3.220 \(\int \frac{e^{-3 \coth ^{-1}(a x)}}{c-a c x} \, dx\)

Optimal. Leaf size=53 \[ \frac{2 \left (a-\frac{1}{x}\right )}{a^2 c \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c} \]

[Out]

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

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Rubi [A]  time = 0.198296, antiderivative size = 53, 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 = {6175, 6178, 1805, 266, 63, 208} \[ \frac{2 \left (a-\frac{1}{x}\right )}{a^2 c \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)),x]

[Out]

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

Rule 6175

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

Rule 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

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

Mathematica [A]  time = 0.0643677, size = 54, normalized size = 1.02 \[ \frac{\frac{2 x \sqrt{1-\frac{1}{a^2 x^2}}}{a x+1}-\frac{\log \left (a x \left (\sqrt{1-\frac{1}{a^2 x^2}}+1\right )\right )}{a}}{c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)),x]

[Out]

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

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Maple [B]  time = 0.141, size = 248, normalized size = 4.7 \begin{align*} -{\frac{1}{ac \left ( ax-1 \right ) } \left ( \ln \left ({ \left ({a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}} \right ){x}^{2}{a}^{3}-\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }{x}^{2}{a}^{2}+2\,\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}}-2\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }xa+a\ln \left ({ \left ({a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}} \right ) -\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) } \right ) \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{{a}^{2}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c),x)

[Out]

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

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Maxima [A]  time = 1.03066, size = 105, normalized size = 1.98 \begin{align*} -a{\left (\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} - \frac{2 \, \sqrt{\frac{a x - 1}{a x + 1}}}{a^{2} c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.57184, size = 150, normalized size = 2.83 \begin{align*} \frac{2 \, \sqrt{\frac{a x - 1}{a x + 1}} - \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) + \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int - \frac{\sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a^{2} x^{2} - 1}\, dx + \int \frac{a x \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{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*x+1))**(3/2)/(-a*c*x+c),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c),x, algorithm="giac")

[Out]

undef