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

Optimal. Leaf size=90 \[ \frac{1}{2} x^2 \sqrt{1-\frac{1}{a^2 x^2}}-\frac{3 x \sqrt{1-\frac{1}{a^2 x^2}}}{a}-\frac{4 \sqrt{1-\frac{1}{a^2 x^2}}}{a \left (a+\frac{1}{x}\right )}+\frac{9 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a^2} \]

[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 = 0.842227, antiderivative size = 90, 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, 266, 51, 63, 208, 264, 651} \[ \frac{1}{2} x^2 \sqrt{1-\frac{1}{a^2 x^2}}-\frac{3 x \sqrt{1-\frac{1}{a^2 x^2}}}{a}-\frac{4 \sqrt{1-\frac{1}{a^2 x^2}}}{a \left (a+\frac{1}{x}\right )}+\frac{9 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[x/E^(3*ArcCoth[a*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 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]

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]

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{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}}}-\frac{4}{a^2 (a+x) \sqrt{1-\frac{x^2}{a^2}}}\right ) \, dx,x,\frac{1}{x}\right )\\ &=-\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{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{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 = 0.0616541, size = 66, normalized size = 0.73 \[ \frac{\frac{a x \sqrt{1-\frac{1}{a^2 x^2}} \left (a^2 x^2-5 a x-14\right )}{a x+1}+9 \log \left (x \left (\sqrt{1-\frac{1}{a^2 x^2}}+1\right )\right )}{2 a^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x/E^(3*ArcCoth[a*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 = 0.128, size = 421, normalized size = 4.7 \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}-10\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }{x}^{2}{a}^{2}+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\,\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}+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+\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}+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}-10\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }-\ln \left ({ \left ({a}^{2}x+\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \right ) a+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 ) \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{{a}^{2}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.995983, size = 204, normalized size = 2.27 \begin{align*} -\frac{1}{2} \, a{\left (\frac{2 \,{\left (7 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}} - 5 \, \sqrt{\frac{a x - 1}{a x + 1}}\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{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}} + \frac{8 \, \sqrt{\frac{a x - 1}{a x + 1}}}{a^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.72874, size = 186, normalized size = 2.07 \begin{align*} \frac{{\left (a^{2} x^{2} - 5 \, a x - 14\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 )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2*x^2 - 5*a*x - 14)*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^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int 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(x*((a*x-1)/(a*x+1))**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef