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

Optimal. Leaf size=105 \[ -\frac{4 \left (a+\frac{1}{x}\right )}{3 a^2 c^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{9 a+\frac{11}{x}}{3 a^2 c^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{c^2}+\frac{3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c^2} \]

[Out]

(-4*(a + x^(-1)))/(3*a^2*c^2*(1 - 1/(a^2*x^2))^(3/2)) - (9*a + 11/x)/(3*a^2*c^2*Sqrt[1 - 1/(a^2*x^2)]) + (Sqrt
[1 - 1/(a^2*x^2)]*x)/c^2 + (3*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/(a*c^2)

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Rubi [A]  time = 0.292632, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{4 \left (a+\frac{1}{x}\right )}{3 a^2 c^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{9 a+\frac{11}{x}}{3 a^2 c^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{c^2}+\frac{3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c^2} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]/(c - c/(a*x))^2,x]

[Out]

(-4*(a + x^(-1)))/(3*a^2*c^2*(1 - 1/(a^2*x^2))^(3/2)) - (9*a + 11/x)/(3*a^2*c^2*Sqrt[1 - 1/(a^2*x^2)]) + (Sqrt
[1 - 1/(a^2*x^2)]*x)/c^2 + (3*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/(a*c^2)

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)}}{\left (c-\frac{c}{a x}\right )^2} \, dx &=-\left (c \operatorname{Subst}\left (\int \frac{\sqrt{1-\frac{x^2}{a^2}}}{x^2 \left (c-\frac{c x}{a}\right )^3} \, dx,x,\frac{1}{x}\right )\right )\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\left (c+\frac{c x}{a}\right )^3}{x^2 \left (1-\frac{x^2}{a^2}\right )^{5/2}} \, dx,x,\frac{1}{x}\right )}{c^5}\\ &=-\frac{4 \left (a+\frac{1}{x}\right )}{3 a^2 c^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{-3 c^3-\frac{9 c^3 x}{a}-\frac{8 c^3 x^2}{a^2}}{x^2 \left (1-\frac{x^2}{a^2}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{3 c^5}\\ &=-\frac{4 \left (a+\frac{1}{x}\right )}{3 a^2 c^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{9 a+\frac{11}{x}}{3 a^2 c^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{\operatorname{Subst}\left (\int \frac{3 c^3+\frac{9 c^3 x}{a}}{x^2 \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{3 c^5}\\ &=-\frac{4 \left (a+\frac{1}{x}\right )}{3 a^2 c^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{9 a+\frac{11}{x}}{3 a^2 c^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{1-\frac{1}{a^2 x^2}} x}{c^2}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{a c^2}\\ &=-\frac{4 \left (a+\frac{1}{x}\right )}{3 a^2 c^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{9 a+\frac{11}{x}}{3 a^2 c^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{1-\frac{1}{a^2 x^2}} x}{c^2}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x}{a^2}}} \, dx,x,\frac{1}{x^2}\right )}{2 a c^2}\\ &=-\frac{4 \left (a+\frac{1}{x}\right )}{3 a^2 c^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{9 a+\frac{11}{x}}{3 a^2 c^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{1-\frac{1}{a^2 x^2}} x}{c^2}+\frac{(3 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^2}\\ &=-\frac{4 \left (a+\frac{1}{x}\right )}{3 a^2 c^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{9 a+\frac{11}{x}}{3 a^2 c^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{1-\frac{1}{a^2 x^2}} x}{c^2}+\frac{3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c^2}\\ \end{align*}

Mathematica [A]  time = 0.0625943, size = 94, normalized size = 0.9 \[ \frac{3 a^3 x^3-16 a^2 x^2+9 a x \sqrt{1-\frac{1}{a^2 x^2}} (a x-1) \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )-5 a x+14}{3 a^2 c^2 x \sqrt{1-\frac{1}{a^2 x^2}} (a x-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(14 - 5*a*x - 16*a^2*x^2 + 3*a^3*x^3 + 9*a*Sqrt[1 - 1/(a^2*x^2)]*x*(-1 + a*x)*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/
(3*a^2*c^2*Sqrt[1 - 1/(a^2*x^2)]*x*(-1 + a*x))

