Integrand size = 21, antiderivative size = 152 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^9} \, dx=-\frac {b c d^3}{56 x^7}+\frac {b c d^2 \left (c^2 d-4 e\right )}{40 x^5}-\frac {b c d \left (c^4 d^2-4 c^2 d e+6 e^2\right )}{24 x^3}+\frac {b c \left (c^2 d-2 e\right ) \left (c^4 d^2-2 c^2 d e+2 e^2\right )}{8 x}+\frac {b \left (c^2 d-e\right )^4 \arctan (c x)}{8 d}-\frac {\left (d+e x^2\right )^4 (a+b \arctan (c x))}{8 d x^8} \]
-1/56*b*c*d^3/x^7+1/40*b*c*d^2*(c^2*d-4*e)/x^5-1/24*b*c*d*(c^4*d^2-4*c^2*d *e+6*e^2)/x^3+1/8*b*c*(c^2*d-2*e)*(c^4*d^2-2*c^2*d*e+2*e^2)/x+1/8*b*(c^2*d -e)^4*arctan(c*x)/d-1/8*(e*x^2+d)^4*(a+b*arctan(c*x))/d/x^8
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.01 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^9} \, dx=-\frac {5 b c d^3 x \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},1,-\frac {5}{2},-c^2 x^2\right )+28 b c d^2 e x^3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-c^2 x^2\right )+35 \left (\left (d^3+4 d^2 e x^2+6 d e^2 x^4+4 e^3 x^6\right ) (a+b \arctan (c x))+2 b c d e^2 x^5 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-c^2 x^2\right )+4 b c e^3 x^7 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-c^2 x^2\right )\right )}{280 x^8} \]
-1/280*(5*b*c*d^3*x*Hypergeometric2F1[-7/2, 1, -5/2, -(c^2*x^2)] + 28*b*c* d^2*e*x^3*Hypergeometric2F1[-5/2, 1, -3/2, -(c^2*x^2)] + 35*((d^3 + 4*d^2* e*x^2 + 6*d*e^2*x^4 + 4*e^3*x^6)*(a + b*ArcTan[c*x]) + 2*b*c*d*e^2*x^5*Hyp ergeometric2F1[-3/2, 1, -1/2, -(c^2*x^2)] + 4*b*c*e^3*x^7*Hypergeometric2F 1[-1/2, 1, 1/2, -(c^2*x^2)]))/x^8
Time = 0.38 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5511, 27, 364, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^9} \, dx\) |
| \(\Big \downarrow \) 5511 |
| \(\displaystyle -b c \int -\frac {\left (e x^2+d\right )^4}{8 d x^8 \left (c^2 x^2+1\right )}dx-\frac {\left (d+e x^2\right )^4 (a+b \arctan (c x))}{8 d x^8}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c \int \frac {\left (e x^2+d\right )^4}{x^8 \left (c^2 x^2+1\right )}dx}{8 d}-\frac {\left (d+e x^2\right )^4 (a+b \arctan (c x))}{8 d x^8}\) |
| \(\Big \downarrow \) 364 |
