Integrand size = 10, antiderivative size = 114 \[ \int e^{\coth ^{-1}(a x)} x^3 \, dx=\frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^3}+\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}{8 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^3}{3 a}+\frac {1}{4} \sqrt {1-\frac {1}{a^2 x^2}} x^4+\frac {3 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{8 a^4} \] Output:
2/3*(1-1/a^2/x^2)^(1/2)*x/a^3+3/8*(1-1/a^2/x^2)^(1/2)*x^2/a^2+1/3*(1-1/a^2 /x^2)^(1/2)*x^3/a+1/4*(1-1/a^2/x^2)^(1/2)*x^4+3/8*arctanh((1-1/a^2/x^2)^(1 /2))/a^4
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.60 \[ \int e^{\coth ^{-1}(a x)} x^3 \, dx=\frac {a \sqrt {1-\frac {1}{a^2 x^2}} x \left (16+9 a x+8 a^2 x^2+6 a^3 x^3\right )+9 \log \left (\left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{24 a^4} \] Input:
Integrate[E^ArcCoth[a*x]*x^3,x]
Output:
(a*Sqrt[1 - 1/(a^2*x^2)]*x*(16 + 9*a*x + 8*a^2*x^2 + 6*a^3*x^3) + 9*Log[(1 + Sqrt[1 - 1/(a^2*x^2)])*x])/(24*a^4)
Time = 0.59 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.400, Rules used = {6719, 539, 25, 27, 539, 25, 27, 539, 25, 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.
| \(\displaystyle \int x^3 e^{\coth ^{-1}(a x)} \, dx\) |
| \(\Big \downarrow \) 6719 |
| \(\displaystyle -\int \frac {\left (1+\frac {1}{a x}\right ) x^5}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle \frac {1}{4} \int -\frac {\left (4 a+\frac {3}{x}\right ) x^4}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}+\frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {1}{4} \int \frac {\left (4 a+\frac {3}{x}\right ) x^4}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {\int \frac {\left (4 a+\frac {3}{x}\right ) x^4}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}}{4 a^2}\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle \frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {-\frac {1}{3} \int -\frac {\left (9 a+\frac {8}{x}\right ) x^3}{a \sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-\frac {4}{3} a x^3 \sqrt {1-\frac {1}{a^2 x^2}}}{4 a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {\frac {1}{3} \int \frac {\left (9 a+\frac {8}{x}\right ) x^3}{a \sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-\frac {4}{3} a x^3 \sqrt {1-\frac {1}{a^2 x^2}}}{4 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {\frac {\int \frac {\left (9 a+\frac {8}{x}\right ) x^3}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}}{3 a}-\frac {4}{3} a x^3 \sqrt {1-\frac {1}{a^2 x^2}}}{4 a^2}\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle \frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {\frac {-\frac {1}{2} \int -\frac {\left (16 a+\frac {9}{x}\right ) x^2}{a \sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-\frac {9}{2} a x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}-\frac {4}{3} a x^3 \sqrt {1-\frac {1}{a^2 x^2}}}{4 a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {\frac {\frac {1}{2} \int \frac {\left (16 a+\frac {9}{x}\right ) x^2}{a \sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-\frac {9}{2} a x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}-\frac {4}{3} a x^3 \sqrt {1-\frac {1}{a^2 x^2}}}{4 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {\frac {\frac {\int \frac {\left (16 a+\frac {9}{x}\right ) x^2}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}}{2 a}-\frac {9}{2} a x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}-\frac {4}{3} a x^3 \sqrt {1-\frac {1}{a^2 x^2}}}{4 a^2}\) |
| \(\Big \downarrow \) 534 |
