3.532 \(\int \frac{e^{-2 \coth ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^5} \, dx\)

Optimal. Leaf size=163 \[ \frac{2 a^4 \left (c-\frac{c}{a x}\right )^{9/2}}{9 c^4}-\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{5/2}}{5 c^2}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}+4 a^4 \sqrt{c-\frac{c}{a x}}-4 \sqrt{2} a^4 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right ) \]

[Out]

4*a^4*Sqrt[c - c/(a*x)] + (2*a^4*(c - c/(a*x))^(3/2))/(3*c) + (2*a^4*(c - c/(a*x))^(5/2))/(5*c^2) - (2*a^4*(c
- c/(a*x))^(7/2))/(7*c^3) + (2*a^4*(c - c/(a*x))^(9/2))/(9*c^4) - 4*Sqrt[2]*a^4*Sqrt[c]*ArcTanh[Sqrt[c - c/(a*
x)]/(Sqrt[2]*Sqrt[c])]

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Rubi [A]  time = 0.458131, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6167, 6133, 25, 514, 446, 88, 50, 63, 208} \[ \frac{2 a^4 \left (c-\frac{c}{a x}\right )^{9/2}}{9 c^4}-\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{5/2}}{5 c^2}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}+4 a^4 \sqrt{c-\frac{c}{a x}}-4 \sqrt{2} a^4 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - c/(a*x)]/(E^(2*ArcCoth[a*x])*x^5),x]

[Out]

4*a^4*Sqrt[c - c/(a*x)] + (2*a^4*(c - c/(a*x))^(3/2))/(3*c) + (2*a^4*(c - c/(a*x))^(5/2))/(5*c^2) - (2*a^4*(c
- c/(a*x))^(7/2))/(7*c^3) + (2*a^4*(c - c/(a*x))^(9/2))/(9*c^4) - 4*Sqrt[2]*a^4*Sqrt[c]*ArcTanh[Sqrt[c - c/(a*
x)]/(Sqrt[2]*Sqrt[c])]

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6133

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

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[(u*(
a + b*x^n)^(m + p))/x^(n*p), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
 NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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^{-2 \coth ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^5} \, dx &=-\int \frac{e^{-2 \tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^5} \, dx\\ &=-\int \frac{\sqrt{c-\frac{c}{a x}} (1-a x)}{x^5 (1+a x)} \, dx\\ &=\frac{a \int \frac{\left (c-\frac{c}{a x}\right )^{3/2}}{x^4 (1+a x)} \, dx}{c}\\ &=\frac{a \int \frac{\left (c-\frac{c}{a x}\right )^{3/2}}{\left (a+\frac{1}{x}\right ) x^5} \, dx}{c}\\ &=-\frac{a \operatorname{Subst}\left (\int \frac{x^3 \left (c-\frac{c x}{a}\right )^{3/2}}{a+x} \, dx,x,\frac{1}{x}\right )}{c}\\ &=-\frac{a \operatorname{Subst}\left (\int \left (a^2 \left (c-\frac{c x}{a}\right )^{3/2}-\frac{a^3 \left (c-\frac{c x}{a}\right )^{3/2}}{a+x}-\frac{a^2 \left (c-\frac{c x}{a}\right )^{5/2}}{c}+\frac{a^2 \left (c-\frac{c x}{a}\right )^{7/2}}{c^2}\right ) \, dx,x,\frac{1}{x}\right )}{c}\\ &=\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{5/2}}{5 c^2}-\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{9/2}}{9 c^4}+\frac{a^4 \operatorname{Subst}\left (\int \frac{\left (c-\frac{c x}{a}\right )^{3/2}}{a+x} \, dx,x,\frac{1}{x}\right )}{c}\\ &=\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{5/2}}{5 c^2}-\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{9/2}}{9 c^4}+\left (2 a^4\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-\frac{c x}{a}}}{a+x} \, dx,x,\frac{1}{x}\right )\\ &=4 a^4 \sqrt{c-\frac{c}{a x}}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{5/2}}{5 c^2}-\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{9/2}}{9 c^4}+\left (4 a^4 c\right ) \operatorname{Subst}\left (\int \frac{1}{(a+x) \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )\\ &=4 a^4 \sqrt{c-\frac{c}{a x}}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{5/2}}{5 c^2}-\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{9/2}}{9 c^4}-\left (8 a^5\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-\frac{c}{a x}}\right )\\ &=4 a^4 \sqrt{c-\frac{c}{a x}}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{5/2}}{5 c^2}-\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}+\frac{2 a^4 \left (c-\frac{c}{a x}\right )^{9/2}}{9 c^4}-4 \sqrt{2} a^4 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.139376, size = 95, normalized size = 0.58 \[ \frac{2 \left (788 a^4 x^4-236 a^3 x^3+138 a^2 x^2-95 a x+35\right ) \sqrt{c-\frac{c}{a x}}}{315 x^4}-4 \sqrt{2} a^4 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c - c/(a*x)]/(E^(2*ArcCoth[a*x])*x^5),x]