| \(\displaystyle \frac {b c \int \left (\frac {d^4}{x^8}-\frac {\left (c^2 d-4 e\right ) d^3}{x^6}+\frac {\left (d^2 c^4-4 d e c^2+6 e^2\right ) d^2}{x^4}+\frac {\left (c^2 d-2 e\right ) \left (-d^2 c^4+2 d e c^2-2 e^2\right ) d}{x^2}+\frac {\left (c^2 d-e\right )^4}{c^2 x^2+1}\right )dx}{8 d}-\frac {\left (d+e x^2\right )^4 (a+b \arctan (c x))}{8 d x^8}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b c \left (\frac {\arctan (c x) \left (c^2 d-e\right )^4}{c}+\frac {d^3 \left (c^2 d-4 e\right )}{5 x^5}-\frac {d^2 \left (c^4 d^2-4 c^2 d e+6 e^2\right )}{3 x^3}+\frac {d \left (c^2 d-2 e\right ) \left (c^4 d^2-2 c^2 d e+2 e^2\right )}{x}-\frac {d^4}{7 x^7}\right )}{8 d}-\frac {\left (d+e x^2\right )^4 (a+b \arctan (c x))}{8 d x^8}\) |
-1/8*((d + e*x^2)^4*(a + b*ArcTan[c*x]))/(d*x^8) + (b*c*(-1/7*d^4/x^7 + (d ^3*(c^2*d - 4*e))/(5*x^5) - (d^2*(c^4*d^2 - 4*c^2*d*e + 6*e^2))/(3*x^3) + (d*(c^2*d - 2*e)*(c^4*d^2 - 2*c^2*d*e + 2*e^2))/x + ((c^2*d - e)^4*ArcTan[ c*x])/c))/(8*d)
3.12.49.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x _)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Sim p[(a + b*ArcTan[c*x]) u, x] - Simp[b*c Int[SimplifyIntegrand[u/(1 + c^2 *x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] && !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] && !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] && !ILt Q[(m - 1)/2, 0]))
Time = 0.35 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.64
| method | result | size |
| parts | \(a \left (-\frac {d^{2} e}{2 x^{6}}-\frac {3 d \,e^{2}}{4 x^{4}}-\frac {d^{3}}{8 x^{8}}-\frac {e^{3}}{2 x^{2}}\right )+b \,c^{8} \left (-\frac {\arctan \left (c x \right ) d^{2} e}{2 c^{8} x^{6}}-\frac {3 \arctan \left (c x \right ) e^{2} d}{4 c^{8} x^{4}}-\frac {\arctan \left (c x \right ) d^{3}}{8 c^{8} x^{8}}-\frac {\arctan \left (c x \right ) e^{3}}{2 c^{8} x^{2}}-\frac {-\frac {c^{6} d^{3}-4 c^{4} d^{2} e +6 e^{2} d \,c^{2}-4 e^{3}}{c x}-\frac {d^{2} \left (c^{2} d -4 e \right )}{5 c \,x^{5}}+\frac {d^{3}}{7 c \,x^{7}}+\frac {d \left (c^{4} d^{2}-4 c^{2} d e +6 e^{2}\right )}{3 c \,x^{3}}+\left (-c^{6} d^{3}+4 c^{4} d^{2} e -6 e^{2} d \,c^{2}+4 e^{3}\right ) \arctan \left (c x \right )}{8 c^{6}}\right )\) | \(249\) |
| derivativedivides | \(c^{8} \left (\frac {a \left (-\frac {3 d \,e^{2}}{4 c^{2} x^{4}}-\frac {d^{2} e}{2 c^{2} x^{6}}-\frac {e^{3}}{2 c^{2} x^{2}}-\frac {d^{3}}{8 c^{2} x^{8}}\right )}{c^{6}}+\frac {b \left (-\frac {3 \arctan \left (c x \right ) d \,e^{2}}{4 c^{2} x^{4}}-\frac {\arctan \left (c x \right ) d^{2} e}{2 c^{2} x^{6}}-\frac {\arctan \left (c x \right ) e^{3}}{2 c^{2} x^{2}}-\frac {\arctan \left (c x \right ) d^{3}}{8 c^{2} x^{8}}-\frac {\left (-c^{6} d^{3}+4 c^{4} d^{2} e -6 e^{2} d \,c^{2}+4 e^{3}\right ) \arctan \left (c x \right )}{8}+\frac {c^{6} d^{3}-4 c^{4} d^{2} e +6 e^{2} d \,c^{2}-4 e^{3}}{8 c x}+\frac {d^{2} \left (c^{2} d -4 e \right )}{40 c \,x^{5}}-\frac {d^{3}}{56 c \,x^{7}}-\frac {d \left (c^{4} d^{2}-4 c^{2} d e +6 e^{2}\right )}{24 c \,x^{3}}\right )}{c^{6}}\right )\) | \(263\) |
| default | \(c^{8} \left (\frac {a \left (-\frac {3 d \,e^{2}}{4 c^{2} x^{4}}-\frac {d^{2} e}{2 c^{2} x^{6}}-\frac {e^{3}}{2 c^{2} x^{2}}-\frac {d^{3}}{8 c^{2} x^{8}}\right )}{c^{6}}+\frac {b \left (-\frac {3 \arctan \left (c x \right ) d \,e^{2}}{4 c^{2} x^{4}}-\frac {\arctan \left (c x \right ) d^{2} e}{2 c^{2} x^{6}}-\frac {\arctan \left (c x \right ) e^{3}}{2 c^{2} x^{2}}-\frac {\arctan \left (c x \right ) d^{3}}{8 c^{2} x^{8}}-\frac {\left (-c^{6} d^{3}+4 c^{4} d^{2} e -6 e^{2} d \,c^{2}+4 e^{3}\right ) \arctan \left (c x \right )}{8}+\frac {c^{6} d^{3}-4 c^{4} d^{2} e +6 e^{2} d \,c^{2}-4 e^{3}}{8 c x}+\frac {d^{2} \left (c^{2} d -4 e \right )}{40 c \,x^{5}}-\frac {d^{3}}{56 c \,x^{7}}-\frac {d \left (c^{4} d^{2}-4 c^{2} d e +6 e^{2}\right )}{24 c \,x^{3}}\right )}{c^{6}}\right )\) | \(263\) |
| parallelrisch | \(\frac {105 x^{8} \arctan \left (c x \right ) b \,c^{8} d^{3}-420 x^{8} \arctan \left (c x \right ) b \,c^{6} d^{2} e +105 b \,c^{7} d^{3} x^{7}+630 x^{8} \arctan \left (c x \right ) b \,c^{4} d \,e^{2}-420 b \,c^{5} d^{2} e \,x^{7}-420 x^{8} \arctan \left (c x \right ) b \,c^{2} e^{3}+420 a \,c^{2} e^{3} x^{8}+630 b \,c^{3} d \,e^{2} x^{7}-35 x^{5} c^{5} d^{3} b -420 b c \,e^{3} x^{7}+140 x^{5} c^{3} d^{2} e b -420 x^{6} \arctan \left (c x \right ) b \,e^{3}-420 a \,e^{3} x^{6}-210 b c \,e^{2} d \,x^{5}+21 x^{3} d^{3} c^{3} b -630 x^{4} \arctan \left (c x \right ) b d \,e^{2}-630 a d \,e^{2} x^{4}-84 b c e \,d^{2} x^{3}-420 x^{2} \arctan \left (c x \right ) b \,d^{2} e -420 a \,d^{2} e \,x^{2}-15 b c \,d^{3} x -105 b \,d^{3} \arctan \left (c x \right )-105 d^{3} a}{840 x^{8}}\) | \(286\) |
