Integrand size = 18, antiderivative size = 29 \[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\arctan \left (\sqrt [4]{2-2 x+x^2}\right )-\text {arctanh}\left (\sqrt [4]{2-2 x+x^2}\right ) \] Output:
arctan((x^2-2*x+2)^(1/4))-arctanh((x^2-2*x+2)^(1/4))
Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\arctan \left (\sqrt [4]{2-2 x+x^2}\right )-\text {arctanh}\left (\sqrt [4]{2-2 x+x^2}\right ) \] Input:
Integrate[1/((-1 + x)*(2 - 2*x + x^2)^(1/4)),x]
Output:
ArcTan[(2 - 2*x + x^2)^(1/4)] - ArcTanh[(2 - 2*x + x^2)^(1/4)]
Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1118, 243, 73, 25, 827, 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.
\(\displaystyle \int \frac {1}{(x-1) \sqrt [4]{x^2-2 x+2}} \, dx\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle \int \frac {1}{\sqrt [4]{(x-1)^2+1} (x-1)}d(x-1)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{(x-1)^2 \sqrt [4]{x}}d(x-1)^2\) |
\(\Big \downarrow \) 73 |
\(\displaystyle 2 \int -\frac {(x-1)^4}{1-(x-1)^8}d\sqrt [4]{x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {(x-1)^4}{1-(x-1)^8}d\sqrt [4]{x}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle 2 \left (\frac {1}{2} \int \frac {1}{(x-1)^4+1}d\sqrt [4]{x}-\frac {1}{2} \int \frac {1}{1-(x-1)^4}d\sqrt [4]{x}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle 2 \left (-\frac {1}{2} \int \frac {1}{1-(x-1)^4}d\sqrt [4]{x}-\frac {1}{2} \arctan (1-x)\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \left (\frac {1}{2} \text {arctanh}(1-x)-\frac {1}{2} \arctan (1-x)\right )\) |
Input:
Int[1/((-1 + x)*(2 - 2*x + x^2)^(1/4)),x]
Output:
2*(-1/2*ArcTan[1 - x] + ArcTanh[1 - x]/2)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, 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[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[1/e Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
Time = 1.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48
method | result | size |
pseudoelliptic | \(\frac {\ln \left (\left (x^{2}-2 x +2\right )^{\frac {1}{4}}-1\right )}{2}-\frac {\ln \left (1+\left (x^{2}-2 x +2\right )^{\frac {1}{4}}\right )}{2}+\arctan \left (\left (x^{2}-2 x +2\right )^{\frac {1}{4}}\right )\) | \(43\) |
trager | \(\frac {\ln \left (-\frac {2 \left (x^{2}-2 x +2\right )^{\frac {3}{4}}-2 \sqrt {x^{2}-2 x +2}-x^{2}+2 \left (x^{2}-2 x +2\right )^{\frac {1}{4}}+2 x -3}{\left (-1+x \right )^{2}}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \left (x^{2}-2 x +2\right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{2}-2 x +2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \left (x^{2}-2 x +2\right )^{\frac {1}{4}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{\left (-1+x \right )^{2}}\right )}{2}\) | \(145\) |
Input:
int(1/(-1+x)/(x^2-2*x+2)^(1/4),x,method=_RETURNVERBOSE)
Output:
1/2*ln((x^2-2*x+2)^(1/4)-1)-1/2*ln(1+(x^2-2*x+2)^(1/4))+arctan((x^2-2*x+2) ^(1/4))
Time = 0.11 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\arctan \left ({\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{4}} - 1\right ) \] Input:
integrate(1/(-1+x)/(x^2-2*x+2)^(1/4),x, algorithm="fricas")
Output:
arctan((x^2 - 2*x + 2)^(1/4)) - 1/2*log((x^2 - 2*x + 2)^(1/4) + 1) + 1/2*l og((x^2 - 2*x + 2)^(1/4) - 1)
\[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\int \frac {1}{\left (x - 1\right ) \sqrt [4]{x^{2} - 2 x + 2}}\, dx \] Input:
integrate(1/(-1+x)/(x**2-2*x+2)**(1/4),x)
Output:
Integral(1/((x - 1)*(x**2 - 2*x + 2)**(1/4)), x)
\[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{4}} {\left (x - 1\right )}} \,d x } \] Input:
integrate(1/(-1+x)/(x^2-2*x+2)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((x^2 - 2*x + 2)^(1/4)*(x - 1)), x)
\[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{4}} {\left (x - 1\right )}} \,d x } \] Input:
integrate(1/(-1+x)/(x^2-2*x+2)^(1/4),x, algorithm="giac")
Output:
integrate(1/((x^2 - 2*x + 2)^(1/4)*(x - 1)), x)
Timed out. \[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\int \frac {1}{\left (x-1\right )\,{\left (x^2-2\,x+2\right )}^{1/4}} \,d x \] Input:
int(1/((x - 1)*(x^2 - 2*x + 2)^(1/4)),x)
Output:
int(1/((x - 1)*(x^2 - 2*x + 2)^(1/4)), x)
\[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\int \frac {1}{\left (x^{2}-2 x +2\right )^{\frac {1}{4}} x -\left (x^{2}-2 x +2\right )^{\frac {1}{4}}}d x \] Input:
int(1/(-1+x)/(x^2-2*x+2)^(1/4),x)
Output:
int(1/((x**2 - 2*x + 2)**(1/4)*x - (x**2 - 2*x + 2)**(1/4)),x)