3.106 \(\int x^3 (d+e x^2)^2 (a+b \text {sech}^{-1}(c x)) \, dx\)

Optimal. Leaf size=278 \[ \frac {1}{4} d^2 x^4 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{3} d e x^6 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{8} e^2 x^8 \left (a+b \text {sech}^{-1}(c x)\right )-\frac {b e \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (1-c^2 x^2\right )^{5/2} \left (8 c^2 d+9 e\right )}{120 c^8}+\frac {b e^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (1-c^2 x^2\right )^{7/2}}{56 c^8}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (1-c^2 x^2\right )^{3/2} \left (6 c^4 d^2+16 c^2 d e+9 e^2\right )}{72 c^8}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (6 c^4 d^2+8 c^2 d e+3 e^2\right )}{24 c^8} \]

[Out]

1/4*d^2*x^4*(a+b*arcsech(c*x))+1/3*d*e*x^6*(a+b*arcsech(c*x))+1/8*e^2*x^8*(a+b*arcsech(c*x))+1/72*b*(6*c^4*d^2
+16*c^2*d*e+9*e^2)*(-c^2*x^2+1)^(3/2)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/c^8-1/120*b*e*(8*c^2*d+9*e)*(-c^2*x^2+1)
^(5/2)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/c^8+1/56*b*e^2*(-c^2*x^2+1)^(7/2)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/c^8-1
/24*b*(6*c^4*d^2+8*c^2*d*e+3*e^2)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/c^8

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Rubi [A]  time = 0.24, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {266, 43, 6301, 12, 1251, 771} \[ \frac {1}{4} d^2 x^4 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{3} d e x^6 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{8} e^2 x^8 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (1-c^2 x^2\right )^{3/2} \left (6 c^4 d^2+16 c^2 d e+9 e^2\right )}{72 c^8}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (6 c^4 d^2+8 c^2 d e+3 e^2\right )}{24 c^8}-\frac {b e \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (1-c^2 x^2\right )^{5/2} \left (8 c^2 d+9 e\right )}{120 c^8}+\frac {b e^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (1-c^2 x^2\right )^{7/2}}{56 c^8} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(d + e*x^2)^2*(a + b*ArcSech[c*x]),x]

[Out]

-(b*(6*c^4*d^2 + 8*c^2*d*e + 3*e^2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(24*c^8) + (b*(6*c^4
*d^2 + 16*c^2*d*e + 9*e^2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*(1 - c^2*x^2)^(3/2))/(72*c^8) - (b*e*(8*c^2*d +
9*e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*(1 - c^2*x^2)^(5/2))/(120*c^8) + (b*e^2*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 +
c*x]*(1 - c^2*x^2)^(7/2))/(56*c^8) + (d^2*x^4*(a + b*ArcSech[c*x]))/4 + (d*e*x^6*(a + b*ArcSech[c*x]))/3 + (e^
2*x^8*(a + b*ArcSech[c*x]))/8

Rule 12

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 6301

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u
= IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSech[c*x], u, x] + Dist[b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)],
Int[SimplifyIntegrand[u/(x*Sqrt[1 - c*x]*Sqrt[1 + c*x]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] &&
((IGtQ[p, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ
[m + 2*p + 3, 0])) || (ILtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps

