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

Optimal. Leaf size=65 \[ -\frac{\left (1-a^2 x^2\right )^{3/2}}{a c (1-a x)^2}-\frac{2 \sqrt{1-a^2 x^2}}{a c}+\frac{2 \sin ^{-1}(a x)}{a c} \]

[Out]

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

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Rubi [A]  time = 0.0956761, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6131, 6128, 793, 665, 216} \[ -\frac{\left (1-a^2 x^2\right )^{3/2}}{a c (1-a x)^2}-\frac{2 \sqrt{1-a^2 x^2}}{a c}+\frac{2 \sin ^{-1}(a x)}{a c} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(c - c/(a*x)),x]

[Out]

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

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 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 665

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

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

Mathematica [A]  time = 0.0382358, size = 52, normalized size = 0.8 \[ \frac{\frac{(a x-3) \sqrt{a x+1}}{\sqrt{1-a x}}-4 \sin ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )}{a c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 96, normalized size = 1.5 \begin{align*} -{\frac{1}{ac}\sqrt{-{a}^{2}{x}^{2}+1}}+2\,{\frac{1}{c\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }+2\,{\frac{1}{{a}^{2}c}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}} \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]

-(-a^2*x^2+1)^(1/2)/a/c+2/c/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+2/a^2/c/(x-1/a)*(-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{a x + 1}{\sqrt{-a^{2} x^{2} + 1}{\left (c - \frac{c}{a x}\right )}}\,{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)/(c-c/a/x),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.09259, size = 154, normalized size = 2.37 \begin{align*} -\frac{3 \, a x + 4 \,{\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 - 3\right )} - 3}{a^{2} c x - a 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)/(c-c/a/x),x, algorithm="fricas")

[Out]

-(3*a*x + 4*(a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + sqrt(-a^2*x^2 + 1)*(a*x - 3) - 3)/(a^2*c*x - a*
c)

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

[Out]

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

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Giac [A]  time = 1.17706, size = 99, normalized size = 1.52 \begin{align*} \frac{2 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{c{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1}}{a c} - \frac{4}{c{\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((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a/x),x, algorithm="giac")

[Out]

2*arcsin(a*x)*sgn(a)/(c*abs(a)) - sqrt(-a^2*x^2 + 1)/(a*c) - 4/(c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1
)*abs(a))