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

Optimal. Leaf size=86 \[ \frac{\left (1-a^2 x^2\right )^{5/2}}{a^2 (a x+1)^3}+\frac{3 \left (1-a^2 x^2\right )^{3/2}}{2 a^2 (a x+1)}+\frac{9 \sqrt{1-a^2 x^2}}{2 a^2}+\frac{9 \sin ^{-1}(a x)}{2 a^2} \]

[Out]

(9*Sqrt[1 - a^2*x^2])/(2*a^2) + (3*(1 - a^2*x^2)^(3/2))/(2*a^2*(1 + a*x)) + (1 - a^2*x^2)^(5/2)/(a^2*(1 + a*x)
^3) + (9*ArcSin[a*x])/(2*a^2)

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Rubi [A]  time = 0.364178, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {6124, 1633, 1593, 12, 793, 665, 216} \[ \frac{\left (1-a^2 x^2\right )^{5/2}}{a^2 (a x+1)^3}+\frac{3 \left (1-a^2 x^2\right )^{3/2}}{2 a^2 (a x+1)}+\frac{9 \sqrt{1-a^2 x^2}}{2 a^2}+\frac{9 \sin ^{-1}(a x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*Sqrt[1 - a^2*x^2])/(2*a^2) + (3*(1 - a^2*x^2)^(3/2))/(2*a^2*(1 + a*x)) + (1 - a^2*x^2)^(5/2)/(a^2*(1 + a*x)
^3) + (9*ArcSin[a*x])/(2*a^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 1633

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 793

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

Rule 665

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

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)} x \, dx &=\int \frac{x (1-a x)^2}{(1+a x) \sqrt{1-a^2 x^2}} \, dx\\ &=a \int \frac{\left (\frac{x}{a}-x^2\right ) \sqrt{1-a^2 x^2}}{(1+a x)^2} \, dx\\ &=a \int \frac{\left (\frac{1}{a}-x\right ) x \sqrt{1-a^2 x^2}}{(1+a x)^2} \, dx\\ &=a^2 \int \frac{x \left (1-a^2 x^2\right )^{3/2}}{a^2 (1+a x)^3} \, dx\\ &=\int \frac{x \left (1-a^2 x^2\right )^{3/2}}{(1+a x)^3} \, dx\\ &=\frac{\left (1-a^2 x^2\right )^{5/2}}{a^2 (1+a x)^3}+\frac{3 \int \frac{\left (1-a^2 x^2\right )^{3/2}}{(1+a x)^2} \, dx}{a}\\ &=\frac{3 \left (1-a^2 x^2\right )^{3/2}}{2 a^2 (1+a x)}+\frac{\left (1-a^2 x^2\right )^{5/2}}{a^2 (1+a x)^3}+\frac{9 \int \frac{\sqrt{1-a^2 x^2}}{1+a x} \, dx}{2 a}\\ &=\frac{9 \sqrt{1-a^2 x^2}}{2 a^2}+\frac{3 \left (1-a^2 x^2\right )^{3/2}}{2 a^2 (1+a x)}+\frac{\left (1-a^2 x^2\right )^{5/2}}{a^2 (1+a x)^3}+\frac{9 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{2 a}\\ &=\frac{9 \sqrt{1-a^2 x^2}}{2 a^2}+\frac{3 \left (1-a^2 x^2\right )^{3/2}}{2 a^2 (1+a x)}+\frac{\left (1-a^2 x^2\right )^{5/2}}{a^2 (1+a x)^3}+\frac{9 \sin ^{-1}(a x)}{2 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0533774, size = 44, normalized size = 0.51 \[ \frac{\sqrt{1-a^2 x^2} \left (-a x+\frac{8}{a x+1}+6\right )+9 \sin ^{-1}(a x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(6 - a*x + 8/(1 + a*x)) + 9*ArcSin[a*x])/(2*a^2)

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Maple [B]  time = 0.043, size = 169, normalized size = 2. \begin{align*} 3\,{\frac{ \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{5/2}}{{a}^{4} \left ( x+{a}^{-1} \right ) ^{2}}}+3\,{\frac{ \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{3/2}}{{a}^{2}}}+{\frac{9\,x}{2\,a}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}+{\frac{9}{2\,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}{{a}^{5} \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(x/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.46216, size = 149, normalized size = 1.73 \begin{align*} -\frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{4} x^{2} + 2 \, a^{3} x + a^{2}} + \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{2 \,{\left (a^{3} x + a^{2}\right )}} + \frac{6 \, \sqrt{-a^{2} x^{2} + 1}}{a^{3} x + a^{2}} + \frac{9 \, \arcsin \left (a x\right )}{2 \, a^{2}} + \frac{3 \, \sqrt{-a^{2} x^{2} + 1}}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-a^2*x^2 + 1)^(3/2)/(a^4*x^2 + 2*a^3*x + a^2) + 1/2*(-a^2*x^2 + 1)^(3/2)/(a^3*x + a^2) + 6*sqrt(-a^2*x^2 + 1
)/(a^3*x + a^2) + 9/2*arcsin(a*x)/a^2 + 3/2*sqrt(-a^2*x^2 + 1)/a^2

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Fricas [A]  time = 1.92971, size = 177, normalized size = 2.06 \begin{align*} \frac{14 \, a x - 18 \,{\left (a x + 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) -{\left (a^{2} x^{2} - 5 \, a x - 14\right )} \sqrt{-a^{2} x^{2} + 1} + 14}{2 \,{\left (a^{3} x + a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(14*a*x - 18*(a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (a^2*x^2 - 5*a*x - 14)*sqrt(-a^2*x^2 + 1)
+ 14)/(a^3*x + a^2)

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

[Out]

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

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Giac [A]  time = 1.17709, size = 105, normalized size = 1.22 \begin{align*} -\frac{1}{2} \, \sqrt{-a^{2} x^{2} + 1}{\left (\frac{x}{a} - \frac{6}{a^{2}}\right )} + \frac{9 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{2 \, a{\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(x/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="giac")

[Out]

-1/2*sqrt(-a^2*x^2 + 1)*(x/a - 6/a^2) + 9/2*arcsin(a*x)*sgn(a)/(a*abs(a)) - 8/(a*((sqrt(-a^2*x^2 + 1)*abs(a) +
a)/(a^2*x) + 1)*abs(a))