Integrand size = 25, antiderivative size = 53 \[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x^4 \left (1+x^3\right )} \, dx=\frac {\sqrt {-1+x^6}}{3 x^3}-\frac {4}{3} \arctan \left (x^3+\sqrt {-1+x^6}\right )+\frac {1}{3} \log \left (x^3+\sqrt {-1+x^6}\right ) \] Output:
1/3*(x^6-1)^(1/2)/x^3-4/3*arctan(x^3+(x^6-1)^(1/2))+1/3*ln(x^3+(x^6-1)^(1/ 2))
Time = 0.10 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.02 \[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x^4 \left (1+x^3\right )} \, dx=\frac {1}{3} \left (\frac {\sqrt {-1+x^6}}{x^3}+4 \arctan \left (x^3-\sqrt {-1+x^6}\right )-\log \left (-x^3+\sqrt {-1+x^6}\right )\right ) \] Input:
Integrate[((-1 + x^3)*Sqrt[-1 + x^6])/(x^4*(1 + x^3)),x]
Output:
(Sqrt[-1 + x^6]/x^3 + 4*ArcTan[x^3 - Sqrt[-1 + x^6]] - Log[-x^3 + Sqrt[-1 + x^6]])/3
Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.51, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1396, 948, 109, 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.
\(\displaystyle \int \frac {\left (x^3-1\right ) \sqrt {x^6-1}}{x^4 \left (x^3+1\right )} \, dx\) |
\(\Big \downarrow \) 1396 |
\(\displaystyle \frac {\sqrt {x^6-1} \int \frac {\left (x^3-1\right )^{3/2}}{x^4 \sqrt {x^3+1}}dx}{\sqrt {x^3-1} \sqrt {x^3+1}}\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {\sqrt {x^6-1} \int \frac {\left (x^3-1\right )^{3/2}}{x^6 \sqrt {x^3+1}}dx^3}{3 \sqrt {x^3-1} \sqrt {x^3+1}}\) |
\(\Big \downarrow \) 109 |
\(\displaystyle \frac {\sqrt {x^6-1} \left (\frac {\sqrt {x^3-1} \sqrt {x^3+1}}{x^3}-\int \frac {2-x^3}{x^3 \sqrt {x^3-1} \sqrt {x^3+1}}dx^3\right )}{3 \sqrt {x^3-1} \sqrt {x^3+1}}\) |
\(\Big \downarrow \) 175 |
\(\displaystyle \frac {\sqrt {x^6-1} \left (\int \frac {1}{\sqrt {x^3-1} \sqrt {x^3+1}}dx^3-2 \int \frac {1}{x^3 \sqrt {x^3-1} \sqrt {x^3+1}}dx^3+\frac {\sqrt {x^3-1} \sqrt {x^3+1}}{x^3}\right )}{3 \sqrt {x^3-1} \sqrt {x^3+1}}\) |
\(\Big \downarrow \) 43 |
\(\displaystyle \frac {\sqrt {x^6-1} \left (-2 \int \frac {1}{x^3 \sqrt {x^3-1} \sqrt {x^3+1}}dx^3+\text {arccosh}\left (x^3\right )+\frac {\sqrt {x^3-1} \sqrt {x^3+1}}{x^3}\right )}{3 \sqrt {x^3-1} \sqrt {x^3+1}}\) |
\(\Big \downarrow \) 103 |
\(\displaystyle \frac {\sqrt {x^6-1} \left (-2 \int \frac {1}{x^6+1}d\left (\sqrt {x^3-1} \sqrt {x^3+1}\right )+\text {arccosh}\left (x^3\right )+\frac {\sqrt {x^3-1} \sqrt {x^3+1}}{x^3}\right )}{3 \sqrt {x^3-1} \sqrt {x^3+1}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\sqrt {x^6-1} \left (\text {arccosh}\left (x^3\right )-2 \arctan \left (\sqrt {x^3-1} \sqrt {x^3+1}\right )+\frac {\sqrt {x^3-1} \sqrt {x^3+1}}{x^3}\right )}{3 \sqrt {x^3-1} \sqrt {x^3+1}}\) |
Input:
Int[((-1 + x^3)*Sqrt[-1 + x^6])/(x^4*(1 + x^3)),x]
Output:
(Sqrt[-1 + x^6]*((Sqrt[-1 + x^3]*Sqrt[1 + x^3])/x^3 + ArcCosh[x^3] - 2*Arc Tan[Sqrt[-1 + x^3]*Sqrt[1 + x^3]]))/(3*Sqrt[-1 + x^3]*Sqrt[1 + x^3])
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*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f *x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
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 = 0.51 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81
method | result | size |
pseudoelliptic | \(\frac {\ln \left (x^{3}+\sqrt {x^{6}-1}\right ) x^{3}+2 \arctan \left (\frac {1}{\sqrt {x^{6}-1}}\right ) x^{3}+\sqrt {x^{6}-1}}{3 x^{3}}\) | \(43\) |
