Integrand size = 21, antiderivative size = 186 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=-\frac {b c d^2}{42 x^6}+\frac {b c d \left (5 c^2 d-14 e\right )}{140 x^4}-\frac {b c \left (15 c^4 d^2-42 c^2 d e+35 e^2\right )}{210 x^2}-\frac {d^2 (a+b \arctan (c x))}{7 x^7}-\frac {2 d e (a+b \arctan (c x))}{5 x^5}-\frac {e^2 (a+b \arctan (c x))}{3 x^3}-\frac {1}{105} b c^3 \left (15 c^4 d^2-42 c^2 d e+35 e^2\right ) \log (x)+\frac {1}{210} b c^3 \left (15 c^4 d^2-42 c^2 d e+35 e^2\right ) \log \left (1+c^2 x^2\right ) \]
-1/42*b*c*d^2/x^6+1/140*b*c*d*(5*c^2*d-14*e)/x^4-1/210*b*c*(15*c^4*d^2-42* c^2*d*e+35*e^2)/x^2-1/7*d^2*(a+b*arctan(c*x))/x^7-2/5*d*e*(a+b*arctan(c*x) )/x^5-1/3*e^2*(a+b*arctan(c*x))/x^3-1/105*b*c^3*(15*c^4*d^2-42*c^2*d*e+35* e^2)*ln(x)+1/210*b*c^3*(15*c^4*d^2-42*c^2*d*e+35*e^2)*ln(c^2*x^2+1)
Time = 0.14 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.95 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=\frac {1}{420} \left (-\frac {60 d^2 (a+b \arctan (c x))}{x^7}-\frac {168 d e (a+b \arctan (c x))}{x^5}-\frac {140 e^2 (a+b \arctan (c x))}{x^3}-70 b c e^2 \left (\frac {1}{x^2}+2 c^2 \log (x)-c^2 \log \left (1+c^2 x^2\right )\right )-42 b c d e \left (\frac {1}{x^4}-\frac {2 c^2}{x^2}-4 c^4 \log (x)+2 c^4 \log \left (1+c^2 x^2\right )\right )-5 b c d^2 \left (\frac {2-3 c^2 x^2+6 c^4 x^4}{x^6}+12 c^6 \log (x)-6 c^6 \log \left (1+c^2 x^2\right )\right )\right ) \]
((-60*d^2*(a + b*ArcTan[c*x]))/x^7 - (168*d*e*(a + b*ArcTan[c*x]))/x^5 - ( 140*e^2*(a + b*ArcTan[c*x]))/x^3 - 70*b*c*e^2*(x^(-2) + 2*c^2*Log[x] - c^2 *Log[1 + c^2*x^2]) - 42*b*c*d*e*(x^(-4) - (2*c^2)/x^2 - 4*c^4*Log[x] + 2*c ^4*Log[1 + c^2*x^2]) - 5*b*c*d^2*((2 - 3*c^2*x^2 + 6*c^4*x^4)/x^6 + 12*c^6 *Log[x] - 6*c^6*Log[1 + c^2*x^2]))/420
Time = 0.45 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5511, 27, 1578, 1195, 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.
