Integrand size = 23, antiderivative size = 193 \[ \int x^2 (d+i c d x)^4 (a+b \arctan (c x)) \, dx=\frac {5 i b d^4 x}{3 c^2}-\frac {88 b d^4 x^2}{105 c}-\frac {5}{9} i b d^4 x^3+\frac {47}{140} b c d^4 x^4+\frac {2}{15} i b c^2 d^4 x^5-\frac {1}{42} b c^3 d^4 x^6+\frac {i d^4 (1+i c x)^5 (a+b \arctan (c x))}{5 c^3}-\frac {i d^4 (1+i c x)^6 (a+b \arctan (c x))}{3 c^3}+\frac {i d^4 (1+i c x)^7 (a+b \arctan (c x))}{7 c^3}+\frac {176 b d^4 \log (i+c x)}{105 c^3} \]
5/3*I*b*d^4*x/c^2-88/105*b*d^4*x^2/c-5/9*I*b*d^4*x^3+47/140*b*c*d^4*x^4+2/ 15*I*b*c^2*d^4*x^5-1/42*b*c^3*d^4*x^6+1/5*I*d^4*(1+I*c*x)^5*(a+b*arctan(c* x))/c^3-1/3*I*d^4*(1+I*c*x)^6*(a+b*arctan(c*x))/c^3+1/7*I*d^4*(1+I*c*x)^7* (a+b*arctan(c*x))/c^3+176/105*b*d^4*ln(c*x+I)/c^3
Time = 0.08 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.43 \[ \int x^2 (d+i c d x)^4 (a+b \arctan (c x)) \, dx=\frac {5 i b d^4 x}{3 c^2}-\frac {88 b d^4 x^2}{105 c}+\frac {1}{3} a d^4 x^3-\frac {5}{9} i b d^4 x^3+i a c d^4 x^4+\frac {47}{140} b c d^4 x^4-\frac {6}{5} a c^2 d^4 x^5+\frac {2}{15} i b c^2 d^4 x^5-\frac {2}{3} i a c^3 d^4 x^6-\frac {1}{42} b c^3 d^4 x^6+\frac {1}{7} a c^4 d^4 x^7-\frac {5 i b d^4 \arctan (c x)}{3 c^3}+\frac {1}{3} b d^4 x^3 \arctan (c x)+i b c d^4 x^4 \arctan (c x)-\frac {6}{5} b c^2 d^4 x^5 \arctan (c x)-\frac {2}{3} i b c^3 d^4 x^6 \arctan (c x)+\frac {1}{7} b c^4 d^4 x^7 \arctan (c x)+\frac {88 b d^4 \log \left (1+c^2 x^2\right )}{105 c^3} \]
(((5*I)/3)*b*d^4*x)/c^2 - (88*b*d^4*x^2)/(105*c) + (a*d^4*x^3)/3 - ((5*I)/ 9)*b*d^4*x^3 + I*a*c*d^4*x^4 + (47*b*c*d^4*x^4)/140 - (6*a*c^2*d^4*x^5)/5 + ((2*I)/15)*b*c^2*d^4*x^5 - ((2*I)/3)*a*c^3*d^4*x^6 - (b*c^3*d^4*x^6)/42 + (a*c^4*d^4*x^7)/7 - (((5*I)/3)*b*d^4*ArcTan[c*x])/c^3 + (b*d^4*x^3*ArcTa n[c*x])/3 + I*b*c*d^4*x^4*ArcTan[c*x] - (6*b*c^2*d^4*x^5*ArcTan[c*x])/5 - ((2*I)/3)*b*c^3*d^4*x^6*ArcTan[c*x] + (b*c^4*d^4*x^7*ArcTan[c*x])/7 + (88* b*d^4*Log[1 + c^2*x^2])/(105*c^3)
Time = 0.39 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {5407, 27, 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.