[Out]

(2*Sqrt[c - c/(a*x)]*(35 - 95*a*x + 138*a^2*x^2 - 236*a^3*x^3 + 788*a^4*x^4))/(315*x^4) - 4*Sqrt[2]*a^4*Sqrt[c
]*ArcTanh[Sqrt[c - c/(a*x)]/(Sqrt[2]*Sqrt[c])]

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Maple [B]  time = 0.186, size = 326, normalized size = 2. \begin{align*} -{\frac{1}{315\,{x}^{5}}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( -1890\,\sqrt{a{x}^{2}-x}{a}^{11/2}\sqrt{{a}^{-1}}{x}^{6}+630\,{a}^{11/2}\sqrt{{a}^{-1}}\sqrt{ \left ( ax-1 \right ) x}{x}^{6}+1260\, \left ( a{x}^{2}-x \right ) ^{3/2}{a}^{9/2}\sqrt{{a}^{-1}}{x}^{4}+945\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}-x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) \sqrt{{a}^{-1}}{x}^{6}{a}^{5}-630\,{a}^{9/2}\sqrt{2}\ln \left ({\frac{2\,\sqrt{2}\sqrt{{a}^{-1}}\sqrt{ \left ( ax-1 \right ) x}a-3\,ax+1}{ax+1}} \right ){x}^{6}-945\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) \sqrt{{a}^{-1}}{x}^{6}{a}^{5}-316\, \left ( a{x}^{2}-x \right ) ^{3/2}{a}^{7/2}\sqrt{{a}^{-1}}{x}^{3}+156\,{a}^{5/2}\sqrt{{a}^{-1}} \left ( a{x}^{2}-x \right ) ^{3/2}{x}^{2}-120\,{a}^{3/2} \left ( a{x}^{2}-x \right ) ^{3/2}x\sqrt{{a}^{-1}}+70\, \left ( a{x}^{2}-x \right ) ^{3/2}\sqrt{a}\sqrt{{a}^{-1}} \right ){\frac{1}{\sqrt{ \left ( ax-1 \right ) x}}}{\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{a}^{-1}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c/a/x)^(1/2)/(a*x+1)*(a*x-1)/x^5,x)

[Out]

-1/315*(c*(a*x-1)/a/x)^(1/2)/x^5*(-1890*(a*x^2-x)^(1/2)*a^(11/2)*(1/a)^(1/2)*x^6+630*a^(11/2)*(1/a)^(1/2)*((a*
x-1)*x)^(1/2)*x^6+1260*(a*x^2-x)^(3/2)*a^(9/2)*(1/a)^(1/2)*x^4+945*ln(1/2*(2*(a*x^2-x)^(1/2)*a^(1/2)+2*a*x-1)/
a^(1/2))*(1/a)^(1/2)*x^6*a^5-630*a^(9/2)*2^(1/2)*ln((2*2^(1/2)*(1/a)^(1/2)*((a*x-1)*x)^(1/2)*a-3*a*x+1)/(a*x+1
))*x^6-945*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*(1/a)^(1/2)*x^6*a^5-316*(a*x^2-x)^(3/2)*a^(7/
2)*(1/a)^(1/2)*x^3+156*a^(5/2)*(1/a)^(1/2)*(a*x^2-x)^(3/2)*x^2-120*a^(3/2)*(a*x^2-x)^(3/2)*x*(1/a)^(1/2)+70*(a
*x^2-x)^(3/2)*a^(1/2)*(1/a)^(1/2))/((a*x-1)*x)^(1/2)/a^(1/2)/(1/a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)*(a*x-1)/(a*x+1)/x^5,x, algorithm="maxima")

