Integrand size = 24, antiderivative size = 67 \[ \int \frac {-1+2 x^4}{\sqrt [4]{-1+x^4} \left (-1+x^8\right )} \, dx=-\frac {x}{2 \sqrt [4]{-1+x^4}}+\frac {3 \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt [4]{2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt [4]{2}} \] Output:
-1/2*x/(x^4-1)^(1/4)+3/8*arctan(2^(1/4)*x/(x^4-1)^(1/4))*2^(3/4)+3/8*arcta nh(2^(1/4)*x/(x^4-1)^(1/4))*2^(3/4)
Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {-1+2 x^4}{\sqrt [4]{-1+x^4} \left (-1+x^8\right )} \, dx=-\frac {x}{2 \sqrt [4]{-1+x^4}}+\frac {3 \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt [4]{2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt [4]{2}} \] Input:
Integrate[(-1 + 2*x^4)/((-1 + x^4)^(1/4)*(-1 + x^8)),x]
Output:
-1/2*x/(-1 + x^4)^(1/4) + (3*ArcTan[(2^(1/4)*x)/(-1 + x^4)^(1/4)])/(4*2^(1 /4)) + (3*ArcTanh[(2^(1/4)*x)/(-1 + x^4)^(1/4)])/(4*2^(1/4))
Time = 0.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1388, 1024, 27, 902, 756, 216, 219}
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.
| \(\displaystyle \int \frac {2 x^4-1}{\sqrt [4]{x^4-1} \left (x^8-1\right )} \, dx\) |
| \(\Big \downarrow \) 1388 |
| \(\displaystyle \int \frac {2 x^4-1}{\left (x^4-1\right )^{5/4} \left (x^4+1\right )}dx\) |
| \(\Big \downarrow \) 1024 |
| \(\displaystyle \frac {1}{2} \int \frac {3}{\sqrt [4]{x^4-1} \left (x^4+1\right )}dx-\frac {x}{2 \sqrt [4]{x^4-1}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3}{2} \int \frac {1}{\sqrt [4]{x^4-1} \left (x^4+1\right )}dx-\frac {x}{2 \sqrt [4]{x^4-1}}\) |
| \(\Big \downarrow \) 902 |
| \(\displaystyle \frac {3}{2} \int \frac {1}{1-\frac {2 x^4}{x^4-1}}d\frac {x}{\sqrt [4]{x^4-1}}-\frac {x}{2 \sqrt [4]{x^4-1}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {3}{2} \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {2} x^2}{\sqrt {x^4-1}}}d\frac {x}{\sqrt [4]{x^4-1}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {2} x^2}{\sqrt {x^4-1}}+1}d\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {x}{2 \sqrt [4]{x^4-1}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {3}{2} \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {2} x^2}{\sqrt {x^4-1}}}d\frac {x}{\sqrt [4]{x^4-1}}+\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{2 \sqrt [4]{2}}\right )-\frac {x}{2 \sqrt [4]{x^4-1}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3}{2} \left (\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{2 \sqrt [4]{2}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{2 \sqrt [4]{2}}\right )-\frac {x}{2 \sqrt [4]{x^4-1}}\) |
Input:
Int[(-1 + 2*x^4)/((-1 + x^4)^(1/4)*(-1 + x^8)),x]
Output:
-1/2*x/(-1 + x^4)^(1/4) + (3*(ArcTan[(2^(1/4)*x)/(-1 + x^4)^(1/4)]/(2*2^(1 /4)) + ArcTanh[(2^(1/4)*x)/(-1 + x^4)^(1/4)]/(2*2^(1/4))))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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[((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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Su bst[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]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f _.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*n*(b*c - a*d)*( p + 1)) Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b *c - a*d)*(p + 1) + d*(b*e - a*f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; Fr eeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
Time = 1.77 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.33
| method | result | size |
| pseudoelliptic | \(\frac {-6 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{4}-1\right )^{\frac {1}{4}}}{2 x}\right ) 2^{\frac {3}{4}} \left (x^{4}-1\right )^{\frac {1}{4}}+3 \ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{4}-1\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{4}-1\right )^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}} \left (x^{4}-1\right )^{\frac {1}{4}}-8 x}{16 \left (x^{4}-1\right )^{\frac {1}{4}}}\) | \(89\) |
| trager | \(-\frac {x}{2 \left (x^{4}-1\right )^{\frac {1}{4}}}-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (\frac {\sqrt {x^{4}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-2 \left (x^{4}-1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}+4 \left (x^{4}-1\right )^{\frac {3}{4}} x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right )}{x^{4}+1}\right )}{16}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {\sqrt {x^{4}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 \left (x^{4}-1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+3 x^{4} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )+4 \left (x^{4}-1\right )^{\frac {3}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )}{x^{4}+1}\right )}{16}\) | \(216\) |
| risch | \(-\frac {x}{2 \left (x^{4}-1\right )^{\frac {1}{4}}}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {\sqrt {x^{4}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 \left (x^{4}-1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+3 x^{4} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )+4 \left (x^{4}-1\right )^{\frac {3}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )}{x^{4}+1}\right )}{16}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {\sqrt {x^{4}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+2 \left (x^{4}-1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}-4 \left (x^{4}-1\right )^{\frac {3}{4}} x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right )}{x^{4}+1}\right )}{16}\) | \(217\) |
Input:
int((2*x^4-1)/(x^4-1)^(1/4)/(x^8-1),x,method=_RETURNVERBOSE)
Output:
1/16*(-6*arctan(1/2*2^(3/4)/x*(x^4-1)^(1/4))*2^(3/4)*(x^4-1)^(1/4)+3*ln((- 2^(1/4)*x-(x^4-1)^(1/4))/(2^(1/4)*x-(x^4-1)^(1/4)))*2^(3/4)*(x^4-1)^(1/4)- 8*x)/(x^4-1)^(1/4)
Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (47) = 94\).
