3.1.27 \(\int \frac {d+e x}{x (d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=117 \[ \frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.10, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {823, 12, 266, 63, 208} \begin {gather*} \frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(x*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(d + e*x)/(5*d^2*(d^2 - e^2*x^2)^(5/2)) + (5*d + 4*e*x)/(15*d^4*(d^2 - e^2*x^2)^(3/2)) + (15*d + 8*e*x)/(15*d^
6*Sqrt[d^2 - e^2*x^2]) - ArcTanh[Sqrt[d^2 - e^2*x^2]/d]/d^6

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {d+e x}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {\int \frac {5 d^3 e^2+4 d^2 e^3 x}{x \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^4 e^2}\\ &=\frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^5 e^4+8 d^4 e^5 x}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^8 e^4}\\ &=\frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {15 d^7 e^6}{x \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{12} e^6}\\ &=\frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^5}\\ &=\frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^5}\\ &=\frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\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 )}{d^5 e^2}\\ &=\frac {d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 131, normalized size = 1.12 \begin {gather*} \frac {23 d^4-8 d^3 e x-27 d^2 e^2 x^2-15 (d-e x)^2 (d+e x) \sqrt {d^2-e^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )+7 d e^3 x^3+8 e^4 x^4}{15 d^6 (d-e x)^2 (d+e x) \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)/(x*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(23*d^4 - 8*d^3*e*x - 27*d^2*e^2*x^2 + 7*d*e^3*x^3 + 8*e^4*x^4 - 15*(d - e*x)^2*(d + e*x)*Sqrt[d^2 - e^2*x^2]*
ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(15*d^6*(d - e*x)^2*(d + e*x)*Sqrt[d^2 - e^2*x^2])

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IntegrateAlgebraic [A]  time = 0.72, size = 122, normalized size = 1.04 \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6}+\frac {\sqrt {d^2-e^2 x^2} \left (23 d^4-8 d^3 e x-27 d^2 e^2 x^2+7 d e^3 x^3+8 e^4 x^4\right )}{15 d^6 (d-e x)^3 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)/(x*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(23*d^4 - 8*d^3*e*x - 27*d^2*e^2*x^2 + 7*d*e^3*x^3 + 8*e^4*x^4))/(15*d^6*(d - e*x)^3*(d +
e*x)^2) + (2*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d])/d^6

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fricas [B]  time = 0.42, size = 244, normalized size = 2.09 \begin {gather*} \frac {23 \, e^{5} x^{5} - 23 \, d e^{4} x^{4} - 46 \, d^{2} e^{3} x^{3} + 46 \, d^{3} e^{2} x^{2} + 23 \, d^{4} e x - 23 \, d^{5} + 15 \, {\left (e^{5} x^{5} - d e^{4} x^{4} - 2 \, d^{2} e^{3} x^{3} + 2 \, d^{3} e^{2} x^{2} + d^{4} e x - d^{5}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (8 \, e^{4} x^{4} + 7 \, d e^{3} x^{3} - 27 \, d^{2} e^{2} x^{2} - 8 \, d^{3} e x + 23 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{6} e^{5} x^{5} - d^{7} e^{4} x^{4} - 2 \, d^{8} e^{3} x^{3} + 2 \, d^{9} e^{2} x^{2} + d^{10} e x - d^{11}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/x/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(23*e^5*x^5 - 23*d*e^4*x^4 - 46*d^2*e^3*x^3 + 46*d^3*e^2*x^2 + 23*d^4*e*x - 23*d^5 + 15*(e^5*x^5 - d*e^4*
x^4 - 2*d^2*e^3*x^3 + 2*d^3*e^2*x^2 + d^4*e*x - d^5)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (8*e^4*x^4 + 7*d*e^3
*x^3 - 27*d^2*e^2*x^2 - 8*d^3*e*x + 23*d^4)*sqrt(-e^2*x^2 + d^2))/(d^6*e^5*x^5 - d^7*e^4*x^4 - 2*d^8*e^3*x^3 +
2*d^9*e^2*x^2 + d^10*e*x - d^11)

