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

Optimal. Leaf size=41 \[ \frac{c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{a}-\frac{c \sqrt{1-a^2 x^2}}{a} \]

[Out]

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

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Rubi [A]  time = 0.0677943, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6131, 6128, 266, 50, 63, 208} \[ \frac{c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{a}-\frac{c \sqrt{1-a^2 x^2}}{a} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*(c - c/(a*x)),x]

[Out]

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

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

Mathematica [A]  time = 0.0396717, size = 42, normalized size = 1.02 \[ -\frac{c \left (\sqrt{1-a^2 x^2}-\log \left (\sqrt{1-a^2 x^2}+1\right )+\log (x)\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 34, normalized size = 0.8 \begin{align*}{\frac{c}{a} \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)*(c-c/a/x),x)

[Out]

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

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Maxima [A]  time = 0.947166, size = 68, normalized size = 1.66 \begin{align*} \frac{c \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right )}{a} - \frac{\sqrt{-a^{2} x^{2} + 1} c}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.10142, size = 85, normalized size = 2.07 \begin{align*} -\frac{c \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) + \sqrt{-a^{2} x^{2} + 1} c}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 16.261, size = 61, normalized size = 1.49 \begin{align*} \begin{cases} \frac{- c \sqrt{- a^{2} x^{2} + 1} + \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}}{a} & \text{for}\: a \neq 0 \\c x + \tilde{\infty } c \log{\left (x \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12334, size = 74, normalized size = 1.8 \begin{align*} -\frac{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 )}}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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