| risch | \(\frac {i b \left (4 e^{3} x^{6}+6 x^{4} e^{2} d +4 e \,d^{2} x^{2}+d^{3}\right ) \ln \left (i c x +1\right )}{16 x^{8}}-\frac {1260 a d \,e^{2} x^{4}+840 a \,d^{2} e \,x^{2}+840 b c \,e^{3} x^{7}-42 x^{3} d^{3} c^{3} b +70 x^{5} c^{5} d^{3} b +420 i b \,d^{2} e \ln \left (-i c x +1\right ) x^{2}+630 i b \,e^{2} d \ln \left (-i c x +1\right ) x^{4}+30 b c \,d^{3} x +210 d^{3} a -105 i \ln \left (-c x -i\right ) b \,c^{8} d^{3} x^{8}+105 i \ln \left (-c x +i\right ) b \,c^{8} d^{3} x^{8}+420 i \ln \left (-c x -i\right ) b \,c^{2} e^{3} x^{8}-420 i \ln \left (-c x +i\right ) b \,c^{2} e^{3} x^{8}+840 a \,e^{3} x^{6}-280 x^{5} c^{3} d^{2} e b -210 b \,c^{7} d^{3} x^{7}+840 b \,c^{5} d^{2} e \,x^{7}-1260 b \,c^{3} d \,e^{2} x^{7}+105 i b \,d^{3} \ln \left (-i c x +1\right )+168 b c e \,d^{2} x^{3}+420 b c \,e^{2} d \,x^{5}-630 i \ln \left (-c x -i\right ) b \,c^{4} d \,e^{2} x^{8}+630 i \ln \left (-c x +i\right ) b \,c^{4} d \,e^{2} x^{8}+420 i b \,e^{3} \ln \left (-i c x +1\right ) x^{6}+420 i \ln \left (-c x -i\right ) b \,c^{6} d^{2} e \,x^{8}-420 i \ln \left (-c x +i\right ) b \,c^{6} d^{2} e \,x^{8}}{1680 x^{8}}\) | \(446\) |
a*(-1/2*d^2*e/x^6-3/4*d*e^2/x^4-1/8*d^3/x^8-1/2*e^3/x^2)+b*c^8*(-1/2*arcta n(c*x)/c^8*d^2*e/x^6-3/4*arctan(c*x)/c^8*e^2*d/x^4-1/8*arctan(c*x)*d^3/c^8 /x^8-1/2*arctan(c*x)/c^8*e^3/x^2-1/8/c^6*(-(c^6*d^3-4*c^4*d^2*e+6*c^2*d*e^ 2-4*e^3)/c/x-1/5/c*d^2*(c^2*d-4*e)/x^5+1/7/c*d^3/x^7+1/3*d/c*(c^4*d^2-4*c^ 2*d*e+6*e^2)/x^3+(-c^6*d^3+4*c^4*d^2*e-6*c^2*d*e^2+4*e^3)*arctan(c*x)))
Time = 0.26 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.50 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^9} \, dx=-\frac {420 \, a e^{3} x^{6} + 630 \, a d e^{2} x^{4} - 105 \, {\left (b c^{7} d^{3} - 4 \, b c^{5} d^{2} e + 6 \, b c^{3} d e^{2} - 4 \, b c e^{3}\right )} x^{7} + 15 \, b c d^{3} x + 420 \, a d^{2} e x^{2} + 35 \, {\left (b c^{5} d^{3} - 4 \, b c^{3} d^{2} e + 6 \, b c d e^{2}\right )} x^{5} + 105 \, a d^{3} - 21 \, {\left (b c^{3} d^{3} - 4 \, b c d^{2} e\right )} x^{3} + 105 \, {\left (4 \, b e^{3} x^{6} - {\left (b c^{8} d^{3} - 4 \, b c^{6} d^{2} e + 6 \, b c^{4} d e^{2} - 4 \, b c^{2} e^{3}\right )} x^{8} + 6 \, b d e^{2} x^{4} + 4 \, b d^{2} e x^{2} + b d^{3}\right )} \arctan \left (c x\right )}{840 \, x^{8}} \]
-1/840*(420*a*e^3*x^6 + 630*a*d*e^2*x^4 - 105*(b*c^7*d^3 - 4*b*c^5*d^2*e + 6*b*c^3*d*e^2 - 4*b*c*e^3)*x^7 + 15*b*c*d^3*x + 420*a*d^2*e*x^2 + 35*(b*c ^5*d^3 - 4*b*c^3*d^2*e + 6*b*c*d*e^2)*x^5 + 105*a*d^3 - 21*(b*c^3*d^3 - 4* b*c*d^2*e)*x^3 + 105*(4*b*e^3*x^6 - (b*c^8*d^3 - 4*b*c^6*d^2*e + 6*b*c^4*d *e^2 - 4*b*c^2*e^3)*x^8 + 6*b*d*e^2*x^4 + 4*b*d^2*e*x^2 + b*d^3)*arctan(c* x))/x^8
Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (139) = 278\).
