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

Optimal. Leaf size=97 \[ \log \left (\sqrt [3]{x^3+1}+x\right )+\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+1}-x}\right )+\frac {\left (x^3+1\right )^{2/3} \left (11 x^3-4\right )}{10 x^5}-\frac {1}{2} \log \left (-\sqrt [3]{x^3+1} x+\left (x^3+1\right )^{2/3}+x^2\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 107, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {580, 583, 12, 377, 200, 31, 634, 618, 204, 628} \begin {gather*} \log \left (\frac {x}{\sqrt [3]{x^3+1}}+1\right )-\sqrt {3} \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{x^3+1}}}{\sqrt {3}}\right )-\frac {2 \left (x^3+1\right )^{2/3}}{5 x^5}+\frac {11 \left (x^3+1\right )^{2/3}}{10 x^2}-\frac {1}{2} \log \left (-\frac {x}{\sqrt [3]{x^3+1}}+\frac {x^2}{\left (x^3+1\right )^{2/3}}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 + x^3)^(2/3))/(5*x^5) + (11*(1 + x^3)^(2/3))/(10*x^2) - Sqrt[3]*ArcTan[(1 - (2*x)/(1 + x^3)^(1/3))/Sqrt
[3]] - Log[1 + x^2/(1 + x^3)^(2/3) - x/(1 + x^3)^(1/3)]/2 + Log[1 + x/(1 + x^3)^(1/3)]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 377

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

Rule 580

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*g*(m + 1)), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {\left (1+x^3\right )^{2/3} \left (2+x^3\right )}{x^6 \left (1+2 x^3\right )} \, dx &=-\frac {2 \left (1+x^3\right )^{2/3}}{5 x^5}+\frac {1}{5} \int \frac {-11-7 x^3}{x^3 \sqrt [3]{1+x^3} \left (1+2 x^3\right )} \, dx\\ &=-\frac {2 \left (1+x^3\right )^{2/3}}{5 x^5}+\frac {11 \left (1+x^3\right )^{2/3}}{10 x^2}-\frac {1}{10} \int -\frac {30}{\sqrt [3]{1+x^3} \left (1+2 x^3\right )} \, dx\\ &=-\frac {2 \left (1+x^3\right )^{2/3}}{5 x^5}+\frac {11 \left (1+x^3\right )^{2/3}}{10 x^2}+3 \int \frac {1}{\sqrt [3]{1+x^3} \left (1+2 x^3\right )} \, dx\\ &=-\frac {2 \left (1+x^3\right )^{2/3}}{5 x^5}+\frac {11 \left (1+x^3\right )^{2/3}}{10 x^2}+3 \operatorname {Subst}\left (\int \frac {1}{1+x^3} \, dx,x,\frac {x}{\sqrt [3]{1+x^3}}\right )\\ &=-\frac {2 \left (1+x^3\right )^{2/3}}{5 x^5}+\frac {11 \left (1+x^3\right )^{2/3}}{10 x^2}+\operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\frac {x}{\sqrt [3]{1+x^3}}\right )+\operatorname {Subst}\left (\int \frac {2-x}{1-x+x^2} \, dx,x,\frac {x}{\sqrt [3]{1+x^3}}\right )\\ &=-\frac {2 \left (1+x^3\right )^{2/3}}{5 x^5}+\frac {11 \left (1+x^3\right )^{2/3}}{10 x^2}+\log \left (1+\frac {x}{\sqrt [3]{1+x^3}}\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\frac {x}{\sqrt [3]{1+x^3}}\right )+\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\frac {x}{\sqrt [3]{1+x^3}}\right )\\ &=-\frac {2 \left (1+x^3\right )^{2/3}}{5 x^5}+\frac {11 \left (1+x^3\right )^{2/3}}{10 x^2}-\frac {1}{2} \log \left (1+\frac {x^2}{\left (1+x^3\right )^{2/3}}-\frac {x}{\sqrt [3]{1+x^3}}\right )+\log \left (1+\frac {x}{\sqrt [3]{1+x^3}}\right )-3 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+\frac {2 x}{\sqrt [3]{1+x^3}}\right )\\ &=-\frac {2 \left (1+x^3\right )^{2/3}}{5 x^5}+\frac {11 \left (1+x^3\right )^{2/3}}{10 x^2}-\sqrt {3} \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )-\frac {1}{2} \log \left (1+\frac {x^2}{\left (1+x^3\right )^{2/3}}-\frac {x}{\sqrt [3]{1+x^3}}\right )+\log \left (1+\frac {x}{\sqrt [3]{1+x^3}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 106, normalized size = 1.09 \begin {gather*} 3 \left (\frac {1}{3} \log \left (\frac {x}{\sqrt [3]{x^3+1}}+1\right )+\frac {\tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{6} \log \left (-\frac {x}{\sqrt [3]{x^3+1}}+\frac {x^2}{\left (x^3+1\right )^{2/3}}+1\right )\right )+\left (x^3+1\right )^{2/3} \left (\frac {11}{10 x^2}-\frac {2}{5 x^5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2/(5*x^5) + 11/(10*x^2))*(1 + x^3)^(2/3) + 3*(ArcTan[(-1 + (2*x)/(1 + x^3)^(1/3))/Sqrt[3]]/Sqrt[3] - Log[1 +
x^2/(1 + x^3)^(2/3) - x/(1 + x^3)^(1/3)]/6 + Log[1 + x/(1 + x^3)^(1/3)]/3)

