3.43 \(\int \frac {(d+e x)^2}{x^6 \sqrt {d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=169 \[ -\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{4 d^4}-\frac {6 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2} \]

[Out]

-3/4*e^5*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^4-1/5*(-e^2*x^2+d^2)^(1/2)/x^5-1/2*e*(-e^2*x^2+d^2)^(1/2)/d/x^4-3/5
*e^2*(-e^2*x^2+d^2)^(1/2)/d^2/x^3-3/4*e^3*(-e^2*x^2+d^2)^(1/2)/d^3/x^2-6/5*e^4*(-e^2*x^2+d^2)^(1/2)/d^4/x

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Rubi [A]  time = 0.20, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1807, 835, 807, 266, 63, 208} \[ -\frac {6 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{4 d^4} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(x^6*Sqrt[d^2 - e^2*x^2]),x]

[Out]

-Sqrt[d^2 - e^2*x^2]/(5*x^5) - (e*Sqrt[d^2 - e^2*x^2])/(2*d*x^4) - (3*e^2*Sqrt[d^2 - e^2*x^2])/(5*d^2*x^3) - (
3*e^3*Sqrt[d^2 - e^2*x^2])/(4*d^3*x^2) - (6*e^4*Sqrt[d^2 - e^2*x^2])/(5*d^4*x) - (3*e^5*ArcTanh[Sqrt[d^2 - e^2
*x^2]/d])/(4*d^4)

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 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])

Rubi steps

\begin {align*} \int \frac {(d+e x)^2}{x^6 \sqrt {d^2-e^2 x^2}} \, dx &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {\int \frac {-10 d^3 e-9 d^2 e^2 x}{x^5 \sqrt {d^2-e^2 x^2}} \, dx}{5 d^2}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}+\frac {\int \frac {36 d^4 e^2+30 d^3 e^3 x}{x^4 \sqrt {d^2-e^2 x^2}} \, dx}{20 d^4}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {\int \frac {-90 d^5 e^3-72 d^4 e^4 x}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{60 d^6}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2}+\frac {\int \frac {144 d^6 e^4+90 d^5 e^5 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{120 d^8}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2}-\frac {6 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}+\frac {\left (3 e^5\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{4 d^3}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2}-\frac {6 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}+\frac {\left (3 e^5\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{8 d^3}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2}-\frac {6 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}-\frac {\left (3 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{4 d^3}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2}-\frac {6 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}-\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{4 d^4}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 79, normalized size = 0.47 \[ -\frac {\sqrt {d^2-e^2 x^2} \left (d^5+3 d^3 e^2 x^2+10 e^5 x^5 \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};1-\frac {e^2 x^2}{d^2}\right )+6 d e^4 x^4\right )}{5 d^5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2/(x^6*Sqrt[d^2 - e^2*x^2]),x]

[Out]

-1/5*(Sqrt[d^2 - e^2*x^2]*(d^5 + 3*d^3*e^2*x^2 + 6*d*e^4*x^4 + 10*e^5*x^5*Hypergeometric2F1[1/2, 3, 3/2, 1 - (
e^2*x^2)/d^2]))/(d^5*x^5)

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fricas [A]  time = 0.77, size = 98, normalized size = 0.58 \[ \frac {15 \, e^{5} x^{5} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (24 \, e^{4} x^{4} + 15 \, d e^{3} x^{3} + 12 \, d^{2} e^{2} x^{2} + 10 \, d^{3} e x + 4 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{20 \, d^{4} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^6/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

1/20*(15*e^5*x^5*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (24*e^4*x^4 + 15*d*e^3*x^3 + 12*d^2*e^2*x^2 + 10*d^3*e*x
+ 4*d^4)*sqrt(-e^2*x^2 + d^2))/(d^4*x^5)

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giac [B]  time = 0.27, size = 365, normalized size = 2.16 \[ \frac {x^{5} {\left (\frac {5 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{10}}{x} + \frac {15 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{8}}{x^{2}} + \frac {40 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{6}}{x^{3}} + \frac {110 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{4}}{x^{4}} + e^{12}\right )} e^{3}}{160 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d^{4}} - \frac {3 \, e^{5} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right )}{4 \, d^{4}} - \frac {{\left (\frac {110 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{16} e^{38}}{x} + \frac {40 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{16} e^{36}}{x^{2}} + \frac {15 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{16} e^{34}}{x^{3}} + \frac {5 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{16} e^{32}}{x^{4}} + \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d^{16} e^{30}}{x^{5}}\right )} e^{\left (-35\right )}}{160 \, d^{20}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^6/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

