3.396 \(\int e^{3 \coth ^{-1}(a x)} (c-\frac{c}{a x})^4 \, dx\)

Optimal. Leaf size=103 \[ \frac{c^4 \sqrt{1-\frac{1}{a^2 x^2}} \left (2 a+\frac{3}{x}\right )}{2 a^2}+\frac{c^4 x \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (3 a+\frac{1}{x}\right )}{3 a}-\frac{c^4 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a}+\frac{3 c^4 \csc ^{-1}(a x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.13326, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {6177, 813, 815, 844, 216, 266, 63, 208} \[ \frac{c^4 \sqrt{1-\frac{1}{a^2 x^2}} \left (2 a+\frac{3}{x}\right )}{2 a^2}+\frac{c^4 x \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (3 a+\frac{1}{x}\right )}{3 a}-\frac{c^4 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a}+\frac{3 c^4 \csc ^{-1}(a x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcCoth[a*x])*(c - c/(a*x))^4,x]

[Out]

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

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 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 815

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

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

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 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 e^{3 \coth ^{-1}(a x)} \left (c-\frac{c}{a x}\right )^4 \, dx &=-\left (c^3 \operatorname{Subst}\left (\int \frac{\left (c-\frac{c x}{a}\right ) \left (1-\frac{x^2}{a^2}\right )^{3/2}}{x^2} \, dx,x,\frac{1}{x}\right )\right )\\ &=\frac{c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (3 a+\frac{1}{x}\right ) x}{3 a}+\frac{1}{2} c^3 \operatorname{Subst}\left (\int \frac{\left (\frac{2 c}{a}+\frac{6 c x}{a^2}\right ) \sqrt{1-\frac{x^2}{a^2}}}{x} \, dx,x,\frac{1}{x}\right )\\ &=\frac{c^4 \sqrt{1-\frac{1}{a^2 x^2}} \left (2 a+\frac{3}{x}\right )}{2 a^2}+\frac{c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (3 a+\frac{1}{x}\right ) x}{3 a}-\frac{1}{4} \left (a^2 c^3\right ) \operatorname{Subst}\left (\int \frac{-\frac{4 c}{a^3}-\frac{6 c x}{a^4}}{x \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{c^4 \sqrt{1-\frac{1}{a^2 x^2}} \left (2 a+\frac{3}{x}\right )}{2 a^2}+\frac{c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (3 a+\frac{1}{x}\right ) x}{3 a}+\frac{\left (3 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{2 a^2}+\frac{c^4 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{a}\\ &=\frac{c^4 \sqrt{1-\frac{1}{a^2 x^2}} \left (2 a+\frac{3}{x}\right )}{2 a^2}+\frac{c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (3 a+\frac{1}{x}\right ) x}{3 a}+\frac{3 c^4 \csc ^{-1}(a x)}{2 a}+\frac{c^4 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x}{a^2}}} \, dx,x,\frac{1}{x^2}\right )}{2 a}\\ &=\frac{c^4 \sqrt{1-\frac{1}{a^2 x^2}} \left (2 a+\frac{3}{x}\right )}{2 a^2}+\frac{c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (3 a+\frac{1}{x}\right ) x}{3 a}+\frac{3 c^4 \csc ^{-1}(a x)}{2 a}-\left (a c^4\right ) \operatorname{Subst}\left (\int \frac{1}{a^2-a^2 x^2} \, dx,x,\sqrt{1-\frac{1}{a^2 x^2}}\right )\\ &=\frac{c^4 \sqrt{1-\frac{1}{a^2 x^2}} \left (2 a+\frac{3}{x}\right )}{2 a^2}+\frac{c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (3 a+\frac{1}{x}\right ) x}{3 a}+\frac{3 c^4 \csc ^{-1}(a x)}{2 a}-\frac{c^4 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.21715, size = 175, normalized size = 1.7 \[ -\frac{c^4 \left (-24 a^5 x^5-32 a^4 x^4+12 a^3 x^3+40 a^2 x^2+42 a^4 x^4 \sqrt{1-\frac{1}{a^2 x^2}} \sin ^{-1}\left (\frac{\sqrt{1-\frac{1}{a x}}}{\sqrt{2}}\right )-15 a^4 x^4 \sqrt{1-\frac{1}{a^2 x^2}} \sin ^{-1}\left (\frac{1}{a x}\right )+24 a^4 x^4 \sqrt{1-\frac{1}{a^2 x^2}} \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )+12 a x-8\right )}{24 a^5 x^4 \sqrt{1-\frac{1}{a^2 x^2}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(3*ArcCoth[a*x])*(c - c/(a*x))^4,x]

