Integrand size = 10, antiderivative size = 64 \[ \int e^{-\coth ^{-1}(a x)} x \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{a}+\frac {1}{2} \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^2} \] Output:
-(1-1/a^2/x^2)^(1/2)*x/a+1/2*(1-1/a^2/x^2)^(1/2)*x^2+1/2*arctanh((1-1/a^2/ x^2)^(1/2))/a^2
Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.77 \[ \int e^{-\coth ^{-1}(a x)} x \, dx=\frac {a \sqrt {1-\frac {1}{a^2 x^2}} x (-2+a x)+\log \left (\left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{2 a^2} \] Input:
Integrate[x/E^ArcCoth[a*x],x]
Output:
(a*Sqrt[1 - 1/(a^2*x^2)]*x*(-2 + a*x) + Log[(1 + Sqrt[1 - 1/(a^2*x^2)])*x] )/(2*a^2)
Time = 0.47 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {6719, 539, 27, 534, 243, 73, 221}
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.
\(\displaystyle \int x e^{-\coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6719 |
\(\displaystyle -\int \frac {\left (1-\frac {1}{a x}\right ) x^3}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}\) |
\(\Big \downarrow \) 539 |
\(\displaystyle \frac {1}{2} \int \frac {\left (2 a-\frac {1}{x}\right ) x^2}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}+\frac {1}{2} x^2 \sqrt {1-\frac {1}{a^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\left (2 a-\frac {1}{x}\right ) x^2}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}}{2 a^2}+\frac {1}{2} x^2 \sqrt {1-\frac {1}{a^2 x^2}}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle \frac {-\int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-2 a x \sqrt {1-\frac {1}{a^2 x^2}}}{2 a^2}+\frac {1}{2} x^2 \sqrt {1-\frac {1}{a^2 x^2}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {-\frac {1}{2} \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x^2}-2 a x \sqrt {1-\frac {1}{a^2 x^2}}}{2 a^2}+\frac {1}{2} x^2 \sqrt {1-\frac {1}{a^2 x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {a^2 \int \frac {1}{a^2-a^2 \sqrt {1-\frac {1}{a^2 x^2}}}d\sqrt {1-\frac {1}{a^2 x^2}}-2 a x \sqrt {1-\frac {1}{a^2 x^2}}}{2 a^2}+\frac {1}{2} x^2 \sqrt {1-\frac {1}{a^2 x^2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )-2 a x \sqrt {1-\frac {1}{a^2 x^2}}}{2 a^2}+\frac {1}{2} x^2 \sqrt {1-\frac {1}{a^2 x^2}}\) |
Input:
Int[x/E^ArcCoth[a*x],x]
Output:
(Sqrt[1 - 1/(a^2*x^2)]*x^2)/2 + (-2*a*Sqrt[1 - 1/(a^2*x^2)]*x + ArcTanh[Sq rt[1 - 1/(a^2*x^2)]])/(2*a^2)
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[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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_.), x_Symbol] :> -Subst[Int[(1 + x/a)^((n + 1)/2)/(x^(m + 2)*(1 - x/a)^((n - 1)/2)*Sqrt[1 - x^2/a^2]), x], x , 1/x] /; FreeQ[a, x] && IntegerQ[(n - 1)/2] && IntegerQ[m]
Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.56
method | result | size |
risch | \(\frac {\left (a x -2\right ) \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}{2 a^{2}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right ) \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{2 a \sqrt {a^{2}}\, \left (a x -1\right )}\) | \(100\) |
default | \(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) \left (-\sqrt {a^{2}}\, \sqrt {a^{2} x^{2}-1}\, a x +2 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}+\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a -2 a \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )\right )}{2 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{2} \sqrt {a^{2}}}\) | \(152\) |
Input:
int(x*((a*x-1)/(a*x+1))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2*(a*x-2)*(a*x+1)/a^2*((a*x-1)/(a*x+1))^(1/2)+1/2/a*ln(a^2*x/(a^2)^(1/2) +(a^2*x^2-1)^(1/2))/(a^2)^(1/2)*((a*x-1)/(a*x+1))^(1/2)*((a*x-1)*(a*x+1))^ (1/2)/(a*x-1)
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.14 \[ \int e^{-\coth ^{-1}(a x)} x \, dx=\frac {{\left (a^{2} x^{2} - a x - 2\right )} \sqrt {\frac {a x - 1}{a x + 1}} + \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{2 \, a^{2}} \] Input:
integrate(x*((a*x-1)/(a*x+1))^(1/2),x, algorithm="fricas")
Output:
1/2*((a^2*x^2 - a*x - 2)*sqrt((a*x - 1)/(a*x + 1)) + log(sqrt((a*x - 1)/(a *x + 1)) + 1) - log(sqrt((a*x - 1)/(a*x + 1)) - 1))/a^2
\[ \int e^{-\coth ^{-1}(a x)} x \, dx=\int x \sqrt {\frac {a x - 1}{a x + 1}}\, dx \] Input:
integrate(x*((a*x-1)/(a*x+1))**(1/2),x)
Output:
Integral(x*sqrt((a*x - 1)/(a*x + 1)), x)
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (54) = 108\).
