Integrand size = 25, antiderivative size = 50 \[ \int \frac {\left (1+x^3\right ) \sqrt {-1+x^6}}{x \left (-1+x^3\right )} \, dx=\frac {1}{3} \sqrt {-1+x^6}+\frac {2}{3} \arctan \left (x^3+\sqrt {-1+x^6}\right )+\frac {2}{3} \log \left (x^3+\sqrt {-1+x^6}\right ) \]
Time = 0.09 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^3\right ) \sqrt {-1+x^6}}{x \left (-1+x^3\right )} \, dx=\frac {1}{3} \left (\sqrt {-1+x^6}-2 \arctan \left (x^3-\sqrt {-1+x^6}\right )-2 \log \left (-x^3+\sqrt {-1+x^6}\right )\right ) \]
Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.54, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1396, 948, 113, 175, 43, 103, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (x^3+1\right ) \sqrt {x^6-1}}{x \left (x^3-1\right )} \, dx\) |
\(\Big \downarrow \) 1396 |
\(\displaystyle \frac {\sqrt {x^6-1} \int \frac {\left (x^3+1\right )^{3/2}}{x \sqrt {x^3-1}}dx}{\sqrt {x^3-1} \sqrt {x^3+1}}\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {\sqrt {x^6-1} \int \frac {\left (x^3+1\right )^{3/2}}{x^3 \sqrt {x^3-1}}dx^3}{3 \sqrt {x^3-1} \sqrt {x^3+1}}\) |
\(\Big \downarrow \) 113 |
\(\displaystyle \frac {\sqrt {x^6-1} \left (\int \frac {2 x^3+1}{x^3 \sqrt {x^3-1} \sqrt {x^3+1}}dx^3+\sqrt {x^3-1} \sqrt {x^3+1}\right )}{3 \sqrt {x^3-1} \sqrt {x^3+1}}\) |
\(\Big \downarrow \) 175 |
\(\displaystyle \frac {\sqrt {x^6-1} \left (2 \int \frac {1}{\sqrt {x^3-1} \sqrt {x^3+1}}dx^3+\int \frac {1}{x^3 \sqrt {x^3-1} \sqrt {x^3+1}}dx^3+\sqrt {x^3-1} \sqrt {x^3+1}\right )}{3 \sqrt {x^3-1} \sqrt {x^3+1}}\) |
\(\Big \downarrow \) 43 |
\(\displaystyle \frac {\sqrt {x^6-1} \left (\int \frac {1}{x^3 \sqrt {x^3-1} \sqrt {x^3+1}}dx^3+2 \text {arccosh}\left (x^3\right )+\sqrt {x^3-1} \sqrt {x^3+1}\right )}{3 \sqrt {x^3-1} \sqrt {x^3+1}}\) |
\(\Big \downarrow \) 103 |
\(\displaystyle \frac {\sqrt {x^6-1} \left (\int \frac {1}{x^6+1}d\left (\sqrt {x^3-1} \sqrt {x^3+1}\right )+2 \text {arccosh}\left (x^3\right )+\sqrt {x^3-1} \sqrt {x^3+1}\right )}{3 \sqrt {x^3-1} \sqrt {x^3+1}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\sqrt {x^6-1} \left (2 \text {arccosh}\left (x^3\right )+\arctan \left (\sqrt {x^3-1} \sqrt {x^3+1}\right )+\sqrt {x^3-1} \sqrt {x^3+1}\right )}{3 \sqrt {x^3-1} \sqrt {x^3+1}}\) |
(Sqrt[-1 + x^6]*(Sqrt[-1 + x^3]*Sqrt[1 + x^3] + 2*ArcCosh[x^3] + ArcTan[Sq rt[-1 + x^3]*Sqrt[1 + x^3]]))/(3*Sqrt[-1 + x^3]*Sqrt[1 + x^3])
3.7.38.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ ArcCosh[b*(x/a)]/(b*Sqrt[d/b]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a *d, 0] && GtQ[a, 0] && GtQ[d/b, 0]
Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_ ))), x_] :> Simp[b*f Subst[Int[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sq rt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[2*b*d *e - f*(b*c + a*d), 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ )))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b Int[(c + d*x)^n*(e + f*x)^p, x] , x] + Simp[(b*g - a*h)/b Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_.), x _Symbol] :> Simp[(a + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p]) Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a* e^2, 0] && !IntegerQ[p] && !(EqQ[q, 1] && EqQ[n, 2])
Time = 1.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.70
method | result | size |
pseudoelliptic | \(\frac {\sqrt {x^{6}-1}}{3}-\frac {\arctan \left (\frac {1}{\sqrt {x^{6}-1}}\right )}{3}+\frac {2 \ln \left (x^{3}+\sqrt {x^{6}-1}\right )}{3}\) | \(35\) |
trager | \(\frac {\sqrt {x^{6}-1}}{3}+\frac {2 \ln \left (x^{3}+\sqrt {x^{6}-1}\right )}{3}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+\sqrt {x^{6}-1}}{x^{3}}\right )}{3}\) | \(52\) |
Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.84 \[ \int \frac {\left (1+x^3\right ) \sqrt {-1+x^6}}{x \left (-1+x^3\right )} \, dx=\frac {1}{3} \, \sqrt {x^{6} - 1} + \frac {2}{3} \, \arctan \left (-x^{3} + \sqrt {x^{6} - 1}\right ) - \frac {2}{3} \, \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) \]
\[ \int \frac {\left (1+x^3\right ) \sqrt {-1+x^6}}{x \left (-1+x^3\right )} \, dx=\int \frac {\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )} \left (x + 1\right ) \left (x^{2} - x + 1\right )}{x \left (x - 1\right ) \left (x^{2} + x + 1\right )}\, dx \]
Integral(sqrt((x - 1)*(x + 1)*(x**2 - x + 1)*(x**2 + x + 1))*(x + 1)*(x**2 - x + 1)/(x*(x - 1)*(x**2 + x + 1)), x)
\[ \int \frac {\left (1+x^3\right ) \sqrt {-1+x^6}}{x \left (-1+x^3\right )} \, dx=\int { \frac {\sqrt {x^{6} - 1} {\left (x^{3} + 1\right )}}{{\left (x^{3} - 1\right )} x} \,d x } \]
\[ \int \frac {\left (1+x^3\right ) \sqrt {-1+x^6}}{x \left (-1+x^3\right )} \, dx=\int { \frac {\sqrt {x^{6} - 1} {\left (x^{3} + 1\right )}}{{\left (x^{3} - 1\right )} x} \,d x } \]
Timed out. \[ \int \frac {\left (1+x^3\right ) \sqrt {-1+x^6}}{x \left (-1+x^3\right )} \, dx=\int \frac {\left (x^3+1\right )\,\sqrt {x^6-1}}{x\,\left (x^3-1\right )} \,d x \]