\(\int \frac {(1+x^6)^{2/3}}{x^7} \, dx\) [1247]

Optimal result

Integrand size = 13, antiderivative size = 90 \[ \int \frac {\left (1+x^6\right )^{2/3}}{x^7} \, dx=-\frac {\left (1+x^6\right )^{2/3}}{6 x^6}+\frac {\arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^6}}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {1}{9} \log \left (-1+\sqrt [3]{1+x^6}\right )-\frac {1}{18} \log \left (1+\sqrt [3]{1+x^6}+\left (1+x^6\right )^{2/3}\right ) \]

[Out]

-1/6*(x^6+1)^(2/3)/x^6+1/9*arctan(1/3*3^(1/2)+2/3*(x^6+1)^(1/3)*3^(1/2))*3^(1/2)+1/9*ln(-1+(x^6+1)^(1/3))-1/18
*ln(1+(x^6+1)^(1/3)+(x^6+1)^(2/3))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.78, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {272, 43, 57, 632, 210, 31} \[ \int \frac {\left (1+x^6\right )^{2/3}}{x^7} \, dx=\frac {\arctan \left (\frac {2 \sqrt [3]{x^6+1}+1}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {\left (x^6+1\right )^{2/3}}{6 x^6}+\frac {1}{6} \log \left (1-\sqrt [3]{x^6+1}\right )-\frac {\log (x)}{3} \]

[In]

Int[(1 + x^6)^(2/3)/x^7,x]

[Out]

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

Rule 31

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

Rule 43

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

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 210

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

Rule 272

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

Rule 632

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} \text {Subst}\left (\int \frac {(1+x)^{2/3}}{x^2} \, dx,x,x^6\right ) \\ & = -\frac {\left (1+x^6\right )^{2/3}}{6 x^6}+\frac {1}{9} \text {Subst}\left (\int \frac {1}{x \sqrt [3]{1+x}} \, dx,x,x^6\right ) \\ & = -\frac {\left (1+x^6\right )^{2/3}}{6 x^6}-\frac {\log (x)}{3}-\frac {1}{6} \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1+x^6}\right )+\frac {1}{6} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{1+x^6}\right ) \\ & = -\frac {\left (1+x^6\right )^{2/3}}{6 x^6}-\frac {\log (x)}{3}+\frac {1}{6} \log \left (1-\sqrt [3]{1+x^6}\right )-\frac {1}{3} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1+x^6}\right ) \\ & = -\frac {\left (1+x^6\right )^{2/3}}{6 x^6}+\frac {\arctan \left (\frac {1+2 \sqrt [3]{1+x^6}}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {\log (x)}{3}+\frac {1}{6} \log \left (1-\sqrt [3]{1+x^6}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int \frac {\left (1+x^6\right )^{2/3}}{x^7} \, dx=\frac {1}{18} \left (-\frac {3 \left (1+x^6\right )^{2/3}}{x^6}+2 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1+x^6}}{\sqrt {3}}\right )+2 \log \left (-1+\sqrt [3]{1+x^6}\right )-\log \left (1+\sqrt [3]{1+x^6}+\left (1+x^6\right )^{2/3}\right )\right ) \]

[In]

Integrate[(1 + x^6)^(2/3)/x^7,x]

[Out]

((-3*(1 + x^6)^(2/3))/x^6 + 2*Sqrt[3]*ArcTan[(1 + 2*(1 + x^6)^(1/3))/Sqrt[3]] + 2*Log[-1 + (1 + x^6)^(1/3)] -
Log[1 + (1 + x^6)^(1/3) + (1 + x^6)^(2/3)])/18

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 6.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84

method	result	size
meijerg	\(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (\frac {\pi \sqrt {3}\, x^{6} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 3\right ], -x^{6}\right )}{9 \Gamma \left (\frac {2}{3}\right )}-\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}-1+6 \ln \left (x \right )\right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}+\frac {\pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right ) x^{6}}\right )}{18 \pi }\)	\(76\)
risch	\(-\frac {\left (x^{6}+1\right )^{\frac {2}{3}}}{6 x^{6}}+\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (-\frac {2 \pi \sqrt {3}\, x^{6} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 2\right ], -x^{6}\right )}{9 \Gamma \left (\frac {2}{3}\right )}+\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+6 \ln \left (x \right )\right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}\right )}{18 \pi }\)	\(76\)
pseudoelliptic	\(\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{6}+1\right )^{\frac {1}{3}}+1\right ) \sqrt {3}}{3}\right ) x^{6}+2 \ln \left (-1+\left (x^{6}+1\right )^{\frac {1}{3}}\right ) x^{6}-\ln \left (1+\left (x^{6}+1\right )^{\frac {1}{3}}+\left (x^{6}+1\right )^{\frac {2}{3}}\right ) x^{6}-3 \left (x^{6}+1\right )^{\frac {2}{3}}}{18 \left (-1+\left (x^{6}+1\right )^{\frac {1}{3}}\right ) \left (1+\left (x^{6}+1\right )^{\frac {1}{3}}+\left (x^{6}+1\right )^{\frac {2}{3}}\right )}\)	\(104\)
trager	\(-\frac {\left (x^{6}+1\right )^{\frac {2}{3}}}{6 x^{6}}+\frac {\ln \left (\frac {184653895 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{6}-538491663 x^{6} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )[phone] x^{6}[phone] \left (x^{6}+1\right )^{\frac {2}{3}} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-184653895 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-1030573260 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {1}{3}}[phone] \left (x^{6}+1\right )^{\frac {2}{3}}-754085602 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-1600004967 \left (x^{6}+1\right )^{\frac {1}{3}}[phone]}{x^{6}}\right )}{9}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\frac {169183873 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{6}[phone] x^{6} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-553961685 x^{6}-1600004967 \left (x^{6}+1\right )^{\frac {2}{3}} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-169183873 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}[phone] \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {1}{3}}-1030573260 \left (x^{6}+1\right )^{\frac {2}{3}}[phone] \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )[phone] \left (x^{6}+1\right )^{\frac {1}{3}}-738615580}{x^{6}}\right )}{9}\)	\(241\)

