Integrand size = 13, antiderivative size = 71 \[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=\frac {1}{8} \sqrt {1-\frac {1}{x^2}} x^2-\frac {1}{3} \left (1-\frac {1}{x^2}\right )^{3/2} x^3+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4-\frac {1}{8} \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right ) \]
-1/3*(1-1/x^2)^(3/2)*x^3+1/4*(1-1/x^2)^(3/2)*x^4-1/8*arctanh((1-1/x^2)^(1/ 2))+1/8*x^2*(1-1/x^2)^(1/2)
Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.73 \[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=\frac {1}{24} \sqrt {1-\frac {1}{x^2}} x \left (8-3 x-8 x^2+6 x^3\right )-\frac {1}{8} \log \left (\left (1+\sqrt {1-\frac {1}{x^2}}\right ) x\right ) \]
Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {6728, 25, 539, 534, 243, 51, 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.
\(\displaystyle \int (1-x)^2 x e^{\coth ^{-1}(x)} \, dx\) |
\(\Big \downarrow \) 6728 |
\(\displaystyle \int -\sqrt {1-\frac {1}{x^2}} \left (1-\frac {1}{x}\right ) x^5d\frac {1}{x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \sqrt {1-\frac {1}{x^2}} \left (1-\frac {1}{x}\right ) x^5d\frac {1}{x}\) |
\(\Big \downarrow \) 539 |
\(\displaystyle \frac {1}{4} \int \sqrt {1-\frac {1}{x^2}} \left (4-\frac {1}{x}\right ) x^4d\frac {1}{x}+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4\) |
\(\Big \downarrow \) 534 |
\(\displaystyle \frac {1}{4} \left (-\int \sqrt {1-\frac {1}{x^2}} x^3d\frac {1}{x}-\frac {4}{3} \left (1-\frac {1}{x^2}\right )^{3/2} x^3\right )+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{4} \left (-\frac {1}{2} \int \sqrt {1-\frac {1}{x^2}} x^2d\frac {1}{x^2}-\frac {4}{3} \left (1-\frac {1}{x^2}\right )^{3/2} x^3\right )+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {x}{\sqrt {1-\frac {1}{x^2}}}d\frac {1}{x^2}+\sqrt {1-\frac {1}{x^2}} x\right )-\frac {4}{3} \left (1-\frac {1}{x^2}\right )^{3/2} x^3\right )+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\sqrt {1-\frac {1}{x^2}} x-\int \frac {1}{1-\sqrt {1-\frac {1}{x^2}}}d\sqrt {1-\frac {1}{x^2}}\right )-\frac {4}{3} \left (1-\frac {1}{x^2}\right )^{3/2} x^3\right )+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\sqrt {1-\frac {1}{x^2}} x-\text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right )\right )-\frac {4}{3} \left (1-\frac {1}{x^2}\right )^{3/2} x^3\right )+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4\) |
((1 - x^(-2))^(3/2)*x^4)/4 + ((-4*(1 - x^(-2))^(3/2)*x^3)/3 + (Sqrt[1 - x^ (-2)]*x - ArcTanh[Sqrt[1 - x^(-2)]])/2)/4
3.3.85.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] + Simp[1/(a*(m + 1)) Int[x^(m + 1)*(a + b*x^2)^p*(a*d*(m + 1) - b*c*(m + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
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 = 0.46 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.99
method | result | size |
default | \(\frac {\left (x -1\right ) \left (6 x \left (x^{2}-1\right )^{\frac {3}{2}}-8 \left (\left (x -1\right ) \left (1+x \right )\right )^{\frac {3}{2}}+3 x \sqrt {x^{2}-1}-3 \ln \left (x +\sqrt {x^{2}-1}\right )\right )}{24 \sqrt {\frac {x -1}{1+x}}\, \sqrt {\left (x -1\right ) \left (1+x \right )}}\) | \(70\) |
