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

Optimal. Leaf size=56 \[ -\frac{2 (1-a x)^2}{a \sqrt{1-a^2 x^2}}-\frac{3 \sqrt{1-a^2 x^2}}{a}-\frac{3 \sin ^{-1}(a x)}{a} \]

[Out]

(-2*(1 - a*x)^2)/(a*Sqrt[1 - a^2*x^2]) - (3*Sqrt[1 - a^2*x^2])/a - (3*ArcSin[a*x])/a

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Rubi [A]  time = 0.0492489, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {6123, 853, 669, 641, 216} \[ -\frac{2 (1-a x)^2}{a \sqrt{1-a^2 x^2}}-\frac{3 \sqrt{1-a^2 x^2}}{a}-\frac{3 \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 - a*x)^2)/(a*Sqrt[1 - a^2*x^2]) - (3*Sqrt[1 - a^2*x^2])/a - (3*ArcSin[a*x])/a

Rule 6123

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

Rule 853

Int[((d_) + (e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a^
m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[m, 0] && IntegerQ[n]

Rule 669

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

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

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]

Rubi steps

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

Mathematica [A]  time = 0.0298534, size = 39, normalized size = 0.7 \[ \frac{\sqrt{1-a^2 x^2} \left (-\frac{4}{a x+1}-1\right )}{a}-\frac{3 \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(-1 - 4/(1 + a*x)))/a - (3*ArcSin[a*x])/a

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Maple [B]  time = 0.041, size = 164, normalized size = 2.9 \begin{align*} -{\frac{1}{{a}^{4} \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}}}}-2\,{\frac{ \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{5/2}}{{a}^{3} \left ( x+{a}^{-1} \right ) ^{2}}}-2\,{\frac{ \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{3/2}}{a}}-3\,\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }x-3\,{\frac{1}{\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}} \right ) } \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)

[Out]

-1/a^4/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-2/a^3/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-2/a*(-a
^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)-3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^
2*(x+1/a)^2+2*a*(x+1/a))^(1/2))

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Maxima [A]  time = 1.43039, size = 85, normalized size = 1.52 \begin{align*} \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{3} x^{2} + 2 \, a^{2} x + a} - \frac{3 \, \arcsin \left (a x\right )}{a} - \frac{6 \, \sqrt{-a^{2} x^{2} + 1}}{a^{2} x + a} \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, algorithm="maxima")

[Out]

(-a^2*x^2 + 1)^(3/2)/(a^3*x^2 + 2*a^2*x + a) - 3*arcsin(a*x)/a - 6*sqrt(-a^2*x^2 + 1)/(a^2*x + a)

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Fricas [A]  time = 2.01886, size = 149, normalized size = 2.66 \begin{align*} -\frac{5 \, a x - 6 \,{\left (a x + 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt{-a^{2} x^{2} + 1}{\left (a x + 5\right )} + 5}{a^{2} x + a} \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, algorithm="fricas")

[Out]

-(5*a*x - 6*(a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + sqrt(-a^2*x^2 + 1)*(a*x + 5) + 5)/(a^2*x + a)

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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}}}{\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)

[Out]

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

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Giac [A]  time = 1.21332, size = 86, normalized size = 1.54 \begin{align*} -\frac{3 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1}}{a} + \frac{8}{{\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, algorithm="giac")

[Out]

-3*arcsin(a*x)*sgn(a)/abs(a) - sqrt(-a^2*x^2 + 1)/a + 8/(((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 1)*abs(a))