\begin {align*} \int x^3 \left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx &=\frac {1}{4} d^2 x^4 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{3} d e x^6 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{8} e^2 x^8 \left (a+b \text {sech}^{-1}(c x)\right )+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {x^3 \left (6 d^2+8 d e x^2+3 e^2 x^4\right )}{24 \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {1}{4} d^2 x^4 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{3} d e x^6 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{8} e^2 x^8 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{24} \left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {x^3 \left (6 d^2+8 d e x^2+3 e^2 x^4\right )}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {1}{4} d^2 x^4 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{3} d e x^6 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{8} e^2 x^8 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{48} \left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {x \left (6 d^2+8 d e x+3 e^2 x^2\right )}{\sqrt {1-c^2 x}} \, dx,x,x^2\right )\\ &=\frac {1}{4} d^2 x^4 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{3} d e x^6 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{8} e^2 x^8 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{48} \left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \left (\frac {6 c^4 d^2+8 c^2 d e+3 e^2}{c^6 \sqrt {1-c^2 x}}+\frac {\left (-6 c^4 d^2-16 c^2 d e-9 e^2\right ) \sqrt {1-c^2 x}}{c^6}+\frac {e \left (8 c^2 d+9 e\right ) \left (1-c^2 x\right )^{3/2}}{c^6}-\frac {3 e^2 \left (1-c^2 x\right )^{5/2}}{c^6}\right ) \, dx,x,x^2\right )\\ &=-\frac {b \left (6 c^4 d^2+8 c^2 d e+3 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{24 c^8}+\frac {b \left (6 c^4 d^2+16 c^2 d e+9 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \left (1-c^2 x^2\right )^{3/2}}{72 c^8}-\frac {b e \left (8 c^2 d+9 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \left (1-c^2 x^2\right )^{5/2}}{120 c^8}+\frac {b e^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \left (1-c^2 x^2\right )^{7/2}}{56 c^8}+\frac {1}{4} d^2 x^4 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{3} d e x^6 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{8} e^2 x^8 \left (a+b \text {sech}^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 168, normalized size = 0.60 \[ \frac {1}{24} \left (6 a d^2 x^4+8 a d e x^6+3 a e^2 x^8-\frac {b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (3 c^6 \left (70 d^2 x^2+56 d e x^4+15 e^2 x^6\right )+c^4 \left (420 d^2+224 d e x^2+54 e^2 x^4\right )+8 c^2 e \left (56 d+9 e x^2\right )+144 e^2\right )}{105 c^8}+b x^4 \text {sech}^{-1}(c x) \left (6 d^2+8 d e x^2+3 e^2 x^4\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(d + e*x^2)^2*(a + b*ArcSech[c*x]),x]

[Out]

(6*a*d^2*x^4 + 8*a*d*e*x^6 + 3*a*e^2*x^8 - (b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(144*e^2 + 8*c^2*e*(56*d + 9
*e*x^2) + c^4*(420*d^2 + 224*d*e*x^2 + 54*e^2*x^4) + 3*c^6*(70*d^2*x^2 + 56*d*e*x^4 + 15*e^2*x^6)))/(105*c^8)
+ b*x^4*(6*d^2 + 8*d*e*x^2 + 3*e^2*x^4)*ArcSech[c*x])/24

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fricas [A]  time = 1.08, size = 227, normalized size = 0.82 \[ \frac {315 \, a c^{7} e^{2} x^{8} + 840 \, a c^{7} d e x^{6} + 630 \, a c^{7} d^{2} x^{4} + 105 \, {\left (3 \, b c^{7} e^{2} x^{8} + 8 \, b c^{7} d e x^{6} + 6 \, b c^{7} d^{2} x^{4}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (45 \, b c^{6} e^{2} x^{7} + 6 \, {\left (28 \, b c^{6} d e + 9 \, b c^{4} e^{2}\right )} x^{5} + 2 \, {\left (105 \, b c^{6} d^{2} + 112 \, b c^{4} d e + 36 \, b c^{2} e^{2}\right )} x^{3} + 4 \, {\left (105 \, b c^{4} d^{2} + 112 \, b c^{2} d e + 36 \, b e^{2}\right )} x\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{2520 \, c^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^2*(a+b*arcsech(c*x)),x, algorithm="fricas")

[Out]

1/2520*(315*a*c^7*e^2*x^8 + 840*a*c^7*d*e*x^6 + 630*a*c^7*d^2*x^4 + 105*(3*b*c^7*e^2*x^8 + 8*b*c^7*d*e*x^6 + 6
*b*c^7*d^2*x^4)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - (45*b*c^6*e^2*x^7 + 6*(28*b*c^6*d*e + 9*
b*c^4*e^2)*x^5 + 2*(105*b*c^6*d^2 + 112*b*c^4*d*e + 36*b*c^2*e^2)*x^3 + 4*(105*b*c^4*d^2 + 112*b*c^2*d*e + 36*
b*e^2)*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/c^7