[Out]

-(c^4*(-8 + 12*a*x + 40*a^2*x^2 + 12*a^3*x^3 - 32*a^4*x^4 - 24*a^5*x^5 + 42*a^4*Sqrt[1 - 1/(a^2*x^2)]*x^4*ArcS
in[Sqrt[1 - 1/(a*x)]/Sqrt[2]] - 15*a^4*Sqrt[1 - 1/(a^2*x^2)]*x^4*ArcSin[1/(a*x)] + 24*a^4*Sqrt[1 - 1/(a^2*x^2)
]*x^4*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]]))/(24*a^5*Sqrt[1 - 1/(a^2*x^2)]*x^4)

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Maple [B]  time = 0.168, size = 233, normalized size = 2.3 \begin{align*} -{\frac{ \left ( ax-1 \right ) ^{2}{c}^{4}}{ \left ( 6\,ax+6 \right ){a}^{4}{x}^{3}} \left ( -6\,\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}{x}^{4}{a}^{4}+6\, \left ({a}^{2}{x}^{2}-1 \right ) ^{3/2}\sqrt{{a}^{2}}{x}^{2}{a}^{2}-9\,\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}{x}^{3}{a}^{3}+6\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}}{\sqrt{{a}^{2}}}} \right ){x}^{3}{a}^{4}-9\,{a}^{3}{x}^{3}\sqrt{{a}^{2}}\arctan \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}-1}}} \right ) +3\,\sqrt{{a}^{2}} \left ({a}^{2}{x}^{2}-1 \right ) ^{3/2}xa-2\, \left ({a}^{2}{x}^{2}-1 \right ) ^{3/2}\sqrt{{a}^{2}} \right ) \left ({\frac{ax-1}{ax+1}} \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}{\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.59112, size = 301, normalized size = 2.92 \begin{align*} -\frac{1}{3} \,{\left (\frac{9 \, c^{4} \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right )}{a^{2}} + \frac{3 \, c^{4} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac{3 \, c^{4} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2}} - \frac{3 \, c^{4} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{7}{2}} + c^{4} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{5}{2}} + 29 \, c^{4} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}} + 15 \, c^{4} \sqrt{\frac{a x - 1}{a x + 1}}}{\frac{2 \,{\left (a x - 1\right )} a^{2}}{a x + 1} - \frac{2 \,{\left (a x - 1\right )}^{3} a^{2}}{{\left (a x + 1\right )}^{3}} - \frac{{\left (a x - 1\right )}^{4} a^{2}}{{\left (a x + 1\right )}^{4}} + a^{2}}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(c-c/a/x)^4,x, algorithm="maxima")

[Out]

-1/3*(9*c^4*arctan(sqrt((a*x - 1)/(a*x + 1)))/a^2 + 3*c^4*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 3*c^4*log(s
qrt((a*x - 1)/(a*x + 1)) - 1)/a^2 - (3*c^4*((a*x - 1)/(a*x + 1))^(7/2) + c^4*((a*x - 1)/(a*x + 1))^(5/2) + 29*
c^4*((a*x - 1)/(a*x + 1))^(3/2) + 15*c^4*sqrt((a*x - 1)/(a*x + 1)))/(2*(a*x - 1)*a^2/(a*x + 1) - 2*(a*x - 1)^3
*a^2/(a*x + 1)^3 - (a*x - 1)^4*a^2/(a*x + 1)^4 + a^2))*a