\(\displaystyle \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx\) |
\(\Big \downarrow \) 5511 |
\(\displaystyle -b c \int -\frac {35 e^2 x^4+42 d e x^2+15 d^2}{105 x^7 \left (c^2 x^2+1\right )}dx-\frac {d^2 (a+b \arctan (c x))}{7 x^7}-\frac {2 d e (a+b \arctan (c x))}{5 x^5}-\frac {e^2 (a+b \arctan (c x))}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{105} b c \int \frac {35 e^2 x^4+42 d e x^2+15 d^2}{x^7 \left (c^2 x^2+1\right )}dx-\frac {d^2 (a+b \arctan (c x))}{7 x^7}-\frac {2 d e (a+b \arctan (c x))}{5 x^5}-\frac {e^2 (a+b \arctan (c x))}{3 x^3}\) |
\(\Big \downarrow \) 1578 |
\(\displaystyle \frac {1}{210} b c \int \frac {35 e^2 x^4+42 d e x^2+15 d^2}{x^8 \left (c^2 x^2+1\right )}dx^2-\frac {d^2 (a+b \arctan (c x))}{7 x^7}-\frac {2 d e (a+b \arctan (c x))}{5 x^5}-\frac {e^2 (a+b \arctan (c x))}{3 x^3}\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \frac {1}{210} b c \int \left (\frac {15 d^2}{x^8}-\frac {3 \left (5 c^2 d-14 e\right ) d}{x^6}+\frac {15 d^2 c^8-42 d e c^6+35 e^2 c^4}{c^2 x^2+1}+\frac {-15 d^2 c^6+42 d e c^4-35 e^2 c^2}{x^2}+\frac {15 d^2 c^4-42 d e c^2+35 e^2}{x^4}\right )dx^2-\frac {d^2 (a+b \arctan (c x))}{7 x^7}-\frac {2 d e (a+b \arctan (c x))}{5 x^5}-\frac {e^2 (a+b \arctan (c x))}{3 x^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {d^2 (a+b \arctan (c x))}{7 x^7}-\frac {2 d e (a+b \arctan (c x))}{5 x^5}-\frac {e^2 (a+b \arctan (c x))}{3 x^3}+\frac {1}{210} b c \left (\frac {3 d \left (5 c^2 d-14 e\right )}{2 x^4}-\frac {15 c^4 d^2-42 c^2 d e+35 e^2}{x^2}-\left (c^2 \log \left (x^2\right ) \left (15 c^4 d^2-42 c^2 d e+35 e^2\right )\right )+c^2 \left (15 c^4 d^2-42 c^2 d e+35 e^2\right ) \log \left (c^2 x^2+1\right )-\frac {5 d^2}{x^6}\right )\) |
-1/7*(d^2*(a + b*ArcTan[c*x]))/x^7 - (2*d*e*(a + b*ArcTan[c*x]))/(5*x^5) - (e^2*(a + b*ArcTan[c*x]))/(3*x^3) + (b*c*((-5*d^2)/x^6 + (3*d*(5*c^2*d - 14*e))/(2*x^4) - (15*c^4*d^2 - 42*c^2*d*e + 35*e^2)/x^2 - c^2*(15*c^4*d^2 - 42*c^2*d*e + 35*e^2)*Log[x^2] + c^2*(15*c^4*d^2 - 42*c^2*d*e + 35*e^2)*L og[1 + c^2*x^2]))/210
3.12.36.3.1 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[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[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] && Int egerQ[(m - 1)/2]
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 = 0.26 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.09
method | result | size |
parts | \(a \left (-\frac {d^{2}}{7 x^{7}}-\frac {e^{2}}{3 x^{3}}-\frac {2 e d}{5 x^{5}}\right )+b \,c^{7} \left (-\frac {\arctan \left (c x \right ) d^{2}}{7 c^{7} x^{7}}-\frac {\arctan \left (c x \right ) e^{2}}{3 c^{7} x^{3}}-\frac {2 \arctan \left (c x \right ) d e}{5 c^{7} x^{5}}-\frac {-\frac {-15 c^{4} d^{2}+42 c^{2} d e -35 e^{2}}{2 c^{2} x^{2}}+\left (15 c^{4} d^{2}-42 c^{2} d e +35 e^{2}\right ) \ln \left (c x \right )+\frac {5 d^{2}}{2 c^{2} x^{6}}-\frac {3 d \left (5 c^{2} d -14 e \right )}{4 c^{2} x^{4}}+\frac {\left (-15 c^{4} d^{2}+42 c^{2} d e -35 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}}{105 c^{4}}\right )\) | \(202\) |
