3.894 \(\int \frac{e^{\tanh ^{-1}(a x)}}{x^3 (c-a^2 c x^2)} \, dx\)

Optimal. Leaf size=99 \[ -\frac{2 a \sqrt{1-a^2 x^2}}{c x}-\frac{3 \sqrt{1-a^2 x^2}}{2 c x^2}+\frac{a x+1}{c x^2 \sqrt{1-a^2 x^2}}-\frac{3 a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 c} \]

[Out]

(1 + a*x)/(c*x^2*Sqrt[1 - a^2*x^2]) - (3*Sqrt[1 - a^2*x^2])/(2*c*x^2) - (2*a*Sqrt[1 - a^2*x^2])/(c*x) - (3*a^2
*ArcTanh[Sqrt[1 - a^2*x^2]])/(2*c)

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Rubi [A]  time = 0.138867, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {6148, 823, 835, 807, 266, 63, 208} \[ -\frac{2 a \sqrt{1-a^2 x^2}}{c x}-\frac{3 \sqrt{1-a^2 x^2}}{2 c x^2}+\frac{a x+1}{c x^2 \sqrt{1-a^2 x^2}}-\frac{3 a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(x^3*(c - a^2*c*x^2)),x]

[Out]

(1 + a*x)/(c*x^2*Sqrt[1 - a^2*x^2]) - (3*Sqrt[1 - a^2*x^2])/(2*c*x^2) - (2*a*Sqrt[1 - a^2*x^2])/(c*x) - (3*a^2
*ArcTanh[Sqrt[1 - a^2*x^2]])/(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 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[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^3 \left (c-a^2 c x^2\right )} \, dx &=\frac{\int \frac{1+a x}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx}{c}\\ &=\frac{1+a x}{c x^2 \sqrt{1-a^2 x^2}}+\frac{\int \frac{3 a^2+2 a^3 x}{x^3 \sqrt{1-a^2 x^2}} \, dx}{a^2 c}\\ &=\frac{1+a x}{c x^2 \sqrt{1-a^2 x^2}}-\frac{3 \sqrt{1-a^2 x^2}}{2 c x^2}-\frac{\int \frac{-4 a^3-3 a^4 x}{x^2 \sqrt{1-a^2 x^2}} \, dx}{2 a^2 c}\\ &=\frac{1+a x}{c x^2 \sqrt{1-a^2 x^2}}-\frac{3 \sqrt{1-a^2 x^2}}{2 c x^2}-\frac{2 a \sqrt{1-a^2 x^2}}{c x}+\frac{\left (3 a^2\right ) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx}{2 c}\\ &=\frac{1+a x}{c x^2 \sqrt{1-a^2 x^2}}-\frac{3 \sqrt{1-a^2 x^2}}{2 c x^2}-\frac{2 a \sqrt{1-a^2 x^2}}{c x}+\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )}{4 c}\\ &=\frac{1+a x}{c x^2 \sqrt{1-a^2 x^2}}-\frac{3 \sqrt{1-a^2 x^2}}{2 c x^2}-\frac{2 a \sqrt{1-a^2 x^2}}{c x}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^2}\right )}{2 c}\\ &=\frac{1+a x}{c x^2 \sqrt{1-a^2 x^2}}-\frac{3 \sqrt{1-a^2 x^2}}{2 c x^2}-\frac{2 a \sqrt{1-a^2 x^2}}{c x}-\frac{3 a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 c}\\ \end{align*}

Mathematica [A]  time = 0.0363257, size = 83, normalized size = 0.84 \[ -\frac{-4 a^3 x^3-3 a^2 x^2+3 a^2 x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+2 a x+1}{2 c x^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]/(x^3*(c - a^2*c*x^2)),x]

[Out]

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

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Maple [A]  time = 0.041, size = 97, normalized size = 1. \begin{align*} -{\frac{1}{c} \left ({\frac{a}{x}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{3\,{a}^{2}}{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }+{a\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}}+{\frac{1}{2\,{x}^{2}}\sqrt{-{a}^{2}{x}^{2}+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^3/(-a^2*c*x^2+c),x)

[Out]

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

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

[Out]

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

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Fricas [A]  time = 1.54449, size = 197, normalized size = 1.99 \begin{align*} \frac{2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} + 3 \,{\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) -{\left (4 \, a^{2} x^{2} - a x - 1\right )} \sqrt{-a^{2} x^{2} + 1}}{2 \,{\left (a c x^{3} - c x^{2}\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^3/(-a^2*c*x^2+c),x, algorithm="fricas")

[Out]

1/2*(2*a^3*x^3 - 2*a^2*x^2 + 3*(a^3*x^3 - a^2*x^2)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (4*a^2*x^2 - a*x - 1)*sqr
t(-a^2*x^2 + 1))/(a*c*x^3 - c*x^2)

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

[Out]

(Integral(a/(-a**2*x**4*sqrt(-a**2*x**2 + 1) + x**2*sqrt(-a**2*x**2 + 1)), x) + Integral(1/(-a**2*x**5*sqrt(-a
**2*x**2 + 1) + x**3*sqrt(-a**2*x**2 + 1)), x))/c

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Giac [B]  time = 1.21046, size = 302, normalized size = 3.05 \begin{align*} -\frac{{\left (a^{3} + \frac{3 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a}{x} - \frac{20 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a x^{2}}\right )} a^{4} x^{2}}{8 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} c{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}{\left | a \right |}} - \frac{3 \, a^{3} \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{2 \, c{\left | a \right |}} - \frac{\frac{4 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a c{\left | a \right |}}{x} + \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} c{\left | a \right |}}{a x^{2}}}{8 \, a^{2} c^{2}} \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^3/(-a^2*c*x^2+c),x, algorithm="giac")

[Out]

-1/8*(a^3 + 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a/x - 20*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a*x^2))*a^4*x^2/((sq
rt(-a^2*x^2 + 1)*abs(a) + a)^2*c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)*abs(a)) - 3/2*a^3*log(1/2*abs(-
2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c*abs(a)) - 1/8*(4*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a*c*abs(a
)/x + (sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c*abs(a)/(a*x^2))/(a^2*c^2)