Time = 0.03 (sec) , antiderivative size = 130, normalized size of antiderivative = 2.03 \[ \int e^{-\coth ^{-1}(a x)} x \, dx=-\frac {1}{2} \, a {\left (\frac {2 \, {\left (3 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {2 \, {\left (a x - 1\right )} a^{3}}{a x + 1} - \frac {{\left (a x - 1\right )}^{2} a^{3}}{{\left (a x + 1\right )}^{2}} - a^{3}} - \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{3}} + \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{3}}\right )} \] Input:
integrate(x*((a*x-1)/(a*x+1))^(1/2),x, algorithm="maxima")
Output:
-1/2*a*(2*(3*((a*x - 1)/(a*x + 1))^(3/2) - sqrt((a*x - 1)/(a*x + 1)))/(2*( a*x - 1)*a^3/(a*x + 1) - (a*x - 1)^2*a^3/(a*x + 1)^2 - a^3) - log(sqrt((a* x - 1)/(a*x + 1)) + 1)/a^3 + log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^3)
Time = 0.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11 \[ \int e^{-\coth ^{-1}(a x)} x \, dx=\frac {1}{2} \, \sqrt {a^{2} x^{2} - 1} {\left (\frac {x \mathrm {sgn}\left (a x + 1\right )}{a} - \frac {2 \, \mathrm {sgn}\left (a x + 1\right )}{a^{2}}\right )} - \frac {\log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right ) \mathrm {sgn}\left (a x + 1\right )}{2 \, a {\left | a \right |}} \] Input:
integrate(x*((a*x-1)/(a*x+1))^(1/2),x, algorithm="giac")
Output:
1/2*sqrt(a^2*x^2 - 1)*(x*sgn(a*x + 1)/a - 2*sgn(a*x + 1)/a^2) - 1/2*log(ab s(-x*abs(a) + sqrt(a^2*x^2 - 1)))*sgn(a*x + 1)/(a*abs(a))
Time = 0.06 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.52 \[ \int e^{-\coth ^{-1}(a x)} x \, dx=\frac {\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a^2}-\frac {\sqrt {\frac {a\,x-1}{a\,x+1}}-3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{a^2+\frac {a^2\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {2\,a^2\,\left (a\,x-1\right )}{a\,x+1}} \] Input:
int(x*((a*x - 1)/(a*x + 1))^(1/2),x)
Output:
atanh(((a*x - 1)/(a*x + 1))^(1/2))/a^2 - (((a*x - 1)/(a*x + 1))^(1/2) - 3* ((a*x - 1)/(a*x + 1))^(3/2))/(a^2 + (a^2*(a*x - 1)^2)/(a*x + 1)^2 - (2*a^2 *(a*x - 1))/(a*x + 1))
Time = 0.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int e^{-\coth ^{-1}(a x)} x \, dx=\frac {\sqrt {a x +1}\, \sqrt {a x -1}\, a x -2 \sqrt {a x +1}\, \sqrt {a x -1}+2 \,\mathrm {log}\left (\frac {\sqrt {a x -1}+\sqrt {a x +1}}{\sqrt {2}}\right )}{2 a^{2}} \] Input:
int(x*((a*x-1)/(a*x+1))^(1/2),x)
Output:
(sqrt(a*x + 1)*sqrt(a*x - 1)*a*x - 2*sqrt(a*x + 1)*sqrt(a*x - 1) + 2*log(( sqrt(a*x - 1) + sqrt(a*x + 1))/sqrt(2)))/(2*a**2)