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

Optimal. Leaf size=70 \[ \frac{a x+1}{c x \sqrt{1-a^2 x^2}}-\frac{2 \sqrt{1-a^2 x^2}}{c x}-\frac{a \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{c} \]

[Out]

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

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Rubi [A]  time = 0.116746, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {6148, 823, 807, 266, 63, 208} \[ \frac{a x+1}{c x \sqrt{1-a^2 x^2}}-\frac{2 \sqrt{1-a^2 x^2}}{c x}-\frac{a \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6148

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

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

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]

Rubi steps

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

Mathematica [A]  time = 0.0259739, size = 67, normalized size = 0.96 \[ \frac{2 a^2 x^2-a x \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+a x-1}{c x \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.042, size = 75, normalized size = 1.1 \begin{align*} -{\frac{1}{c} \left ({\frac{1}{x}\sqrt{-{a}^{2}{x}^{2}+1}}+a{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) +{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00868, size = 140, normalized size = 2. \begin{align*} -\frac{\frac{a^{2} \log \left (\sqrt{-a^{2} x^{2} + 1} + 1\right )}{c} - \frac{a^{2} \log \left (\sqrt{-a^{2} x^{2} + 1} - 1\right )}{c} - \frac{2 \, a^{2}}{\sqrt{-a^{2} x^{2} + 1} c}}{2 \, a} + \frac{2 \, a^{2} x^{2} - 1}{\sqrt{a x + 1} \sqrt{-a x + 1} c x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.58153, size = 157, normalized size = 2.24 \begin{align*} \frac{a^{2} x^{2} - a x +{\left (a^{2} x^{2} - a x\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) - \sqrt{-a^{2} x^{2} + 1}{\left (2 \, a x - 1\right )}}{a c x^{2} - c x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{- a^{2} x^{3} \sqrt{- a^{2} x^{2} + 1} + x \sqrt{- a^{2} x^{2} + 1}}\, dx + \int \frac{1}{- a^{2} x^{4} \sqrt{- a^{2} x^{2} + 1} + x^{2} \sqrt{- a^{2} x^{2} + 1}}\, dx}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(-a**2*x**3*sqrt(-a**2*x**2 + 1) + x*sqrt(-a**2*x**2 + 1)), x) + Integral(1/(-a**2*x**4*sqrt(-a**2
*x**2 + 1) + x**2*sqrt(-a**2*x**2 + 1)), x))/c

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Giac [B]  time = 1.21332, size = 215, normalized size = 3.07 \begin{align*} -\frac{a^{2} \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{c{\left | a \right |}} - \frac{{\left (a^{2} - \frac{5 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{x}\right )} a^{2} x}{2 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} c{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{2 \, c x{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c*abs(a)) - 1/2*(a^2 - 5*(sqrt(-a^2*x^2 +
1)*abs(a) + a)/x)*a^2*x/((sqrt(-a^2*x^2 + 1)*abs(a) + a)*c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)*abs(a
)) - 1/2*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(c*x*abs(a))