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IntegrateAlgebraic [A]  time = 0.19, size = 97, normalized size = 1.00 \begin {gather*} \frac {\left (1+x^3\right )^{2/3} \left (-4+11 x^3\right )}{10 x^5}+\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{-x+2 \sqrt [3]{1+x^3}}\right )+\log \left (x+\sqrt [3]{1+x^3}\right )-\frac {1}{2} \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)*(2 + x^3))/(x^6*(1 + 2*x^3)),x]

[Out]

((1 + x^3)^(2/3)*(-4 + 11*x^3))/(10*x^5) + Sqrt[3]*ArcTan[(Sqrt[3]*x)/(-x + 2*(1 + x^3)^(1/3))] + Log[x + (1 +
x^3)^(1/3)] - Log[x^2 - x*(1 + x^3)^(1/3) + (1 + x^3)^(2/3)]/2

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fricas [A]  time = 0.88, size = 124, normalized size = 1.28 \begin {gather*} -\frac {10 \, \sqrt {3} x^{5} \arctan \left (\frac {4 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} + 2 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {2}{3}} x + \sqrt {3} {\left (x^{3} + 1\right )}}{7 \, x^{3} - 1}\right ) - 5 \, x^{5} \log \left (\frac {2 \, x^{3} + 3 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} + 3 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} x + 1}{2 \, x^{3} + 1}\right ) - {\left (11 \, x^{3} - 4\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{10 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(10*sqrt(3)*x^5*arctan((4*sqrt(3)*(x^3 + 1)^(1/3)*x^2 + 2*sqrt(3)*(x^3 + 1)^(2/3)*x + sqrt(3)*(x^3 + 1))
/(7*x^3 - 1)) - 5*x^5*log((2*x^3 + 3*(x^3 + 1)^(1/3)*x^2 + 3*(x^3 + 1)^(2/3)*x + 1)/(2*x^3 + 1)) - (11*x^3 - 4
)*(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 (x^{3} + 2\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{{\left (2 \, x^{3} + 1\right )} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 4.27, size = 258, normalized size = 2.66

method	result	size
risch	\(\frac {11 x^{6}+7 x^{3}-4}{10 x^{5} \left (x^{3}+1\right )^{\frac {1}{3}}}+\ln \left (-\frac {3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x +6 \left (x^{3}+1\right )^{\frac {1}{3}} \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{3}-x \left (x^{3}+1\right )^{\frac {2}{3}}+x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}-1}{2 x^{3}+1}\right )+3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \ln \left (\frac {9 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{3}+3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x -3 \left (x^{3}+1\right )^{\frac {1}{3}} \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{3}+2 x \left (x^{3}+1\right )^{\frac {2}{3}}+x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}+3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{2 x^{3}+1}\right )\)	\(258\)
trager	\(\frac {\left (x^{3}+1\right )^{\frac {2}{3}} \left (11 x^{3}-4\right )}{10 x^{5}}+\ln \left (\frac {-18 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{3}+18 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x +63 \left (x^{3}+1\right )^{\frac {1}{3}} \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+21 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{3}-15 x \left (x^{3}+1\right )^{\frac {2}{3}}+6 x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}-5 x^{3}+18 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}-3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-10}{2 x^{3}+1}\right )+3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \ln \left (-\frac {45 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{3}+18 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x -45 \left (x^{3}+1\right )^{\frac {1}{3}} \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+24 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{3}+21 x \left (x^{3}+1\right )^{\frac {2}{3}}+6 x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}-4 x^{3}-45 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}+21 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-2}{2 x^{3}+1}\right )\)	\(327\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(11*x^6+7*x^3-4)/x^5/(x^3+1)^(1/3)+ln(-(3*RootOf(9*_Z^2+3*_Z+1)*(x^3+1)^(2/3)*x+6*(x^3+1)^(1/3)*RootOf(9*
_Z^2+3*_Z+1)*x^2+3*RootOf(9*_Z^2+3*_Z+1)*x^3-x*(x^3+1)^(2/3)+x^2*(x^3+1)^(1/3)-1)/(2*x^3+1))+3*RootOf(9*_Z^2+3
*_Z+1)*ln((9*RootOf(9*_Z^2+3*_Z+1)^2*x^3+3*RootOf(9*_Z^2+3*_Z+1)*(x^3+1)^(2/3)*x-3*(x^3+1)^(1/3)*RootOf(9*_Z^2
+3*_Z+1)*x^2+3*RootOf(9*_Z^2+3*_Z+1)*x^3+2*x*(x^3+1)^(2/3)+x^2*(x^3+1)^(1/3)+3*RootOf(9*_Z^2+3*_Z+1))/(2*x^3+1
))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^3+1\right )}^{2/3}\,\left (x^3+2\right )}{x^6\,\left (2\,x^3+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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