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Maple [B]  time = 0.139, size = 339, normalized size = 3.2 \begin{align*}{\frac{1}{3\,a \left ( ax-1 \right ) ^{2}{c}^{2}} \left ( 9\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ){x}^{3}{a}^{4}+9\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }{x}^{3}{a}^{3}-27\,\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}-6\,\sqrt{{a}^{2}} \left ( \left ( ax-1 \right ) \left ( ax+1 \right ) \right ) ^{3/2}xa-27\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }{x}^{2}{a}^{2}+27\,\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}+5\, \left ( \left ( ax-1 \right ) \left ( ax+1 \right ) \right ) ^{3/2}\sqrt{{a}^{2}}+27\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }xa-9\,a\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) -9\,\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)^2,x)

[Out]

1/3*(9*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*x^3*a^4+9*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/
2)*x^3*a^3-27*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*x^2*a^3-6*(a^2)^(1/2)*((a*x-1)*(a*x+
1))^(3/2)*x*a-27*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^2*a^2+27*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))
/(a^2)^(1/2))*x*a^2+5*((a*x-1)*(a*x+1))^(3/2)*(a^2)^(1/2)+27*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x*a-9*a*ln((a
^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))-9*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/a/(a^2)^(1/2)/(a
*x-1)^2/c^2/((a*x-1)*(a*x+1))^(1/2)/((a*x-1)/(a*x+1))^(1/2)

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Maxima [A]  time = 1.02613, size = 185, normalized size = 1.76 \begin{align*} \frac{1}{3} \, a{\left (\frac{\frac{11 \,{\left (a x - 1\right )}}{a x + 1} - \frac{18 \,{\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + 1}{a^{2} c^{2} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{5}{2}} - a^{2} c^{2} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}} + \frac{9 \, \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2} c^{2}} - \frac{9 \, \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2} c^{2}}\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)^2,x, algorithm="maxima")

[Out]

1/3*a*((11*(a*x - 1)/(a*x + 1) - 18*(a*x - 1)^2/(a*x + 1)^2 + 1)/(a^2*c^2*((a*x - 1)/(a*x + 1))^(5/2) - a^2*c^
2*((a*x - 1)/(a*x + 1))^(3/2)) + 9*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/(a^2*c^2) - 9*log(sqrt((a*x - 1)/(a*x +
1)) - 1)/(a^2*c^2))

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Fricas [A]  time = 1.60619, size = 309, normalized size = 2.94 \begin{align*} \frac{9 \,{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) - 9 \,{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right ) +{\left (3 \, a^{3} x^{3} - 16 \, a^{2} x^{2} - 5 \, a x + 14\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{3 \,{\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\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)^2,x, algorithm="fricas")

[Out]

1/3*(9*(a^2*x^2 - 2*a*x + 1)*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 9*(a^2*x^2 - 2*a*x + 1)*log(sqrt((a*x - 1)/(
a*x + 1)) - 1) + (3*a^3*x^3 - 16*a^2*x^2 - 5*a*x + 14)*sqrt((a*x - 1)/(a*x + 1)))/(a^3*c^2*x^2 - 2*a^2*c^2*x +
a*c^2)

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

[Out]

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

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Giac [A]  time = 1.16618, size = 200, normalized size = 1.9 \begin{align*} \frac{1}{3} \, a{\left (\frac{9 \, \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2} c^{2}} - \frac{9 \, \log \left ({\left | \sqrt{\frac{a x - 1}{a x + 1}} - 1 \right |}\right )}{a^{2} c^{2}} - \frac{{\left (a x + 1\right )}{\left (\frac{12 \,{\left (a x - 1\right )}}{a x + 1} + 1\right )}}{{\left (a x - 1\right )} a^{2} c^{2} \sqrt{\frac{a x - 1}{a x + 1}}} - \frac{6 \, \sqrt{\frac{a x - 1}{a x + 1}}}{a^{2} c^{2}{\left (\frac{a x - 1}{a x + 1} - 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)^2,x, algorithm="giac")

[Out]

1/3*a*(9*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/(a^2*c^2) - 9*log(abs(sqrt((a*x - 1)/(a*x + 1)) - 1))/(a^2*c^2) -
(a*x + 1)*(12*(a*x - 1)/(a*x + 1) + 1)/((a*x - 1)*a^2*c^2*sqrt((a*x - 1)/(a*x + 1))) - 6*sqrt((a*x - 1)/(a*x +
1))/(a^2*c^2*((a*x - 1)/(a*x + 1) - 1)))