Integrand size = 13, antiderivative size = 55 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1-x)^2} \, dx=-\frac {4 \left (1+\frac {1}{x}\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {3+\frac {5}{x}}{3 \sqrt {1-\frac {1}{x^2}}}+\text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right ) \]
Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1-x)^2} \, dx=\frac {\sqrt {1-\frac {1}{x^2}} (5-7 x) x}{3 (-1+x)^2}+\log \left (\left (1+\sqrt {1-\frac {1}{x^2}}\right ) x\right ) \]
Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {6728, 25, 570, 532, 25, 532, 27, 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.
\(\displaystyle \int \frac {x e^{\coth ^{-1}(x)}}{(1-x)^2} \, dx\) |
\(\Big \downarrow \) 6728 |
\(\displaystyle \int -\frac {\sqrt {1-\frac {1}{x^2}} x}{\left (1-\frac {1}{x}\right )^3}d\frac {1}{x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\sqrt {1-\frac {1}{x^2}} x}{\left (1-\frac {1}{x}\right )^3}d\frac {1}{x}\) |
\(\Big \downarrow \) 570 |
\(\displaystyle -\int \frac {\left (1+\frac {1}{x}\right )^3 x}{\left (1-\frac {1}{x^2}\right )^{5/2}}d\frac {1}{x}\) |
\(\Big \downarrow \) 532 |
\(\displaystyle \frac {1}{3} \int -\frac {\left (3+\frac {5}{x}\right ) x}{\left (1-\frac {1}{x^2}\right )^{3/2}}d\frac {1}{x}-\frac {4 \left (\frac {1}{x}+1\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{3} \int \frac {\left (3+\frac {5}{x}\right ) x}{\left (1-\frac {1}{x^2}\right )^{3/2}}d\frac {1}{x}-\frac {4 \left (\frac {1}{x}+1\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}\) |
\(\Big \downarrow \) 532 |
\(\displaystyle \frac {1}{3} \left (\int -\frac {3 x}{\sqrt {1-\frac {1}{x^2}}}d\frac {1}{x}-\frac {\frac {5}{x}+3}{\sqrt {1-\frac {1}{x^2}}}\right )-\frac {4 \left (\frac {1}{x}+1\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (-3 \int \frac {x}{\sqrt {1-\frac {1}{x^2}}}d\frac {1}{x}-\frac {\frac {5}{x}+3}{\sqrt {1-\frac {1}{x^2}}}\right )-\frac {4 \left (\frac {1}{x}+1\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{3} \left (-\frac {3}{2} \int \frac {x}{\sqrt {1-\frac {1}{x^2}}}d\frac {1}{x^2}-\frac {\frac {5}{x}+3}{\sqrt {1-\frac {1}{x^2}}}\right )-\frac {4 \left (\frac {1}{x}+1\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (3 \int \frac {1}{1-\sqrt {1-\frac {1}{x^2}}}d\sqrt {1-\frac {1}{x^2}}-\frac {\frac {5}{x}+3}{\sqrt {1-\frac {1}{x^2}}}\right )-\frac {4 \left (\frac {1}{x}+1\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (3 \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right )-\frac {\frac {5}{x}+3}{\sqrt {1-\frac {1}{x^2}}}\right )-\frac {4 \left (\frac {1}{x}+1\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}\) |
(-4*(1 + x^(-1)))/(3*(1 - x^(-2))^(3/2)) + (-((3 + 5/x)/Sqrt[1 - x^(-2)]) + 3*ArcTanh[Sqrt[1 - x^(-2)]])/3
3.3.93.3.1 Defintions of rubi rules used
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_))^(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_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[x^m *(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*p]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^(2*n)/a^n Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I LtQ[n, -1] && !(IGtQ[m, 0] && ILtQ[m + n, 0] && !GtQ[p, 1])
