Integrand size = 12, antiderivative size = 142 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=-\frac {\sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{4 a}+\frac {1}{2} \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4} x^2-\frac {\arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2} \]
-1/4*(1-1/a/x)^(1/4)*(1+1/a/x)^(3/4)*x/a+1/2*(1-1/a/x)^(5/4)*(1+1/a/x)^(3/ 4)*x^2-1/4*arctan((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a^2+1/4*arctanh((1+1/a/ x)^(1/4)/(1-1/a/x)^(1/4))/a^2
Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.46 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\frac {-\frac {2 e^{\frac {3}{2} \coth ^{-1}(a x)} \left (-5+e^{2 \coth ^{-1}(a x)}\right )}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^2}-\arctan \left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )+\text {arctanh}\left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )}{4 a^2} \]
((-2*E^((3*ArcCoth[a*x])/2)*(-5 + E^(2*ArcCoth[a*x])))/(-1 + E^(2*ArcCoth[ a*x]))^2 - ArcTan[E^(ArcCoth[a*x]/2)] + ArcTanh[E^(ArcCoth[a*x]/2)])/(4*a^ 2)
Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6721, 107, 105, 104, 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.
| \(\displaystyle \int x e^{-\frac {1}{2} \coth ^{-1}(a x)} \, dx\) |
| \(\Big \downarrow \) 6721 |
| \(\displaystyle -\int \frac {\sqrt [4]{1-\frac {1}{a x}} x^3}{\sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}\) |
| \(\Big \downarrow \) 107 |
| \(\displaystyle \frac {\int \frac {\sqrt [4]{1-\frac {1}{a x}} x^2}{\sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}}{4 a}+\frac {1}{2} x^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 105 |
| \(\displaystyle \frac {x \left (-\sqrt [4]{1-\frac {1}{a x}}\right ) \left (\frac {1}{a x}+1\right )^{3/4}-\frac {\int \frac {x}{\left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}}{2 a}}{4 a}+\frac {1}{2} x^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {x \left (-\sqrt [4]{1-\frac {1}{a x}}\right ) \left (\frac {1}{a x}+1\right )^{3/4}-\frac {2 \int -\frac {1}{\left (1-\frac {1}{x^4}\right ) x^2}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}}{a}}{4 a}+\frac {1}{2} x^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {2 \int \frac {1}{\left (1-\frac {1}{x^4}\right ) x^2}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}}{a}-x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{4 a}+\frac {1}{2} x^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {x \left (-\sqrt [4]{1-\frac {1}{a x}}\right ) \left (\frac {1}{a x}+1\right )^{3/4}-\frac {2 \left (\frac {1}{2} \int \frac {1}{1+\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}-\frac {1}{2} \int \frac {1}{1-\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}}{4 a}+\frac {1}{2} x^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {x \left (-\sqrt [4]{1-\frac {1}{a x}}\right ) \left (\frac {1}{a x}+1\right )^{3/4}-\frac {2 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}}{4 a}+\frac {1}{2} x^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (\frac {1}{a x}+1\right )^{3/4}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {x \left (-\sqrt [4]{1-\frac {1}{a x}}\right ) \left (\frac {1}{a x}+1\right )^{3/4}-\frac {2 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )\right )}{a}}{4 a}+\frac {1}{2} x^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (\frac {1}{a x}+1\right )^{3/4}\) |
((1 - 1/(a*x))^(5/4)*(1 + 1/(a*x))^(3/4)*x^2)/2 + (-((1 - 1/(a*x))^(1/4)*( 1 + 1/(a*x))^(3/4)*x) - (2*(ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4) ]/2 - ArcTanh[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)]/2))/a)/(4*a)
3.1.89.3.1 Defintions of rubi rules used
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] || !SumSimplerQ[p, 1]) && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x ] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
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])
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_)^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]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> -Subst[Int[(1 + x /a)^(n/2)/(x^(m + 2)*(1 - x/a)^(n/2)), x], x, 1/x] /; FreeQ[{a, n}, x] && !IntegerQ[n] && IntegerQ[m]
\[\int x \left (\frac {a x -1}{a x +1}\right )^{\frac {1}{4}}d x\]
Time = 0.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.65 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\frac {2 \, {\left (2 \, a^{2} x^{2} - a x - 3\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 2 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) + \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) - \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{8 \, a^{2}} \]
1/8*(2*(2*a^2*x^2 - a*x - 3)*((a*x - 1)/(a*x + 1))^(1/4) + 2*arctan(((a*x - 1)/(a*x + 1))^(1/4)) + log(((a*x - 1)/(a*x + 1))^(1/4) + 1) - log(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^2
\[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\int x \sqrt [4]{\frac {a x - 1}{a x + 1}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.06 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=-\frac {1}{8} \, a {\left (\frac {4 \, {\left (5 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} - \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{\frac {2 \, {\left (a x - 1\right )} a^{3}}{a x + 1} - \frac {{\left (a x - 1\right )}^{2} a^{3}}{{\left (a x + 1\right )}^{2}} - a^{3}} - \frac {2 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} - \frac {\log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} + \frac {\log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{3}}\right )} \]
-1/8*a*(4*(5*((a*x - 1)/(a*x + 1))^(5/4) - ((a*x - 1)/(a*x + 1))^(1/4))/(2 *(a*x - 1)*a^3/(a*x + 1) - (a*x - 1)^2*a^3/(a*x + 1)^2 - a^3) - 2*arctan(( (a*x - 1)/(a*x + 1))^(1/4))/a^3 - log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^3 + log(((a*x - 1)/(a*x + 1))^(1/4) - 1)/a^3)
Time = 0.31 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.99 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\frac {1}{8} \, a {\left (\frac {2 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} + \frac {\log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} - \frac {\log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{3}} + \frac {4 \, {\left (\frac {5 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a x + 1} - \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{a^{3} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{2}}\right )} \]
1/8*a*(2*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^3 + log(((a*x - 1)/(a*x + 1 ))^(1/4) + 1)/a^3 - log(abs(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^3 + 4*(5*( a*x - 1)*((a*x - 1)/(a*x + 1))^(1/4)/(a*x + 1) - ((a*x - 1)/(a*x + 1))^(1/ 4))/(a^3*((a*x - 1)/(a*x + 1) - 1)^2))
Time = 0.07 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.85 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\frac {\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4\,a^2}-\frac {\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{2}-\frac {5\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}}{2}}{a^2+\frac {a^2\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {2\,a^2\,\left (a\,x-1\right )}{a\,x+1}}+\frac {\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4\,a^2} \]