| \(\displaystyle \frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {\frac {\frac {9 \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-16 a x \sqrt {1-\frac {1}{a^2 x^2}}}{2 a}-\frac {9}{2} a x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}-\frac {4}{3} a x^3 \sqrt {1-\frac {1}{a^2 x^2}}}{4 a^2}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {\frac {\frac {\frac {9}{2} \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x^2}-16 a x \sqrt {1-\frac {1}{a^2 x^2}}}{2 a}-\frac {9}{2} a x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}-\frac {4}{3} a x^3 \sqrt {1-\frac {1}{a^2 x^2}}}{4 a^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {\frac {\frac {-9 a^2 \int \frac {1}{a^2-a^2 \sqrt {1-\frac {1}{a^2 x^2}}}d\sqrt {1-\frac {1}{a^2 x^2}}-16 a x \sqrt {1-\frac {1}{a^2 x^2}}}{2 a}-\frac {9}{2} a x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}-\frac {4}{3} a x^3 \sqrt {1-\frac {1}{a^2 x^2}}}{4 a^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{4} x^4 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {\frac {\frac {-9 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )-16 a x \sqrt {1-\frac {1}{a^2 x^2}}}{2 a}-\frac {9}{2} a x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 a}-\frac {4}{3} a x^3 \sqrt {1-\frac {1}{a^2 x^2}}}{4 a^2}\) |
Input:
Int[E^ArcCoth[a*x]*x^3,x]
Output:
(Sqrt[1 - 1/(a^2*x^2)]*x^4)/4 - ((-4*a*Sqrt[1 - 1/(a^2*x^2)]*x^3)/3 + ((-9 *a*Sqrt[1 - 1/(a^2*x^2)]*x^2)/2 + (-16*a*Sqrt[1 - 1/(a^2*x^2)]*x - 9*ArcTa nh[Sqrt[1 - 1/(a^2*x^2)]])/(2*a))/(3*a))/(4*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 = 0.12 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.03
| method | result | size |
| risch | \(\frac {\left (6 a^{3} x^{3}+8 a^{2} x^{2}+9 a x +16\right ) \left (a x -1\right )}{24 a^{4} \sqrt {\frac {a x -1}{a x +1}}}+\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right ) \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{8 a^{3} \sqrt {a^{2}}\, \sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right )}\) | \(117\) |
| default | \(-\frac {\left (a x -1\right ) \left (-6 \sqrt {a^{2}}\, \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} a x -8 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-15 \sqrt {a^{2}}\, \sqrt {a^{2} x^{2}-1}\, a x +15 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a -24 a \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )-24 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\right )}{24 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{4} \sqrt {a^{2}}}\) | \(193\) |
Input:
int(1/((a*x-1)/(a*x+1))^(1/2)*x^3,x,method=_RETURNVERBOSE)
Output:
1/24*(6*a^3*x^3+8*a^2*x^2+9*a*x+16)*(a*x-1)/a^4/((a*x-1)/(a*x+1))^(1/2)+3/ 8/a^3*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 = 92, normalized size of antiderivative = 0.81 \[ \int e^{\coth ^{-1}(a x)} x^3 \, dx=\frac {{\left (6 \, a^{4} x^{4} + 14 \, a^{3} x^{3} + 17 \, a^{2} x^{2} + 25 \, a x + 16\right )} \sqrt {\frac {a x - 1}{a x + 1}} + 9 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 9 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{24 \, a^{4}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^3,x, algorithm="fricas")
Output:
1/24*((6*a^4*x^4 + 14*a^3*x^3 + 17*a^2*x^2 + 25*a*x + 16)*sqrt((a*x - 1)/( a*x + 1)) + 9*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 9*log(sqrt((a*x - 1)/(a *x + 1)) - 1))/a^4
\[ \int e^{\coth ^{-1}(a x)} x^3 \, dx=\int \frac {x^{3}}{\sqrt {\frac {a x - 1}{a x + 1}}}\, dx \] Input:
integrate(1/((a*x-1)/(a*x+1))**(1/2)*x**3,x)
Output:
Integral(x**3/sqrt((a*x - 1)/(a*x + 1)), x)
Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (94) = 188\).