derivativedivides | \(c^{7} \left (\frac {a \left (-\frac {2 d e}{5 c^{3} x^{5}}-\frac {d^{2}}{7 c^{3} x^{7}}-\frac {e^{2}}{3 c^{3} x^{3}}\right )}{c^{4}}+\frac {b \left (-\frac {2 \arctan \left (c x \right ) d e}{5 c^{3} x^{5}}-\frac {\arctan \left (c x \right ) d^{2}}{7 c^{3} x^{7}}-\frac {\arctan \left (c x \right ) e^{2}}{3 c^{3} x^{3}}-\frac {\left (-15 c^{4} d^{2}+42 c^{2} d e -35 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{210}+\frac {-15 c^{4} d^{2}+42 c^{2} d e -35 e^{2}}{210 c^{2} x^{2}}-\frac {\left (15 c^{4} d^{2}-42 c^{2} d e +35 e^{2}\right ) \ln \left (c x \right )}{105}-\frac {d^{2}}{42 c^{2} x^{6}}+\frac {d \left (5 c^{2} d -14 e \right )}{140 c^{2} x^{4}}\right )}{c^{4}}\right )\) | \(213\) |
default | \(c^{7} \left (\frac {a \left (-\frac {2 d e}{5 c^{3} x^{5}}-\frac {d^{2}}{7 c^{3} x^{7}}-\frac {e^{2}}{3 c^{3} x^{3}}\right )}{c^{4}}+\frac {b \left (-\frac {2 \arctan \left (c x \right ) d e}{5 c^{3} x^{5}}-\frac {\arctan \left (c x \right ) d^{2}}{7 c^{3} x^{7}}-\frac {\arctan \left (c x \right ) e^{2}}{3 c^{3} x^{3}}-\frac {\left (-15 c^{4} d^{2}+42 c^{2} d e -35 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{210}+\frac {-15 c^{4} d^{2}+42 c^{2} d e -35 e^{2}}{210 c^{2} x^{2}}-\frac {\left (15 c^{4} d^{2}-42 c^{2} d e +35 e^{2}\right ) \ln \left (c x \right )}{105}-\frac {d^{2}}{42 c^{2} x^{6}}+\frac {d \left (5 c^{2} d -14 e \right )}{140 c^{2} x^{4}}\right )}{c^{4}}\right )\) | \(213\) |
parallelrisch | \(-\frac {60 \ln \left (x \right ) b \,c^{7} d^{2} x^{7}-30 \ln \left (c^{2} x^{2}+1\right ) b \,c^{7} d^{2} x^{7}-30 b \,c^{7} d^{2} x^{7}-168 \ln \left (x \right ) b \,c^{5} d e \,x^{7}+84 \ln \left (c^{2} x^{2}+1\right ) b \,c^{5} d e \,x^{7}+84 b \,c^{5} d e \,x^{7}+140 \ln \left (x \right ) b \,c^{3} e^{2} x^{7}-70 \ln \left (c^{2} x^{2}+1\right ) b \,c^{3} e^{2} x^{7}-70 b \,c^{3} e^{2} x^{7}+30 b \,c^{5} d^{2} x^{5}-84 b \,c^{3} d e \,x^{5}+70 b c \,e^{2} x^{5}-15 x^{3} b \,c^{3} d^{2}+140 x^{4} \arctan \left (c x \right ) b \,e^{2}+140 a \,e^{2} x^{4}+42 b c e d \,x^{3}+168 x^{2} \arctan \left (c x \right ) b d e +168 a d e \,x^{2}+10 b c \,d^{2} x +60 b \,d^{2} \arctan \left (c x \right )+60 d^{2} a}{420 x^{7}}\) | \(268\) |
risch | \(\frac {i b \left (35 x^{4} e^{2}+42 x^{2} e d +15 d^{2}\right ) \ln \left (i c x +1\right )}{210 x^{7}}-\frac {60 \ln \left (x \right ) b \,c^{7} d^{2} x^{7}-30 \ln \left (c^{2} x^{2}+1\right ) b \,c^{7} d^{2} x^{7}-168 \ln \left (x \right ) b \,c^{5} d e \,x^{7}+84 \ln \left (c^{2} x^{2}+1\right ) b \,c^{5} d e \,x^{7}+140 \ln \left (x \right ) b \,c^{3} e^{2} x^{7}-70 \ln \left (c^{2} x^{2}+1\right ) b \,c^{3} e^{2} x^{7}+30 b \,c^{5} d^{2} x^{5}-84 b \,c^{3} d e \,x^{5}+70 i b \,e^{2} \ln \left (-i c x +1\right ) x^{4}-15 x^{3} b \,c^{3} d^{2}+70 b c \,e^{2} x^{5}+30 i b \,d^{2} \ln \left (-i c x +1\right )+140 a \,e^{2} x^{4}+42 b c e d \,x^{3}+84 i b d e \ln \left (-i c x +1\right ) x^{2}+168 a d e \,x^{2}+10 b c \,d^{2} x +60 d^{2} a}{420 x^{7}}\) | \(285\) |
