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

Optimal. Leaf size=103 \[ \frac {c^2 (2-3 a x) \sqrt {1-a^2 x^2}}{2 a^2 x}+\frac {3 c^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{2 a}-\frac {c^2 (3 a x+2) \left (1-a^2 x^2\right )^{3/2}}{6 a^4 x^3}+\frac {c^2 \sin ^{-1}(a x)}{a} \]

[Out]

-1/6*c^2*(3*a*x+2)*(-a^2*x^2+1)^(3/2)/a^4/x^3+c^2*arcsin(a*x)/a+3/2*c^2*arctanh((-a^2*x^2+1)^(1/2))/a+1/2*c^2*
(-3*a*x+2)*(-a^2*x^2+1)^(1/2)/a^2/x

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Rubi [A]  time = 0.16, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {6157, 6148, 811, 813, 844, 216, 266, 63, 208} \[ -\frac {c^2 (3 a x+2) \left (1-a^2 x^2\right )^{3/2}}{6 a^4 x^3}+\frac {c^2 (2-3 a x) \sqrt {1-a^2 x^2}}{2 a^2 x}+\frac {3 c^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{2 a}+\frac {c^2 \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(2 - 3*a*x)*Sqrt[1 - a^2*x^2])/(2*a^2*x) - (c^2*(2 + 3*a*x)*(1 - a^2*x^2)^(3/2))/(6*a^4*x^3) + (c^2*ArcSi
n[a*x])/a + (3*c^2*ArcTanh[Sqrt[1 - a^2*x^2]])/(2*a)

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 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 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 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 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 6148

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

Rule 6157

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

Rubi steps

\begin {align*} \int e^{\tanh ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx &=\frac {c^2 \int \frac {e^{\tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^2}{x^4} \, dx}{a^4}\\ &=\frac {c^2 \int \frac {(1+a x) \left (1-a^2 x^2\right )^{3/2}}{x^4} \, dx}{a^4}\\ &=-\frac {c^2 (2+3 a x) \left (1-a^2 x^2\right )^{3/2}}{6 a^4 x^3}-\frac {c^2 \int \frac {\left (4 a^2+6 a^3 x\right ) \sqrt {1-a^2 x^2}}{x^2} \, dx}{4 a^4}\\ &=\frac {c^2 (2-3 a x) \sqrt {1-a^2 x^2}}{2 a^2 x}-\frac {c^2 (2+3 a x) \left (1-a^2 x^2\right )^{3/2}}{6 a^4 x^3}+\frac {c^2 \int \frac {-12 a^3+8 a^4 x}{x \sqrt {1-a^2 x^2}} \, dx}{8 a^4}\\ &=\frac {c^2 (2-3 a x) \sqrt {1-a^2 x^2}}{2 a^2 x}-\frac {c^2 (2+3 a x) \left (1-a^2 x^2\right )^{3/2}}{6 a^4 x^3}+c^2 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx-\frac {\left (3 c^2\right ) \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx}{2 a}\\ &=\frac {c^2 (2-3 a x) \sqrt {1-a^2 x^2}}{2 a^2 x}-\frac {c^2 (2+3 a x) \left (1-a^2 x^2\right )^{3/2}}{6 a^4 x^3}+\frac {c^2 \sin ^{-1}(a x)}{a}-\frac {\left (3 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )}{4 a}\\ &=\frac {c^2 (2-3 a x) \sqrt {1-a^2 x^2}}{2 a^2 x}-\frac {c^2 (2+3 a x) \left (1-a^2 x^2\right )^{3/2}}{6 a^4 x^3}+\frac {c^2 \sin ^{-1}(a x)}{a}+\frac {\left (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 )}{2 a^3}\\ &=\frac {c^2 (2-3 a x) \sqrt {1-a^2 x^2}}{2 a^2 x}-\frac {c^2 (2+3 a x) \left (1-a^2 x^2\right )^{3/2}}{6 a^4 x^3}+\frac {c^2 \sin ^{-1}(a x)}{a}+\frac {3 c^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{2 a}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 70, normalized size = 0.68 \[ \frac {c^2 \left (-\frac {5 \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};a^2 x^2\right )}{x^3}-3 a^3 \left (1-a^2 x^2\right )^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};1-a^2 x^2\right )\right )}{15 a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^2*((-5*Hypergeometric2F1[-3/2, -3/2, -1/2, a^2*x^2])/x^3 - 3*a^3*(1 - a^2*x^2)^(5/2)*Hypergeometric2F1[2, 5
/2, 7/2, 1 - a^2*x^2]))/(15*a^4)

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fricas [A]  time = 0.40, size = 131, normalized size = 1.27 \[ -\frac {12 \, a^{3} c^{2} x^{3} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + 9 \, a^{3} c^{2} x^{3} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + 6 \, a^{3} c^{2} x^{3} + {\left (6 \, a^{3} c^{2} x^{3} - 8 \, a^{2} c^{2} x^{2} + 3 \, a c^{2} x + 2 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, a^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*a^3*c^2*x^3*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + 9*a^3*c^2*x^3*log((sqrt(-a^2*x^2 + 1) - 1)/x) +
6*a^3*c^2*x^3 + (6*a^3*c^2*x^3 - 8*a^2*c^2*x^2 + 3*a*c^2*x + 2*c^2)*sqrt(-a^2*x^2 + 1))/(a^4*x^3)