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )} x^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^2*(a+b*arcsech(c*x)),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^2*(b*arcsech(c*x) + a)*x^3, x)

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maple [A]  time = 0.07, size = 212, normalized size = 0.76 \[ \frac {\frac {a \left (\frac {1}{8} e^{2} c^{8} x^{8}+\frac {1}{3} c^{8} d e \,x^{6}+\frac {1}{4} c^{8} d^{2} x^{4}\right )}{c^{4}}+\frac {b \left (\frac {\mathrm {arcsech}\left (c x \right ) e^{2} c^{8} x^{8}}{8}+\frac {\mathrm {arcsech}\left (c x \right ) c^{8} d e \,x^{6}}{3}+\frac {\mathrm {arcsech}\left (c x \right ) c^{8} x^{4} d^{2}}{4}-\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (45 c^{6} e^{2} x^{6}+168 c^{6} d e \,x^{4}+210 c^{6} d^{2} x^{2}+54 c^{4} e^{2} x^{4}+224 c^{4} d e \,x^{2}+420 d^{2} c^{4}+72 c^{2} e^{2} x^{2}+448 c^{2} d e +144 e^{2}\right )}{2520}\right )}{c^{4}}}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x^2+d)^2*(a+b*arcsech(c*x)),x)

[Out]

1/c^4*(a/c^4*(1/8*e^2*c^8*x^8+1/3*c^8*d*e*x^6+1/4*c^8*d^2*x^4)+b/c^4*(1/8*arcsech(c*x)*e^2*c^8*x^8+1/3*arcsech
(c*x)*c^8*d*e*x^6+1/4*arcsech(c*x)*c^8*x^4*d^2-1/2520*(-(c*x-1)/c/x)^(1/2)*c*x*((c*x+1)/c/x)^(1/2)*(45*c^6*e^2
*x^6+168*c^6*d*e*x^4+210*c^6*d^2*x^2+54*c^4*e^2*x^4+224*c^4*d*e*x^2+420*c^4*d^2+72*c^2*e^2*x^2+448*c^2*d*e+144
*e^2)))

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maxima [A]  time = 0.33, size = 245, normalized size = 0.88 \[ \frac {1}{8} \, a e^{2} x^{8} + \frac {1}{3} \, a d e x^{6} + \frac {1}{4} \, a d^{2} x^{4} + \frac {1}{12} \, {\left (3 \, x^{4} \operatorname {arsech}\left (c x\right ) + \frac {c^{2} x^{3} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} - 3 \, x \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{3}}\right )} b d^{2} + \frac {1}{45} \, {\left (15 \, x^{6} \operatorname {arsech}\left (c x\right ) - \frac {3 \, c^{4} x^{5} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {5}{2}} - 10 \, c^{2} x^{3} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} + 15 \, x \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{5}}\right )} b d e + \frac {1}{280} \, {\left (35 \, x^{8} \operatorname {arsech}\left (c x\right ) + \frac {5 \, c^{6} x^{7} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {7}{2}} - 21 \, c^{4} x^{5} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {5}{2}} + 35 \, c^{2} x^{3} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} - 35 \, x \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{7}}\right )} b e^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^2*(a+b*arcsech(c*x)),x, algorithm="maxima")

[Out]

