3.1093 \(\int \frac{e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=115 \[ a^3 \left (-c^{3/2}\right ) \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )+a^3 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right )-\frac{a c (a x+1) \sqrt{c-a^2 c x^2}}{x^2}-\frac{\left (c-a^2 c x^2\right )^{3/2}}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.2824, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6151, 1807, 811, 844, 217, 203, 266, 63, 208} \[ a^3 \left (-c^{3/2}\right ) \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )+a^3 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right )-\frac{a c (a x+1) \sqrt{c-a^2 c x^2}}{x^2}-\frac{\left (c-a^2 c x^2\right )^{3/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^(3/2))/x^4,x]

[Out]

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

Rule 6151

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

Rule 1807

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

Rule 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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^{2 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2}}{x^4} \, dx &=c \int \frac{(1+a x)^2 \sqrt{c-a^2 c x^2}}{x^4} \, dx\\ &=-\frac{\left (c-a^2 c x^2\right )^{3/2}}{3 x^3}-\frac{1}{3} \int \frac{\left (-6 a c-3 a^2 c x\right ) \sqrt{c-a^2 c x^2}}{x^3} \, dx\\ &=-\frac{a c (1+a x) \sqrt{c-a^2 c x^2}}{x^2}-\frac{\left (c-a^2 c x^2\right )^{3/2}}{3 x^3}+\frac{\int \frac{-12 a^3 c^3-12 a^4 c^3 x}{x \sqrt{c-a^2 c x^2}} \, dx}{12 c}\\ &=-\frac{a c (1+a x) \sqrt{c-a^2 c x^2}}{x^2}-\frac{\left (c-a^2 c x^2\right )^{3/2}}{3 x^3}-\left (a^3 c^2\right ) \int \frac{1}{x \sqrt{c-a^2 c x^2}} \, dx-\left (a^4 c^2\right ) \int \frac{1}{\sqrt{c-a^2 c x^2}} \, dx\\ &=-\frac{a c (1+a x) \sqrt{c-a^2 c x^2}}{x^2}-\frac{\left (c-a^2 c x^2\right )^{3/2}}{3 x^3}-\frac{1}{2} \left (a^3 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c-a^2 c x}} \, dx,x,x^2\right )-\left (a^4 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+a^2 c x^2} \, dx,x,\frac{x}{\sqrt{c-a^2 c x^2}}\right )\\ &=-\frac{a c (1+a x) \sqrt{c-a^2 c x^2}}{x^2}-\frac{\left (c-a^2 c x^2\right )^{3/2}}{3 x^3}-a^3 c^{3/2} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )+(a c) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2 c}} \, dx,x,\sqrt{c-a^2 c x^2}\right )\\ &=-\frac{a c (1+a x) \sqrt{c-a^2 c x^2}}{x^2}-\frac{\left (c-a^2 c x^2\right )^{3/2}}{3 x^3}-a^3 c^{3/2} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )+a^3 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.15585, size = 127, normalized size = 1.1 \[ a^3 c^{3/2} \log \left (\sqrt{c} \sqrt{c-a^2 c x^2}+c\right )+a^3 c^{3/2} \tan ^{-1}\left (\frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{c} \left (a^2 x^2-1\right )}\right )-a^3 c^{3/2} \log (x)-\frac{c \left (2 a^2 x^2+3 a x+1\right ) \sqrt{c-a^2 c x^2}}{3 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^(3/2))/x^4,x]

[Out]

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

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Maple [B]  time = 0.049, size = 339, normalized size = 3. \begin{align*} -{\frac{4\,{a}^{2}}{3\,cx} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{4\,{a}^{4}x}{3} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-2\,{a}^{4}cx\sqrt{-{a}^{2}c{x}^{2}+c}-2\,{\frac{{a}^{4}{c}^{2}}{\sqrt{{a}^{2}c}}\arctan \left ({\frac{\sqrt{{a}^{2}c}x}{\sqrt{-{a}^{2}c{x}^{2}+c}}} \right ) }-{\frac{{a}^{3}}{3} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{a}^{3}{c}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{-{a}^{2}c{x}^{2}+c} \right ) } \right ) -{a}^{3}\sqrt{-{a}^{2}c{x}^{2}+c}c-{\frac{2\,{a}^{3}}{3} \left ( -c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}+{a}^{4}c\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }x+{{a}^{4}{c}^{2}\arctan \left ({x\sqrt{{a}^{2}c}{\frac{1}{\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}-{\frac{a}{c{x}^{2}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{1}{3\,c{x}^{3}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.83923, size = 601, normalized size = 5.23 \begin{align*} \left [\frac{6 \, a^{3} c^{\frac{3}{2}} x^{3} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} a \sqrt{c} x}{a^{2} c x^{2} - c}\right ) + 3 \, a^{3} c^{\frac{3}{2}} x^{3} \log \left (-\frac{a^{2} c x^{2} - 2 \, \sqrt{-a^{2} c x^{2} + c} \sqrt{c} - 2 \, c}{x^{2}}\right ) - 2 \,{\left (2 \, a^{2} c x^{2} + 3 \, a c x + c\right )} \sqrt{-a^{2} c x^{2} + c}}{6 \, x^{3}}, \frac{6 \, a^{3} \sqrt{-c} c x^{3} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} \sqrt{-c}}{a^{2} c x^{2} - c}\right ) + 3 \, a^{3} \sqrt{-c} c x^{3} \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt{-a^{2} c x^{2} + c} a \sqrt{-c} x - c\right ) - 2 \,{\left (2 \, a^{2} c x^{2} + 3 \, a c x + c\right )} \sqrt{-a^{2} c x^{2} + c}}{6 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(6*a^3*c^(3/2)*x^3*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) + 3*a^3*c^(3/2)*x^3*log(-(a^2
*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) - 2*(2*a^2*c*x^2 + 3*a*c*x + c)*sqrt(-a^2*c*x^2 + c))/x^3,
1/6*(6*a^3*sqrt(-c)*c*x^3*arctan(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/(a^2*c*x^2 - c)) + 3*a^3*sqrt(-c)*c*x^3*log(2*
a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) - 2*(2*a^2*c*x^2 + 3*a*c*x + c)*sqrt(-a^2*c*x^2 + c))/x^3
]

