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

Optimal. Leaf size=116 \[ \frac {3 a c^4 \left (1-a^2 x^2\right )^{3/2}}{x}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{2 x^2}+\frac {5}{2} a^2 c^4 (a x+1) \sqrt {1-a^2 x^2}-\frac {5}{2} a^2 c^4 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+\frac {5}{2} a^2 c^4 \sin ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {6128, 1807, 815, 844, 216, 266, 63, 208} \[ \frac {3 a c^4 \left (1-a^2 x^2\right )^{3/2}}{x}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{2 x^2}+\frac {5}{2} a^2 c^4 (a x+1) \sqrt {1-a^2 x^2}-\frac {5}{2} a^2 c^4 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+\frac {5}{2} a^2 c^4 \sin ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 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 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 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)^4}{x^3} \, dx &=c \int \frac {(c-a c x)^3 \sqrt {1-a^2 x^2}}{x^3} \, dx\\ &=-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{2 x^2}-\frac {1}{2} c \int \frac {\sqrt {1-a^2 x^2} \left (6 a c^3-5 a^2 c^3 x+2 a^3 c^3 x^2\right )}{x^2} \, dx\\ &=-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{2 x^2}+\frac {3 a c^4 \left (1-a^2 x^2\right )^{3/2}}{x}+\frac {1}{2} c \int \frac {\left (5 a^2 c^3+10 a^3 c^3 x\right ) \sqrt {1-a^2 x^2}}{x} \, dx\\ &=\frac {5}{2} a^2 c^4 (1+a x) \sqrt {1-a^2 x^2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{2 x^2}+\frac {3 a c^4 \left (1-a^2 x^2\right )^{3/2}}{x}-\frac {c \int \frac {-10 a^4 c^3-10 a^5 c^3 x}{x \sqrt {1-a^2 x^2}} \, dx}{4 a^2}\\ &=\frac {5}{2} a^2 c^4 (1+a x) \sqrt {1-a^2 x^2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{2 x^2}+\frac {3 a c^4 \left (1-a^2 x^2\right )^{3/2}}{x}+\frac {1}{2} \left (5 a^2 c^4\right ) \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx+\frac {1}{2} \left (5 a^3 c^4\right ) \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {5}{2} a^2 c^4 (1+a x) \sqrt {1-a^2 x^2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{2 x^2}+\frac {3 a c^4 \left (1-a^2 x^2\right )^{3/2}}{x}+\frac {5}{2} a^2 c^4 \sin ^{-1}(a x)+\frac {1}{4} \left (5 a^2 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=\frac {5}{2} a^2 c^4 (1+a x) \sqrt {1-a^2 x^2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{2 x^2}+\frac {3 a c^4 \left (1-a^2 x^2\right )^{3/2}}{x}+\frac {5}{2} a^2 c^4 \sin ^{-1}(a x)-\frac {1}{2} \left (5 c^4\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 {5}{2} a^2 c^4 (1+a x) \sqrt {1-a^2 x^2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{2 x^2}+\frac {3 a c^4 \left (1-a^2 x^2\right )^{3/2}}{x}+\frac {5}{2} a^2 c^4 \sin ^{-1}(a x)-\frac {5}{2} a^2 c^4 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 106, normalized size = 0.91 \[ \frac {1}{4} c^4 \left (-10 a^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+5 a^2 \sin ^{-1}(a x)-10 a^2 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )+\frac {2 (a x+1)^2 \left (a^3 x^3-8 a^2 x^2+8 a x-1\right )}{x^2 \sqrt {1-a^2 x^2}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^4*((2*(1 + a*x)^2*(-1 + 8*a*x - 8*a^2*x^2 + a^3*x^3))/(x^2*Sqrt[1 - a^2*x^2]) + 5*a^2*ArcSin[a*x] - 10*a^2*
ArcSin[Sqrt[1 - a*x]/Sqrt[2]] - 10*a^2*ArcTanh[Sqrt[1 - a^2*x^2]]))/4

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fricas [A]  time = 0.49, size = 125, normalized size = 1.08 \[ -\frac {10 \, a^{2} c^{4} x^{2} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - 5 \, a^{2} c^{4} x^{2} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - 6 \, a^{2} c^{4} x^{2} + {\left (a^{3} c^{4} x^{3} - 6 \, a^{2} c^{4} x^{2} - 6 \, a c^{4} x + c^{4}\right )} \sqrt {-a^{2} x^{2} + 1}}{2 \, x^{2}} \]

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)^4/x^3,x, algorithm="fricas")

[Out]

-1/2*(10*a^2*c^4*x^2*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - 5*a^2*c^4*x^2*log((sqrt(-a^2*x^2 + 1) - 1)/x) -
6*a^2*c^4*x^2 + (a^3*c^4*x^3 - 6*a^2*c^4*x^2 - 6*a*c^4*x + c^4)*sqrt(-a^2*x^2 + 1))/x^2

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giac [B]  time = 0.23, size = 224, normalized size = 1.93 \[ \frac {5 \, a^{3} c^{4} \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{2 \, {\left | a \right |}} - \frac {5 \, a^{3} c^{4} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{2 \, {\left | a \right |}} + \frac {{\left (a^{3} c^{4} - \frac {12 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a c^{4}}{x}\right )} a^{4} x^{2}}{8 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} {\left | a \right |}} - \frac {1}{2} \, {\left (a^{3} c^{4} x - 6 \, a^{2} c^{4}\right )} \sqrt {-a^{2} x^{2} + 1} + \frac {\frac {12 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a c^{4} {\left | a \right |}}{x} - \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c^{4} {\left | a \right |}}{a x^{2}}}{8 \, a^{2}} \]

