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

Optimal. Leaf size=108 \[ -\frac {4}{9} \log \left (\sqrt [3]{x^3+1}-x\right )+\frac {4 \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+1}+x}\right )}{3 \sqrt {3}}+\frac {2}{9} \log \left (\sqrt [3]{x^3+1} x+\left (x^3+1\right )^{2/3}+x^2\right )+\frac {\left (x^3+1\right )^{2/3} \left (10 x^6-3 x^3-3\right )}{15 x^5} \]

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Rubi [A]  time = 0.03, antiderivative size = 97, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1487, 451, 277, 239} \begin {gather*} -\frac {2}{3} \log \left (\sqrt [3]{x^3+1}-x\right )+\frac {4 \tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {\left (x^3+1\right )^{5/3}}{5 x^5}+\frac {2 \left (x^3+1\right )^{5/3}}{3 x^2}-\frac {2 \left (x^3+1\right )^{2/3}}{3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 + x^3)^(2/3))/(3*x^2) - (1 + x^3)^(5/3)/(5*x^5) + (2*(1 + x^3)^(5/3))/(3*x^2) + (4*ArcTan[(1 + (2*x)/(1
+ x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[3]) - (2*Log[-x + (1 + x^3)^(1/3)])/3

Rule 239

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + (2*Rt[b, 3]*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 277

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

Rule 451

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

Rule 1487

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> 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 + 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, c, d, e, f, m, q}, x] && EqQ[n2,
2*n] && 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+2 x^6\right )}{x^6} \, dx &=\frac {2 \left (1+x^3\right )^{5/3}}{3 x^2}+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3} \left (3+4 x^3\right )}{x^6} \, dx\\ &=-\frac {\left (1+x^3\right )^{5/3}}{5 x^5}+\frac {2 \left (1+x^3\right )^{5/3}}{3 x^2}+\frac {4}{3} \int \frac {\left (1+x^3\right )^{2/3}}{x^3} \, dx\\ &=-\frac {2 \left (1+x^3\right )^{2/3}}{3 x^2}-\frac {\left (1+x^3\right )^{5/3}}{5 x^5}+\frac {2 \left (1+x^3\right )^{5/3}}{3 x^2}+\frac {4}{3} \int \frac {1}{\sqrt [3]{1+x^3}} \, dx\\ &=-\frac {2 \left (1+x^3\right )^{2/3}}{3 x^2}-\frac {\left (1+x^3\right )^{5/3}}{5 x^5}+\frac {2 \left (1+x^3\right )^{5/3}}{3 x^2}+\frac {4 \tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {2}{3} \log \left (-x+\sqrt [3]{1+x^3}\right )\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 35, normalized size = 0.32 \begin {gather*} 2 x \, _2F_1\left (-\frac {2}{3},\frac {1}{3};\frac {4}{3};-x^3\right )-\frac {\left (x^3+1\right )^{5/3}}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(1 + x^3)^(5/3)/x^5 + 2*x*Hypergeometric2F1[-2/3, 1/3, 4/3, -x^3]

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IntegrateAlgebraic [A]  time = 0.21, size = 108, normalized size = 1.00 \begin {gather*} \frac {\left (1+x^3\right )^{2/3} \left (-3-3 x^3+10 x^6\right )}{15 x^5}+\frac {4 \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{1+x^3}}\right )}{3 \sqrt {3}}-\frac {4}{9} \log \left (-x+\sqrt [3]{1+x^3}\right )+\frac {2}{9} \log \left (x^2+x \sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + x^3)^(2/3)*(-3 - 3*x^3 + 10*x^6))/(15*x^5) + (4*ArcTan[(Sqrt[3]*x)/(x + 2*(1 + x^3)^(1/3))])/(3*Sqrt[3])
- (4*Log[-x + (1 + x^3)^(1/3)])/9 + (2*Log[x^2 + x*(1 + x^3)^(1/3) + (1 + x^3)^(2/3)])/9

