Integrand size = 20, antiderivative size = 55 \[ \int \frac {\left (1+2 x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\frac {1}{3} \left (1+x^3\right ) \sqrt {-1+x^6}-\frac {2}{3} \arctan \left (x^3+\sqrt {-1+x^6}\right )-\frac {1}{3} \log \left (x^3+\sqrt {-1+x^6}\right ) \]
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.09 \[ \int \frac {\left (1+2 x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\frac {1}{3} \left (\left (1+x^3\right ) \sqrt {-1+x^6}+2 \arctan \left (\frac {\sqrt {-1+x^6}}{-1+x^3}\right )-2 \text {arctanh}\left (\frac {\sqrt {-1+x^6}}{-1+x^3}\right )\right ) \]
((1 + x^3)*Sqrt[-1 + x^6] + 2*ArcTan[Sqrt[-1 + x^6]/(-1 + x^3)] - 2*ArcTan h[Sqrt[-1 + x^6]/(-1 + x^3)])/3
Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {1803, 535, 27, 538, 224, 219, 243, 73, 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.
\(\displaystyle \int \frac {\left (2 x^3+1\right ) \sqrt {x^6-1}}{x} \, dx\) |
\(\Big \downarrow \) 1803 |
\(\displaystyle \frac {1}{3} \int \frac {\left (2 x^3+1\right ) \sqrt {x^6-1}}{x^3}dx^3\) |
\(\Big \downarrow \) 535 |
\(\displaystyle \frac {1}{3} \left (\left (x^3+1\right ) \sqrt {x^6-1}-\frac {1}{2} \int \frac {2 \left (x^3+1\right )}{x^3 \sqrt {x^6-1}}dx^3\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (\left (x^3+1\right ) \sqrt {x^6-1}-\int \frac {x^3+1}{x^3 \sqrt {x^6-1}}dx^3\right )\) |
\(\Big \downarrow \) 538 |
\(\displaystyle \frac {1}{3} \left (-\int \frac {1}{\sqrt {x^6-1}}dx^3-\int \frac {1}{x^3 \sqrt {x^6-1}}dx^3+\sqrt {x^6-1} \left (x^3+1\right )\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{3} \left (-\int \frac {1}{1-x^6}d\frac {x^3}{\sqrt {x^6-1}}-\int \frac {1}{x^3 \sqrt {x^6-1}}dx^3+\sqrt {x^6-1} \left (x^3+1\right )\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (-\int \frac {1}{x^3 \sqrt {x^6-1}}dx^3-\text {arctanh}\left (\frac {x^3}{\sqrt {x^6-1}}\right )+\sqrt {x^6-1} \left (x^3+1\right )\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{3} \left (-\frac {1}{2} \int \frac {1}{x^3 \sqrt {x^6-1}}dx^6-\text {arctanh}\left (\frac {x^3}{\sqrt {x^6-1}}\right )+\sqrt {x^6-1} \left (x^3+1\right )\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (-\int \frac {1}{\sqrt {x^6-1}+1}d\sqrt {x^6-1}-\text {arctanh}\left (\frac {x^3}{\sqrt {x^6-1}}\right )+\sqrt {x^6-1} \left (x^3+1\right )\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{3} \left (-\arctan \left (\sqrt {x^6-1}\right )-\text {arctanh}\left (\frac {x^3}{\sqrt {x^6-1}}\right )+\sqrt {x^6-1} \left (x^3+1\right )\right )\) |
3.8.14.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_), x_Symbol] :> Sim p[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^p/(2*p*(2*p + 1))), x] + Simp[a/(2*p + 1) Int[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^(p - 1)/x), x], x] /; Free Q[{a, b, c, d}, x] && GtQ[p, 0] && IntegerQ[2*p]
Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp [c Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d Int[1/Sqrt[a + b*x^2], x] , x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x )^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.94 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85
method | result | size |