1/8*a*e^2*x^8 + 1/3*a*d*e*x^6 + 1/4*a*d^2*x^4 + 1/12*(3*x^4*arcsech(c*x) + (c^2*x^3*(1/(c^2*x^2) - 1)^(3/2) -
3*x*sqrt(1/(c^2*x^2) - 1))/c^3)*b*d^2 + 1/45*(15*x^6*arcsech(c*x) - (3*c^4*x^5*(1/(c^2*x^2) - 1)^(5/2) - 10*c^
2*x^3*(1/(c^2*x^2) - 1)^(3/2) + 15*x*sqrt(1/(c^2*x^2) - 1))/c^5)*b*d*e + 1/280*(35*x^8*arcsech(c*x) + (5*c^6*x
^7*(1/(c^2*x^2) - 1)^(7/2) - 21*c^4*x^5*(1/(c^2*x^2) - 1)^(5/2) + 35*c^2*x^3*(1/(c^2*x^2) - 1)^(3/2) - 35*x*sq
rt(1/(c^2*x^2) - 1))/c^7)*b*e^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,{\left (e\,x^2+d\right )}^2\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(d + e*x^2)^2*(a + b*acosh(1/(c*x))),x)

[Out]

int(x^3*(d + e*x^2)^2*(a + b*acosh(1/(c*x))), x)

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sympy [A]  time = 16.46, size = 332, normalized size = 1.19 \[ \begin {cases} \frac {a d^{2} x^{4}}{4} + \frac {a d e x^{6}}{3} + \frac {a e^{2} x^{8}}{8} + \frac {b d^{2} x^{4} \operatorname {asech}{\left (c x \right )}}{4} + \frac {b d e x^{6} \operatorname {asech}{\left (c x \right )}}{3} + \frac {b e^{2} x^{8} \operatorname {asech}{\left (c x \right )}}{8} - \frac {b d^{2} x^{2} \sqrt {- c^{2} x^{2} + 1}}{12 c^{2}} - \frac {b d e x^{4} \sqrt {- c^{2} x^{2} + 1}}{15 c^{2}} - \frac {b e^{2} x^{6} \sqrt {- c^{2} x^{2} + 1}}{56 c^{2}} - \frac {b d^{2} \sqrt {- c^{2} x^{2} + 1}}{6 c^{4}} - \frac {4 b d e x^{2} \sqrt {- c^{2} x^{2} + 1}}{45 c^{4}} - \frac {3 b e^{2} x^{4} \sqrt {- c^{2} x^{2} + 1}}{140 c^{4}} - \frac {8 b d e \sqrt {- c^{2} x^{2} + 1}}{45 c^{6}} - \frac {b e^{2} x^{2} \sqrt {- c^{2} x^{2} + 1}}{35 c^{6}} - \frac {2 b e^{2} \sqrt {- c^{2} x^{2} + 1}}{35 c^{8}} & \text {for}\: c \neq 0 \\\left (a + \infty b\right ) \left (\frac {d^{2} x^{4}}{4} + \frac {d e x^{6}}{3} + \frac {e^{2} x^{8}}{8}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x**2+d)**2*(a+b*asech(c*x)),x)

[Out]

Piecewise((a*d**2*x**4/4 + a*d*e*x**6/3 + a*e**2*x**8/8 + b*d**2*x**4*asech(c*x)/4 + b*d*e*x**6*asech(c*x)/3 +
b*e**2*x**8*asech(c*x)/8 - b*d**2*x**2*sqrt(-c**2*x**2 + 1)/(12*c**2) - b*d*e*x**4*sqrt(-c**2*x**2 + 1)/(15*c
**2) - b*e**2*x**6*sqrt(-c**2*x**2 + 1)/(56*c**2) - b*d**2*sqrt(-c**2*x**2 + 1)/(6*c**4) - 4*b*d*e*x**2*sqrt(-
c**2*x**2 + 1)/(45*c**4) - 3*b*e**2*x**4*sqrt(-c**2*x**2 + 1)/(140*c**4) - 8*b*d*e*sqrt(-c**2*x**2 + 1)/(45*c*
*6) - b*e**2*x**2*sqrt(-c**2*x**2 + 1)/(35*c**6) - 2*b*e**2*sqrt(-c**2*x**2 + 1)/(35*c**8), Ne(c, 0)), ((a + o
o*b)*(d**2*x**4/4 + d*e*x**6/3 + e**2*x**8/8), True))

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