3.304 \(\int \frac {e^{\tanh ^{-1}(a x)} (c-a c x)^2}{x^6} \, dx\)

Optimal. Leaf size=129 \[ -\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}+\frac {a c^2 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}-\frac {2 a^2 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}-\frac {1}{8} a^5 c^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+\frac {a^3 c^2 \sqrt {1-a^2 x^2}}{8 x^2} \]

[Out]

-1/5*c^2*(-a^2*x^2+1)^(3/2)/x^5+1/4*a*c^2*(-a^2*x^2+1)^(3/2)/x^4-2/15*a^2*c^2*(-a^2*x^2+1)^(3/2)/x^3-1/8*a^5*c
^2*arctanh((-a^2*x^2+1)^(1/2))+1/8*a^3*c^2*(-a^2*x^2+1)^(1/2)/x^2

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Rubi [A]  time = 0.14, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6128, 835, 807, 266, 47, 63, 208} \[ \frac {a^3 c^2 \sqrt {1-a^2 x^2}}{8 x^2}-\frac {2 a^2 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}+\frac {a c^2 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}-\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}-\frac {1}{8} a^5 c^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*(c - a*c*x)^2)/x^6,x]

[Out]

(a^3*c^2*Sqrt[1 - a^2*x^2])/(8*x^2) - (c^2*(1 - a^2*x^2)^(3/2))/(5*x^5) + (a*c^2*(1 - a^2*x^2)^(3/2))/(4*x^4)
- (2*a^2*c^2*(1 - a^2*x^2)^(3/2))/(15*x^3) - (a^5*c^2*ArcTanh[Sqrt[1 - a^2*x^2]])/8

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]

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 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 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 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]

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)} (c-a c x)^2}{x^6} \, dx &=c \int \frac {(c-a c x) \sqrt {1-a^2 x^2}}{x^6} \, dx\\ &=-\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}-\frac {1}{5} c \int \frac {\left (5 a c-2 a^2 c x\right ) \sqrt {1-a^2 x^2}}{x^5} \, dx\\ &=-\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}+\frac {a c^2 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}+\frac {1}{20} c \int \frac {\left (8 a^2 c-5 a^3 c x\right ) \sqrt {1-a^2 x^2}}{x^4} \, dx\\ &=-\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}+\frac {a c^2 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}-\frac {2 a^2 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}-\frac {1}{4} \left (a^3 c^2\right ) \int \frac {\sqrt {1-a^2 x^2}}{x^3} \, dx\\ &=-\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}+\frac {a c^2 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}-\frac {2 a^2 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}-\frac {1}{8} \left (a^3 c^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1-a^2 x}}{x^2} \, dx,x,x^2\right )\\ &=\frac {a^3 c^2 \sqrt {1-a^2 x^2}}{8 x^2}-\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}+\frac {a c^2 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}-\frac {2 a^2 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}+\frac {1}{16} \left (a^5 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=\frac {a^3 c^2 \sqrt {1-a^2 x^2}}{8 x^2}-\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}+\frac {a c^2 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}-\frac {2 a^2 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}-\frac {1}{8} \left (a^3 c^2\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^2 \sqrt {1-a^2 x^2}}{8 x^2}-\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{5 x^5}+\frac {a c^2 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}-\frac {2 a^2 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 x^3}-\frac {1}{8} a^5 c^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 107, normalized size = 0.83 \[ -\frac {c^2 \left (16 a^6 x^6-15 a^5 x^5-8 a^4 x^4+45 a^3 x^3-32 a^2 x^2+15 a^5 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-30 a x+24\right )}{120 x^5 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*(c - a*c*x)^2)/x^6,x]

[Out]

-1/120*(c^2*(24 - 30*a*x - 32*a^2*x^2 + 45*a^3*x^3 - 8*a^4*x^4 - 15*a^5*x^5 + 16*a^6*x^6 + 15*a^5*x^5*Sqrt[1 -
a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]]))/(x^5*Sqrt[1 - a^2*x^2])

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fricas [A]  time = 0.44, size = 95, normalized size = 0.74 \[ \frac {15 \, a^{5} c^{2} x^{5} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + {\left (16 \, a^{4} c^{2} x^{4} - 15 \, a^{3} c^{2} x^{3} + 8 \, a^{2} c^{2} x^{2} + 30 \, a c^{2} x - 24 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1}}{120 \, x^{5}} \]

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)^2/x^6,x, algorithm="fricas")

[Out]

1/120*(15*a^5*c^2*x^5*log((sqrt(-a^2*x^2 + 1) - 1)/x) + (16*a^4*c^2*x^4 - 15*a^3*c^2*x^3 + 8*a^2*c^2*x^2 + 30*
a*c^2*x - 24*c^2)*sqrt(-a^2*x^2 + 1))/x^5

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giac [B]  time = 0.30, size = 297, normalized size = 2.30 \[ \frac {{\left (6 \, a^{6} c^{2} - \frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{4} c^{2}}{x} + \frac {10 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} a^{2} c^{2}}{x^{2}} - \frac {60 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4} c^{2}}{a^{2} x^{4}}\right )} a^{10} x^{5}}{960 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{5} {\left | a \right |}} - \frac {a^{6} c^{2} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{8 \, {\left | a \right |}} + \frac {\frac {60 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{8} c^{2}}{x} - \frac {10 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} a^{4} c^{2}}{x^{3}} + \frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4} a^{2} c^{2}}{x^{4}} - \frac {6 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{5} c^{2}}{x^{5}}}{960 \, a^{4} {\left | a \right |}} \]