Time = 2.98 (sec) , antiderivative size = 210, normalized size of antiderivative = 3.13 \[ \int \frac {-1+2 x^4}{\sqrt [4]{-1+x^4} \left (-1+x^8\right )} \, dx=-\frac {6 \cdot 2^{\frac {3}{4}} {\left (x^{4} - 1\right )} \arctan \left (\frac {2 \, {\left (2^{\frac {3}{4}} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 2^{\frac {1}{4}} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x\right )}}{x^{4} + 1}\right ) - 3 \cdot 2^{\frac {3}{4}} {\left (x^{4} - 1\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} - 1} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{4} - 1\right )} + 4 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x}{x^{4} + 1}\right ) + 3 \cdot 2^{\frac {3}{4}} {\left (x^{4} - 1\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} - 1} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{4} - 1\right )} + 4 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x}{x^{4} + 1}\right ) + 16 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x}{32 \, {\left (x^{4} - 1\right )}} \] Input:
integrate((2*x^4-1)/(x^4-1)^(1/4)/(x^8-1),x, algorithm="fricas")
Output:
-1/32*(6*2^(3/4)*(x^4 - 1)*arctan(2*(2^(3/4)*(x^4 - 1)^(1/4)*x^3 + 2^(1/4) *(x^4 - 1)^(3/4)*x)/(x^4 + 1)) - 3*2^(3/4)*(x^4 - 1)*log((4*sqrt(2)*(x^4 - 1)^(1/4)*x^3 + 4*2^(1/4)*sqrt(x^4 - 1)*x^2 + 2^(3/4)*(3*x^4 - 1) + 4*(x^4 - 1)^(3/4)*x)/(x^4 + 1)) + 3*2^(3/4)*(x^4 - 1)*log((4*sqrt(2)*(x^4 - 1)^( 1/4)*x^3 - 4*2^(1/4)*sqrt(x^4 - 1)*x^2 - 2^(3/4)*(3*x^4 - 1) + 4*(x^4 - 1) ^(3/4)*x)/(x^4 + 1)) + 16*(x^4 - 1)^(3/4)*x)/(x^4 - 1)
\[ \int \frac {-1+2 x^4}{\sqrt [4]{-1+x^4} \left (-1+x^8\right )} \, dx=\int \frac {2 x^{4} - 1}{\sqrt [4]{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )}\, dx \] Input:
integrate((2*x**4-1)/(x**4-1)**(1/4)/(x**8-1),x)
Output:
Integral((2*x**4 - 1)/(((x - 1)*(x + 1)*(x**2 + 1))**(1/4)*(x - 1)*(x + 1) *(x**2 + 1)*(x**4 + 1)), x)
\[ \int \frac {-1+2 x^4}{\sqrt [4]{-1+x^4} \left (-1+x^8\right )} \, dx=\int { \frac {2 \, x^{4} - 1}{{\left (x^{8} - 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate((2*x^4-1)/(x^4-1)^(1/4)/(x^8-1),x, algorithm="maxima")
Output:
integrate((2*x^4 - 1)/((x^8 - 1)*(x^4 - 1)^(1/4)), x)
\[ \int \frac {-1+2 x^4}{\sqrt [4]{-1+x^4} \left (-1+x^8\right )} \, dx=\int { \frac {2 \, x^{4} - 1}{{\left (x^{8} - 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate((2*x^4-1)/(x^4-1)^(1/4)/(x^8-1),x, algorithm="giac")
Output:
integrate((2*x^4 - 1)/((x^8 - 1)*(x^4 - 1)^(1/4)), x)
Timed out. \[ \int \frac {-1+2 x^4}{\sqrt [4]{-1+x^4} \left (-1+x^8\right )} \, dx=\int \frac {2\,x^4-1}{{\left (x^4-1\right )}^{1/4}\,\left (x^8-1\right )} \,d x \] Input:
int((2*x^4 - 1)/((x^4 - 1)^(1/4)*(x^8 - 1)),x)
Output:
int((2*x^4 - 1)/((x^4 - 1)^(1/4)*(x^8 - 1)), x)
\[ \int \frac {-1+2 x^4}{\sqrt [4]{-1+x^4} \left (-1+x^8\right )} \, dx=2 \left (\int \frac {x^{4}}{\left (x^{4}-1\right )^{\frac {1}{4}} x^{8}-\left (x^{4}-1\right )^{\frac {1}{4}}}d x \right )-\left (\int \frac {1}{\left (x^{4}-1\right )^{\frac {1}{4}} x^{8}-\left (x^{4}-1\right )^{\frac {1}{4}}}d x \right ) \] Input:
int((2*x^4-1)/(x^4-1)^(1/4)/(x^8-1),x)
Output:
2*int(x**4/((x**4 - 1)**(1/4)*x**8 - (x**4 - 1)**(1/4)),x) - int(1/((x**4 - 1)**(1/4)*x**8 - (x**4 - 1)**(1/4)),x)