\(\displaystyle \int x^2 (d+i c d x)^4 (a+b \arctan (c x)) \, dx\) |
\(\Big \downarrow \) 5407 |
\(\displaystyle -b c \int -\frac {d^4 (i-c x)^4 \left (-15 c^2 x^2-5 i c x+1\right )}{105 c^3 (c x+i)}dx+\frac {i d^4 (1+i c x)^7 (a+b \arctan (c x))}{7 c^3}-\frac {i d^4 (1+i c x)^6 (a+b \arctan (c x))}{3 c^3}+\frac {i d^4 (1+i c x)^5 (a+b \arctan (c x))}{5 c^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b d^4 \int \frac {(i-c x)^4 \left (-15 c^2 x^2-5 i c x+1\right )}{c x+i}dx}{105 c^2}+\frac {i d^4 (1+i c x)^7 (a+b \arctan (c x))}{7 c^3}-\frac {i d^4 (1+i c x)^6 (a+b \arctan (c x))}{3 c^3}+\frac {i d^4 (1+i c x)^5 (a+b \arctan (c x))}{5 c^3}\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \frac {b d^4 \int \left (-15 c^5 x^5+70 i c^4 x^4+141 c^3 x^3-175 i c^2 x^2-176 c x+\frac {176}{c x+i}+175 i\right )dx}{105 c^2}+\frac {i d^4 (1+i c x)^7 (a+b \arctan (c x))}{7 c^3}-\frac {i d^4 (1+i c x)^6 (a+b \arctan (c x))}{3 c^3}+\frac {i d^4 (1+i c x)^5 (a+b \arctan (c x))}{5 c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i d^4 (1+i c x)^7 (a+b \arctan (c x))}{7 c^3}-\frac {i d^4 (1+i c x)^6 (a+b \arctan (c x))}{3 c^3}+\frac {i d^4 (1+i c x)^5 (a+b \arctan (c x))}{5 c^3}+\frac {b d^4 \left (-\frac {5}{2} c^5 x^6+14 i c^4 x^5+\frac {141 c^3 x^4}{4}-\frac {175}{3} i c^2 x^3-88 c x^2+\frac {176 \log (c x+i)}{c}+175 i x\right )}{105 c^2}\) |
((I/5)*d^4*(1 + I*c*x)^5*(a + b*ArcTan[c*x]))/c^3 - ((I/3)*d^4*(1 + I*c*x) ^6*(a + b*ArcTan[c*x]))/c^3 + ((I/7)*d^4*(1 + I*c*x)^7*(a + b*ArcTan[c*x]) )/c^3 + (b*d^4*((175*I)*x - 88*c*x^2 - ((175*I)/3)*c^2*x^3 + (141*c^3*x^4) /4 + (14*I)*c^4*x^5 - (5*c^5*x^6)/2 + (176*Log[I + c*x])/c))/(105*c^2)
3.1.32.3.1 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[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x _))^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x)^q, x]}, Simp[(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, q}, x] && NeQ[q, -1] && IntegerQ [2*m] && ((IGtQ[m, 0] && IGtQ[q, 0]) || (ILtQ[m + q + 1, 0] && LtQ[m*q, 0]) )
Time = 1.38 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.94
method | result | size |
parts | \(d^{4} a \left (\frac {1}{7} c^{4} x^{7}-\frac {2}{3} i c^{3} x^{6}-\frac {6}{5} c^{2} x^{5}+i c \,x^{4}+\frac {1}{3} x^{3}\right )+\frac {d^{4} b \left (\frac {\arctan \left (c x \right ) c^{7} x^{7}}{7}-\frac {2 i \arctan \left (c x \right ) c^{6} x^{6}}{3}-\frac {6 c^{5} x^{5} \arctan \left (c x \right )}{5}+i \arctan \left (c x \right ) c^{4} x^{4}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {5 i c x}{3}-\frac {c^{6} x^{6}}{42}+\frac {2 i c^{5} x^{5}}{15}+\frac {47 c^{4} x^{4}}{140}-\frac {5 i c^{3} x^{3}}{9}-\frac {88 c^{2} x^{2}}{105}+\frac {88 \ln \left (c^{2} x^{2}+1\right )}{105}-\frac {5 i \arctan \left (c x \right )}{3}\right )}{c^{3}}\) | \(182\) |