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Fricas [A]  time = 1.72577, size = 358, normalized size = 3.48 \begin{align*} -\frac{18 \, a^{3} c^{4} x^{3} \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right ) + 6 \, a^{3} c^{4} x^{3} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) - 6 \, a^{3} c^{4} x^{3} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right ) -{\left (6 \, a^{4} c^{4} x^{4} + 14 \, a^{3} c^{4} x^{3} + 11 \, a^{2} c^{4} x^{2} + a c^{4} x - 2 \, c^{4}\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{6 \, a^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(c-c/a/x)^4,x, algorithm="fricas")

[Out]

-1/6*(18*a^3*c^4*x^3*arctan(sqrt((a*x - 1)/(a*x + 1))) + 6*a^3*c^4*x^3*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 6*
a^3*c^4*x^3*log(sqrt((a*x - 1)/(a*x + 1)) - 1) - (6*a^4*c^4*x^4 + 14*a^3*c^4*x^3 + 11*a^2*c^4*x^2 + a*c^4*x -
2*c^4)*sqrt((a*x - 1)/(a*x + 1)))/(a^4*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{c^{4} \left (\int - \frac{4 a}{\frac{a x^{4} \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a x + 1} - \frac{x^{3} \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a x + 1}}\, dx + \int \frac{6 a^{2}}{\frac{a x^{3} \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a x + 1} - \frac{x^{2} \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a x + 1}}\, dx + \int - \frac{4 a^{3}}{\frac{a x^{2} \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a x + 1} - \frac{x \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a x + 1}}\, dx + \int \frac{a^{4}}{\frac{a x \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a x + 1} - \frac{\sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a x + 1}}\, dx + \int \frac{1}{\frac{a x^{5} \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a x + 1} - \frac{x^{4} \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a x + 1}}\, dx\right )}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))**(3/2)*(c-c/a/x)**4,x)

[Out]

c**4*(Integral(-4*a/(a*x**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - x**3*sqrt(a*x/(a*x + 1) - 1/(a*x + 1
))/(a*x + 1)), x) + Integral(6*a**2/(a*x**3*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - x**2*sqrt(a*x/(a*x +
1) - 1/(a*x + 1))/(a*x + 1)), x) + Integral(-4*a**3/(a*x**2*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - x*s
qrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1)), x) + Integral(a**4/(a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x +
1) - sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1)), x) + Integral(1/(a*x**5*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(
a*x + 1) - x**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1)), x))/a**4

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Giac [B]  time = 1.20557, size = 311, normalized size = 3.02 \begin{align*} -\frac{1}{3} \,{\left (\frac{9 \, c^{4} \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right )}{a^{2}} + \frac{3 \, c^{4} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac{3 \, c^{4} \log \left ({\left | \sqrt{\frac{a x - 1}{a x + 1}} - 1 \right |}\right )}{a^{2}} + \frac{6 \, c^{4} \sqrt{\frac{a x - 1}{a x + 1}}}{a^{2}{\left (\frac{a x - 1}{a x + 1} - 1\right )}} - \frac{\frac{20 \,{\left (a x - 1\right )} c^{4} \sqrt{\frac{a x - 1}{a x + 1}}}{a x + 1} + \frac{3 \,{\left (a x - 1\right )}^{2} c^{4} \sqrt{\frac{a x - 1}{a x + 1}}}{{\left (a x + 1\right )}^{2}} + 9 \, c^{4} \sqrt{\frac{a x - 1}{a x + 1}}}{a^{2}{\left (\frac{a x - 1}{a x + 1} + 1\right )}^{3}}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(c-c/a/x)^4,x, algorithm="giac")

[Out]

-1/3*(9*c^4*arctan(sqrt((a*x - 1)/(a*x + 1)))/a^2 + 3*c^4*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 3*c^4*log(a
bs(sqrt((a*x - 1)/(a*x + 1)) - 1))/a^2 + 6*c^4*sqrt((a*x - 1)/(a*x + 1))/(a^2*((a*x - 1)/(a*x + 1) - 1)) - (20
*(a*x - 1)*c^4*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1) + 3*(a*x - 1)^2*c^4*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1)^2 +
9*c^4*sqrt((a*x - 1)/(a*x + 1)))/(a^2*((a*x - 1)/(a*x + 1) + 1)^3))*a