Integrand size = 13, antiderivative size = 51 \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=\frac {2 \sqrt {1-\frac {1}{x^2}}}{1-\frac {1}{x}}-\sqrt {1-\frac {1}{x^2}} x-2 \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right ) \] Output:
2*(1-1/x^2)^(1/2)/(1-1/x)-(1-1/x^2)^(1/2)*x-2*arctanh((1-1/x^2)^(1/2))
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=-\frac {\sqrt {1-\frac {1}{x^2}} (-3+x) x}{-1+x}-2 \log \left (\left (1+\sqrt {1-\frac {1}{x^2}}\right ) x\right ) \] Input:
Integrate[(E^ArcCoth[x]*x)/(1 - x),x]
Output:
-((Sqrt[1 - x^(-2)]*(-3 + x)*x)/(-1 + x)) - 2*Log[(1 + Sqrt[1 - x^(-2)])*x ]
Time = 0.44 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {6728, 564, 534, 243, 73, 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.
\(\displaystyle \int \frac {x e^{\coth ^{-1}(x)}}{1-x} \, dx\) |
\(\Big \downarrow \) 6728 |
\(\displaystyle \int \frac {\sqrt {1-\frac {1}{x^2}} x^2}{\left (1-\frac {1}{x}\right )^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 564 |
\(\displaystyle \int \frac {\left (1+\frac {2}{x}\right ) x^2}{\sqrt {1-\frac {1}{x^2}}}d\frac {1}{x}+\frac {2 \sqrt {1-\frac {1}{x^2}}}{1-\frac {1}{x}}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle 2 \int \frac {x}{\sqrt {1-\frac {1}{x^2}}}d\frac {1}{x}-\sqrt {1-\frac {1}{x^2}} x+\frac {2 \sqrt {1-\frac {1}{x^2}}}{1-\frac {1}{x}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \int \frac {x}{\sqrt {1-\frac {1}{x^2}}}d\frac {1}{x^2}-\sqrt {1-\frac {1}{x^2}} x+\frac {2 \sqrt {1-\frac {1}{x^2}}}{1-\frac {1}{x}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -2 \int \frac {1}{1-\sqrt {1-\frac {1}{x^2}}}d\sqrt {1-\frac {1}{x^2}}-\sqrt {1-\frac {1}{x^2}} x+\frac {2 \sqrt {1-\frac {1}{x^2}}}{1-\frac {1}{x}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -2 \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right )-\sqrt {1-\frac {1}{x^2}} x+\frac {2 \sqrt {1-\frac {1}{x^2}}}{1-\frac {1}{x}}\) |
Input:
Int[(E^ArcCoth[x]*x)/(1 - x),x]
Output:
(2*Sqrt[1 - x^(-2)])/(1 - x^(-1)) - Sqrt[1 - x^(-2)]*x - 2*ArcTanh[Sqrt[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]))* 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_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d Int[ x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b ^(n + 2)*(c + d*x))), x] - Simp[d^(2*n + 2)/b^(n + 1) Int[(x^m/Sqrt[a + b *x^2])*ExpandToSum[((2^(-n - 1)*(-c)^(m - n - 1))/(d^m*x^m) - (-c + d*x)^(- n - 1))/(c + d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^ 2, 0] && ILtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_))^(p_.), x_S ymbol] :> Simp[-d^n Subst[Int[(d + c*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^(m + p + 2)), x], x, 1/x], x] /; FreeQ[{a, c, d}, x] && EqQ[a*c + d, 0] && In tegerQ[p] && IntegerQ[n] && IntegerQ[m]
Time = 0.14 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.27
method | result | size |
risch | \(-\frac {x^{2}-2 x -3}{\sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}-\frac {2 \ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (x -1\right ) \left (1+x \right )}}{\sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}\) | \(65\) |
trager | \(-\frac {\left (1+x \right ) \left (-3+x \right ) \sqrt {-\frac {1-x}{1+x}}}{x -1}+2 \ln \left (-\sqrt {-\frac {1-x}{1+x}}\, x -\sqrt {-\frac {1-x}{1+x}}+x \right )\) | \(67\) |