1/160*x^5*(5*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^10/x + 15*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^8/x^2 + 40*(d*e + s
qrt(-x^2*e^2 + d^2)*e)^3*e^6/x^3 + 110*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^4/x^4 + e^12)*e^3/((d*e + sqrt(-x^2*
e^2 + d^2)*e)^5*d^4) - 3/4*e^5*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^4 - 1/160*(110*
(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^16*e^38/x + 40*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^16*e^36/x^2 + 15*(d*e + sqr
t(-x^2*e^2 + d^2)*e)^3*d^16*e^34/x^3 + 5*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d^16*e^32/x^4 + (d*e + sqrt(-x^2*e^2
+ d^2)*e)^5*d^16*e^30/x^5)*e^(-35)/d^20

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maple [A]  time = 0.02, size = 164, normalized size = 0.97 \[ -\frac {3 e^{5} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{4 \sqrt {d^{2}}\, d^{3}}-\frac {6 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}{5 d^{4} x}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{3}}{4 d^{3} x^{2}}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{2}}{5 d^{2} x^{3}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e}{2 d \,x^{4}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{5 x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/x^6/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-3/5*e^2*(-e^2*x^2+d^2)^(1/2)/d^2/x^3-6/5*e^4*(-e^2*x^2+d^2)^(1/2)/d^4/x-1/2*e*(-e^2*x^2+d^2)^(1/2)/d/x^4-3/4*
e^3*(-e^2*x^2+d^2)^(1/2)/d^3/x^2-3/4/d^3*e^5/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-1/5*
(-e^2*x^2+d^2)^(1/2)/x^5

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maxima [A]  time = 0.98, size = 158, normalized size = 0.93 \[ -\frac {3 \, e^{5} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{4 \, d^{4}} - \frac {6 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}}{5 \, d^{4} x} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{3}}{4 \, d^{3} x^{2}} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{2}}{5 \, d^{2} x^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} e}{2 \, d x^{4}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^6/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

-3/4*e^5*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^4 - 6/5*sqrt(-e^2*x^2 + d^2)*e^4/(d^4*x) - 3/4*
sqrt(-e^2*x^2 + d^2)*e^3/(d^3*x^2) - 3/5*sqrt(-e^2*x^2 + d^2)*e^2/(d^2*x^3) - 1/2*sqrt(-e^2*x^2 + d^2)*e/(d*x^
4) - 1/5*sqrt(-e^2*x^2 + d^2)/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^2}{x^6\,\sqrt {d^2-e^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(x^6*(d^2 - e^2*x^2)^(1/2)),x)

[Out]

int((d + e*x)^2/(x^6*(d^2 - e^2*x^2)^(1/2)), x)

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sympy [C]  time = 8.96, size = 510, normalized size = 3.02 \[ d^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{5 d^{2} x^{4}} - \frac {4 e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{15 d^{4} x^{2}} - \frac {8 e^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{15 d^{6}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{5 d^{2} x^{4}} - \frac {4 i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{15 d^{4} x^{2}} - \frac {8 i e^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{15 d^{6}} & \text {otherwise} \end {cases}\right ) + 2 d e \left (\begin {cases} - \frac {1}{4 e x^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {e}{8 d^{2} x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {3 e^{3}}{8 d^{4} x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {3 e^{4} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{8 d^{5}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i}{4 e x^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e}{8 d^{2} x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {3 i e^{3}}{8 d^{4} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {3 i e^{4} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{8 d^{5}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2} x^{2}} - \frac {2 e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{4}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2} x^{2}} - \frac {2 i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{4}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/x**6/(-e**2*x**2+d**2)**(1/2),x)

[Out]

d**2*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(5*d**2*x**4) - 4*e**3*sqrt(d**2/(e**2*x**2) - 1)/(15*d**4*x**2)
- 8*e**5*sqrt(d**2/(e**2*x**2) - 1)/(15*d**6), Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/
(5*d**2*x**4) - 4*I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(15*d**4*x**2) - 8*I*e**5*sqrt(-d**2/(e**2*x**2) + 1)/(15
*d**6), True)) + 2*d*e*Piecewise((-1/(4*e*x**5*sqrt(d**2/(e**2*x**2) - 1)) - e/(8*d**2*x**3*sqrt(d**2/(e**2*x*
*2) - 1)) + 3*e**3/(8*d**4*x*sqrt(d**2/(e**2*x**2) - 1)) - 3*e**4*acosh(d/(e*x))/(8*d**5), Abs(d**2/(e**2*x**2
)) > 1), (I/(4*e*x**5*sqrt(-d**2/(e**2*x**2) + 1)) + I*e/(8*d**2*x**3*sqrt(-d**2/(e**2*x**2) + 1)) - 3*I*e**3/
(8*d**4*x*sqrt(-d**2/(e**2*x**2) + 1)) + 3*I*e**4*asin(d/(e*x))/(8*d**5), True)) + e**2*Piecewise((-e*sqrt(d**
2/(e**2*x**2) - 1)/(3*d**2*x**2) - 2*e**3*sqrt(d**2/(e**2*x**2) - 1)/(3*d**4), Abs(d**2/(e**2*x**2)) > 1), (-I
*e*sqrt(-d**2/(e**2*x**2) + 1)/(3*d**2*x**2) - 2*I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(3*d**4), True))

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