Time = 0.56 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.03 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^9} \, dx=- \frac {a d^{3}}{8 x^{8}} - \frac {a d^{2} e}{2 x^{6}} - \frac {3 a d e^{2}}{4 x^{4}} - \frac {a e^{3}}{2 x^{2}} + \frac {b c^{8} d^{3} \operatorname {atan}{\left (c x \right )}}{8} + \frac {b c^{7} d^{3}}{8 x} - \frac {b c^{6} d^{2} e \operatorname {atan}{\left (c x \right )}}{2} - \frac {b c^{5} d^{3}}{24 x^{3}} - \frac {b c^{5} d^{2} e}{2 x} + \frac {3 b c^{4} d e^{2} \operatorname {atan}{\left (c x \right )}}{4} + \frac {b c^{3} d^{3}}{40 x^{5}} + \frac {b c^{3} d^{2} e}{6 x^{3}} + \frac {3 b c^{3} d e^{2}}{4 x} - \frac {b c^{2} e^{3} \operatorname {atan}{\left (c x \right )}}{2} - \frac {b c d^{3}}{56 x^{7}} - \frac {b c d^{2} e}{10 x^{5}} - \frac {b c d e^{2}}{4 x^{3}} - \frac {b c e^{3}}{2 x} - \frac {b d^{3} \operatorname {atan}{\left (c x \right )}}{8 x^{8}} - \frac {b d^{2} e \operatorname {atan}{\left (c x \right )}}{2 x^{6}} - \frac {3 b d e^{2} \operatorname {atan}{\left (c x \right )}}{4 x^{4}} - \frac {b e^{3} \operatorname {atan}{\left (c x \right )}}{2 x^{2}} \]
-a*d**3/(8*x**8) - a*d**2*e/(2*x**6) - 3*a*d*e**2/(4*x**4) - a*e**3/(2*x** 2) + b*c**8*d**3*atan(c*x)/8 + b*c**7*d**3/(8*x) - b*c**6*d**2*e*atan(c*x) /2 - b*c**5*d**3/(24*x**3) - b*c**5*d**2*e/(2*x) + 3*b*c**4*d*e**2*atan(c* x)/4 + b*c**3*d**3/(40*x**5) + b*c**3*d**2*e/(6*x**3) + 3*b*c**3*d*e**2/(4 *x) - b*c**2*e**3*atan(c*x)/2 - b*c*d**3/(56*x**7) - b*c*d**2*e/(10*x**5) - b*c*d*e**2/(4*x**3) - b*c*e**3/(2*x) - b*d**3*atan(c*x)/(8*x**8) - b*d** 2*e*atan(c*x)/(2*x**6) - 3*b*d*e**2*atan(c*x)/(4*x**4) - b*e**3*atan(c*x)/ (2*x**2)
Time = 0.29 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.43 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^9} \, dx=\frac {1}{840} \, {\left ({\left (105 \, c^{7} \arctan \left (c x\right ) + \frac {105 \, c^{6} x^{6} - 35 \, c^{4} x^{4} + 21 \, c^{2} x^{2} - 15}{x^{7}}\right )} c - \frac {105 \, \arctan \left (c x\right )}{x^{8}}\right )} b d^{3} - \frac {1}{30} \, {\left ({\left (15 \, c^{5} \arctan \left (c x\right ) + \frac {15 \, c^{4} x^{4} - 5 \, c^{2} x^{2} + 3}{x^{5}}\right )} c + \frac {15 \, \arctan \left (c x\right )}{x^{6}}\right )} b d^{2} e + \frac {1}{4} \, {\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac {3 \, \arctan \left (c x\right )}{x^{4}}\right )} b d e^{2} - \frac {1}{2} \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b e^{3} - \frac {a e^{3}}{2 \, x^{2}} - \frac {3 \, a d e^{2}}{4 \, x^{4}} - \frac {a d^{2} e}{2 \, x^{6}} - \frac {a d^{3}}{8 \, x^{8}} \]