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Sympy [C]  time = 10.0809, size = 359, normalized size = 3.12 \begin{align*} a^{2} c \left (\begin{cases} - \frac{i a^{2} \sqrt{c} x}{\sqrt{a^{2} x^{2} - 1}} + i a \sqrt{c} \operatorname{acosh}{\left (a x \right )} + \frac{i \sqrt{c}}{x \sqrt{a^{2} x^{2} - 1}} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac{a^{2} \sqrt{c} x}{\sqrt{- a^{2} x^{2} + 1}} - a \sqrt{c} \operatorname{asin}{\left (a x \right )} - \frac{\sqrt{c}}{x \sqrt{- a^{2} x^{2} + 1}} & \text{otherwise} \end{cases}\right ) + 2 a c \left (\begin{cases} \frac{a^{2} \sqrt{c} \operatorname{acosh}{\left (\frac{1}{a x} \right )}}{2} + \frac{a \sqrt{c}}{2 x \sqrt{-1 + \frac{1}{a^{2} x^{2}}}} - \frac{\sqrt{c}}{2 a x^{3} \sqrt{-1 + \frac{1}{a^{2} x^{2}}}} & \text{for}\: \frac{1}{\left |{a^{2} x^{2}}\right |} > 1 \\- \frac{i a^{2} \sqrt{c} \operatorname{asin}{\left (\frac{1}{a x} \right )}}{2} - \frac{i a \sqrt{c} \sqrt{1 - \frac{1}{a^{2} x^{2}}}}{2 x} & \text{otherwise} \end{cases}\right ) + c \left (\begin{cases} \frac{a^{3} \sqrt{c} \sqrt{-1 + \frac{1}{a^{2} x^{2}}}}{3} - \frac{a \sqrt{c} \sqrt{-1 + \frac{1}{a^{2} x^{2}}}}{3 x^{2}} & \text{for}\: \frac{1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac{i a^{3} \sqrt{c} \sqrt{1 - \frac{1}{a^{2} x^{2}}}}{3} - \frac{i a \sqrt{c} \sqrt{1 - \frac{1}{a^{2} x^{2}}}}{3 x^{2}} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*c*Piecewise((-I*a**2*sqrt(c)*x/sqrt(a**2*x**2 - 1) + I*a*sqrt(c)*acosh(a*x) + I*sqrt(c)/(x*sqrt(a**2*x**2
- 1)), Abs(a**2*x**2) > 1), (a**2*sqrt(c)*x/sqrt(-a**2*x**2 + 1) - a*sqrt(c)*asin(a*x) - sqrt(c)/(x*sqrt(-a**
2*x**2 + 1)), True)) + 2*a*c*Piecewise((a**2*sqrt(c)*acosh(1/(a*x))/2 + a*sqrt(c)/(2*x*sqrt(-1 + 1/(a**2*x**2)
)) - sqrt(c)/(2*a*x**3*sqrt(-1 + 1/(a**2*x**2))), 1/Abs(a**2*x**2) > 1), (-I*a**2*sqrt(c)*asin(1/(a*x))/2 - I*
a*sqrt(c)*sqrt(1 - 1/(a**2*x**2))/(2*x), True)) + c*Piecewise((a**3*sqrt(c)*sqrt(-1 + 1/(a**2*x**2))/3 - a*sqr
t(c)*sqrt(-1 + 1/(a**2*x**2))/(3*x**2), 1/Abs(a**2*x**2) > 1), (I*a**3*sqrt(c)*sqrt(1 - 1/(a**2*x**2))/3 - I*a
*sqrt(c)*sqrt(1 - 1/(a**2*x**2))/(3*x**2), True))

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Giac [B]  time = 1.18994, size = 350, normalized size = 3.04 \begin{align*} -\frac{2 \, a^{3} c^{2} \arctan \left (-\frac{\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} - \frac{a^{4} \sqrt{-c} c \log \left ({\left | -\sqrt{-a^{2} c} x + \sqrt{-a^{2} c x^{2} + c} \right |}\right )}{{\left | a \right |}} - \frac{2 \,{\left (3 \,{\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )}^{5} a^{3} c^{2}{\left | a \right |} + 6 \,{\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )}^{2} a^{4} \sqrt{-c} c^{3} - 3 \,{\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )} a^{3} c^{4}{\left | a \right |} - 2 \, a^{4} \sqrt{-c} c^{4}\right )}}{3 \,{\left ({\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )}^{2} - c\right )}^{3}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^3*c^2*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c) - a^4*sqrt(-c)*c*log(abs(-sqrt(-
a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/abs(a) - 2/3*(3*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^5*a^3*c^2*abs(a) + 6
*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2*a^4*sqrt(-c)*c^3 - 3*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))*a^3*c^
4*abs(a) - 2*a^4*sqrt(-c)*c^4)/(((sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2 - c)^3*abs(a))