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

Optimal. Leaf size=153 \[ \frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 d^7 x}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^7}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}} \]

[Out]

1/5*(e*x+d)/d^2/x/(-e^2*x^2+d^2)^(5/2)+1/15*(5*e*x+6*d)/d^4/x/(-e^2*x^2+d^2)^(3/2)-e*arctanh((-e^2*x^2+d^2)^(1
/2)/d)/d^7+1/5*(5*e*x+8*d)/d^6/x/(-e^2*x^2+d^2)^(1/2)-16/5*(-e^2*x^2+d^2)^(1/2)/d^7/x

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Rubi [A]  time = 0.13, antiderivative size = 153, 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, 807, 266, 63, 208} \[ \frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 d^7 x}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d + e*x)/(5*d^2*x*(d^2 - e^2*x^2)^(5/2)) + (6*d + 5*e*x)/(15*d^4*x*(d^2 - e^2*x^2)^(3/2)) + (8*d + 5*e*x)/(5*
d^6*x*Sqrt[d^2 - e^2*x^2]) - (16*Sqrt[d^2 - e^2*x^2])/(5*d^7*x) - (e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^7

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 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^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {\int \frac {6 d^3 e^2+5 d^2 e^3 x}{x^2 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^4 e^2}\\ &=\frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {24 d^5 e^4+15 d^4 e^5 x}{x^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^8 e^4}\\ &=\frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {48 d^7 e^6+15 d^6 e^7 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{12} e^6}\\ &=\frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 d^7 x}+\frac {e \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^6}\\ &=\frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 d^7 x}+\frac {e \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^6}\\ &=\frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 d^7 x}-\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^6 e}\\ &=\frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 d^7 x}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^7}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 147, normalized size = 0.96 \[ \frac {-15 d^5+38 d^4 e x+52 d^3 e^2 x^2-87 d^2 e^3 x^3-15 e x (d-e x)^2 (d+e x) \sqrt {d^2-e^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )-33 d e^4 x^4+48 e^5 x^5}{15 d^7 x (d-e x)^2 (d+e x) \sqrt {d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*d^5 + 38*d^4*e*x + 52*d^3*e^2*x^2 - 87*d^2*e^3*x^3 - 33*d*e^4*x^4 + 48*e^5*x^5 - 15*e*x*(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^7*x*(d - e*x)^2*(d + e*x)*Sqrt[d^2 - e^2*x^2])

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fricas [A]  time = 1.06, size = 270, normalized size = 1.76 \[ \frac {23 \, e^{6} x^{6} - 23 \, d e^{5} x^{5} - 46 \, d^{2} e^{4} x^{4} + 46 \, d^{3} e^{3} x^{3} + 23 \, d^{4} e^{2} x^{2} - 23 \, d^{5} e x + 15 \, {\left (e^{6} x^{6} - d e^{5} x^{5} - 2 \, d^{2} e^{4} x^{4} + 2 \, d^{3} e^{3} x^{3} + d^{4} e^{2} x^{2} - d^{5} e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (48 \, e^{5} x^{5} - 33 \, d e^{4} x^{4} - 87 \, d^{2} e^{3} x^{3} + 52 \, d^{3} e^{2} x^{2} + 38 \, d^{4} e x - 15 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{7} e^{5} x^{6} - d^{8} e^{4} x^{5} - 2 \, d^{9} e^{3} x^{4} + 2 \, d^{10} e^{2} x^{3} + d^{11} e x^{2} - d^{12} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(23*e^6*x^6 - 23*d*e^5*x^5 - 46*d^2*e^4*x^4 + 46*d^3*e^3*x^3 + 23*d^4*e^2*x^2 - 23*d^5*e*x + 15*(e^6*x^6
- d*e^5*x^5 - 2*d^2*e^4*x^4 + 2*d^3*e^3*x^3 + d^4*e^2*x^2 - d^5*e*x)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (48*
e^5*x^5 - 33*d*e^4*x^4 - 87*d^2*e^3*x^3 + 52*d^3*e^2*x^2 + 38*d^4*e*x - 15*d^5)*sqrt(-e^2*x^2 + d^2))/(d^7*e^5
*x^6 - d^8*e^4*x^5 - 2*d^9*e^3*x^4 + 2*d^10*e^2*x^3 + d^11*e*x^2 - d^12*x)

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giac [A]  time = 0.29, size = 189, normalized size = 1.24 \[ -\frac {\sqrt {-x^{2} e^{2} + d^{2}} {\left ({\left ({\left (3 \, {\left (x {\left (\frac {11 \, x e^{6}}{d^{7}} + \frac {5 \, e^{5}}{d^{6}}\right )} - \frac {25 \, e^{4}}{d^{5}}\right )} x - \frac {35 \, e^{3}}{d^{4}}\right )} x + \frac {45 \, e^{2}}{d^{3}}\right )} x + \frac {23 \, e}{d^{2}}\right )}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} - \frac {e \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right )}{d^{7}} + \frac {x e^{3}}{2 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{7}} - \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-1\right )}}{2 \, d^{7} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*sqrt(-x^2*e^2 + d^2)*(((3*(x*(11*x*e^6/d^7 + 5*e^5/d^6) - 25*e^4/d^5)*x - 35*e^3/d^4)*x + 45*e^2/d^3)*x
+ 23*e/d^2)/(x^2*e^2 - d^2)^3 - e*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^7 + 1/2*x*e^
3/((d*e + sqrt(-x^2*e^2 + d^2)*e)*d^7) - 1/2*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-1)/(d^7*x)