[In]

int((x^6+1)^(2/3)/x^7,x,method=_RETURNVERBOSE)

[Out]

-1/18/Pi*3^(1/2)*GAMMA(2/3)*(1/9*Pi*3^(1/2)/GAMMA(2/3)*x^6*hypergeom([1,1,4/3],[2,3],-x^6)-2/3*(-1/6*Pi*3^(1/2
)-3/2*ln(3)-1+6*ln(x))*Pi*3^(1/2)/GAMMA(2/3)+Pi*3^(1/2)/GAMMA(2/3)/x^6)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {\left (1+x^6\right )^{2/3}}{x^7} \, dx=\frac {2 \, \sqrt {3} x^{6} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{6} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - x^{6} \log \left ({\left (x^{6} + 1\right )}^{\frac {2}{3}} + {\left (x^{6} + 1\right )}^{\frac {1}{3}} + 1\right ) + 2 \, x^{6} \log \left ({\left (x^{6} + 1\right )}^{\frac {1}{3}} - 1\right ) - 3 \, {\left (x^{6} + 1\right )}^{\frac {2}{3}}}{18 \, x^{6}} \]

[In]

integrate((x^6+1)^(2/3)/x^7,x, algorithm="fricas")

[Out]

1/18*(2*sqrt(3)*x^6*arctan(2/3*sqrt(3)*(x^6 + 1)^(1/3) + 1/3*sqrt(3)) - x^6*log((x^6 + 1)^(2/3) + (x^6 + 1)^(1
/3) + 1) + 2*x^6*log((x^6 + 1)^(1/3) - 1) - 3*(x^6 + 1)^(2/3))/x^6

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.79 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.36 \[ \int \frac {\left (1+x^6\right )^{2/3}}{x^7} \, dx=- \frac {\Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{6}}} \right )}}{6 x^{2} \Gamma \left (\frac {4}{3}\right )} \]

[In]

integrate((x**6+1)**(2/3)/x**7,x)

[Out]

-gamma(1/3)*hyper((-2/3, 1/3), (4/3,), exp_polar(I*pi)/x**6)/(6*x**2*gamma(4/3))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {\left (1+x^6\right )^{2/3}}{x^7} \, dx=\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{6} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {{\left (x^{6} + 1\right )}^{\frac {2}{3}}}{6 \, x^{6}} - \frac {1}{18} \, \log \left ({\left (x^{6} + 1\right )}^{\frac {2}{3}} + {\left (x^{6} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{9} \, \log \left ({\left (x^{6} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

[In]

integrate((x^6+1)^(2/3)/x^7,x, algorithm="maxima")

[Out]

1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^6 + 1)^(1/3) + 1)) - 1/6*(x^6 + 1)^(2/3)/x^6 - 1/18*log((x^6 + 1)^(2/3) +
(x^6 + 1)^(1/3) + 1) + 1/9*log((x^6 + 1)^(1/3) - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {\left (1+x^6\right )^{2/3}}{x^7} \, dx=\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{6} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {{\left (x^{6} + 1\right )}^{\frac {2}{3}}}{6 \, x^{6}} - \frac {1}{18} \, \log \left ({\left (x^{6} + 1\right )}^{\frac {2}{3}} + {\left (x^{6} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{9} \, \log \left ({\left (x^{6} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

[In]

integrate((x^6+1)^(2/3)/x^7,x, algorithm="giac")

[Out]

1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^6 + 1)^(1/3) + 1)) - 1/6*(x^6 + 1)^(2/3)/x^6 - 1/18*log((x^6 + 1)^(2/3) +
(x^6 + 1)^(1/3) + 1) + 1/9*log((x^6 + 1)^(1/3) - 1)

Mupad [B] (verification not implemented)

Time = 5.64 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {\left (1+x^6\right )^{2/3}}{x^7} \, dx=\frac {\ln \left (\frac {{\left (x^6+1\right )}^{1/3}}{9}-\frac {1}{9}\right )}{9}+\ln \left (\frac {{\left (x^6+1\right )}^{1/3}}{9}-9\,{\left (-\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )}^2\right )\,\left (-\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )-\ln \left (\frac {{\left (x^6+1\right )}^{1/3}}{9}-9\,{\left (\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )}^2\right )\,\left (\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )-\frac {{\left (x^6+1\right )}^{2/3}}{6\,x^6} \]

[In]

int((x^6 + 1)^(2/3)/x^7,x)

[Out]

log((x^6 + 1)^(1/3)/9 - 1/9)/9 + log((x^6 + 1)^(1/3)/9 - 9*((3^(1/2)*1i)/18 - 1/18)^2)*((3^(1/2)*1i)/18 - 1/18
) - log((x^6 + 1)^(1/3)/9 - 9*((3^(1/2)*1i)/18 + 1/18)^2)*((3^(1/2)*1i)/18 + 1/18) - (x^6 + 1)^(2/3)/(6*x^6)