a*(-1/7*d^2/x^7-1/3*e^2/x^3-2/5*e*d/x^5)+b*c^7*(-1/7*arctan(c*x)*d^2/c^7/x ^7-1/3*arctan(c*x)/c^7*e^2/x^3-2/5*arctan(c*x)/c^7*d*e/x^5-1/105/c^4*(-1/2 *(-15*c^4*d^2+42*c^2*d*e-35*e^2)/c^2/x^2+(15*c^4*d^2-42*c^2*d*e+35*e^2)*ln (c*x)+5/2/c^2*d^2/x^6-3/4*d/c^2*(5*c^2*d-14*e)/x^4+1/2*(-15*c^4*d^2+42*c^2 *d*e-35*e^2)*ln(c^2*x^2+1)))
Time = 0.27 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.04 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=\frac {2 \, {\left (15 \, b c^{7} d^{2} - 42 \, b c^{5} d e + 35 \, b c^{3} e^{2}\right )} x^{7} \log \left (c^{2} x^{2} + 1\right ) - 4 \, {\left (15 \, b c^{7} d^{2} - 42 \, b c^{5} d e + 35 \, b c^{3} e^{2}\right )} x^{7} \log \left (x\right ) - 140 \, a e^{2} x^{4} - 2 \, {\left (15 \, b c^{5} d^{2} - 42 \, b c^{3} d e + 35 \, b c e^{2}\right )} x^{5} - 10 \, b c d^{2} x - 168 \, a d e x^{2} + 3 \, {\left (5 \, b c^{3} d^{2} - 14 \, b c d e\right )} x^{3} - 60 \, a d^{2} - 4 \, {\left (35 \, b e^{2} x^{4} + 42 \, b d e x^{2} + 15 \, b d^{2}\right )} \arctan \left (c x\right )}{420 \, x^{7}} \]
1/420*(2*(15*b*c^7*d^2 - 42*b*c^5*d*e + 35*b*c^3*e^2)*x^7*log(c^2*x^2 + 1) - 4*(15*b*c^7*d^2 - 42*b*c^5*d*e + 35*b*c^3*e^2)*x^7*log(x) - 140*a*e^2*x ^4 - 2*(15*b*c^5*d^2 - 42*b*c^3*d*e + 35*b*c*e^2)*x^5 - 10*b*c*d^2*x - 168 *a*d*e*x^2 + 3*(5*b*c^3*d^2 - 14*b*c*d*e)*x^3 - 60*a*d^2 - 4*(35*b*e^2*x^4 + 42*b*d*e*x^2 + 15*b*d^2)*arctan(c*x))/x^7
Time = 0.72 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.55 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=\begin {cases} - \frac {a d^{2}}{7 x^{7}} - \frac {2 a d e}{5 x^{5}} - \frac {a e^{2}}{3 x^{3}} - \frac {b c^{7} d^{2} \log {\left (x \right )}}{7} + \frac {b c^{7} d^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{14} - \frac {b c^{5} d^{2}}{14 x^{2}} + \frac {2 b c^{5} d e \log {\left (x \right )}}{5} - \frac {b c^{5} d e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{5} + \frac {b c^{3} d^{2}}{28 x^{4}} + \frac {b c^{3} d e}{5 x^{2}} - \frac {b c^{3} e^{2} \log {\left (x \right )}}{3} + \frac {b c^{3} e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6} - \frac {b c d^{2}}{42 x^{6}} - \frac {b c d e}{10 x^{4}} - \frac {b c e^{2}}{6 x^{2}} - \frac {b d^{2} \operatorname {atan}{\left (c x \right )}}{7 x^{7}} - \frac {2 b d e \operatorname {atan}{\left (c x \right )}}{5 x^{5}} - \frac {b e^{2} \operatorname {atan}{\left (c x \right )}}{3 x^{3}} & \text {for}\: c \neq 0 \\a \left (- \frac {d^{2}}{7 x^{7}} - \frac {2 d e}{5 x^{5}} - \frac {e^{2}}{3 x^{3}}\right ) & \text {otherwise} \end {cases} \]
