3.2.5 \(\int \frac {x^2 (d^2-e^2 x^2)^{5/2}}{d+e x} \, dx\)

Optimal. Leaf size=140 \[ \frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3}+\frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2} \]

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Rubi [A]  time = 0.15, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1639, 12, 785, 780, 195, 217, 203} \begin {gather*} \frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(d^2 - e^2*x^2)^(5/2))/(d + e*x),x]

[Out]

(d^5*x*Sqrt[d^2 - e^2*x^2])/(16*e^2) + (d^3*x*(d^2 - e^2*x^2)^(3/2))/(24*e^2) + (d*(6*d - 5*e*x)*(d^2 - e^2*x^
2)^(5/2))/(30*e^3) - (d^2 - e^2*x^2)^(7/2)/(7*e^3) + (d^7*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*e^3)

Rule 12

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 785

Int[(x_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^m*e^m, Int[(x*(a + c*x^2)^(m
+ p))/(a*e + c*d*x)^m, x], x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[
m, 0] && EqQ[m, -1] &&  !ILtQ[p - 1/2, 0]

Rule 1639

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx &=-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}-\frac {\int \frac {7 d e^3 x \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx}{7 e^4}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}-\frac {d \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx}{e}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}-\frac {\int x \left (d^2 e-d e^2 x\right ) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{e^2}\\ &=\frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^3 \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{6 e^2}\\ &=\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^5 \int \sqrt {d^2-e^2 x^2} \, dx}{8 e^2}\\ &=\frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{16 e^2}\\ &=\frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2}\\ &=\frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 113, normalized size = 0.81 \begin {gather*} \frac {105 d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\sqrt {d^2-e^2 x^2} \left (96 d^6-105 d^5 e x+48 d^4 e^2 x^2+490 d^3 e^3 x^3-384 d^2 e^4 x^4-280 d e^5 x^5+240 e^6 x^6\right )}{1680 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(d^2 - e^2*x^2)^(5/2))/(d + e*x),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(96*d^6 - 105*d^5*e*x + 48*d^4*e^2*x^2 + 490*d^3*e^3*x^3 - 384*d^2*e^4*x^4 - 280*d*e^5*x^
5 + 240*e^6*x^6) + 105*d^7*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(1680*e^3)

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IntegrateAlgebraic [A]  time = 0.39, size = 136, normalized size = 0.97 \begin {gather*} \frac {d^7 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{16 e^4}+\frac {\sqrt {d^2-e^2 x^2} \left (96 d^6-105 d^5 e x+48 d^4 e^2 x^2+490 d^3 e^3 x^3-384 d^2 e^4 x^4-280 d e^5 x^5+240 e^6 x^6\right )}{1680 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^2*(d^2 - e^2*x^2)^(5/2))/(d + e*x),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(96*d^6 - 105*d^5*e*x + 48*d^4*e^2*x^2 + 490*d^3*e^3*x^3 - 384*d^2*e^4*x^4 - 280*d*e^5*x^
5 + 240*e^6*x^6))/(1680*e^3) + (d^7*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(16*e^4)

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fricas [A]  time = 0.41, size = 117, normalized size = 0.84 \begin {gather*} -\frac {210 \, d^{7} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (240 \, e^{6} x^{6} - 280 \, d e^{5} x^{5} - 384 \, d^{2} e^{4} x^{4} + 490 \, d^{3} e^{3} x^{3} + 48 \, d^{4} e^{2} x^{2} - 105 \, d^{5} e x + 96 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{1680 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-e^2*x^2+d^2)^(5/2)/(e*x+d),x, algorithm="fricas")

[Out]

-1/1680*(210*d^7*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (240*e^6*x^6 - 280*d*e^5*x^5 - 384*d^2*e^4*x^4 +
490*d^3*e^3*x^3 + 48*d^4*e^2*x^2 - 105*d^5*e*x + 96*d^6)*sqrt(-e^2*x^2 + d^2))/e^3

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-e^2*x^2+d^2)^(5/2)/(e*x+d),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/2*(12*d^7*exp(1)^4*exp(2)^2-8*d^7*exp(
2)^4-4*d^7*exp(1)^6*exp(2))*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+ex
p(2)^2))/sqrt(-exp(1)^4+exp(2)^2)/exp(1)^8/exp(1)+1/16*d^7*sign(d)*asin(x*exp(2)/d/exp(1))/exp(1)/exp(2)+2*(((
(((5760*exp(1)^13*1/80640/exp(1)^10*x-6720*exp(1)^12*d*1/80640/exp(1)^10)*x-9216*exp(1)^11*d^2*1/80640/exp(1)^
10)*x+11760*exp(1)^10*d^3*1/80640/exp(1)^10)*x+1152*exp(1)^9*d^4*1/80640/exp(1)^10)*x-2520*exp(1)^8*d^5*1/8064
0/exp(1)^10)*x+2304*exp(1)^7*d^6*1/80640/exp(1)^10)*sqrt(d^2-x^2*exp(2))

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maple [B]  time = 0.01, size = 282, normalized size = 2.01 \begin {gather*} \frac {3 d^{7} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{8 \sqrt {e^{2}}\, e^{2}}-\frac {5 d^{7} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 \sqrt {e^{2}}\, e^{2}}-\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{5} x}{16 e^{2}}+\frac {3 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{5} x}{8 e^{2}}-\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{3} x}{24 e^{2}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d^{3} x}{4 e^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d x}{6 e^{2}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} d^{2}}{5 e^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{7 e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(-e^2*x^2+d^2)^(5/2)/(e*x+d),x)