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giac [A]  time = 0.27, size = 122, normalized size = 1.04 \begin {gather*} -\frac {\sqrt {-x^{2} e^{2} + d^{2}} {\left ({\left ({\left ({\left (x {\left (\frac {8 \, x e^{5}}{d^{6}} + \frac {15 \, e^{4}}{d^{5}}\right )} - \frac {20 \, e^{3}}{d^{4}}\right )} x - \frac {35 \, e^{2}}{d^{3}}\right )} x + \frac {15 \, e}{d^{2}}\right )} x + \frac {23}{d}\right )}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} - \frac {\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 )}{d^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/x/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-1/15*sqrt(-x^2*e^2 + d^2)*((((x*(8*x*e^5/d^6 + 15*e^4/d^5) - 20*e^3/d^4)*x - 35*e^2/d^3)*x + 15*e/d^2)*x + 23
/d)/(x^2*e^2 - d^2)^3 - log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^6

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maple [A]  time = 0.01, size = 163, normalized size = 1.39 \begin {gather*} \frac {e x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{2}}+\frac {1}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d}+\frac {4 e x}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{4}}+\frac {1}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{3}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}\, d^{5}}+\frac {8 e x}{15 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{6}}+\frac {1}{\sqrt {-e^{2} x^{2}+d^{2}}\, d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)/x/(-e^2*x^2+d^2)^(7/2),x)

[Out]

1/5*e*x/d^2/(-e^2*x^2+d^2)^(5/2)+4/15*e/d^4*x/(-e^2*x^2+d^2)^(3/2)+8/15*e/d^6*x/(-e^2*x^2+d^2)^(1/2)+1/5/d/(-e
^2*x^2+d^2)^(5/2)+1/3/d^3/(-e^2*x^2+d^2)^(3/2)+1/d^5/(-e^2*x^2+d^2)^(1/2)-1/d^5/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^
(1/2)*(-e^2*x^2+d^2)^(1/2))/x)

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maxima [A]  time = 0.45, size = 157, normalized size = 1.34 \begin {gather*} \frac {e x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}} + \frac {1}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d} + \frac {4 \, e x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}} + \frac {1}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}} + \frac {8 \, e x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{6}} - \frac {\log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{6}} + \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/x/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/5*e*x/((-e^2*x^2 + d^2)^(5/2)*d^2) + 1/5/((-e^2*x^2 + d^2)^(5/2)*d) + 4/15*e*x/((-e^2*x^2 + d^2)^(3/2)*d^4)
+ 1/3/((-e^2*x^2 + d^2)^(3/2)*d^3) + 8/15*e*x/(sqrt(-e^2*x^2 + d^2)*d^6) - log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2
+ d^2)*d/abs(x))/d^6 + 1/(sqrt(-e^2*x^2 + d^2)*d^5)

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mupad [B]  time = 3.08, size = 127, normalized size = 1.09 \begin {gather*} \frac {\frac {d^2-e^2\,x^2}{3\,d^3}+\frac {{\left (d^2-e^2\,x^2\right )}^2}{d^5}+\frac {1}{5\,d}}{{\left (d^2-e^2\,x^2\right )}^{5/2}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d^2-e^2\,x^2}}{d}\right )}{d^6}+\frac {e\,x\,\left (15\,d^4-20\,d^2\,e^2\,x^2+8\,e^4\,x^4\right )}{15\,d^6\,{\left (d^2-e^2\,x^2\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)/(x*(d^2 - e^2*x^2)^(7/2)),x)

[Out]

((d^2 - e^2*x^2)/(3*d^3) + (d^2 - e^2*x^2)^2/d^5 + 1/(5*d))/(d^2 - e^2*x^2)^(5/2) - atanh((d^2 - e^2*x^2)^(1/2
)/d)/d^6 + (e*x*(15*d^4 + 8*e^4*x^4 - 20*d^2*e^2*x^2))/(15*d^6*(d^2 - e^2*x^2)^(5/2))

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sympy [C]  time = 41.14, size = 2378, normalized size = 20.32

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/x/(-e**2*x**2+d**2)**(7/2),x)