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maple [A]  time = 0.02, size = 195, normalized size = 1.27 \[ \frac {6 e^{2} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{3}}+\frac {e}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{2}}-\frac {1}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d x}+\frac {8 e^{2} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{5}}+\frac {e}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{4}}-\frac {e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}\, d^{6}}+\frac {16 e^{2} x}{5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{7}}+\frac {e}{\sqrt {-e^{2} x^{2}+d^{2}}\, d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*e/d^2/(-e^2*x^2+d^2)^(5/2)+1/3*e/d^4/(-e^2*x^2+d^2)^(3/2)+e/d^6/(-e^2*x^2+d^2)^(1/2)-e/d^6/(d^2)^(1/2)*ln(
(2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-1/d/x/(-e^2*x^2+d^2)^(5/2)+6/5*e^2/d^3*x/(-e^2*x^2+d^2)^(5/2)+8/
5*e^2/d^5*x/(-e^2*x^2+d^2)^(3/2)+16/5*e^2/d^7*x/(-e^2*x^2+d^2)^(1/2)

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maxima [A]  time = 0.46, size = 189, normalized size = 1.24 \[ \frac {6 \, e^{2} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}} + \frac {e}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}} + \frac {8 \, e^{2} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{5}} + \frac {e}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}} - \frac {1}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x} + \frac {16 \, e^{2} x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{7}} - \frac {e \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{7}} + \frac {e}{\sqrt {-e^{2} x^{2} + d^{2}} d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6/5*e^2*x/((-e^2*x^2 + d^2)^(5/2)*d^3) + 1/5*e/((-e^2*x^2 + d^2)^(5/2)*d^2) + 8/5*e^2*x/((-e^2*x^2 + d^2)^(3/2
)*d^5) + 1/3*e/((-e^2*x^2 + d^2)^(3/2)*d^4) - 1/((-e^2*x^2 + d^2)^(5/2)*d*x) + 16/5*e^2*x/(sqrt(-e^2*x^2 + d^2
)*d^7) - e*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^7 + e/(sqrt(-e^2*x^2 + d^2)*d^6)

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mupad [B]  time = 3.31, size = 141, normalized size = 0.92 \[ \frac {\frac {e}{5\,d^2}+\frac {e\,{\left (d^2-e^2\,x^2\right )}^2}{d^6}+\frac {e\,\left (d^2-e^2\,x^2\right )}{3\,d^4}}{{\left (d^2-e^2\,x^2\right )}^{5/2}}-\frac {e\,\mathrm {atanh}\left (\frac {\sqrt {d^2-e^2\,x^2}}{d}\right )}{d^7}-\frac {d^6-6\,d^4\,e^2\,x^2+8\,d^2\,e^4\,x^4-\frac {16\,e^6\,x^6}{5}}{d^7\,x\,{\left (d^2-e^2\,x^2\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e/(5*d^2) + (e*(d^2 - e^2*x^2)^2)/d^6 + (e*(d^2 - e^2*x^2))/(3*d^4))/(d^2 - e^2*x^2)^(5/2) - (e*atanh((d^2 -
e^2*x^2)^(1/2)/d))/d^7 - (d^6 - (16*e^6*x^6)/5 - 6*d^4*e^2*x^2 + 8*d^2*e^4*x^4)/(d^7*x*(d^2 - e^2*x^2)^(5/2))

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sympy [C]  time = 31.22, size = 2404, normalized size = 15.71 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*Piecewise((5*d**6*e*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*
e**6*x**6) - 30*d**4*e**3*x**2*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4
+ 5*d**8*e**6*x**6) + 40*d**2*e**5*x**4*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e
**4*x**4 + 5*d**8*e**6*x**6) - 16*e**7*x**6*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**
10*e**4*x**4 + 5*d**8*e**6*x**6), Abs(d**2/(e**2*x**2)) > 1), (5*I*d**6*e*sqrt(-d**2/(e**2*x**2) + 1)/(-5*d**1
4 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) - 30*I*d**4*e**3*x**2*sqrt(-d**2/(e**2*x**2) +
1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) + 40*I*d**2*e**5*x**4*sqrt(-d**2/(
e**2*x**2) + 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) - 16*I*e**7*x**6*sqrt(
-d**2/(e**2*x**2) + 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6), True)) + e*Pie
cewise((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*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) - 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*l
og(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*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(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*a
sin(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)/(3
0*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)/(3
0*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))

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