3.315 \(\int \frac{e^{\tanh ^{-1}(a x)} (c-a c x)^3}{x^6} \, dx\)

Optimal. Leaf size=129 \[ \frac{a^3 c^3 \sqrt{1-a^2 x^2}}{4 x^2}-\frac{7 a^2 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}+\frac{a c^3 \left (1-a^2 x^2\right )^{3/2}}{2 x^4}-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}-\frac{1}{4} a^5 c^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

[Out]

(a^3*c^3*Sqrt[1 - a^2*x^2])/(4*x^2) - (c^3*(1 - a^2*x^2)^(3/2))/(5*x^5) + (a*c^3*(1 - a^2*x^2)^(3/2))/(2*x^4)
- (7*a^2*c^3*(1 - a^2*x^2)^(3/2))/(15*x^3) - (a^5*c^3*ArcTanh[Sqrt[1 - a^2*x^2]])/4

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Rubi [A]  time = 0.190251, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {6128, 1807, 835, 807, 266, 47, 63, 208} \[ \frac{a^3 c^3 \sqrt{1-a^2 x^2}}{4 x^2}-\frac{7 a^2 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}+\frac{a c^3 \left (1-a^2 x^2\right )^{3/2}}{2 x^4}-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}-\frac{1}{4} a^5 c^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*(c - a*c*x)^3)/x^6,x]

[Out]

(a^3*c^3*Sqrt[1 - a^2*x^2])/(4*x^2) - (c^3*(1 - a^2*x^2)^(3/2))/(5*x^5) + (a*c^3*(1 - a^2*x^2)^(3/2))/(2*x^4)
- (7*a^2*c^3*(1 - a^2*x^2)^(3/2))/(15*x^3) - (a^5*c^3*ArcTanh[Sqrt[1 - a^2*x^2]])/4

Rule 6128

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

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 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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^{\tanh ^{-1}(a x)} (c-a c x)^3}{x^6} \, dx &=c \int \frac{(c-a c x)^2 \sqrt{1-a^2 x^2}}{x^6} \, dx\\ &=-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}-\frac{1}{5} c \int \frac{\left (10 a c^2-7 a^2 c^2 x\right ) \sqrt{1-a^2 x^2}}{x^5} \, dx\\ &=-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}+\frac{a c^3 \left (1-a^2 x^2\right )^{3/2}}{2 x^4}+\frac{1}{20} c \int \frac{\left (28 a^2 c^2-10 a^3 c^2 x\right ) \sqrt{1-a^2 x^2}}{x^4} \, dx\\ &=-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}+\frac{a c^3 \left (1-a^2 x^2\right )^{3/2}}{2 x^4}-\frac{7 a^2 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}-\frac{1}{2} \left (a^3 c^3\right ) \int \frac{\sqrt{1-a^2 x^2}}{x^3} \, dx\\ &=-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}+\frac{a c^3 \left (1-a^2 x^2\right )^{3/2}}{2 x^4}-\frac{7 a^2 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}-\frac{1}{4} \left (a^3 c^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-a^2 x}}{x^2} \, dx,x,x^2\right )\\ &=\frac{a^3 c^3 \sqrt{1-a^2 x^2}}{4 x^2}-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}+\frac{a c^3 \left (1-a^2 x^2\right )^{3/2}}{2 x^4}-\frac{7 a^2 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}+\frac{1}{8} \left (a^5 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )\\ &=\frac{a^3 c^3 \sqrt{1-a^2 x^2}}{4 x^2}-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}+\frac{a c^3 \left (1-a^2 x^2\right )^{3/2}}{2 x^4}-\frac{7 a^2 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}-\frac{1}{4} \left (a^3 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^2}\right )\\ &=\frac{a^3 c^3 \sqrt{1-a^2 x^2}}{4 x^2}-\frac{c^3 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}+\frac{a c^3 \left (1-a^2 x^2\right )^{3/2}}{2 x^4}-\frac{7 a^2 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}-\frac{1}{4} a^5 c^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0332235, size = 107, normalized size = 0.83 \[ -\frac{c^3 \left (28 a^6 x^6-15 a^5 x^5-44 a^4 x^4+45 a^3 x^3+4 a^2 x^2+15 a^5 x^5 \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-30 a x+12\right )}{60 x^5 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*(c - a*c*x)^3)/x^6,x]

[Out]

