3.55 \(\int \frac{e^{-3 \tanh ^{-1}(a x)}}{x} \, dx\)

Optimal. Leaf size=45 \[ \frac{4 \sqrt{1-a^2 x^2}}{a x+1}-\tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+\sin ^{-1}(a x) \]

[Out]

(4*Sqrt[1 - a^2*x^2])/(1 + a*x) + ArcSin[a*x] - ArcTanh[Sqrt[1 - a^2*x^2]]

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Rubi [A]  time = 0.708101, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {6124, 6742, 216, 266, 63, 208, 651} \[ \frac{4 \sqrt{1-a^2 x^2}}{a x+1}-\tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+\sin ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*x),x]

[Out]

(4*Sqrt[1 - a^2*x^2])/(1 + a*x) + ArcSin[a*x] - ArcTanh[Sqrt[1 - a^2*x^2]]

Rule 6124

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/
2)*Sqrt[1 - a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(n - 1)/2]

Rule 6742

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

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

Mathematica [A]  time = 0.0369465, size = 49, normalized size = 1.09 \[ \frac{4 \sqrt{1-a^2 x^2}}{a x+1}-\log \left (\sqrt{1-a^2 x^2}+1\right )+\sin ^{-1}(a x)+\log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*x),x]

[Out]

(4*Sqrt[1 - a^2*x^2])/(1 + a*x) + ArcSin[a*x] + Log[x] - Log[1 + Sqrt[1 - a^2*x^2]]

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Maple [B]  time = 0.049, size = 200, normalized size = 4.4 \begin{align*}{\frac{1}{{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{2}{3} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}+a\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }x+{a\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{3} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+\sqrt{-{a}^{2}{x}^{2}+1}-{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) +{\frac{1}{{a}^{3} \left ( x+{a}^{-1} \right ) ^{3}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)+2/3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)+a*(-a^2*(x+1/a)^2+2*
a*(x+1/a))^(1/2)*x+a/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))+1/3*(-a^2*x^2+1)^(3/
2)+(-a^2*x^2+1)^(1/2)-arctanh(1/(-a^2*x^2+1)^(1/2))+1/a^3/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.06162, size = 193, normalized size = 4.29 \begin{align*} \frac{4 \, a x - 2 \,{\left (a x + 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (a x + 1\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) + 4 \, \sqrt{-a^{2} x^{2} + 1} + 4}{a x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a*x - 2*(a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (a*x + 1)*log((sqrt(-a^2*x^2 + 1) - 1)/x) + 4*sq
rt(-a^2*x^2 + 1) + 4)/(a*x + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.16404, size = 116, normalized size = 2.58 \begin{align*} \frac{a \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}} - \frac{a \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{{\left | a \right |}} - \frac{8 \, a}{{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} + 1\right )}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*arcsin(a*x)*sgn(a)/abs(a) - a*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - 8*a/(((
sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 1)*abs(a))