derivativedivides | \(\frac {d^{4} a \left (\frac {1}{7} c^{7} x^{7}-\frac {2}{3} i c^{6} x^{6}-\frac {6}{5} c^{5} x^{5}+i c^{4} x^{4}+\frac {1}{3} c^{3} x^{3}\right )+d^{4} b \left (\frac {\arctan \left (c x \right ) c^{7} x^{7}}{7}-\frac {2 i \arctan \left (c x \right ) c^{6} x^{6}}{3}-\frac {6 c^{5} x^{5} \arctan \left (c x \right )}{5}+i \arctan \left (c x \right ) c^{4} x^{4}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {5 i c x}{3}-\frac {c^{6} x^{6}}{42}+\frac {2 i c^{5} x^{5}}{15}+\frac {47 c^{4} x^{4}}{140}-\frac {5 i c^{3} x^{3}}{9}-\frac {88 c^{2} x^{2}}{105}+\frac {88 \ln \left (c^{2} x^{2}+1\right )}{105}-\frac {5 i \arctan \left (c x \right )}{3}\right )}{c^{3}}\) | \(188\) |
default | \(\frac {d^{4} a \left (\frac {1}{7} c^{7} x^{7}-\frac {2}{3} i c^{6} x^{6}-\frac {6}{5} c^{5} x^{5}+i c^{4} x^{4}+\frac {1}{3} c^{3} x^{3}\right )+d^{4} b \left (\frac {\arctan \left (c x \right ) c^{7} x^{7}}{7}-\frac {2 i \arctan \left (c x \right ) c^{6} x^{6}}{3}-\frac {6 c^{5} x^{5} \arctan \left (c x \right )}{5}+i \arctan \left (c x \right ) c^{4} x^{4}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {5 i c x}{3}-\frac {c^{6} x^{6}}{42}+\frac {2 i c^{5} x^{5}}{15}+\frac {47 c^{4} x^{4}}{140}-\frac {5 i c^{3} x^{3}}{9}-\frac {88 c^{2} x^{2}}{105}+\frac {88 \ln \left (c^{2} x^{2}+1\right )}{105}-\frac {5 i \arctan \left (c x \right )}{3}\right )}{c^{3}}\) | \(188\) |
parallelrisch | \(-\frac {-180 c^{7} b \,d^{4} \arctan \left (c x \right ) x^{7}-2100 i b \,d^{4} x c -180 a \,c^{7} d^{4} x^{7}-168 i x^{5} b \,c^{5} d^{4}+30 b \,c^{6} d^{4} x^{6}-1260 i x^{4} \arctan \left (c x \right ) b \,c^{4} d^{4}+1512 c^{5} b \,d^{4} \arctan \left (c x \right ) x^{5}+2100 i b \,d^{4} \arctan \left (c x \right )+1512 a \,c^{5} d^{4} x^{5}+840 i x^{6} \arctan \left (c x \right ) b \,c^{6} d^{4}-423 b \,c^{4} d^{4} x^{4}+700 i x^{3} b \,c^{3} d^{4}-420 x^{3} \arctan \left (c x \right ) b \,d^{4} c^{3}-420 a \,c^{3} d^{4} x^{3}+1056 b \,c^{2} d^{4} x^{2}-1260 i x^{4} a \,c^{4} d^{4}+840 i x^{6} a \,c^{6} d^{4}-1056 b \,d^{4} \ln \left (c^{2} x^{2}+1\right )}{1260 c^{3}}\) | \(249\) |
risch | \(-\frac {5 i b \,d^{4} x^{3}}{9}-\frac {i d^{4} b \left (15 c^{4} x^{7}-70 i c^{3} x^{6}-126 c^{2} x^{5}+105 i c \,x^{4}+35 x^{3}\right ) \ln \left (i c x +1\right )}{210}+\frac {a \,c^{4} d^{4} x^{7}}{7}+\frac {i d^{4} c^{4} b \,x^{7} \ln \left (-i c x +1\right )}{14}+\frac {d^{4} c^{3} x^{6} b \ln \left (-i c x +1\right )}{3}-\frac {2 i a \,c^{3} d^{4} x^{6}}{3}-\frac {b \,c^{3} d^{4} x^{6}}{42}+\frac {5 i b \,d^{4} x}{3 c^{2}}-\frac {6 a \,c^{2} d^{4} x^{5}}{5}+\frac {2 i b \,c^{2} d^{4} x^{5}}{15}-\frac {d^{4} c \,x^{4} b \ln \left (-i c x +1\right )}{2}+i a c \,d^{4} x^{4}+\frac {47 b c \,d^{4} x^{4}}{140}-\frac {3 i d^{4} c^{2} b \,x^{5} \ln \left (-i c x +1\right )}{5}+\frac {a \,d^{4} x^{3}}{3}-\frac {88 b \,d^{4} x^{2}}{105 c}-\frac {5 i d^{4} b \arctan \left (c x \right )}{3 c^{3}}+\frac {i d^{4} b \,x^{3} \ln \left (-i c x +1\right )}{6}+\frac {88 d^{4} b \ln \left (c^{2} x^{2}+1\right )}{105 c^{3}}\) | \(311\) |