-(c^3*(12 - 30*a*x + 4*a^2*x^2 + 45*a^3*x^3 - 44*a^4*x^4 - 15*a^5*x^5 + 28*a^6*x^6 + 15*a^5*x^5*Sqrt[1 - a^2*x
^2]*ArcTanh[Sqrt[1 - a^2*x^2]]))/(60*x^5*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.043, size = 190, normalized size = 1.5 \begin{align*} -{c}^{3} \left ( 2\,a \left ( -1/4\,{\frac{\sqrt{-{a}^{2}{x}^{2}+1}}{{x}^{4}}}+3/4\,{a}^{2} \left ( -1/2\,{\frac{\sqrt{-{a}^{2}{x}^{2}+1}}{{x}^{2}}}-1/2\,{a}^{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) \right ) \right ) -{\frac{{a}^{4}}{x}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{1}{5\,{x}^{5}}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{4\,{a}^{2}}{5} \left ( -{\frac{1}{3\,{x}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{2\,{a}^{2}}{3\,x}\sqrt{-{a}^{2}{x}^{2}+1}} \right ) }-2\,{a}^{3} \left ( -1/2\,{\frac{\sqrt{-{a}^{2}{x}^{2}+1}}{{x}^{2}}}-1/2\,{a}^{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3/x^6,x)

[Out]

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

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Maxima [A]  time = 1.42988, size = 196, normalized size = 1.52 \begin{align*} -\frac{1}{4} \, a^{5} c^{3} \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) + \frac{7 \, \sqrt{-a^{2} x^{2} + 1} a^{4} c^{3}}{15 \, x} - \frac{\sqrt{-a^{2} x^{2} + 1} a^{3} c^{3}}{4 \, x^{2}} - \frac{4 \, \sqrt{-a^{2} x^{2} + 1} a^{2} c^{3}}{15 \, x^{3}} + \frac{\sqrt{-a^{2} x^{2} + 1} a c^{3}}{2 \, x^{4}} - \frac{\sqrt{-a^{2} x^{2} + 1} c^{3}}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3/x^6,x, algorithm="maxima")

[Out]

-1/4*a^5*c^3*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) + 7/15*sqrt(-a^2*x^2 + 1)*a^4*c^3/x - 1/4*sqrt(-a^2*x
^2 + 1)*a^3*c^3/x^2 - 4/15*sqrt(-a^2*x^2 + 1)*a^2*c^3/x^3 + 1/2*sqrt(-a^2*x^2 + 1)*a*c^3/x^4 - 1/5*sqrt(-a^2*x
^2 + 1)*c^3/x^5

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Fricas [A]  time = 1.6267, size = 207, normalized size = 1.6 \begin{align*} \frac{15 \, a^{5} c^{3} x^{5} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) +{\left (28 \, a^{4} c^{3} x^{4} - 15 \, a^{3} c^{3} x^{3} - 16 \, a^{2} c^{3} x^{2} + 30 \, a c^{3} x - 12 \, c^{3}\right )} \sqrt{-a^{2} x^{2} + 1}}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3/x^6,x, algorithm="fricas")

[Out]

1/60*(15*a^5*c^3*x^5*log((sqrt(-a^2*x^2 + 1) - 1)/x) + (28*a^4*c^3*x^4 - 15*a^3*c^3*x^3 - 16*a^2*c^3*x^2 + 30*
a*c^3*x - 12*c^3)*sqrt(-a^2*x^2 + 1))/x^5

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Sympy [C]  time = 10.1219, size = 476, normalized size = 3.69 \begin{align*} - a^{4} c^{3} \left (\begin{cases} - \frac{i \sqrt{a^{2} x^{2} - 1}}{x} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac{\sqrt{- a^{2} x^{2} + 1}}{x} & \text{otherwise} \end{cases}\right ) + 2 a^{3} c^{3} \left (\begin{cases} - \frac{a^{2} \operatorname{acosh}{\left (\frac{1}{a x} \right )}}{2} - \frac{a \sqrt{-1 + \frac{1}{a^{2} x^{2}}}}{2 x} & \text{for}\: \frac{1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac{i a^{2} \operatorname{asin}{\left (\frac{1}{a x} \right )}}{2} - \frac{i a}{2 x \sqrt{1 - \frac{1}{a^{2} x^{2}}}} + \frac{i}{2 a x^{3} \sqrt{1 - \frac{1}{a^{2} x^{2}}}} & \text{otherwise} \end{cases}\right ) - 2 a c^{3} \left (\begin{cases} - \frac{3 a^{4} \operatorname{acosh}{\left (\frac{1}{a x} \right )}}{8} + \frac{3 a^{3}}{8 x \sqrt{-1 + \frac{1}{a^{2} x^{2}}}} - \frac{a}{8 x^{3} \sqrt{-1 + \frac{1}{a^{2} x^{2}}}} - \frac{1}{4 a x^{5} \sqrt{-1 + \frac{1}{a^{2} x^{2}}}} & \text{for}\: \frac{1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac{3 i a^{4} \operatorname{asin}{\left (\frac{1}{a x} \right )}}{8} - \frac{3 i a^{3}}{8 x \sqrt{1 - \frac{1}{a^{2} x^{2}}}} + \frac{i a}{8 x^{3} \sqrt{1 - \frac{1}{a^{2} x^{2}}}} + \frac{i}{4 a x^{5} \sqrt{1 - \frac{1}{a^{2} x^{2}}}} & \text{otherwise} \end{cases}\right ) + c^{3} \left (\begin{cases} - \frac{8 a^{5} \sqrt{-1 + \frac{1}{a^{2} x^{2}}}}{15} - \frac{4 a^{3} \sqrt{-1 + \frac{1}{a^{2} x^{2}}}}{15 x^{2}} - \frac{a \sqrt{-1 + \frac{1}{a^{2} x^{2}}}}{5 x^{4}} & \text{for}\: \frac{1}{\left |{a^{2} x^{2}}\right |} > 1 \\- \frac{8 i a^{5} \sqrt{1 - \frac{1}{a^{2} x^{2}}}}{15} - \frac{4 i a^{3} \sqrt{1 - \frac{1}{a^{2} x^{2}}}}{15 x^{2}} - \frac{i a \sqrt{1 - \frac{1}{a^{2} x^{2}}}}{5 x^{4}} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a*c*x+c)**3/x**6,x)

