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

Optimal. Leaf size=61 \[ -\frac{a x+1}{a c \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{a c}+\frac{\sin ^{-1}(a x)}{a c} \]

[Out]

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

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Rubi [A]  time = 0.128842, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {6157, 6148, 797, 641, 216, 637} \[ -\frac{a x+1}{a c \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{a c}+\frac{\sin ^{-1}(a x)}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6157

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

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 797

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c, Int[(f + g*x)*(a + c*x^2)^(p
+ 1), x], x] - Dist[a/c, Int[(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && EqQ[a*g^2 + f^2*
c, 0]

Rule 641

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

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]

Rule 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rubi steps

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

Mathematica [A]  time = 0.0280983, size = 53, normalized size = 0.87 \[ \frac{a^2 x^2+\sqrt{1-a^2 x^2} \sin ^{-1}(a x)-a x-2}{a c \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.039, size = 94, normalized size = 1.5 \begin{align*} -{\frac{1}{ac}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{1}{c}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{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^2/x^2),x)

[Out]

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

[Out]

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

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Fricas [A]  time = 2.07212, size = 154, normalized size = 2.52 \begin{align*} -\frac{2 \, a x + 2 \,{\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 - 2\right )} - 2}{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^2/x^2),x, algorithm="fricas")

[Out]

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

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

[Out]

a**2*Integral(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 = 1.17708, size = 97, normalized size = 1.59 \begin{align*} \frac{\arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{c{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1}}{a c} - \frac{2}{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^2/x^2),x, algorithm="giac")

[Out]

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