[Out]

-1/7*(-e^2*x^2+d^2)^(7/2)/e^3-1/6*(-e^2*x^2+d^2)^(5/2)*d/e^2*x-5/24*(-e^2*x^2+d^2)^(3/2)*d^3/e^2*x-5/16*(-e^2*
x^2+d^2)^(1/2)*d^5/e^2*x-5/16/(e^2)^(1/2)*d^7/e^2*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+1/5*d^2/e^3*(2*(x
+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)+1/4*d^3/e^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x+3/8*d^5/e^2*(2*(x+d/e)*d*e-(x
+d/e)^2*e^2)^(1/2)*x+3/8*d^7/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)

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maxima [C]  time = 1.04, size = 198, normalized size = 1.41 \begin {gather*} -\frac {3 i \, d^{7} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e^{3}} - \frac {5 \, d^{7} \arcsin \left (\frac {e x}{d}\right )}{16 \, e^{3}} + \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{5} x}{8 \, e^{2}} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5} x}{16 \, e^{2}} + \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{6}}{4 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} x}{24 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x}{6 \, e^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{5 \, e^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}}{7 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-e^2*x^2+d^2)^(5/2)/(e*x+d),x, algorithm="maxima")

[Out]

-3/8*I*d^7*arcsin(e*x/d + 2)/e^3 - 5/16*d^7*arcsin(e*x/d)/e^3 + 3/8*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^5*x/e^2
- 5/16*sqrt(-e^2*x^2 + d^2)*d^5*x/e^2 + 3/4*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^6/e^3 + 1/24*(-e^2*x^2 + d^2)^(3
/2)*d^3*x/e^2 - 1/6*(-e^2*x^2 + d^2)^(5/2)*d*x/e^2 + 1/5*(-e^2*x^2 + d^2)^(5/2)*d^2/e^3 - 1/7*(-e^2*x^2 + d^2)
^(7/2)/e^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(d^2 - e^2*x^2)^(5/2))/(d + e*x),x)

[Out]

int((x^2*(d^2 - e^2*x^2)^(5/2))/(d + e*x), x)

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sympy [C]  time = 16.66, size = 653, normalized size = 4.66 \begin {gather*} d^{3} \left (\begin {cases} - \frac {i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{3}} + \frac {i d^{3} x}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {3 i d x^{3}}{8 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{5}}{4 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{3}} - \frac {d^{3} x}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {3 d x^{3}}{8 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) - d^{2} e \left (\begin {cases} - \frac {2 d^{4} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac {d^{2} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{2}} + \frac {x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {x^{4} \sqrt {d^{2}}}{4} & \text {otherwise} \end {cases}\right ) - d e^{2} \left (\begin {cases} - \frac {i d^{6} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{16 e^{5}} + \frac {i d^{5} x}{16 e^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i d^{3} x^{3}}{48 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {5 i d x^{5}}{24 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{7}}{6 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{6} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{16 e^{5}} - \frac {d^{5} x}{16 e^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {d^{3} x^{3}}{48 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {5 d x^{5}}{24 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{7}}{6 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} - \frac {8 d^{6} \sqrt {d^{2} - e^{2} x^{2}}}{105 e^{6}} - \frac {4 d^{4} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{105 e^{4}} - \frac {d^{2} x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{35 e^{2}} + \frac {x^{6} \sqrt {d^{2} - e^{2} x^{2}}}{7} & \text {for}\: e \neq 0 \\\frac {x^{6} \sqrt {d^{2}}}{6} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(-e**2*x**2+d**2)**(5/2)/(e*x+d),x)

[Out]

d**3*Piecewise((-I*d**4*acosh(e*x/d)/(8*e**3) + I*d**3*x/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - 3*I*d*x**3/(8*sq
rt(-1 + e**2*x**2/d**2)) + I*e**2*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**4*asin(e
*x/d)/(8*e**3) - d**3*x/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + 3*d*x**3/(8*sqrt(1 - e**2*x**2/d**2)) - e**2*x**5/
(4*d*sqrt(1 - e**2*x**2/d**2)), True)) - d**2*e*Piecewise((-2*d**4*sqrt(d**2 - e**2*x**2)/(15*e**4) - d**2*x**
2*sqrt(d**2 - e**2*x**2)/(15*e**2) + x**4*sqrt(d**2 - e**2*x**2)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True)) - d*
e**2*Piecewise((-I*d**6*acosh(e*x/d)/(16*e**5) + I*d**5*x/(16*e**4*sqrt(-1 + e**2*x**2/d**2)) - I*d**3*x**3/(4
8*e**2*sqrt(-1 + e**2*x**2/d**2)) - 5*I*d*x**5/(24*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**7/(6*d*sqrt(-1 + e**
2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**6*asin(e*x/d)/(16*e**5) - d**5*x/(16*e**4*sqrt(1 - e**2*x**2/d**2
)) + d**3*x**3/(48*e**2*sqrt(1 - e**2*x**2/d**2)) + 5*d*x**5/(24*sqrt(1 - e**2*x**2/d**2)) - e**2*x**7/(6*d*sq
rt(1 - e**2*x**2/d**2)), True)) + e**3*Piecewise((-8*d**6*sqrt(d**2 - e**2*x**2)/(105*e**6) - 4*d**4*x**2*sqrt
(d**2 - e**2*x**2)/(105*e**4) - d**2*x**4*sqrt(d**2 - e**2*x**2)/(35*e**2) + x**6*sqrt(d**2 - e**2*x**2)/7, Ne
(e, 0)), (x**6*sqrt(d**2)/6, True))

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