trager | \(\frac {\sqrt {x^{6}-1}}{3 x^{3}}+\frac {\ln \left (x^{3}+\sqrt {x^{6}-1}\right )}{3}-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+\sqrt {x^{6}-1}}{x^{3}}\right )}{3}\) | \(55\) |
risch | \(\frac {\sqrt {x^{6}-1}}{3 x^{3}}+\frac {\sqrt {-\operatorname {signum}\left (x^{6}-1\right )}\, \arcsin \left (x^{3}\right )}{3 \sqrt {\operatorname {signum}\left (x^{6}-1\right )}}-\frac {\sqrt {-\operatorname {signum}\left (x^{6}-1\right )}\, \left (\left (-2 \ln \left (2\right )+6 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{6}+1}}{2}\right )\right )}{3 \sqrt {\pi }\, \sqrt {\operatorname {signum}\left (x^{6}-1\right )}}\) | \(98\) |
Input:
int((x^3-1)*(x^6-1)^(1/2)/x^4/(x^3+1),x,method=_RETURNVERBOSE)
Output:
1/3*(ln(x^3+(x^6-1)^(1/2))*x^3+2*arctan(1/(x^6-1)^(1/2))*x^3+(x^6-1)^(1/2) )/x^3
Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.08 \[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x^4 \left (1+x^3\right )} \, dx=-\frac {4 \, x^{3} \arctan \left (-x^{3} + \sqrt {x^{6} - 1}\right ) + x^{3} \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) - x^{3} - \sqrt {x^{6} - 1}}{3 \, x^{3}} \] Input:
integrate((x^3-1)*(x^6-1)^(1/2)/x^4/(x^3+1),x, algorithm="fricas")
Output:
-1/3*(4*x^3*arctan(-x^3 + sqrt(x^6 - 1)) + x^3*log(-x^3 + sqrt(x^6 - 1)) - x^3 - sqrt(x^6 - 1))/x^3
\[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x^4 \left (1+x^3\right )} \, dx=\int \frac {\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )} \left (x - 1\right ) \left (x^{2} + x + 1\right )}{x^{4} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \] Input:
integrate((x**3-1)*(x**6-1)**(1/2)/x**4/(x**3+1),x)
Output:
Integral(sqrt((x - 1)*(x + 1)*(x**2 - x + 1)*(x**2 + x + 1))*(x - 1)*(x**2 + x + 1)/(x**4*(x + 1)*(x**2 - x + 1)), x)
\[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x^4 \left (1+x^3\right )} \, dx=\int { \frac {\sqrt {x^{6} - 1} {\left (x^{3} - 1\right )}}{{\left (x^{3} + 1\right )} x^{4}} \,d x } \] Input:
integrate((x^3-1)*(x^6-1)^(1/2)/x^4/(x^3+1),x, algorithm="maxima")
Output:
integrate(sqrt(x^6 - 1)*(x^3 - 1)/((x^3 + 1)*x^4), x)
\[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x^4 \left (1+x^3\right )} \, dx=\int { \frac {\sqrt {x^{6} - 1} {\left (x^{3} - 1\right )}}{{\left (x^{3} + 1\right )} x^{4}} \,d x } \] Input:
integrate((x^3-1)*(x^6-1)^(1/2)/x^4/(x^3+1),x, algorithm="giac")
Output:
integrate(sqrt(x^6 - 1)*(x^3 - 1)/((x^3 + 1)*x^4), x)
Timed out. \[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x^4 \left (1+x^3\right )} \, dx=\int \frac {\left (x^3-1\right )\,\sqrt {x^6-1}}{x^4\,\left (x^3+1\right )} \,d x \] Input:
int(((x^3 - 1)*(x^6 - 1)^(1/2))/(x^4*(x^3 + 1)),x)
Output:
int(((x^3 - 1)*(x^6 - 1)^(1/2))/(x^4*(x^3 + 1)), x)
Time = 0.17 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.55 \[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x^4 \left (1+x^3\right )} \, dx=\frac {-2 \mathit {atan} \left (\frac {\sqrt {x^{6}-1}\, x^{6}-2 \sqrt {x^{6}-1}}{2 x^{6}-2}\right ) x^{3}+2 \sqrt {x^{6}-1}+\mathrm {log}\left (\sqrt {x^{6}-1}+x^{3}\right ) x^{3}-\mathrm {log}\left (\sqrt {x^{6}-1}-x^{3}\right ) x^{3}}{6 x^{3}} \] Input:
int((x^3-1)*(x^6-1)^(1/2)/x^4/(x^3+1),x)
Output:
( - 2*atan((sqrt(x**6 - 1)*x**6 - 2*sqrt(x**6 - 1))/(2*x**6 - 2))*x**3 + 2 *sqrt(x**6 - 1) + log(sqrt(x**6 - 1) + x**3)*x**3 - log(sqrt(x**6 - 1) - x **3)*x**3)/(6*x**3)