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))
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
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
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[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]
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]
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]
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])
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))
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))
\[ \int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx=\int \frac {x^{2} \cdot \left (10 x^{6} - 1\right )}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )} \left (2 x^{3} - 1\right ) \left (2 x^{3} + 1\right )}\, dx \] 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)
\[ \int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx=\int { \frac {10 \, x^{8} - x^{2}}{{\left (4 \, x^{6} - 1\right )} \sqrt {x^{6} - 1}} \,d x } \] 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)
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
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)
\[ \int \frac {-x^2+10 x^8}{\sqrt {-1+x^6} \left (-1+4 x^6\right )} \, dx=10 \left (\int \frac {\sqrt {x^{6}-1}\, x^{8}}{4 x^{12}-5 x^{6}+1}d x \right )-\left (\int \frac {\sqrt {x^{6}-1}\, x^{2}}{4 x^{12}-5 x^{6}+1}d x \right ) \] 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)