risch | \(\frac {\left (6 x^{3}-8 x^{2}-3 x +8\right ) \left (x -1\right )}{24 \sqrt {\frac {x -1}{1+x}}}-\frac {\ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (x -1\right ) \left (1+x \right )}}{8 \sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}\) | \(70\) |
trager | \(\frac {\left (1+x \right ) \left (6 x^{3}-8 x^{2}-3 x +8\right ) \sqrt {-\frac {1-x}{1+x}}}{24}-\frac {\ln \left (\sqrt {-\frac {1-x}{1+x}}\, x +\sqrt {-\frac {1-x}{1+x}}+x \right )}{8}\) | \(71\) |
1/24*(x-1)*(6*x*(x^2-1)^(3/2)-8*((x-1)*(1+x))^(3/2)+3*x*(x^2-1)^(1/2)-3*ln (x+(x^2-1)^(1/2)))/((x-1)/(1+x))^(1/2)/((x-1)*(1+x))^(1/2)
Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=\frac {1}{24} \, {\left (6 \, x^{4} - 2 \, x^{3} - 11 \, x^{2} + 5 \, x + 8\right )} \sqrt {\frac {x - 1}{x + 1}} - \frac {1}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]
1/24*(6*x^4 - 2*x^3 - 11*x^2 + 5*x + 8)*sqrt((x - 1)/(x + 1)) - 1/8*log(sq rt((x - 1)/(x + 1)) + 1) + 1/8*log(sqrt((x - 1)/(x + 1)) - 1)
\[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=\int \frac {x \left (x - 1\right )^{2}}{\sqrt {\frac {x - 1}{x + 1}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (55) = 110\).
Time = 0.21 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.94 \[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=-\frac {3 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {7}{2}} + 53 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {5}{2}} - 11 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}} + 3 \, \sqrt {\frac {x - 1}{x + 1}}}{12 \, {\left (\frac {4 \, {\left (x - 1\right )}}{x + 1} - \frac {6 \, {\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} + \frac {4 \, {\left (x - 1\right )}^{3}}{{\left (x + 1\right )}^{3}} - \frac {{\left (x - 1\right )}^{4}}{{\left (x + 1\right )}^{4}} - 1\right )}} - \frac {1}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]
-1/12*(3*((x - 1)/(x + 1))^(7/2) + 53*((x - 1)/(x + 1))^(5/2) - 11*((x - 1 )/(x + 1))^(3/2) + 3*sqrt((x - 1)/(x + 1)))/(4*(x - 1)/(x + 1) - 6*(x - 1) ^2/(x + 1)^2 + 4*(x - 1)^3/(x + 1)^3 - (x - 1)^4/(x + 1)^4 - 1) - 1/8*log( sqrt((x - 1)/(x + 1)) + 1) + 1/8*log(sqrt((x - 1)/(x + 1)) - 1)
Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01 \[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=\frac {1}{24} \, {\left ({\left (2 \, x {\left (\frac {3 \, x}{\mathrm {sgn}\left (x + 1\right )} - \frac {4}{\mathrm {sgn}\left (x + 1\right )}\right )} - \frac {3}{\mathrm {sgn}\left (x + 1\right )}\right )} x + \frac {8}{\mathrm {sgn}\left (x + 1\right )}\right )} \sqrt {x^{2} - 1} + \frac {\log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{8 \, \mathrm {sgn}\left (x + 1\right )} \]
1/24*((2*x*(3*x/sgn(x + 1) - 4/sgn(x + 1)) - 3/sgn(x + 1))*x + 8/sgn(x + 1 ))*sqrt(x^2 - 1) + 1/8*log(abs(-x + sqrt(x^2 - 1)))/sgn(x + 1)
Time = 4.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.66 \[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=\frac {\frac {\sqrt {\frac {x-1}{x+1}}}{4}-\frac {11\,{\left (\frac {x-1}{x+1}\right )}^{3/2}}{12}+\frac {53\,{\left (\frac {x-1}{x+1}\right )}^{5/2}}{12}+\frac {{\left (\frac {x-1}{x+1}\right )}^{7/2}}{4}}{\frac {6\,{\left (x-1\right )}^2}{{\left (x+1\right )}^2}-\frac {4\,\left (x-1\right )}{x+1}-\frac {4\,{\left (x-1\right )}^3}{{\left (x+1\right )}^3}+\frac {{\left (x-1\right )}^4}{{\left (x+1\right )}^4}+1}-\frac {\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )}{4} \]