d^4*a*(1/7*c^4*x^7-2/3*I*c^3*x^6-6/5*c^2*x^5+I*c*x^4+1/3*x^3)+d^4*b/c^3*(1 /7*arctan(c*x)*c^7*x^7-2/3*I*arctan(c*x)*c^6*x^6-6/5*c^5*x^5*arctan(c*x)+I *arctan(c*x)*c^4*x^4+1/3*c^3*x^3*arctan(c*x)+5/3*I*c*x-1/42*c^6*x^6+2/15*I *c^5*x^5+47/140*c^4*x^4-5/9*I*c^3*x^3-88/105*c^2*x^2+88/105*ln(c^2*x^2+1)- 5/3*I*arctan(c*x))
Time = 0.25 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.13 \[ \int x^2 (d+i c d x)^4 (a+b \arctan (c x)) \, dx=\frac {180 \, a c^{7} d^{4} x^{7} - 30 \, {\left (28 i \, a + b\right )} c^{6} d^{4} x^{6} - 168 \, {\left (9 \, a - i \, b\right )} c^{5} d^{4} x^{5} - 9 \, {\left (-140 i \, a - 47 \, b\right )} c^{4} d^{4} x^{4} + 140 \, {\left (3 \, a - 5 i \, b\right )} c^{3} d^{4} x^{3} - 1056 \, b c^{2} d^{4} x^{2} + 2100 i \, b c d^{4} x + 2106 \, b d^{4} \log \left (\frac {c x + i}{c}\right ) + 6 \, b d^{4} \log \left (\frac {c x - i}{c}\right ) - 6 \, {\left (-15 i \, b c^{7} d^{4} x^{7} - 70 \, b c^{6} d^{4} x^{6} + 126 i \, b c^{5} d^{4} x^{5} + 105 \, b c^{4} d^{4} x^{4} - 35 i \, b c^{3} d^{4} x^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{1260 \, c^{3}} \]
1/1260*(180*a*c^7*d^4*x^7 - 30*(28*I*a + b)*c^6*d^4*x^6 - 168*(9*a - I*b)* c^5*d^4*x^5 - 9*(-140*I*a - 47*b)*c^4*d^4*x^4 + 140*(3*a - 5*I*b)*c^3*d^4* x^3 - 1056*b*c^2*d^4*x^2 + 2100*I*b*c*d^4*x + 2106*b*d^4*log((c*x + I)/c) + 6*b*d^4*log((c*x - I)/c) - 6*(-15*I*b*c^7*d^4*x^7 - 70*b*c^6*d^4*x^6 + 1 26*I*b*c^5*d^4*x^5 + 105*b*c^4*d^4*x^4 - 35*I*b*c^3*d^4*x^3)*log(-(c*x + I )/(c*x - I)))/c^3
Time = 3.02 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.90 \[ \int x^2 (d+i c d x)^4 (a+b \arctan (c x)) \, dx=\frac {a c^{4} d^{4} x^{7}}{7} - \frac {88 b d^{4} x^{2}}{105 c} + \frac {5 i b d^{4} x}{3 c^{2}} + \frac {b d^{4} \left (\frac {\log {\left (2299 b c d^{4} x - 2299 i b d^{4} \right )}}{210} + \frac {769 \log {\left (2299 b c d^{4} x + 2299 i b d^{4} \right )}}{560}\right )}{c^{3}} + x^{6} \left (- \frac {2 i a c^{3} d^{4}}{3} - \frac {b c^{3} d^{4}}{42}\right ) + x^{5} \left (- \frac {6 a c^{2} d^{4}}{5} + \frac {2 i b c^{2} d^{4}}{15}\right ) + x^{4} \left (i a c d^{4} + \frac {47 b c d^{4}}{140}\right ) + x^{3} \left (\frac {a d^{4}}{3} - \frac {5 i b d^{4}}{9}\right ) + \left (- \frac {i b c^{4} d^{4} x^{7}}{14} - \frac {b c^{3} d^{4} x^{6}}{3} + \frac {3 i b c^{2} d^{4} x^{5}}{5} + \frac {b c d^{4} x^{4}}{2} - \frac {i b d^{4} x^{3}}{6}\right ) \log {\left (i c x + 1 \right )} + \frac {\left (120 i b c^{7} d^{4} x^{7} + 560 b c^{6} d^{4} x^{6} - 1008 i b c^{5} d^{4} x^{5} - 840 b c^{4} d^{4} x^{4} + 280 i b c^{3} d^{4} x^{3} + 501 b d^{4}\right ) \log {\left (- i c x + 1 \right )}}{1680 c^{3}} \]