[Out]

-a**4*c**3*Piecewise((-I*sqrt(a**2*x**2 - 1)/x, Abs(a**2*x**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True)) + 2*a**3
*c**3*Piecewise((-a**2*acosh(1/(a*x))/2 - a*sqrt(-1 + 1/(a**2*x**2))/(2*x), 1/Abs(a**2*x**2) > 1), (I*a**2*asi
n(1/(a*x))/2 - I*a/(2*x*sqrt(1 - 1/(a**2*x**2))) + I/(2*a*x**3*sqrt(1 - 1/(a**2*x**2))), True)) - 2*a*c**3*Pie
cewise((-3*a**4*acosh(1/(a*x))/8 + 3*a**3/(8*x*sqrt(-1 + 1/(a**2*x**2))) - a/(8*x**3*sqrt(-1 + 1/(a**2*x**2)))
- 1/(4*a*x**5*sqrt(-1 + 1/(a**2*x**2))), 1/Abs(a**2*x**2) > 1), (3*I*a**4*asin(1/(a*x))/8 - 3*I*a**3/(8*x*sqr
t(1 - 1/(a**2*x**2))) + I*a/(8*x**3*sqrt(1 - 1/(a**2*x**2))) + I/(4*a*x**5*sqrt(1 - 1/(a**2*x**2))), True)) +
c**3*Piecewise((-8*a**5*sqrt(-1 + 1/(a**2*x**2))/15 - 4*a**3*sqrt(-1 + 1/(a**2*x**2))/(15*x**2) - a*sqrt(-1 +
1/(a**2*x**2))/(5*x**4), 1/Abs(a**2*x**2) > 1), (-8*I*a**5*sqrt(1 - 1/(a**2*x**2))/15 - 4*I*a**3*sqrt(1 - 1/(a
**2*x**2))/(15*x**2) - I*a*sqrt(1 - 1/(a**2*x**2))/(5*x**4), True))

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Giac [B]  time = 1.1874, size = 401, normalized size = 3.11 \begin{align*} \frac{{\left (3 \, a^{6} c^{3} - \frac{15 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{4} c^{3}}{x} + \frac{25 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} a^{2} c^{3}}{x^{2}} - \frac{90 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4} c^{3}}{a^{2} x^{4}}\right )} a^{10} x^{5}}{480 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{5}{\left | a \right |}} - \frac{a^{6} c^{3} \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{4 \,{\left | a \right |}} + \frac{\frac{90 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{8} c^{3}}{x} - \frac{25 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3} a^{4} c^{3}}{x^{3}} + \frac{15 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4} a^{2} c^{3}}{x^{4}} - \frac{3 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{5} c^{3}}{x^{5}}}{480 \, a^{4}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3/x^6,x, algorithm="giac")

[Out]

1/480*(3*a^6*c^3 - 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^4*c^3/x + 25*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a^2*c^3
/x^2 - 90*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*c^3/(a^2*x^4))*a^10*x^5/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^5*abs(a))
- 1/4*a^6*c^3*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) + 1/480*(90*(sqrt(-a^2*x^2
+ 1)*abs(a) + a)*a^8*c^3/x - 25*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*a^4*c^3/x^3 + 15*(sqrt(-a^2*x^2 + 1)*abs(a)
+ a)^4*a^2*c^3/x^4 - 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^5*c^3/x^5)/(a^4*abs(a))