3.4.44 \(\int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx\) [344]

3.4.44.1 Optimal result

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))
 

3.4.44.2 Mathematica [A] (verified)

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)]
 

3.4.44.3 Rubi [A] (warning: unable to verify)

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)
 

3.4.44.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 73

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]
 

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 243

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]
 

rule 827

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]
 

rule 1118

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]
 

3.4.44.4 Maple [A] (verified)

Time = 2.80 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48

method	result	size
pseudoelliptic	\(\frac {\ln \left (\left (x^{2}-2 x +2\right )^{\frac {1}{4}}-1\right )}{2}-\frac {\ln \left (\left (x^{2}-2 x +2\right )^{\frac {1}{4}}+1\right )}{2}+\arctan \left (\left (x^{2}-2 x +2\right )^{\frac {1}{4}}\right )\)	\(43\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \left (x^{2}-2 x +2\right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{2}-2 x +2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \left (x^{2}-2 x +2\right )^{\frac {1}{4}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{\left (x -1\right )^{2}}\right )}{2}-\frac {\ln \left (-\frac {2 \left (x^{2}-2 x +2\right )^{\frac {3}{4}}+2 \sqrt {x^{2}-2 x +2}+x^{2}+2 \left (x^{2}-2 x +2\right )^{\frac {1}{4}}-2 x +3}{\left (x -1\right )^{2}}\right )}{2}\)	\(143\)

input

int(1/(x-1)/(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((x^2-2*x+2)^(1/4)+1)+arctan((x^2-2*x+2) 
^(1/4))
 

3.4.44.5 Fricas [A] (verification not implemented)

Time = 0.26 (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)
 

3.4.44.6 Sympy [F]

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

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)
 

3.4.44.7 Maxima [F]

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

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)
 

3.4.44.8 Giac [F]

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

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)
 

3.4.44.9 Mupad [F(-1)]

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)