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giac [B]  time = 0.57, size = 263, normalized size = 2.55 \[ \frac {{\left (c^{2} + \frac {3 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c^{2}}{a^{2} x} - \frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c^{2}}{a^{4} x^{2}}\right )} a^{6} x^{3}}{24 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} {\left | a \right |}} + \frac {c^{2} \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{{\left | a \right |}} + \frac {3 \, c^{2} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{2 \, {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{2}}{a} + \frac {\frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c^{2}}{x} - \frac {3 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c^{2}}{a^{2} x^{2}} - \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c^{2}}{a^{4} x^{3}}}{24 \, a^{2} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(c^2 + 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^2/(a^2*x) - 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c^2/(a^4*x^2)
)*a^6*x^3/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*abs(a)) + c^2*arcsin(a*x)*sgn(a)/abs(a) + 3/2*c^2*log(1/2*abs(-2*
sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - sqrt(-a^2*x^2 + 1)*c^2/a + 1/24*(15*(sqrt(-a^2*x^2 + 1
)*abs(a) + a)*c^2/x - 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c^2/(a^2*x^2) - (sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*c^
2/(a^4*x^3))/(a^2*abs(a))

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maple [A]  time = 0.04, size = 141, normalized size = 1.37 \[ -\frac {c^{2} \sqrt {-a^{2} x^{2}+1}}{a}+\frac {c^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}+\frac {3 c^{2} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a}+\frac {4 c^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{2} x}-\frac {c^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{4} x^{3}}-\frac {c^{2} \sqrt {-a^{2} x^{2}+1}}{2 a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.41, size = 189, normalized size = 1.83 \[ \frac {c^{2} \arcsin \left (a x\right )}{a} + \frac {2 \, c^{2} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right )}{a} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{2}}{a} - \frac {{\left (a^{2} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\sqrt {-a^{2} x^{2} + 1}}{x^{2}}\right )} c^{2}}{2 \, a^{3}} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1} c^{2}}{a^{2} x} - \frac {{\left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a^{2}}{x} + \frac {\sqrt {-a^{2} x^{2} + 1}}{x^{3}}\right )} c^{2}}{3 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 136, normalized size = 1.32 \[ \frac {c^2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}-\frac {c^2\,\sqrt {1-a^2\,x^2}}{a}+\frac {4\,c^2\,\sqrt {1-a^2\,x^2}}{3\,a^2\,x}-\frac {c^2\,\sqrt {1-a^2\,x^2}}{2\,a^3\,x^2}-\frac {c^2\,\sqrt {1-a^2\,x^2}}{3\,a^4\,x^3}-\frac {c^2\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 9.86, size = 354, normalized size = 3.44 \[ a c^{2} \left (\begin {cases} \frac {x^{2}}{2} & \text {for}\: a^{2} = 0 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {otherwise} \end {cases}\right ) + c^{2} \left (\begin {cases} \sqrt {\frac {1}{a^{2}}} \operatorname {asin}{\left (x \sqrt {a^{2}} \right )} & \text {for}\: a^{2} > 0 \\\sqrt {- \frac {1}{a^{2}}} \operatorname {asinh}{\left (x \sqrt {- a^{2}} \right )} & \text {for}\: a^{2} < 0 \end {cases}\right ) - \frac {2 c^{2} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right )}{a} - \frac {2 c^{2} \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right )}{a^{2}} + \frac {c^{2} \left (\begin {cases} - \frac {a^{2} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{2} - \frac {a \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{2 x} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {i a^{2} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{2} - \frac {i a}{2 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i}{2 a x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right )}{a^{3}} + \frac {c^{2} \left (\begin {cases} - \frac {2 i a^{2} \sqrt {a^{2} x^{2} - 1}}{3 x} - \frac {i \sqrt {a^{2} x^{2} - 1}}{3 x^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {2 a^{2} \sqrt {- a^{2} x^{2} + 1}}{3 x} - \frac {\sqrt {- a^{2} x^{2} + 1}}{3 x^{3}} & \text {otherwise} \end {cases}\right )}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c**2*Piecewise((x**2/2, Eq(a**2, 0)), (-sqrt(-a**2*x**2 + 1)/a**2, True)) + c**2*Piecewise((sqrt(a**(-2))*as
in(x*sqrt(a**2)), a**2 > 0), (sqrt(-1/a**2)*asinh(x*sqrt(-a**2)), a**2 < 0)) - 2*c**2*Piecewise((-acosh(1/(a*x
)), 1/Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), True))/a - 2*c**2*Piecewise((-I*sqrt(a**2*x**2 - 1)/x, Abs(a**2*x
**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True))/a**2 + 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**3 + 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), T
rue))/a**4

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