default | \(\frac {\left (x^{2}-1\right )^{\frac {3}{2}}-2 x^{2} \sqrt {x^{2}-1}-2 \ln \left (x +\sqrt {x^{2}-1}\right ) x^{2}+4 x \sqrt {x^{2}-1}+4 \ln \left (x +\sqrt {x^{2}-1}\right ) x -2 \sqrt {x^{2}-1}-2 \ln \left (x +\sqrt {x^{2}-1}\right )}{\left (x -1\right ) \sqrt {\left (x -1\right ) \left (1+x \right )}\, \sqrt {\frac {x -1}{1+x}}}\) | \(106\) |
Input:
int(1/((x-1)/(1+x))^(1/2)*x/(1-x),x,method=_RETURNVERBOSE)
Output:
-(x^2-2*x-3)/((x-1)/(1+x))^(1/2)/(1+x)-2*ln(x+(x^2-1)^(1/2))/((x-1)/(1+x)) ^(1/2)/(1+x)*((x-1)*(1+x))^(1/2)
Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=-\frac {2 \, {\left (x - 1\right )} \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - 2 \, {\left (x - 1\right )} \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) + {\left (x^{2} - 2 \, x - 3\right )} \sqrt {\frac {x - 1}{x + 1}}}{x - 1} \] Input:
integrate(1/((x-1)/(1+x))^(1/2)*x/(1-x),x, algorithm="fricas")
Output:
-(2*(x - 1)*log(sqrt((x - 1)/(x + 1)) + 1) - 2*(x - 1)*log(sqrt((x - 1)/(x + 1)) - 1) + (x^2 - 2*x - 3)*sqrt((x - 1)/(x + 1)))/(x - 1)
\[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=- \int \frac {x}{x \sqrt {\frac {x}{x + 1} - \frac {1}{x + 1}} - \sqrt {\frac {x}{x + 1} - \frac {1}{x + 1}}}\, dx \] Input:
integrate(1/((x-1)/(1+x))**(1/2)*x/(1-x),x)
Output:
-Integral(x/(x*sqrt(x/(x + 1) - 1/(x + 1)) - sqrt(x/(x + 1) - 1/(x + 1))), x)
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.45 \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=\frac {2 \, {\left (\frac {2 \, {\left (x - 1\right )}}{x + 1} - 1\right )}}{\left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}} - \sqrt {\frac {x - 1}{x + 1}}} - 2 \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) + 2 \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \] Input:
integrate(1/((x-1)/(1+x))^(1/2)*x/(1-x),x, algorithm="maxima")
Output:
2*(2*(x - 1)/(x + 1) - 1)/(((x - 1)/(x + 1))^(3/2) - sqrt((x - 1)/(x + 1)) ) - 2*log(sqrt((x - 1)/(x + 1)) + 1) + 2*log(sqrt((x - 1)/(x + 1)) - 1)
Time = 0.14 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.27 \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=\frac {2 \, \log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{\mathrm {sgn}\left (x + 1\right )} - \frac {\sqrt {x^{2} - 1}}{\mathrm {sgn}\left (x + 1\right )} - \frac {4}{{\left (x - \sqrt {x^{2} - 1} - 1\right )} \mathrm {sgn}\left (x + 1\right )} - 2 \, \mathrm {sgn}\left (x + 1\right ) \] Input:
integrate(1/((x-1)/(1+x))^(1/2)*x/(1-x),x, algorithm="giac")
Output:
2*log(abs(-x + sqrt(x^2 - 1)))/sgn(x + 1) - sqrt(x^2 - 1)/sgn(x + 1) - 4/( (x - sqrt(x^2 - 1) - 1)*sgn(x + 1)) - 2*sgn(x + 1)
Time = 13.64 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=-\frac {2\,x+8\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )\,\sqrt {\frac {x-1}{x+1}}-6}{2\,\sqrt {\frac {x-1}{x+1}}} \] Input:
int(-x/(((x - 1)/(x + 1))^(1/2)*(x - 1)),x)
Output:
-(2*x + 8*atanh(((x - 1)/(x + 1))^(1/2))*((x - 1)/(x + 1))^(1/2) - 6)/(2*( (x - 1)/(x + 1))^(1/2))
Time = 0.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=\frac {-8 \sqrt {x -1}\, \mathrm {log}\left (\frac {\sqrt {x -1}+\sqrt {x +1}}{\sqrt {2}}\right )+5 \sqrt {x -1}-2 \sqrt {x +1}\, x +6 \sqrt {x +1}}{2 \sqrt {x -1}} \] Input:
int(1/((x-1)/(1+x))^(1/2)*x/(1-x),x)
Output:
( - 8*sqrt(x - 1)*log((sqrt(x - 1) + sqrt(x + 1))/sqrt(2)) + 5*sqrt(x - 1) - 2*sqrt(x + 1)*x + 6*sqrt(x + 1))/(2*sqrt(x - 1))