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

Optimal. Leaf size=35 \[ c \sqrt{1-a^2 x^2}-c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

[Out]

c*Sqrt[1 - a^2*x^2] - c*ArcTanh[Sqrt[1 - a^2*x^2]]

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Rubi [A]  time = 0.0530402, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {6128, 266, 50, 63, 208} \[ c \sqrt{1-a^2 x^2}-c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*(c - a*c*x))/x,x]

[Out]

c*Sqrt[1 - a^2*x^2] - c*ArcTanh[Sqrt[1 - a^2*x^2]]

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 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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [B]  time = 0.0778286, size = 79, normalized size = 2.26 \[ c \left (-\frac{a^2 x^2}{\sqrt{1-a^2 x^2}}+\frac{1}{\sqrt{1-a^2 x^2}}-\tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+\sin ^{-1}(a x)+2 \sin ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.035, size = 32, normalized size = 0.9 \begin{align*} -c \left ( -\sqrt{-{a}^{2}{x}^{2}+1}+{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) \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)*(-a*c*x+c)/x,x)

[Out]

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

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Maxima [A]  time = 1.41313, size = 59, normalized size = 1.69 \begin{align*} -c \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) + \sqrt{-a^{2} x^{2} + 1} 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)*(-a*c*x+c)/x,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.72293, size = 78, normalized size = 2.23 \begin{align*} c \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) + \sqrt{-a^{2} x^{2} + 1} 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)*(-a*c*x+c)/x,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 20.4257, size = 66, normalized size = 1.89 \begin{align*} \frac{a^{2} c \left (\begin{cases} - x^{2} & \text{for}\: a^{2} = 0 \\\frac{2 \sqrt{- a^{2} x^{2} + 1}}{a^{2}} & \text{otherwise} \end{cases}\right )}{2} - \frac{c \left (- \log{\left (-1 + \frac{1}{\sqrt{- a^{2} x^{2} + 1}} \right )} + \log{\left (1 + \frac{1}{\sqrt{- a^{2} x^{2} + 1}} \right )}\right )}{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)*(-a*c*x+c)/x,x)

[Out]

a**2*c*Piecewise((-x**2, Eq(a**2, 0)), (2*sqrt(-a**2*x**2 + 1)/a**2, True))/2 - c*(-log(-1 + 1/sqrt(-a**2*x**2
+ 1)) + log(1 + 1/sqrt(-a**2*x**2 + 1)))/2

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Giac [A]  time = 1.14362, size = 70, normalized size = 2. \begin{align*} \frac{1}{2} \, c{\left (2 \, \sqrt{-a^{2} x^{2} + 1} - \log \left (\sqrt{-a^{2} x^{2} + 1} + 1\right ) + \log \left (-\sqrt{-a^{2} x^{2} + 1} + 1\right )\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)*(-a*c*x+c)/x,x, algorithm="giac")

[Out]

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