[Out]

d*Piecewise((46*I*d**6*sqrt(-1 + e**2*x**2/d**2)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*
e**6*x**6) + 15*d**6*log(e**2*x**2/d**2)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**
6) - 30*d**6*log(e*x/d)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) + 30*I*d**6*as
in(d/(e*x))/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) - 70*I*d**4*e**2*x**2*sqrt
(-1 + e**2*x**2/d**2)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) - 45*d**4*e**2*x
**2*log(e**2*x**2/d**2)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) + 90*d**4*e**2
*x**2*log(e*x/d)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) - 90*I*d**4*e**2*x**2
*asin(d/(e*x))/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) + 30*I*d**2*e**4*x**4*s
qrt(-1 + e**2*x**2/d**2)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) + 45*d**2*e**
4*x**4*log(e**2*x**2/d**2)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) - 90*d**2*e
**4*x**4*log(e*x/d)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) + 90*I*d**2*e**4*x
**4*asin(d/(e*x))/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) - 15*e**6*x**6*log(e
**2*x**2/d**2)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) + 30*e**6*x**6*log(e*x/
d)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) - 30*I*e**6*x**6*asin(d/(e*x))/(30*
d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6), Abs(e**2*x**2/d**2) > 1), (46*d**6*sqrt(1
- e**2*x**2/d**2)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) + 15*d**6*log(e**2*
x**2/d**2)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) - 30*d**6*log(sqrt(1 - e**2
*x**2/d**2) + 1)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) + 15*I*pi*d**6/(30*d*
*13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) - 70*d**4*e**2*x**2*sqrt(1 - e**2*x**2/d**2)
/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) - 45*d**4*e**2*x**2*log(e**2*x**2/d**
2)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) + 90*d**4*e**2*x**2*log(sqrt(1 - e*
*2*x**2/d**2) + 1)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) - 45*I*pi*d**4*e**2
*x**2/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) + 30*d**2*e**4*x**4*sqrt(1 - e**
2*x**2/d**2)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) + 45*d**2*e**4*x**4*log(e
**2*x**2/d**2)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) - 90*d**2*e**4*x**4*log
(sqrt(1 - e**2*x**2/d**2) + 1)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) + 45*I*
pi*d**2*e**4*x**4/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) - 15*e**6*x**6*log(e
**2*x**2/d**2)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) + 30*e**6*x**6*log(sqrt
(1 - e**2*x**2/d**2) + 1)/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6) - 15*I*pi*e*
*6*x**6/(30*d**13 - 90*d**11*e**2*x**2 + 90*d**9*e**4*x**4 - 30*d**7*e**6*x**6), True)) + e*Piecewise((-15*I*d
**4*x/(15*d**11*sqrt(-1 + e**2*x**2/d**2) - 30*d**9*e**2*x**2*sqrt(-1 + e**2*x**2/d**2) + 15*d**7*e**4*x**4*sq
rt(-1 + e**2*x**2/d**2)) + 20*I*d**2*e**2*x**3/(15*d**11*sqrt(-1 + e**2*x**2/d**2) - 30*d**9*e**2*x**2*sqrt(-1
+ e**2*x**2/d**2) + 15*d**7*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)) - 8*I*e**4*x**5/(15*d**11*sqrt(-1 + e**2*x**
2/d**2) - 30*d**9*e**2*x**2*sqrt(-1 + e**2*x**2/d**2) + 15*d**7*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2
*x**2/d**2) > 1), (15*d**4*x/(15*d**11*sqrt(1 - e**2*x**2/d**2) - 30*d**9*e**2*x**2*sqrt(1 - e**2*x**2/d**2) +
15*d**7*e**4*x**4*sqrt(1 - e**2*x**2/d**2)) - 20*d**2*e**2*x**3/(15*d**11*sqrt(1 - e**2*x**2/d**2) - 30*d**9*
e**2*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d**7*e**4*x**4*sqrt(1 - e**2*x**2/d**2)) + 8*e**4*x**5/(15*d**11*sqrt(
1 - e**2*x**2/d**2) - 30*d**9*e**2*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d**7*e**4*x**4*sqrt(1 - e**2*x**2/d**2))
, True))

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