[Out]

integrate((a*x - 1)*sqrt(c - c/(a*x))/((a*x + 1)*x^5), x)

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Fricas [A]  time = 1.65622, size = 517, normalized size = 3.17 \begin{align*} \left [\frac{2 \,{\left (315 \, \sqrt{2} a^{4} \sqrt{c} x^{4} \log \left (\frac{2 \, \sqrt{2} a \sqrt{c} x \sqrt{\frac{a c x - c}{a x}} - 3 \, a c x + c}{a x + 1}\right ) +{\left (788 \, a^{4} x^{4} - 236 \, a^{3} x^{3} + 138 \, a^{2} x^{2} - 95 \, a x + 35\right )} \sqrt{\frac{a c x - c}{a x}}\right )}}{315 \, x^{4}}, \frac{2 \,{\left (630 \, \sqrt{2} a^{4} \sqrt{-c} x^{4} \arctan \left (\frac{\sqrt{2} \sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{2 \, c}\right ) +{\left (788 \, a^{4} x^{4} - 236 \, a^{3} x^{3} + 138 \, a^{2} x^{2} - 95 \, a x + 35\right )} \sqrt{\frac{a c x - c}{a x}}\right )}}{315 \, x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)*(a*x-1)/(a*x+1)/x^5,x, algorithm="fricas")

[Out]

[2/315*(315*sqrt(2)*a^4*sqrt(c)*x^4*log((2*sqrt(2)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) - 3*a*c*x + c)/(a*x + 1
)) + (788*a^4*x^4 - 236*a^3*x^3 + 138*a^2*x^2 - 95*a*x + 35)*sqrt((a*c*x - c)/(a*x)))/x^4, 2/315*(630*sqrt(2)*
a^4*sqrt(-c)*x^4*arctan(1/2*sqrt(2)*sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) + (788*a^4*x^4 - 236*a^3*x^3 + 138*a^2
*x^2 - 95*a*x + 35)*sqrt((a*c*x - c)/(a*x)))/x^4]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)**(1/2)*(a*x-1)/(a*x+1)/x**5,x)

[Out]

Integral(sqrt(-c*(-1 + 1/(a*x)))*(a*x - 1)/(x**5*(a*x + 1)), x)

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Giac [B]  time = 2.70279, size = 586, normalized size = 3.6 \begin{align*} \frac{4 \, \sqrt{2} a^{5} c \arctan \left (\frac{\sqrt{2}{\left ({\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )} a + \sqrt{c}{\left | a \right |}\right )}}{2 \, a \sqrt{-c}}\right )}{\sqrt{-c}{\left | a \right |} \mathrm{sgn}\left (x\right )} + \frac{2 \,{\left (1260 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{8} a^{9} c - 1260 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{7} a^{8} c^{\frac{3}{2}}{\left | a \right |} + 2100 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{6} a^{9} c^{2} - 3150 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{5} a^{8} c^{\frac{5}{2}}{\left | a \right |} + 3528 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{4} a^{9} c^{3} - 2625 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{3} a^{8} c^{\frac{7}{2}}{\left | a \right |} + 1215 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{2} a^{9} c^{4} - 315 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )} a^{8} c^{\frac{9}{2}}{\left | a \right |} + 35 \, a^{9} c^{5}\right )}}{315 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{9} a^{4}{\left | a \right |} \mathrm{sgn}\left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)*(a*x-1)/(a*x+1)/x^5,x, algorithm="giac")

[Out]

4*sqrt(2)*a^5*c*arctan(1/2*sqrt(2)*((sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*a + sqrt(c)*abs(a))/(a*sqrt(-c))
)/(sqrt(-c)*abs(a)*sgn(x)) + 2/315*(1260*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^8*a^9*c - 1260*(sqrt(a^2*c)
*x - sqrt(a^2*c*x^2 - a*c*x))^7*a^8*c^(3/2)*abs(a) + 2100*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^6*a^9*c^2
- 3150*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^5*a^8*c^(5/2)*abs(a) + 3528*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 -
 a*c*x))^4*a^9*c^3 - 2625*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^3*a^8*c^(7/2)*abs(a) + 1215*(sqrt(a^2*c)*x
 - sqrt(a^2*c*x^2 - a*c*x))^2*a^9*c^4 - 315*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*a^8*c^(9/2)*abs(a) + 35*
a^9*c^5)/((sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^9*a^4*abs(a)*sgn(x))