pseudoelliptic | \(\frac {x^{3} \sqrt {x^{6}-1}}{3}-\frac {\ln \left (x^{3}+\sqrt {x^{6}-1}\right )}{3}+\frac {\sqrt {x^{6}-1}}{3}+\frac {\arctan \left (\frac {1}{\sqrt {x^{6}-1}}\right )}{3}\) | \(47\) |
trager | \(\left (\frac {x^{3}}{3}+\frac {1}{3}\right ) \sqrt {x^{6}-1}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+\sqrt {x^{6}-1}}{x^{3}}\right )}{3}+\frac {\ln \left (-x^{3}+\sqrt {x^{6}-1}\right )}{3}\) | \(62\) |
meijerg | \(-\frac {\sqrt {\operatorname {signum}\left (x^{6}-1\right )}\, \left (4 \sqrt {\pi }-4 \sqrt {\pi }\, \sqrt {-x^{6}+1}+4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{6}+1}}{2}\right )-2 \left (2-2 \ln \left (2\right )+6 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }\right )}{12 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (x^{6}-1\right )}}+\frac {i \sqrt {\operatorname {signum}\left (x^{6}-1\right )}\, \left (-2 i \sqrt {\pi }\, x^{3} \sqrt {-x^{6}+1}-2 i \sqrt {\pi }\, \arcsin \left (x^{3}\right )\right )}{6 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (x^{6}-1\right )}}\) | \(136\) |
1/3*x^3*(x^6-1)^(1/2)-1/3*ln(x^3+(x^6-1)^(1/2))+1/3*(x^6-1)^(1/2)+1/3*arct an(1/(x^6-1)^(1/2))
Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {\left (1+2 x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\frac {1}{3} \, \sqrt {x^{6} - 1} {\left (x^{3} + 1\right )} - \frac {2}{3} \, \arctan \left (-x^{3} + \sqrt {x^{6} - 1}\right ) + \frac {1}{3} \, \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) \]
1/3*sqrt(x^6 - 1)*(x^3 + 1) - 2/3*arctan(-x^3 + sqrt(x^6 - 1)) + 1/3*log(- x^3 + sqrt(x^6 - 1))
Time = 4.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02 \[ \int \frac {\left (1+2 x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\frac {x^{3} \sqrt {x^{6} - 1}}{3} + \frac {\begin {cases} \sqrt {x^{6} - 1} - \operatorname {acos}{\left (\frac {1}{x^{3}} \right )} & \text {for}\: x^{3} > -1 \wedge x^{3} < 1 \end {cases}}{3} - \frac {\log {\left (2 x^{3} + 2 \sqrt {x^{6} - 1} \right )}}{3} \]
x**3*sqrt(x**6 - 1)/3 + Piecewise((sqrt(x**6 - 1) - acos(x**(-3)), (x**3 > -1) & (x**3 < 1)))/3 - log(2*x**3 + 2*sqrt(x**6 - 1))/3
Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.40 \[ \int \frac {\left (1+2 x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\frac {1}{3} \, \sqrt {x^{6} - 1} - \frac {\sqrt {x^{6} - 1}}{3 \, x^{3} {\left (\frac {x^{6} - 1}{x^{6}} - 1\right )}} - \frac {1}{3} \, \arctan \left (\sqrt {x^{6} - 1}\right ) - \frac {1}{6} \, \log \left (\frac {\sqrt {x^{6} - 1}}{x^{3}} + 1\right ) + \frac {1}{6} \, \log \left (\frac {\sqrt {x^{6} - 1}}{x^{3}} - 1\right ) \]
1/3*sqrt(x^6 - 1) - 1/3*sqrt(x^6 - 1)/(x^3*((x^6 - 1)/x^6 - 1)) - 1/3*arct an(sqrt(x^6 - 1)) - 1/6*log(sqrt(x^6 - 1)/x^3 + 1) + 1/6*log(sqrt(x^6 - 1) /x^3 - 1)
\[ \int \frac {\left (1+2 x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\int { \frac {\sqrt {x^{6} - 1} {\left (2 \, x^{3} + 1\right )}}{x} \,d x } \]
Time = 6.38 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98 \[ \int \frac {\left (1+2 x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\frac {\sqrt {x^6-1}}{3}-\frac {\ln \left (\sqrt {x^6-1}+x^3\right )}{3}+\frac {x^3\,\sqrt {x^6-1}}{3}-\frac {\ln \left (\frac {\sqrt {x^6-1}+1{}\mathrm {i}}{x^3}\right )\,1{}\mathrm {i}}{3} \]