### 3.19 $$\int e^{3 \coth ^{-1}(a x)} x \, dx$$

Optimal. Leaf size=92 $\frac{1}{2} x^2 \sqrt{1-\frac{1}{a^2 x^2}}+\frac{3 x \sqrt{1-\frac{1}{a^2 x^2}}}{a}-\frac{4 \sqrt{1-\frac{1}{a^2 x^2}}}{a \left (a-\frac{1}{x}\right )}+\frac{9 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a^2}$

[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)

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Rubi [A]  time = 0.871573, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.8, Rules used = {6169, 6742, 651, 266, 51, 63, 208, 264} $\frac{1}{2} x^2 \sqrt{1-\frac{1}{a^2 x^2}}+\frac{3 x \sqrt{1-\frac{1}{a^2 x^2}}}{a}-\frac{4 \sqrt{1-\frac{1}{a^2 x^2}}}{a \left (a-\frac{1}{x}\right )}+\frac{9 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a^2}$

Antiderivative was successfully veriﬁed.

[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 6169

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 6742

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

Rule 651

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 266

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 51

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 264

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]

Rubi steps

\begin{align*} \int e^{3 \coth ^{-1}(a x)} x \, dx &=-\operatorname{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 )\\ &=-\operatorname{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 \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{a^2}-\frac{4 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{a^2}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{a}-\operatorname{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} \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-\frac{x}{a^2}}} \, dx,x,\frac{1}{x^2}\right )-\frac{2 \operatorname{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 \operatorname{Subst}\left (\int \frac{1}{a^2-a^2 x^2} \, dx,x,\sqrt{1-\frac{1}{a^2 x^2}}\right )-\frac{\operatorname{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 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a^2}+\frac{1}{2} \operatorname{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 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0632278, size = 66, normalized size = 0.72 $\frac{\frac{a x \sqrt{1-\frac{1}{a^2 x^2}} \left (a^2 x^2+5 a x-14\right )}{a x-1}+9 \log \left (x \left (\sqrt{1-\frac{1}{a^2 x^2}}+1\right )\right )}{2 a^2}$

Warning: Unable to verify antiderivative.

[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
)

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Maple [B]  time = 0.165, size = 421, normalized size = 4.6 \begin{align*}{\frac{1}{2\,{a}^{2} \left ( ax+1 \right ) } \left ( \sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}{x}^{3}{a}^{3}-2\,\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}{x}^{2}{a}^{2}-\ln \left ({ \left ({a}^{2}x+\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \right ){x}^{2}{a}^{3}+10\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }{x}^{2}{a}^{2}+10\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ){x}^{2}{a}^{3}+\sqrt{{a}^{2}}\sqrt{{a}^{2}{x}^{2}-1}xa+2\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}}{\sqrt{{a}^{2}}}} \right ) x{a}^{2}-4\, \left ( \left ( ax-1 \right ) \left ( ax+1 \right ) \right ) ^{3/2}\sqrt{{a}^{2}}-20\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }xa-20\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) x{a}^{2}-\ln \left ({ \left ({a}^{2}x+\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \right ) a+10\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }+10\,a\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) \right ){\frac{1}{\sqrt{{a}^{2}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}} \left ({\frac{ax-1}{ax+1}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a^2*((a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^3*a^3-2*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^2*a^2-ln((a^2*x+(a^2*x^2-1)^(
1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^2*a^3+10*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^2*a^2+10*ln((a^2*x+(a^2)^(1/2)
*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*x^2*a^3+(a^2)^(1/2)*(a^2*x^2-1)^(1/2)*x*a+2*ln((a^2*x+(a^2*x^2-1)^(1/2)
*(a^2)^(1/2))/(a^2)^(1/2))*x*a^2-4*((a*x-1)*(a*x+1))^(3/2)*(a^2)^(1/2)-20*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*
x*a-20*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*x*a^2-ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/
2))/(a^2)^(1/2))*a+10*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)+10*a*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/
(a^2)^(1/2)))/(a^2)^(1/2)/((a*x-1)*(a*x+1))^(1/2)/(a*x+1)/((a*x-1)/(a*x+1))^(3/2)

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Maxima [A]  time = 1.00739, size = 196, normalized size = 2.13 \begin{align*} \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 )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.95602, size = 243, normalized size = 2.64 \begin{align*} \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 )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.18944, size = 189, normalized size = 2.05 \begin{align*} \frac{1}{2} \, a{\left (\frac{9 \, \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{3}} - \frac{9 \, \log \left ({\left | \sqrt{\frac{a x - 1}{a x + 1}} - 1 \right |}\right )}{a^{3}} - \frac{8}{a^{3} \sqrt{\frac{a x - 1}{a x + 1}}} - \frac{2 \,{\left (\frac{5 \,{\left (a x - 1\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{a x + 1} - 7 \, \sqrt{\frac{a x - 1}{a x + 1}}\right )}}{a^{3}{\left (\frac{a x - 1}{a x + 1} - 1\right )}^{2}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(9*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^3 - 9*log(abs(sqrt((a*x - 1)/(a*x + 1)) - 1))/a^3 - 8/(a^3*sqrt(
(a*x - 1)/(a*x + 1))) - 2*(5*(a*x - 1)*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1) - 7*sqrt((a*x - 1)/(a*x + 1)))/(a^3
*((a*x - 1)/(a*x + 1) - 1)^2))