)/x^12,x, algorithm="fricas")

[Out]

-1/440*(227*x^9 + 142*x^6 - 45*x^3 + 40)*(x^3 + 1)^(2/3)/x^11

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{6} - x^{3} + 1\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{x^{12}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)^(2/3)*(2*x^6-x^3+1)/x^12,x, algorithm="giac")

[Out]

integrate((2*x^6 - x^3 + 1)*(x^3 + 1)^(2/3)/x^12, x)

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maple [A]  time = 0.07, size = 30, normalized size = 0.91

method	result	size
trager	\(-\frac {\left (227 x^{9}+142 x^{6}-45 x^{3}+40\right ) \left (x^{3}+1\right )^{\frac {2}{3}}}{440 x^{11}}\)	\(30\)
risch	\(-\frac {227 x^{12}+369 x^{9}+97 x^{6}-5 x^{3}+40}{440 x^{11} \left (x^{3}+1\right )^{\frac {1}{3}}}\)	\(35\)
gosper	\(-\frac {\left (x^{2}-x +1\right ) \left (1+x \right ) \left (227 x^{6}-85 x^{3}+40\right ) \left (x^{3}+1\right )^{\frac {2}{3}}}{440 x^{11}}\)	\(36\)
meijerg	\(-\frac {2 \left (x^{3}+1\right )^{\frac {5}{3}}}{5 x^{5}}+\frac {\left (-\frac {3}{5} x^{6}+\frac {2}{5} x^{3}+1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}}{8 x^{8}}-\frac {\left (\frac {9}{20} x^{9}-\frac {3}{10} x^{6}+\frac {1}{4} x^{3}+1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}}{11 x^{11}}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+1)^(2/3)*(2*x^6-x^3+1)/x^12,x,method=_RETURNVERBOSE)

[Out]

-1/440*(227*x^9+142*x^6-45*x^3+40)/x^11*(x^3+1)^(2/3)

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maxima [A]  time = 0.45, size = 37, normalized size = 1.12 \begin {gather*} -\frac {4 \, {\left (x^{3} + 1\right )}^{\frac {5}{3}}}{5 \, x^{5}} + \frac {3 \, {\left (x^{3} + 1\right )}^{\frac {8}{3}}}{8 \, x^{8}} - \frac {{\left (x^{3} + 1\right )}^{\frac {11}{3}}}{11 \, x^{11}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)^(2/3)*(2*x^6-x^3+1)/x^12,x, algorithm="maxima")

[Out]

-4/5*(x^3 + 1)^(5/3)/x^5 + 3/8*(x^3 + 1)^(8/3)/x^8 - 1/11*(x^3 + 1)^(11/3)/x^11

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mupad [B]  time = 0.41, size = 49, normalized size = 1.48 \begin {gather*} \frac {9\,{\left (x^3+1\right )}^{2/3}}{88\,x^8}-\frac {71\,{\left (x^3+1\right )}^{2/3}}{220\,x^5}-\frac {227\,{\left (x^3+1\right )}^{2/3}}{440\,x^2}-\frac {{\left (x^3+1\right )}^{2/3}}{11\,x^{11}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3 + 1)^(2/3)*(2*x^6 - x^3 + 1))/x^12,x)

[Out]

(9*(x^3 + 1)^(2/3))/(88*x^8) - (71*(x^3 + 1)^(2/3))/(220*x^5) - (227*(x^3 + 1)^(2/3))/(440*x^2) - (x^3 + 1)^(2
/3)/(11*x^11)

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sympy [B]  time = 3.94, size = 260, normalized size = 7.88 \begin {gather*} \frac {2 \left (1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 \Gamma \left (- \frac {2}{3}\right )} + \frac {2 \left (1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {11}{3}\right )}{3 \Gamma \left (- \frac {2}{3}\right )} - \frac {\left (x^{3} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{3 x^{2} \Gamma \left (- \frac {2}{3}\right )} + \frac {2 \left (1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 x^{3} \Gamma \left (- \frac {2}{3}\right )} - \frac {4 \left (1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {11}{3}\right )}{9 x^{3} \Gamma \left (- \frac {2}{3}\right )} + \frac {2 \left (x^{3} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{9 x^{5} \Gamma \left (- \frac {2}{3}\right )} + \frac {10 \left (1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {11}{3}\right )}{27 x^{6} \Gamma \left (- \frac {2}{3}\right )} + \frac {5 \left (x^{3} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{9 x^{8} \Gamma \left (- \frac {2}{3}\right )} + \frac {40 \left (1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {11}{3}\right )}{27 x^{9} \Gamma \left (- \frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+1)**(2/3)*(2*x**6-x**3+1)/x**12,x)

[Out]

2*(1 + x**(-3))**(2/3)*gamma(-5/3)/(3*gamma(-2/3)) + 2*(1 + x**(-3))**(2/3)*gamma(-11/3)/(3*gamma(-2/3)) - (x*
*3 + 1)**(2/3)*gamma(-8/3)/(3*x**2*gamma(-2/3)) + 2*(1 + x**(-3))**(2/3)*gamma(-5/3)/(3*x**3*gamma(-2/3)) - 4*
(1 + x**(-3))**(2/3)*gamma(-11/3)/(9*x**3*gamma(-2/3)) + 2*(x**3 + 1)**(2/3)*gamma(-8/3)/(9*x**5*gamma(-2/3))
+ 10*(1 + x**(-3))**(2/3)*gamma(-11/3)/(27*x**6*gamma(-2/3)) + 5*(x**3 + 1)**(2/3)*gamma(-8/3)/(9*x**8*gamma(-
2/3)) + 40*(1 + x**(-3))**(2/3)*gamma(-11/3)/(27*x**9*gamma(-2/3))

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