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

Optimal. Leaf size=70 \[ -\frac{2 \left (a+\frac{1}{x}\right )}{a^2 c \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{c}+\frac{2 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c} \]

[Out]

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

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Rubi [A]  time = 0.195578, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {6177, 852, 1805, 807, 266, 63, 208} \[ -\frac{2 \left (a+\frac{1}{x}\right )}{a^2 c \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{c}+\frac{2 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]/(c - c/(a*x)),x]

[Out]

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

Rule 6177

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

Rule 852

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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

Mathematica [A]  time = 0.0896123, size = 63, normalized size = 0.9 \[ \frac{a x \sqrt{1-\frac{1}{a^2 x^2}} (a x-3)+2 (a x-1) \log \left (x \left (\sqrt{1-\frac{1}{a^2 x^2}}+1\right )\right )}{a c (a x-1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.135, size = 250, normalized size = 3.6 \begin{align*}{\frac{1}{ \left ( ax-1 \right ) ac} \left ( 2\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ){x}^{2}{a}^{3}+2\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }{x}^{2}{a}^{2}-4\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) x{a}^{2}- \left ( \left ( ax-1 \right ) \left ( ax+1 \right ) \right ) ^{{\frac{3}{2}}}\sqrt{{a}^{2}}-4\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }xa+2\,a\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) +2\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00315, size = 157, normalized size = 2.24 \begin{align*} -2 \, a{\left (\frac{\frac{2 \,{\left (a x - 1\right )}}{a x + 1} - 1}{a^{2} c \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}} - a^{2} c \sqrt{\frac{a x - 1}{a x + 1}}} - \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2} c} + \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2} c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.53907, size = 223, normalized size = 3.19 \begin{align*} \frac{2 \,{\left (a x - 1\right )} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) - 2 \,{\left (a x - 1\right )} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right ) +{\left (a^{2} x^{2} - 2 \, a x - 3\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{a^{2} c x - a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a \int \frac{x}{a x \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}} - \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}\, dx}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.15889, size = 173, normalized size = 2.47 \begin{align*} 2 \, a{\left (\frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2} c} - \frac{\log \left ({\left | \sqrt{\frac{a x - 1}{a x + 1}} - 1 \right |}\right )}{a^{2} c} - \frac{\frac{2 \,{\left (a x - 1\right )}}{a x + 1} - 1}{a^{2} c{\left (\frac{{\left (a x - 1\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{a x + 1} - \sqrt{\frac{a x - 1}{a x + 1}}\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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