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

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

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

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))
 

Mathematica [A] (verified)

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

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

Output:

(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
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.80, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2027, 1045, 398, 224, 219, 291, 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.

\(\displaystyle \int \frac {10 x^8-x^2}{\sqrt {x^6-1} \left (4 x^6-1\right )} \, dx\)

\(\Big \downarrow \) 2027

\(\displaystyle \int \frac {x^2 \left (10 x^6-1\right )}{\sqrt {x^6-1} \left (4 x^6-1\right )}dx\)

\(\Big \downarrow \) 1045

\(\displaystyle \frac {1}{3} \int \frac {1-10 x^6}{\left (1-4 x^6\right ) \sqrt {x^6-1}}dx^3\)

\(\Big \downarrow \) 398

\(\displaystyle \frac {1}{3} \left (\frac {5}{2} \int \frac {1}{\sqrt {x^6-1}}dx^3-\frac {3}{2} \int \frac {1}{\left (1-4 x^6\right ) \sqrt {x^6-1}}dx^3\right )\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {1}{3} \left (\frac {5}{2} \int \frac {1}{1-x^6}d\frac {x^3}{\sqrt {x^6-1}}-\frac {3}{2} \int \frac {1}{\left (1-4 x^6\right ) \sqrt {x^6-1}}dx^3\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {5}{2} \text {arctanh}\left (\frac {x^3}{\sqrt {x^6-1}}\right )-\frac {3}{2} \int \frac {1}{\left (1-4 x^6\right ) \sqrt {x^6-1}}dx^3\right )\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {1}{3} \left (\frac {5}{2} \text {arctanh}\left (\frac {x^3}{\sqrt {x^6-1}}\right )-\frac {3}{2} \int \frac {1}{3 x^6+1}d\frac {x^3}{\sqrt {x^6-1}}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{3} \left (\frac {5}{2} \text {arctanh}\left (\frac {x^3}{\sqrt {x^6-1}}\right )-\frac {1}{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} x^3}{\sqrt {x^6-1}}\right )\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 216
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])
 

rule 219
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])
 

rule 224
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 291
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 398
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) 
, x_Symbol] :> Simp[f/b   Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ 
b   Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} 
, x]
 

rule 1045
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]}, Si 
mp[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] && IntegerQ[m]
 

rule 2027
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ 
(p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & 
& PosQ[s - r] &&  !(EqQ[p, 1] && EqQ[u, 1])
 
Maple [A] (verified)

Time = 0.38 (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 \(-\frac {5 \ln \left (x^{3}-\sqrt {x^{6}-1}\right )}{6}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) x^{6}+6 x^{3} \sqrt {x^{6}-1}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right )}{\left (2 x^{3}-1\right ) \left (2 x^{3}+1\right )}\right )}{12}\) \(77\)

Input:

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

Output:

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)

Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.80 \[ \int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {4}{3} \, \sqrt {3} \sqrt {x^{6} - 1} x^{3} - \frac {1}{3} \, \sqrt {3} {\left (4 \, x^{6} - 1\right )}\right ) - \frac {5}{6} \, \log \left (-x^{3} + \sqrt {x^{6} - 1}\right ) \] Input:

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

Output:

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]

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

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

Output:

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]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:rootof minimal polynomial must be u 
nitary Error: Bad Argument Valuerootof minimal polynomial must be unitary 
Error: Ba
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx=10 \left (\int \frac {\sqrt {x^{6}-1}\, x^{8}}{4 x^{12}-5 x^{6}+1}d x \right )-\left (\int \frac {\sqrt {x^{6}-1}\, x^{2}}{4 x^{12}-5 x^{6}+1}d x \right ) \] Input:

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

Output:

10*int((sqrt(x**6 - 1)*x**8)/(4*x**12 - 5*x**6 + 1),x) - int((sqrt(x**6 - 
1)*x**2)/(4*x**12 - 5*x**6 + 1),x)