1/840*((105*c^7*arctan(c*x) + (105*c^6*x^6 - 35*c^4*x^4 + 21*c^2*x^2 - 15) /x^7)*c - 105*arctan(c*x)/x^8)*b*d^3 - 1/30*((15*c^5*arctan(c*x) + (15*c^4 *x^4 - 5*c^2*x^2 + 3)/x^5)*c + 15*arctan(c*x)/x^6)*b*d^2*e + 1/4*((3*c^3*a rctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*b*d*e^2 - 1/2*((c *arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*b*e^3 - 1/2*a*e^3/x^2 - 3/4*a*d*e ^2/x^4 - 1/2*a*d^2*e/x^6 - 1/8*a*d^3/x^8
\[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^9} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{9}} \,d x } \]
Time = 0.73 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.98 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^9} \, dx=\frac {b\,c^2\,\mathrm {atan}\left (\frac {b\,c^2\,x\,\left (2\,e-c^2\,d\right )\,\left (c^4\,d^2-2\,c^2\,d\,e+2\,e^2\right )}{b\,c^7\,d^3-4\,b\,c^5\,d^2\,e+6\,b\,c^3\,d\,e^2-4\,b\,c\,e^3}\right )\,\left (2\,e-c^2\,d\right )\,\left (c^4\,d^2-2\,c^2\,d\,e+2\,e^2\right )}{8}-\frac {\mathrm {atan}\left (c\,x\right )\,\left (\frac {b\,d^3}{8}+\frac {b\,d^2\,e\,x^2}{2}+\frac {3\,b\,d\,e^2\,x^4}{4}+\frac {b\,e^3\,x^6}{2}\right )}{x^8}-\frac {a\,d^3-x^3\,\left (\frac {b\,c^3\,d^3}{5}-\frac {4\,b\,c\,d^2\,e}{5}\right )-x^7\,\left (b\,c^7\,d^3-4\,b\,c^5\,d^2\,e+6\,b\,c^3\,d\,e^2-4\,b\,c\,e^3\right )+x^5\,\left (\frac {b\,c^5\,d^3}{3}-\frac {4\,b\,c^3\,d^2\,e}{3}+2\,b\,c\,d\,e^2\right )+4\,a\,e^3\,x^6+\frac {b\,c\,d^3\,x}{7}+4\,a\,d^2\,e\,x^2+6\,a\,d\,e^2\,x^4}{8\,x^8} \]
(b*c^2*atan((b*c^2*x*(2*e - c^2*d)*(2*e^2 + c^4*d^2 - 2*c^2*d*e))/(b*c^7*d ^3 - 4*b*c*e^3 + 6*b*c^3*d*e^2 - 4*b*c^5*d^2*e))*(2*e - c^2*d)*(2*e^2 + c^ 4*d^2 - 2*c^2*d*e))/8 - (atan(c*x)*((b*d^3)/8 + (b*e^3*x^6)/2 + (b*d^2*e*x ^2)/2 + (3*b*d*e^2*x^4)/4))/x^8 - (a*d^3 - x^3*((b*c^3*d^3)/5 - (4*b*c*d^2 *e)/5) - x^7*(b*c^7*d^3 - 4*b*c*e^3 + 6*b*c^3*d*e^2 - 4*b*c^5*d^2*e) + x^5 *((b*c^5*d^3)/3 + 2*b*c*d*e^2 - (4*b*c^3*d^2*e)/3) + 4*a*e^3*x^6 + (b*c*d^ 3*x)/7 + 4*a*d^2*e*x^2 + 6*a*d*e^2*x^4)/(8*x^8)