\(\int e^{3 \coth ^{-1}(a x)} x \, dx\) [19]

Optimal result

Integrand size = 10, antiderivative size = 92 \[ \int e^{3 \coth ^{-1}(a x)} x \, dx=-\frac {4 \sqrt {1-\frac {1}{a^2 x^2}}}{a \left (a-\frac {1}{x}\right )}+\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} x}{a}+\frac {1}{2} \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {9 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^2} \]

[Out]

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)

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {6304, 6874, 665, 272, 44, 65, 214, 270} \[ \int e^{3 \coth ^{-1}(a x)} x \, dx=\frac {9 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^2}+\frac {1}{2} x^2 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {3 x \sqrt {1-\frac {1}{a^2 x^2}}}{a}-\frac {4 \sqrt {1-\frac {1}{a^2 x^2}}}{a \left (a-\frac {1}{x}\right )} \]

[In]

Int[E^(3*ArcCoth[a*x])*x,x]

[Out]

(-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)

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a + c*x^2)^(p + 1)/
(2*c*d*(p + 1))), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 6304

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]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\left (1+\frac {x}{a}\right )^2}{x^3 \left (1-\frac {x}{a}\right ) \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right ) \\ & = -\text {Subst}\left (\int \left (\frac {4}{a^2 (a-x) \sqrt {1-\frac {x^2}{a^2}}}+\frac {1}{x^3 \sqrt {1-\frac {x^2}{a^2}}}+\frac {3}{a x^2 \sqrt {1-\frac {x^2}{a^2}}}+\frac {4}{a^2 x \sqrt {1-\frac {x^2}{a^2}}}\right ) \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {4 \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a^2}-\frac {4 \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a^2}-\frac {3 \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a}-\text {Subst}\left (\int \frac {1}{x^3 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right ) \\ & = -\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} \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )-\frac {2 \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{a^2} \\ & = -\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+4 \text {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{4 a^2} \\ & = -\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 {4 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a^2}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {1}{a^2 x^2}}\right ) \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.72 \[ \int e^{3 \coth ^{-1}(a x)} x \, dx=\frac {\frac {a \sqrt {1-\frac {1}{a^2 x^2}} x \left (-14+5 a x+a^2 x^2\right )}{-1+a x}+9 \log \left (\left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{2 a^2} \]

[In]

Integrate[E^(3*ArcCoth[a*x])*x,x]

[Out]

((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
)

Maple [A] (verified)

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	\(\frac {\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{3} x^{3}-2 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{2} x^{2}-\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}+10 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{2} x^{2}+10 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}+\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a x +2 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{2} x -4 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-20 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a x -20 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right ) a^{2} x -\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a +10 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}+10 a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right )}{2 a^{2} \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\)	\(421\)

[In]

int(1/((a*x-1)/(a*x+1))^(3/2)*x,x,method=_RETURNVERBOSE)

[Out]

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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.12 \[ \int e^{3 \coth ^{-1}(a x)} x \, dx=\frac {9 \, {\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 9 \, {\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (a^{3} x^{3} + 6 \, a^{2} x^{2} - 9 \, a x - 14\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{2 \, {\left (a^{3} x - a^{2}\right )}} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*x,x, algorithm="fricas")

[Out]

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)

Sympy [F]

\[ \int e^{3 \coth ^{-1}(a x)} x \, dx=\int \frac {x}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/((a*x-1)/(a*x+1))**(3/2)*x,x)

[Out]

Integral(x/((a*x - 1)/(a*x + 1))**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.58 \[ \int e^{3 \coth ^{-1}(a x)} x \, dx=\frac {1}{2} \, a {\left (\frac {2 \, {\left (\frac {15 \, {\left (a x - 1\right )}}{a x + 1} - \frac {9 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 4\right )}}{a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 2 \, a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + a^{3} \sqrt {\frac {a x - 1}{a x + 1}}} + \frac {9 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{3}} - \frac {9 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{3}}\right )} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*x,x, algorithm="maxima")

[Out]

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(s
qrt((a*x - 1)/(a*x + 1)) - 1)/a^3)

Giac [F]

\[ \int e^{3 \coth ^{-1}(a x)} x \, dx=\int { \frac {x}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*x,x, algorithm="giac")

[Out]

undef

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.27 \[ \int e^{3 \coth ^{-1}(a x)} x \, dx=\frac {9\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a^2}-\frac {\frac {9\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {15\,\left (a\,x-1\right )}{a\,x+1}+4}{a^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}-2\,a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}+a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}} \]

[In]

int(x/((a*x - 1)/(a*x + 1))^(3/2),x)

[Out]

(9*atanh(((a*x - 1)/(a*x + 1))^(1/2)))/a^2 - ((9*(a*x - 1)^2)/(a*x + 1)^2 - (15*(a*x - 1))/(a*x + 1) + 4)/(a^2
*((a*x - 1)/(a*x + 1))^(1/2) - 2*a^2*((a*x - 1)/(a*x + 1))^(3/2) + a^2*((a*x - 1)/(a*x + 1))^(5/2))