Integrand size = 10, antiderivative size = 92 \[ \int e^{3 \coth ^{-1}(a x)} x \, dx=-\frac {4 \sqrt {1-\frac {1}{a^2 x^2}}}{a \left (a-\frac {1}{x}\right )}+\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} x}{a}+\frac {1}{2} \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {9 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^2} \]
9/2*arctanh((1-1/a^2/x^2)^(1/2))/a^2-4*(1-1/a^2/x^2)^(1/2)/a/(a-1/x)+3*x*( 1-1/a^2/x^2)^(1/2)/a+1/2*x^2*(1-1/a^2/x^2)^(1/2)
Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.72 \[ \int e^{3 \coth ^{-1}(a x)} x \, dx=\frac {\frac {a \sqrt {1-\frac {1}{a^2 x^2}} x \left (-14+5 a x+a^2 x^2\right )}{-1+a x}+9 \log \left (\left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{2 a^2} \]
((a*Sqrt[1 - 1/(a^2*x^2)]*x*(-14 + 5*a*x + a^2*x^2))/(-1 + a*x) + 9*Log[(1 + Sqrt[1 - 1/(a^2*x^2)])*x])/(2*a^2)
Time = 0.59 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6719, 2353, 2009}
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.
\(\displaystyle \int x e^{3 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6719 |
\(\displaystyle -\int \frac {\left (1+\frac {1}{a x}\right )^2 x^3}{\sqrt {1-\frac {1}{a^2 x^2}} \left (1-\frac {1}{a x}\right )}d\frac {1}{x}\) |
\(\Big \downarrow \) 2353 |
\(\displaystyle -\int \left (\frac {x^3}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 x^2}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 x}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4}{a^2 \sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )}\right )d\frac {1}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {9 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^2}+\frac {1}{2} x^2 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {3 x \sqrt {1-\frac {1}{a^2 x^2}}}{a}-\frac {4 \sqrt {1-\frac {1}{a^2 x^2}}}{a \left (a-\frac {1}{x}\right )}\) |
(-4*Sqrt[1 - 1/(a^2*x^2)])/(a*(a - x^(-1))) + (3*Sqrt[1 - 1/(a^2*x^2)]*x)/ a + (Sqrt[1 - 1/(a^2*x^2)]*x^2)/2 + (9*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/(2* a^2)
3.1.19.3.1 Defintions of rubi rules used
Int[(Px_)*((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2) ^(p_), x_Symbol] :> Int[ExpandIntegrand[Px*(e*x)^m*(c + d*x)^n*(a + b*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && PolyQ[Px, x] && (Integer Q[p] || (IntegerQ[2*p] && IntegerQ[m] && ILtQ[n, 0]))
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 = 0.14 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.54
method | result | size |
risch | \(\frac {\left (a x +6\right ) \left (a x -1\right )}{2 a^{2} \sqrt {\frac {a x -1}{a x +1}}}+\frac {\left (\frac {9 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right )}{2 a \sqrt {a^{2}}}-\frac {4 \sqrt {\left (x -\frac {1}{a}\right )^{2} a^{2}+2 \left (x -\frac {1}{a}\right ) a}}{a^{3} \left (x -\frac {1}{a}\right )}\right ) \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}\) | \(142\) |
default | \(\frac {\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{3} x^{3}-2 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{2} x^{2}-\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}+10 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{2} x^{2}+10 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}+\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a x +2 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{2} x -4 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-20 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a x -20 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right ) a^{2} x -\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a +10 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}+10 a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right )}{2 a^{2} \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\) | \(421\) |
1/2*(a*x+6)*(a*x-1)/a^2/((a*x-1)/(a*x+1))^(1/2)+(9/2/a*ln(a^2*x/(a^2)^(1/2 )+(a^2*x^2-1)^(1/2))/(a^2)^(1/2)-4/a^3/(x-1/a)*((x-1/a)^2*a^2+2*(x-1/a)*a) ^(1/2))/(a*x+1)/((a*x-1)/(a*x+1))^(1/2)*((a*x-1)*(a*x+1))^(1/2)
Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.12 \[ \int e^{3 \coth ^{-1}(a x)} x \, dx=\frac {9 \, {\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 9 \, {\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (a^{3} x^{3} + 6 \, a^{2} x^{2} - 9 \, a x - 14\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{2 \, {\left (a^{3} x - a^{2}\right )}} \]
1/2*(9*(a*x - 1)*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 9*(a*x - 1)*log(sqrt ((a*x - 1)/(a*x + 1)) - 1) + (a^3*x^3 + 6*a^2*x^2 - 9*a*x - 14)*sqrt((a*x - 1)/(a*x + 1)))/(a^3*x - a^2)
\[ \int e^{3 \coth ^{-1}(a x)} x \, dx=\int \frac {x}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.58 \[ \int e^{3 \coth ^{-1}(a x)} x \, dx=\frac {1}{2} \, a {\left (\frac {2 \, {\left (\frac {15 \, {\left (a x - 1\right )}}{a x + 1} - \frac {9 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 4\right )}}{a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 2 \, a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + a^{3} \sqrt {\frac {a x - 1}{a x + 1}}} + \frac {9 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{3}} - \frac {9 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{3}}\right )} \]
1/2*a*(2*(15*(a*x - 1)/(a*x + 1) - 9*(a*x - 1)^2/(a*x + 1)^2 - 4)/(a^3*((a *x - 1)/(a*x + 1))^(5/2) - 2*a^3*((a*x - 1)/(a*x + 1))^(3/2) + a^3*sqrt((a *x - 1)/(a*x + 1))) + 9*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^3 - 9*log(sqr t((a*x - 1)/(a*x + 1)) - 1)/a^3)
\[ \int e^{3 \coth ^{-1}(a x)} x \, dx=\int { \frac {x}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \]
Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.27 \[ \int e^{3 \coth ^{-1}(a x)} x \, dx=\frac {9\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a^2}-\frac {\frac {9\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {15\,\left (a\,x-1\right )}{a\,x+1}+4}{a^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}-2\,a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}+a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}} \]