\(\int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} (-1+4 x^6)} \, dx\) [851]

Optimal result

Integrand size = 30, antiderivative size = 64 \[ \int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx=-\frac {\arctan \left (\frac {1}{\sqrt {3}}-\frac {4 x^6}{\sqrt {3}}-\frac {4 x^3 \sqrt {-1+x^6}}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {5}{6} \log \left (x^3+\sqrt {-1+x^6}\right ) \]

[Out]

1/6*arctan(-1/3*3^(1/2)+4/3*x^6*3^(1/2)+4/3*x^3*(x^6-1)^(1/2)*3^(1/2))*3^(1/2)+5/6*ln(x^3+(x^6-1)^(1/2))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.73, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {1607, 589, 537, 223, 212, 385, 210} \[ \int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx=\frac {5}{6} \text {arctanh}\left (\frac {x^3}{\sqrt {x^6-1}}\right )-\frac {\arctan \left (\frac {\sqrt {3} x^3}{\sqrt {x^6-1}}\right )}{2 \sqrt {3}} \]

[In]

Int[(-x^2 + 10*x^8)/(Sqrt[-1 + x^6]*(-1 + 4*x^6)),x]

[Out]

-1/2*ArcTan[(Sqrt[3]*x^3)/Sqrt[-1 + x^6]]/Sqrt[3] + (5*ArcTanh[x^3/Sqrt[-1 + x^6]])/6

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 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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 589

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^
q*(e + f*x^(n/k))^r, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, c, d, e, f, p, q, r}, x] && IGtQ[n, 0] && Inte
gerQ[m]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 \left (-1+10 x^6\right )}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx \\ & = \frac {1}{3} \text {Subst}\left (\int \frac {-1+10 x^2}{\sqrt {-1+x^2} \left (-1+4 x^2\right )} \, dx,x,x^3\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x^2} \left (-1+4 x^2\right )} \, dx,x,x^3\right )+\frac {5}{6} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x^2}} \, dx,x,x^3\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{-1-3 x^2} \, dx,x,\frac {x^3}{\sqrt {-1+x^6}}\right )+\frac {5}{6} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^3}{\sqrt {-1+x^6}}\right ) \\ & = -\frac {\arctan \left (\frac {\sqrt {3} x^3}{\sqrt {-1+x^6}}\right )}{2 \sqrt {3}}+\frac {5}{6} \text {arctanh}\left (\frac {x^3}{\sqrt {-1+x^6}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.89 \[ \int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx=\frac {1}{6} \left (\sqrt {3} \arctan \left (\frac {1-4 x^6+4 x^3 \sqrt {-1+x^6}}{\sqrt {3}}\right )-5 \log \left (-x^3+\sqrt {-1+x^6}\right )\right ) \]

[In]

Integrate[(-x^2 + 10*x^8)/(Sqrt[-1 + x^6]*(-1 + 4*x^6)),x]

[Out]

(Sqrt[3]*ArcTan[(1 - 4*x^6 + 4*x^3*Sqrt[-1 + x^6])/Sqrt[3]] - 5*Log[-x^3 + Sqrt[-1 + x^6]])/6

Maple [A] (verified)

Time = 1.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.97

method	result	size
pseudoelliptic	\(\frac {5 \ln \left (x^{3}+\sqrt {x^{6}-1}\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (x^{3}+2\right ) \sqrt {3}}{3 \sqrt {x^{6}-1}}\right )}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (x^{3}-2\right ) \sqrt {3}}{3 \sqrt {x^{6}-1}}\right )}{12}\)	\(62\)
trager	\(-\frac {5 \ln \left (x^{3}-\sqrt {x^{6}-1}\right )}{6}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) \ln \left (-\frac {-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) x^{6}+6 x^{3} \sqrt {x^{6}-1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right )}{\left (2 x^{3}-1\right ) \left (2 x^{3}+1\right )}\right )}{12}\)	\(79\)

[In]

int((10*x^8-x^2)/(x^6-1)^(1/2)/(4*x^6-1),x,method=_RETURNVERBOSE)

[Out]

5/6*ln(x^3+(x^6-1)^(1/2))+1/12*3^(1/2)*arctan(1/3*(x^3+2)*3^(1/2)/(x^6-1)^(1/2))+1/12*3^(1/2)*arctan(1/3*(x^3-
2)*3^(1/2)/(x^6-1)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.80 \[ \int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {4}{3} \, \sqrt {3} \sqrt {x^{6} - 1} x^{3} - \frac {1}{3} \, \sqrt {3} {\left (4 \, x^{6} - 1\right )}\right ) - \frac {5}{6} \, \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) \]

[In]

integrate((10*x^8-x^2)/(x^6-1)^(1/2)/(4*x^6-1),x, algorithm="fricas")

[Out]

1/6*sqrt(3)*arctan(4/3*sqrt(3)*sqrt(x^6 - 1)*x^3 - 1/3*sqrt(3)*(4*x^6 - 1)) - 5/6*log(-x^3 + sqrt(x^6 - 1))

Sympy [F]

\[ \int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx=\int \frac {x^{2} \cdot \left (10 x^{6} - 1\right )}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )} \left (2 x^{3} - 1\right ) \left (2 x^{3} + 1\right )}\, dx \]

[In]

integrate((10*x**8-x**2)/(x**6-1)**(1/2)/(4*x**6-1),x)

[Out]

Integral(x**2*(10*x**6 - 1)/(sqrt((x - 1)*(x + 1)*(x**2 - x + 1)*(x**2 + x + 1))*(2*x**3 - 1)*(2*x**3 + 1)), x
)

Maxima [F]

\[ \int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx=\int { \frac {10 \, x^{8} - x^{2}}{{\left (4 \, x^{6} - 1\right )} \sqrt {x^{6} - 1}} \,d x } \]

[In]

integrate((10*x^8-x^2)/(x^6-1)^(1/2)/(4*x^6-1),x, algorithm="maxima")

[Out]

integrate((10*x^8 - x^2)/((4*x^6 - 1)*sqrt(x^6 - 1)), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((10*x^8-x^2)/(x^6-1)^(1/2)/(4*x^6-1),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:rootof minimal polynomial must be unitary Error: Bad Argument Valuerootof minimal polynomial must be unitar
y Error: Ba

Mupad [F(-1)]

Timed out. \[ \int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx=\int -\frac {x^2-10\,x^8}{\sqrt {x^6-1}\,\left (4\,x^6-1\right )} \,d x \]

[In]

int(-(x^2 - 10*x^8)/((x^6 - 1)^(1/2)*(4*x^6 - 1)),x)

[Out]

int(-(x^2 - 10*x^8)/((x^6 - 1)^(1/2)*(4*x^6 - 1)), x)