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)^4/x^3,x, algorithm="giac")

[Out]

5/2*a^3*c^4*arcsin(a*x)*sgn(a)/abs(a) - 5/2*a^3*c^4*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x
)))/abs(a) + 1/8*(a^3*c^4 - 12*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a*c^4/x)*a^4*x^2/((sqrt(-a^2*x^2 + 1)*abs(a) +
a)^2*abs(a)) - 1/2*(a^3*c^4*x - 6*a^2*c^4)*sqrt(-a^2*x^2 + 1) + 1/8*(12*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a*c^4*
abs(a)/x - (sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c^4*abs(a)/(a*x^2))/a^2

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maple [A]  time = 0.04, size = 138, normalized size = 1.19 \[ -\frac {c^{4} a^{3} x \sqrt {-a^{2} x^{2}+1}}{2}+\frac {5 c^{4} a^{3} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}+3 c^{4} a^{2} \sqrt {-a^{2} x^{2}+1}-\frac {5 c^{4} a^{2} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}+\frac {3 c^{4} a \sqrt {-a^{2} x^{2}+1}}{x}-\frac {c^{4} \sqrt {-a^{2} x^{2}+1}}{2 x^{2}} \]

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)^4/x^3,x)

[Out]

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

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maxima [A]  time = 0.52, size = 129, normalized size = 1.11 \[ -\frac {1}{2} \, \sqrt {-a^{2} x^{2} + 1} a^{3} c^{4} x + \frac {5}{2} \, a^{2} c^{4} \arcsin \left (a x\right ) - \frac {5}{2} \, a^{2} c^{4} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + 3 \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{4} + \frac {3 \, \sqrt {-a^{2} x^{2} + 1} a c^{4}}{x} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{4}}{2 \, x^{2}} \]

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)^4/x^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.78, size = 133, normalized size = 1.15 \[ 3\,a^2\,c^4\,\sqrt {1-a^2\,x^2}-\frac {c^4\,\sqrt {1-a^2\,x^2}}{2\,x^2}+\frac {3\,a\,c^4\,\sqrt {1-a^2\,x^2}}{x}-\frac {a^3\,c^4\,x\,\sqrt {1-a^2\,x^2}}{2}+\frac {5\,a^3\,c^4\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,\sqrt {-a^2}}+\frac {a^2\,c^4\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 7.72, size = 357, normalized size = 3.08 \[ a^{5} c^{4} \left (\begin {cases} - \frac {i x \sqrt {a^{2} x^{2} - 1}}{2 a^{2}} - \frac {i \operatorname {acosh}{\left (a x \right )}}{2 a^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {x^{3}}{2 \sqrt {- a^{2} x^{2} + 1}} - \frac {x}{2 a^{2} \sqrt {- a^{2} x^{2} + 1}} + \frac {\operatorname {asin}{\left (a x \right )}}{2 a^{3}} & \text {otherwise} \end {cases}\right ) - 3 a^{4} c^{4} \left (\begin {cases} \frac {x^{2}}{2} & \text {for}\: a^{2} = 0 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {otherwise} \end {cases}\right ) + 2 a^{3} c^{4} \left (\begin {cases} \sqrt {\frac {1}{a^{2}}} \operatorname {asin}{\left (x \sqrt {a^{2}} \right )} & \text {for}\: a^{2} > 0 \\\sqrt {- \frac {1}{a^{2}}} \operatorname {asinh}{\left (x \sqrt {- a^{2}} \right )} & \text {for}\: a^{2} < 0 \end {cases}\right ) + 2 a^{2} c^{4} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right ) - 3 a c^{4} \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right ) + c^{4} \left (\begin {cases} - \frac {a^{2} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{2} - \frac {a \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{2 x} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {i a^{2} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{2} - \frac {i a}{2 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i}{2 a x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right ) \]

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)**4/x**3,x)

[Out]

a**5*c**4*Piecewise((-I*x*sqrt(a**2*x**2 - 1)/(2*a**2) - I*acosh(a*x)/(2*a**3), Abs(a**2*x**2) > 1), (x**3/(2*
sqrt(-a**2*x**2 + 1)) - x/(2*a**2*sqrt(-a**2*x**2 + 1)) + asin(a*x)/(2*a**3), True)) - 3*a**4*c**4*Piecewise((
x**2/2, Eq(a**2, 0)), (-sqrt(-a**2*x**2 + 1)/a**2, True)) + 2*a**3*c**4*Piecewise((sqrt(a**(-2))*asin(x*sqrt(a
**2)), a**2 > 0), (sqrt(-1/a**2)*asinh(x*sqrt(-a**2)), a**2 < 0)) + 2*a**2*c**4*Piecewise((-acosh(1/(a*x)), 1/
Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), True)) - 3*a*c**4*Piecewise((-I*sqrt(a**2*x**2 - 1)/x, Abs(a**2*x**2) >
1), (-sqrt(-a**2*x**2 + 1)/x, True)) + c**4*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))

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