3.490 \(\int \frac{e^{-\tanh ^{-1}(a x)}}{(c-\frac{c}{a x})^2} \, dx\)

Optimal. Leaf size=63 \[ \frac{\sqrt{1-a^2 x^2}}{a c^2 (1-a x)}+\frac{\sqrt{1-a^2 x^2}}{a c^2}-\frac{\sin ^{-1}(a x)}{a c^2} \]

[Out]

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

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Rubi [A]  time = 0.179248, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6131, 6128, 1639, 12, 793, 216} \[ \frac{\sqrt{1-a^2 x^2}}{a c^2 (1-a x)}+\frac{\sqrt{1-a^2 x^2}}{a c^2}-\frac{\sin ^{-1}(a x)}{a c^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*(c - c/(a*x))^2),x]

[Out]

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

Rule 6131

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 + (c*x)/d)
^p*E^(n*ArcTanh[a*x]))/x^p, x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]

Rule 6128

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

Rule 1639

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

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

Mathematica [A]  time = 0.0949117, size = 47, normalized size = 0.75 \[ \frac{\sqrt{1-a^2 x^2} \left (\frac{1}{c^2}-\frac{1}{c^2 (a x-1)}\right )}{a}-\frac{\sin ^{-1}(a x)}{a c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcTanh[a*x]*(c - c/(a*x))^2),x]

[Out]

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

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Maple [B]  time = 0.046, size = 198, normalized size = 3.1 \begin{align*}{\frac{1}{2\,{a}^{3}{c}^{2}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}} \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{5}{4\,a{c}^{2}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}-{\frac{5}{4\,{c}^{2}}\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}{4\,a{c}^{2}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}+{\frac{1}{4\,{c}^{2}}\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}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a^3/c^2/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)+5/4/a/c^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-5/4/c^2/
(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))+1/4/a/c^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1
/2)+1/4/c^2/(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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} x^{2} + 1}}{{\left (a x + 1\right )}{\left (c - \frac{c}{a x}\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.10926, size = 158, normalized size = 2.51 \begin{align*} \frac{2 \, a x + 2 \,{\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 - 2\right )} - 2}{a^{2} c^{2} x - a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14473, size = 97, normalized size = 1.54 \begin{align*} -\frac{\arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{c^{2}{\left | a \right |}} + \frac{\sqrt{-a^{2} x^{2} + 1}}{a c^{2}} + \frac{2}{c^{2}{\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)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^2,x, algorithm="giac")

[Out]

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