Piecewise((-a*d**2/(7*x**7) - 2*a*d*e/(5*x**5) - a*e**2/(3*x**3) - b*c**7* d**2*log(x)/7 + b*c**7*d**2*log(x**2 + c**(-2))/14 - b*c**5*d**2/(14*x**2) + 2*b*c**5*d*e*log(x)/5 - b*c**5*d*e*log(x**2 + c**(-2))/5 + b*c**3*d**2/ (28*x**4) + b*c**3*d*e/(5*x**2) - b*c**3*e**2*log(x)/3 + b*c**3*e**2*log(x **2 + c**(-2))/6 - b*c*d**2/(42*x**6) - b*c*d*e/(10*x**4) - b*c*e**2/(6*x* *2) - b*d**2*atan(c*x)/(7*x**7) - 2*b*d*e*atan(c*x)/(5*x**5) - b*e**2*atan (c*x)/(3*x**3), Ne(c, 0)), (a*(-d**2/(7*x**7) - 2*d*e/(5*x**5) - e**2/(3*x **3)), True))
Time = 0.21 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.06 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=\frac {1}{84} \, {\left ({\left (6 \, c^{6} \log \left (c^{2} x^{2} + 1\right ) - 6 \, c^{6} \log \left (x^{2}\right ) - \frac {6 \, c^{4} x^{4} - 3 \, c^{2} x^{2} + 2}{x^{6}}\right )} c - \frac {12 \, \arctan \left (c x\right )}{x^{7}}\right )} b d^{2} - \frac {1}{10} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d e + \frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b e^{2} - \frac {a e^{2}}{3 \, x^{3}} - \frac {2 \, a d e}{5 \, x^{5}} - \frac {a d^{2}}{7 \, x^{7}} \]
1/84*((6*c^6*log(c^2*x^2 + 1) - 6*c^6*log(x^2) - (6*c^4*x^4 - 3*c^2*x^2 + 2)/x^6)*c - 12*arctan(c*x)/x^7)*b*d^2 - 1/10*((2*c^4*log(c^2*x^2 + 1) - 2* c^4*log(x^2) - (2*c^2*x^2 - 1)/x^4)*c + 4*arctan(c*x)/x^5)*b*d*e + 1/6*((c ^2*log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3)*b*e^2 - 1/3*a*e^2/x^3 - 2/5*a*d*e/x^5 - 1/7*a*d^2/x^7
\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{8}} \,d x } \]
Time = 0.71 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.25 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=-\frac {60\,a\,d^2+60\,b\,d^2\,\mathrm {atan}\left (c\,x\right )+140\,a\,e^2\,x^4-15\,b\,c^3\,d^2\,x^3+30\,b\,c^5\,d^2\,x^5+10\,b\,c\,d^2\,x+168\,a\,d\,e\,x^2+70\,b\,c\,e^2\,x^5+140\,b\,e^2\,x^4\,\mathrm {atan}\left (c\,x\right )+60\,b\,c^7\,d^2\,x^7\,\ln \left (x\right )+140\,b\,c^3\,e^2\,x^7\,\ln \left (x\right )-84\,b\,c^3\,d\,e\,x^5+42\,b\,c\,d\,e\,x^3-30\,b\,c^7\,d^2\,x^7\,\ln \left (c^2\,x^2+1\right )-70\,b\,c^3\,e^2\,x^7\,\ln \left (c^2\,x^2+1\right )+168\,b\,d\,e\,x^2\,\mathrm {atan}\left (c\,x\right )-168\,b\,c^5\,d\,e\,x^7\,\ln \left (x\right )+84\,b\,c^5\,d\,e\,x^7\,\ln \left (c^2\,x^2+1\right )}{420\,x^7} \]
-(60*a*d^2 + 60*b*d^2*atan(c*x) + 140*a*e^2*x^4 - 15*b*c^3*d^2*x^3 + 30*b* c^5*d^2*x^5 + 10*b*c*d^2*x + 168*a*d*e*x^2 + 70*b*c*e^2*x^5 + 140*b*e^2*x^ 4*atan(c*x) + 60*b*c^7*d^2*x^7*log(x) + 140*b*c^3*e^2*x^7*log(x) - 84*b*c^ 3*d*e*x^5 + 42*b*c*d*e*x^3 - 30*b*c^7*d^2*x^7*log(c^2*x^2 + 1) - 70*b*c^3* e^2*x^7*log(c^2*x^2 + 1) + 168*b*d*e*x^2*atan(c*x) - 168*b*c^5*d*e*x^7*log (x) + 84*b*c^5*d*e*x^7*log(c^2*x^2 + 1))/(420*x^7)