\(\int \frac {(-1+x^3) \sqrt {-1+x^6}}{x} \, dx\) [764]

Optimal result

Integrand size = 18, antiderivative size = 59 \[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\frac {1}{6} \left (-2+x^3\right ) \sqrt {-1+x^6}-\frac {2}{3} \arctan \left (\frac {1+x^3}{\sqrt {-1+x^6}}\right )-\frac {1}{3} \text {arctanh}\left (\frac {1+x^3}{\sqrt {-1+x^6}}\right ) \]

[Out]

1/6*(x^3-2)*(x^6-1)^(1/2)-2/3*arctan((x^3+1)/(x^6-1)^(1/2))-1/3*arctanh((x^3+1)/(x^6-1)^(1/2))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {1489, 829, 858, 223, 212, 272, 65, 209} \[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\frac {1}{3} \arctan \left (\sqrt {x^6-1}\right )-\frac {1}{6} \text {arctanh}\left (\frac {x^3}{\sqrt {x^6-1}}\right )-\frac {1}{6} \sqrt {x^6-1} \left (2-x^3\right ) \]

[In]

Int[((-1 + x^3)*Sqrt[-1 + x^6])/x,x]

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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]

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 829

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
+ 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*((a + c*x^2)^p/(c*e^2*(m + 2*p + 1)*(m +
2*p + 2))), x] + Dist[2*(p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 858

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

Rule 1489

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[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]]

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.02 \[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\frac {1}{6} \left (\left (-2+x^3\right ) \sqrt {-1+x^6}-4 \arctan \left (\frac {\sqrt {-1+x^6}}{-1+x^3}\right )-2 \text {arctanh}\left (\frac {\sqrt {-1+x^6}}{-1+x^3}\right )\right ) \]

[In]

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

[Out]

((-2 + x^3)*Sqrt[-1 + x^6] - 4*ArcTan[Sqrt[-1 + x^6]/(-1 + x^3)] - 2*ArcTanh[Sqrt[-1 + x^6]/(-1 + x^3)])/6

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80

method	result	size
pseudoelliptic	\(\frac {x^{3} \sqrt {x^{6}-1}}{6}-\frac {\ln \left (x^{3}+\sqrt {x^{6}-1}\right )}{6}-\frac {\sqrt {x^{6}-1}}{3}-\frac {\arctan \left (\frac {1}{\sqrt {x^{6}-1}}\right )}{3}\)	\(47\)
trager	\(\left (\frac {x^{3}}{6}-\frac {1}{3}\right ) \sqrt {x^{6}-1}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+\sqrt {x^{6}-1}}{x^{3}}\right )}{3}+\frac {\ln \left (-x^{3}+\sqrt {x^{6}-1}\right )}{6}\)	\(62\)
meijerg	\(\frac {\sqrt {\operatorname {signum}\left (x^{6}-1\right )}\, \left (4 \sqrt {\pi }-4 \sqrt {\pi }\, \sqrt {-x^{6}+1}+4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{6}+1}}{2}\right )-2 \left (2-2 \ln \left (2\right )+6 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }\right )}{12 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (x^{6}-1\right )}}+\frac {i \sqrt {\operatorname {signum}\left (x^{6}-1\right )}\, \left (-2 i \sqrt {\pi }\, x^{3} \sqrt {-x^{6}+1}-2 i \sqrt {\pi }\, \arcsin \left (x^{3}\right )\right )}{12 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (x^{6}-1\right )}}\)	\(136\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80 \[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\frac {1}{6} \, \sqrt {x^{6} - 1} {\left (x^{3} - 2\right )} + \frac {2}{3} \, \arctan \left (-x^{3} + \sqrt {x^{6} - 1}\right ) + \frac {1}{6} \, \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 4.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.95 \[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\frac {x^{3} \sqrt {x^{6} - 1}}{6} - \frac {\begin {cases} \sqrt {x^{6} - 1} - \operatorname {acos}{\left (\frac {1}{x^{3}} \right )} & \text {for}\: x^{3} > -1 \wedge x^{3} < 1 \end {cases}}{3} - \frac {\log {\left (2 x^{3} + 2 \sqrt {x^{6} - 1} \right )}}{6} \]

[In]

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

[Out]

x**3*sqrt(x**6 - 1)/6 - Piecewise((sqrt(x**6 - 1) - acos(x**(-3)), (x**3 > -1) & (x**3 < 1)))/3 - log(2*x**3 +
2*sqrt(x**6 - 1))/6

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.31 \[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x} \, dx=-\frac {1}{3} \, \sqrt {x^{6} - 1} - \frac {\sqrt {x^{6} - 1}}{6 \, x^{3} {\left (\frac {x^{6} - 1}{x^{6}} - 1\right )}} + \frac {1}{3} \, \arctan \left (\sqrt {x^{6} - 1}\right ) - \frac {1}{12} \, \log \left (\frac {\sqrt {x^{6} - 1}}{x^{3}} + 1\right ) + \frac {1}{12} \, \log \left (\frac {\sqrt {x^{6} - 1}}{x^{3}} - 1\right ) \]

[In]

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

[Out]

-1/3*sqrt(x^6 - 1) - 1/6*sqrt(x^6 - 1)/(x^3*((x^6 - 1)/x^6 - 1)) + 1/3*arctan(sqrt(x^6 - 1)) - 1/12*log(sqrt(x
^6 - 1)/x^3 + 1) + 1/12*log(sqrt(x^6 - 1)/x^3 - 1)

Giac [F]

\[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\int { \frac {\sqrt {x^{6} - 1} {\left (x^{3} - 1\right )}}{x} \,d x } \]

[In]

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

[Out]

integrate(sqrt(x^6 - 1)*(x^3 - 1)/x, x)

Mupad [B] (verification not implemented)

Time = 6.31 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92 \[ \int \frac {\left (-1+x^3\right ) \sqrt {-1+x^6}}{x} \, dx=\frac {x^3\,\sqrt {x^6-1}}{6}-\frac {\sqrt {x^6-1}}{3}-\frac {\ln \left (\sqrt {x^6-1}+x^3\right )}{6}+\frac {\ln \left (\frac {\sqrt {x^6-1}+1{}\mathrm {i}}{x^3}\right )\,1{}\mathrm {i}}{3} \]

[In]

int(((x^3 - 1)*(x^6 - 1)^(1/2))/x,x)

[Out]

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