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 = 0.48 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.16
method | result | size |
trager | \(-\frac {\left (1+x \right ) \left (7 x -5\right ) \sqrt {-\frac {1-x}{1+x}}}{3 \left (x -1\right )^{2}}+\ln \left (\sqrt {-\frac {1-x}{1+x}}\, x +\sqrt {-\frac {1-x}{1+x}}+x \right )\) | \(64\) |
risch | \(-\frac {7 x^{2}+2 x -5}{3 \left (x -1\right ) \sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}+\frac {\ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (x -1\right ) \left (1+x \right )}}{\sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}\) | \(71\) |
default | \(-\frac {3 x \left (x^{2}-1\right )^{\frac {3}{2}}-3 \sqrt {x^{2}-1}\, x^{3}-3 \ln \left (x +\sqrt {x^{2}-1}\right ) x^{3}-2 \left (x^{2}-1\right )^{\frac {3}{2}}+9 x^{2} \sqrt {x^{2}-1}+9 \ln \left (x +\sqrt {x^{2}-1}\right ) x^{2}-9 x \sqrt {x^{2}-1}-9 \ln \left (x +\sqrt {x^{2}-1}\right ) x +3 \sqrt {x^{2}-1}+3 \ln \left (x +\sqrt {x^{2}-1}\right )}{3 \left (x -1\right )^{2} \sqrt {\left (x -1\right ) \left (1+x \right )}\, \sqrt {\frac {x -1}{1+x}}}\) | \(146\) |
-1/3*(1+x)*(7*x-5)/(x-1)^2*(-(1-x)/(1+x))^(1/2)+ln((-(1-x)/(1+x))^(1/2)*x+ (-(1-x)/(1+x))^(1/2)+x)
Time = 0.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.53 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1-x)^2} \, dx=\frac {3 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - 3 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) - {\left (7 \, x^{2} + 2 \, x - 5\right )} \sqrt {\frac {x - 1}{x + 1}}}{3 \, {\left (x^{2} - 2 \, x + 1\right )}} \]
1/3*(3*(x^2 - 2*x + 1)*log(sqrt((x - 1)/(x + 1)) + 1) - 3*(x^2 - 2*x + 1)* log(sqrt((x - 1)/(x + 1)) - 1) - (7*x^2 + 2*x - 5)*sqrt((x - 1)/(x + 1)))/ (x^2 - 2*x + 1)
\[ \int \frac {e^{\coth ^{-1}(x)} x}{(1-x)^2} \, dx=\int \frac {x}{\sqrt {\frac {x - 1}{x + 1}} \left (x - 1\right )^{2}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1-x)^2} \, dx=-\frac {\frac {6 \, {\left (x - 1\right )}}{x + 1} + 1}{3 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}}} + \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]
-1/3*(6*(x - 1)/(x + 1) + 1)/((x - 1)/(x + 1))^(3/2) + log(sqrt((x - 1)/(x + 1)) + 1) - log(sqrt((x - 1)/(x + 1)) - 1)
Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1-x)^2} \, dx=-\frac {\log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{\mathrm {sgn}\left (x + 1\right )} + \frac {2 \, {\left (9 \, {\left (x - \sqrt {x^{2} - 1}\right )}^{2} - 12 \, x + 12 \, \sqrt {x^{2} - 1} + 7\right )}}{3 \, {\left (x - \sqrt {x^{2} - 1} - 1\right )}^{3} \mathrm {sgn}\left (x + 1\right )} + \frac {7}{3} \, \mathrm {sgn}\left (x + 1\right ) \]
-log(abs(-x + sqrt(x^2 - 1)))/sgn(x + 1) + 2/3*(9*(x - sqrt(x^2 - 1))^2 - 12*x + 12*sqrt(x^2 - 1) + 7)/((x - sqrt(x^2 - 1) - 1)^3*sgn(x + 1)) + 7/3* sgn(x + 1)
Time = 4.62 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1-x)^2} \, dx=2\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )-\frac {\frac {2\,\left (x-1\right )}{x+1}+\frac {1}{3}}{{\left (\frac {x-1}{x+1}\right )}^{3/2}} \]