Time = 0.04 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.78 \[ \int e^{\coth ^{-1}(a x)} x^3 \, dx=\frac {1}{24} \, a {\left (\frac {2 \, {\left (9 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} - 49 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + 31 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 39 \, \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {4 \, {\left (a x - 1\right )} a^{5}}{a x + 1} - \frac {6 \, {\left (a x - 1\right )}^{2} a^{5}}{{\left (a x + 1\right )}^{2}} + \frac {4 \, {\left (a x - 1\right )}^{3} a^{5}}{{\left (a x + 1\right )}^{3}} - \frac {{\left (a x - 1\right )}^{4} a^{5}}{{\left (a x + 1\right )}^{4}} - a^{5}} + \frac {9 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{5}} - \frac {9 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{5}}\right )} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^3,x, algorithm="maxima")
Output:
1/24*a*(2*(9*((a*x - 1)/(a*x + 1))^(7/2) - 49*((a*x - 1)/(a*x + 1))^(5/2) + 31*((a*x - 1)/(a*x + 1))^(3/2) - 39*sqrt((a*x - 1)/(a*x + 1)))/(4*(a*x - 1)*a^5/(a*x + 1) - 6*(a*x - 1)^2*a^5/(a*x + 1)^2 + 4*(a*x - 1)^3*a^5/(a*x + 1)^3 - (a*x - 1)^4*a^5/(a*x + 1)^4 - a^5) + 9*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^5 - 9*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^5)
Time = 0.15 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.97 \[ \int e^{\coth ^{-1}(a x)} x^3 \, dx=\frac {1}{24} \, \sqrt {a^{2} x^{2} - 1} {\left ({\left (2 \, x {\left (\frac {3 \, x}{a \mathrm {sgn}\left (a x + 1\right )} + \frac {4}{a^{2} \mathrm {sgn}\left (a x + 1\right )}\right )} + \frac {9}{a^{3} \mathrm {sgn}\left (a x + 1\right )}\right )} x + \frac {16}{a^{4} \mathrm {sgn}\left (a x + 1\right )}\right )} - \frac {3 \, \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right )}{8 \, a^{3} {\left | a \right |} \mathrm {sgn}\left (a x + 1\right )} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^3,x, algorithm="giac")
Output:
1/24*sqrt(a^2*x^2 - 1)*((2*x*(3*x/(a*sgn(a*x + 1)) + 4/(a^2*sgn(a*x + 1))) + 9/(a^3*sgn(a*x + 1)))*x + 16/(a^4*sgn(a*x + 1))) - 3/8*log(abs(-x*abs(a ) + sqrt(a^2*x^2 - 1)))/(a^3*abs(a)*sgn(a*x + 1))
Time = 0.11 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.50 \[ \int e^{\coth ^{-1}(a x)} x^3 \, dx=\frac {\frac {13\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{4}-\frac {31\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{12}+\frac {49\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{12}-\frac {3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{4}}{a^4+\frac {6\,a^4\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {4\,a^4\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}+\frac {a^4\,{\left (a\,x-1\right )}^4}{{\left (a\,x+1\right )}^4}-\frac {4\,a^4\,\left (a\,x-1\right )}{a\,x+1}}+\frac {3\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{4\,a^4} \] Input:
int(x^3/((a*x - 1)/(a*x + 1))^(1/2),x)
Output:
((13*((a*x - 1)/(a*x + 1))^(1/2))/4 - (31*((a*x - 1)/(a*x + 1))^(3/2))/12 + (49*((a*x - 1)/(a*x + 1))^(5/2))/12 - (3*((a*x - 1)/(a*x + 1))^(7/2))/4) /(a^4 + (6*a^4*(a*x - 1)^2)/(a*x + 1)^2 - (4*a^4*(a*x - 1)^3)/(a*x + 1)^3 + (a^4*(a*x - 1)^4)/(a*x + 1)^4 - (4*a^4*(a*x - 1))/(a*x + 1)) + (3*atanh( ((a*x - 1)/(a*x + 1))^(1/2)))/(4*a^4)
Time = 0.16 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.85 \[ \int e^{\coth ^{-1}(a x)} x^3 \, dx=\frac {6 \sqrt {a x +1}\, \sqrt {a x -1}\, a^{3} x^{3}+8 \sqrt {a x +1}\, \sqrt {a x -1}\, a^{2} x^{2}+9 \sqrt {a x +1}\, \sqrt {a x -1}\, a x +16 \sqrt {a x +1}\, \sqrt {a x -1}+18 \,\mathrm {log}\left (\frac {\sqrt {a x -1}+\sqrt {a x +1}}{\sqrt {2}}\right )}{24 a^{4}} \] Input:
int(1/((a*x-1)/(a*x+1))^(1/2)*x^3,x)
Output:
(6*sqrt(a*x + 1)*sqrt(a*x - 1)*a**3*x**3 + 8*sqrt(a*x + 1)*sqrt(a*x - 1)*a **2*x**2 + 9*sqrt(a*x + 1)*sqrt(a*x - 1)*a*x + 16*sqrt(a*x + 1)*sqrt(a*x - 1) + 18*log((sqrt(a*x - 1) + sqrt(a*x + 1))/sqrt(2)))/(24*a**4)