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fricas [A]  time = 0.92, size = 117, normalized size = 1.08 \begin {gather*} \frac {20 \, \sqrt {3} x^{5} \arctan \left (-\frac {25382 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} - 13720 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {2}{3}} x + \sqrt {3} {\left (5831 \, x^{3} + 7200\right )}}{58653 \, x^{3} + 8000}\right ) - 10 \, x^{5} \log \left (3 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} - 3 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} x + 1\right ) + 3 \, {\left (10 \, x^{6} - 3 \, x^{3} - 3\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{45 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/45*(20*sqrt(3)*x^5*arctan(-(25382*sqrt(3)*(x^3 + 1)^(1/3)*x^2 - 13720*sqrt(3)*(x^3 + 1)^(2/3)*x + sqrt(3)*(5
831*x^3 + 7200))/(58653*x^3 + 8000)) - 10*x^5*log(3*(x^3 + 1)^(1/3)*x^2 - 3*(x^3 + 1)^(2/3)*x + 1) + 3*(10*x^6
- 3*x^3 - 3)*(x^3 + 1)^(2/3))/x^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 2.98, size = 28, normalized size = 0.26

method	result	size
meijerg	\(2 x \hypergeom \left (\left [-\frac {2}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{3}\right )-\frac {\left (x^{3}+1\right )^{\frac {5}{3}}}{5 x^{5}}\)	\(28\)
risch	\(\frac {10 x^{9}+7 x^{6}-6 x^{3}-3}{15 x^{5} \left (x^{3}+1\right )^{\frac {1}{3}}}+\frac {4 x \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{3}\right )}{3}\)	\(45\)
trager	\(\frac {\left (x^{3}+1\right )^{\frac {2}{3}} \left (10 x^{6}-3 x^{3}-3\right )}{15 x^{5}}-\frac {4 \ln \left (317 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{3}-555 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x +2358 \left (x^{3}+1\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-2120 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}-1803 x \left (x^{3}+1\right )^{\frac {2}{3}}-555 x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}+2675 x^{3}-317 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}-99 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+1070\right )}{9}+\frac {4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (-535 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{3}+555 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x +1803 \left (x^{3}+1\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-1823 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}-2358 x \left (x^{3}+1\right )^{\frac {2}{3}}+555 x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}+1268 x^{3}+535 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}-1922 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+951\right )}{9}\)	\(287\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x*hypergeom([-2/3,1/3],[4/3],-x^3)-1/5*(x^3+1)^(5/3)/x^5

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maxima [A]  time = 0.41, size = 106, normalized size = 0.98 \begin {gather*} -\frac {4}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) + \frac {2 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{3 \, x^{2} {\left (\frac {x^{3} + 1}{x^{3}} - 1\right )}} - \frac {{\left (x^{3} + 1\right )}^{\frac {5}{3}}}{5 \, x^{5}} + \frac {2}{9} \, \log \left (\frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) - \frac {4}{9} \, \log \left (\frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 1)^(1/3)/x + 1)) + 2/3*(x^3 + 1)^(2/3)/(x^2*((x^3 + 1)/x^3 - 1)) - 1
/5*(x^3 + 1)^(5/3)/x^5 + 2/9*log((x^3 + 1)^(1/3)/x + (x^3 + 1)^(2/3)/x^2 + 1) - 4/9*log((x^3 + 1)^(1/3)/x - 1)

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mupad [B]  time = 1.12, size = 38, normalized size = 0.35 \begin {gather*} 2\,x\,{{}}_2{\mathrm {F}}_1\left (-\frac {2}{3},\frac {1}{3};\ \frac {4}{3};\ -x^3\right )-\frac {{\left (x^3+1\right )}^{2/3}+x^3\,{\left (x^3+1\right )}^{2/3}}{5\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x*hypergeom([-2/3, 1/3], 4/3, -x^3) - ((x^3 + 1)^(2/3) + x^3*(x^3 + 1)^(2/3))/(5*x^5)

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sympy [C]  time = 2.39, size = 85, normalized size = 0.79 \begin {gather*} \frac {2 x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {\left (1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 \Gamma \left (- \frac {2}{3}\right )} + \frac {\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x*gamma(1/3)*hyper((-2/3, 1/3), (4/3,), x**3*exp_polar(I*pi))/(3*gamma(4/3)) + (1 + x**(-3))**(2/3)*gamma(-5
/3)/(3*gamma(-2/3)) + (1 + x**(-3))**(2/3)*gamma(-5/3)/(3*x**3*gamma(-2/3))

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