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)^2/x^6,x, algorithm="giac")

[Out]

1/960*(6*a^6*c^2 - 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^4*c^2/x + 10*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a^2*c^2
/x^2 - 60*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*c^2/(a^2*x^4))*a^10*x^5/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^5*abs(a))
- 1/8*a^6*c^2*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) + 1/960*(60*(sqrt(-a^2*x^2
+ 1)*abs(a) + a)*a^8*c^2/x - 10*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*a^4*c^2/x^3 + 15*(sqrt(-a^2*x^2 + 1)*abs(a)
+ a)^4*a^2*c^2/x^4 - 6*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^5*c^2/x^5)/(a^4*abs(a))

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maple [A]  time = 0.04, size = 168, normalized size = 1.30 \[ c^{2} \left (-\frac {a^{2} \left (-\frac {\sqrt {-a^{2} x^{2}+1}}{3 x^{3}}-\frac {2 a^{2} \sqrt {-a^{2} x^{2}+1}}{3 x}\right )}{5}+a^{3} \left (-\frac {\sqrt {-a^{2} x^{2}+1}}{2 x^{2}}-\frac {a^{2} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}\right )-a \left (-\frac {\sqrt {-a^{2} x^{2}+1}}{4 x^{4}}+\frac {3 a^{2} \left (-\frac {\sqrt {-a^{2} x^{2}+1}}{2 x^{2}}-\frac {a^{2} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}\right )}{4}\right )-\frac {\sqrt {-a^{2} x^{2}+1}}{5 x^{5}}\right ) \]

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)^2/x^6,x)

[Out]

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

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maxima [A]  time = 0.41, size = 145, normalized size = 1.12 \[ -\frac {1}{8} \, a^{5} c^{2} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {2 \, \sqrt {-a^{2} x^{2} + 1} a^{4} c^{2}}{15 \, x} - \frac {\sqrt {-a^{2} x^{2} + 1} a^{3} c^{2}}{8 \, x^{2}} + \frac {\sqrt {-a^{2} x^{2} + 1} a^{2} c^{2}}{15 \, x^{3}} + \frac {\sqrt {-a^{2} x^{2} + 1} a c^{2}}{4 \, x^{4}} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{2}}{5 \, x^{5}} \]

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)^2/x^6,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.79, size = 136, normalized size = 1.05 \[ \frac {a\,c^2\,\sqrt {1-a^2\,x^2}}{4\,x^4}-\frac {c^2\,\sqrt {1-a^2\,x^2}}{5\,x^5}+\frac {a^2\,c^2\,\sqrt {1-a^2\,x^2}}{15\,x^3}-\frac {a^3\,c^2\,\sqrt {1-a^2\,x^2}}{8\,x^2}+\frac {2\,a^4\,c^2\,\sqrt {1-a^2\,x^2}}{15\,x}+\frac {a^5\,c^2\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^5*c^2*atan((1 - a^2*x^2)^(1/2)*1i)*1i)/8 - (c^2*(1 - a^2*x^2)^(1/2))/(5*x^5) + (a*c^2*(1 - a^2*x^2)^(1/2))/
(4*x^4) + (a^2*c^2*(1 - a^2*x^2)^(1/2))/(15*x^3) - (a^3*c^2*(1 - a^2*x^2)^(1/2))/(8*x^2) + (2*a^4*c^2*(1 - a^2
*x^2)^(1/2))/(15*x)

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sympy [C]  time = 8.61, size = 522, normalized size = 4.05 \[ a^{3} c^{2} \left (\begin {cases} - \frac {a^{2} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{2} - \frac {a \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{2 x} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {i a^{2} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{2} - \frac {i a}{2 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i}{2 a x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right ) - a^{2} c^{2} \left (\begin {cases} - \frac {2 i a^{2} \sqrt {a^{2} x^{2} - 1}}{3 x} - \frac {i \sqrt {a^{2} x^{2} - 1}}{3 x^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {2 a^{2} \sqrt {- a^{2} x^{2} + 1}}{3 x} - \frac {\sqrt {- a^{2} x^{2} + 1}}{3 x^{3}} & \text {otherwise} \end {cases}\right ) - a c^{2} \left (\begin {cases} - \frac {3 a^{4} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{8} + \frac {3 a^{3}}{8 x \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {a}{8 x^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{4 a x^{5} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {3 i a^{4} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{8} - \frac {3 i a^{3}}{8 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i a}{8 x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i}{4 a x^{5} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right ) + c^{2} \left (\begin {cases} - \frac {8 a^{5} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{15} - \frac {4 a^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{15 x^{2}} - \frac {a \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{5 x^{4}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\- \frac {8 i a^{5} \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{15} - \frac {4 i a^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{15 x^{2}} - \frac {i a \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{5 x^{4}} & \text {otherwise} \end {cases}\right ) \]

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)**2/x**6,x)

[Out]

a**3*c**2*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
*asin(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)) - a**2*c**
2*Piecewise((-2*I*a**2*sqrt(a**2*x**2 - 1)/(3*x) - I*sqrt(a**2*x**2 - 1)/(3*x**3), Abs(a**2*x**2) > 1), (-2*a*
*2*sqrt(-a**2*x**2 + 1)/(3*x) - sqrt(-a**2*x**2 + 1)/(3*x**3), True)) - a*c**2*Piecewise((-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*sqrt(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**2*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/A
bs(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*s
qrt(1 - 1/(a**2*x**2))/(5*x**4), True))

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