a*c**4*d**4*x**7/7 - 88*b*d**4*x**2/(105*c) + 5*I*b*d**4*x/(3*c**2) + b*d* *4*(log(2299*b*c*d**4*x - 2299*I*b*d**4)/210 + 769*log(2299*b*c*d**4*x + 2 299*I*b*d**4)/560)/c**3 + x**6*(-2*I*a*c**3*d**4/3 - b*c**3*d**4/42) + x** 5*(-6*a*c**2*d**4/5 + 2*I*b*c**2*d**4/15) + x**4*(I*a*c*d**4 + 47*b*c*d**4 /140) + x**3*(a*d**4/3 - 5*I*b*d**4/9) + (-I*b*c**4*d**4*x**7/14 - b*c**3* d**4*x**6/3 + 3*I*b*c**2*d**4*x**5/5 + b*c*d**4*x**4/2 - I*b*d**4*x**3/6)* log(I*c*x + 1) + (120*I*b*c**7*d**4*x**7 + 560*b*c**6*d**4*x**6 - 1008*I*b *c**5*d**4*x**5 - 840*b*c**4*d**4*x**4 + 280*I*b*c**3*d**4*x**3 + 501*b*d* *4)*log(-I*c*x + 1)/(1680*c**3)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (153) = 306\).
Time = 0.28 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.65 \[ \int x^2 (d+i c d x)^4 (a+b \arctan (c x)) \, dx=\frac {1}{7} \, a c^{4} d^{4} x^{7} - \frac {2}{3} i \, a c^{3} d^{4} x^{6} - \frac {6}{5} \, a c^{2} d^{4} x^{5} + \frac {1}{84} \, {\left (12 \, x^{7} \arctan \left (c x\right ) - c {\left (\frac {2 \, c^{4} x^{6} - 3 \, c^{2} x^{4} + 6 \, x^{2}}{c^{6}} - \frac {6 \, \log \left (c^{2} x^{2} + 1\right )}{c^{8}}\right )}\right )} b c^{4} d^{4} + i \, a c d^{4} x^{4} - \frac {2}{45} i \, {\left (15 \, x^{6} \arctan \left (c x\right ) - c {\left (\frac {3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac {15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} b c^{3} d^{4} - \frac {3}{10} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c^{2} d^{4} + \frac {1}{3} \, a d^{4} x^{3} + \frac {1}{3} i \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b c d^{4} + \frac {1}{6} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d^{4} \]
1/7*a*c^4*d^4*x^7 - 2/3*I*a*c^3*d^4*x^6 - 6/5*a*c^2*d^4*x^5 + 1/84*(12*x^7 *arctan(c*x) - c*((2*c^4*x^6 - 3*c^2*x^4 + 6*x^2)/c^6 - 6*log(c^2*x^2 + 1) /c^8))*b*c^4*d^4 + I*a*c*d^4*x^4 - 2/45*I*(15*x^6*arctan(c*x) - c*((3*c^4* x^5 - 5*c^2*x^3 + 15*x)/c^6 - 15*arctan(c*x)/c^7))*b*c^3*d^4 - 3/10*(4*x^5 *arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*b*c^2*d ^4 + 1/3*a*d^4*x^3 + 1/3*I*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3 *arctan(c*x)/c^5))*b*c*d^4 + 1/6*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2 *x^2 + 1)/c^4))*b*d^4
\[ \int x^2 (d+i c d x)^4 (a+b \arctan (c x)) \, dx=\int { {\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )} x^{2} \,d x } \]
Time = 0.69 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.06 \[ \int x^2 (d+i c d x)^4 (a+b \arctan (c x)) \, dx=\frac {c^4\,d^4\,\left (180\,a\,x^7+180\,b\,x^7\,\mathrm {atan}\left (c\,x\right )\right )}{1260}+\frac {d^4\,\left (420\,a\,x^3+420\,b\,x^3\,\mathrm {atan}\left (c\,x\right )-b\,x^3\,700{}\mathrm {i}\right )}{1260}-\frac {\frac {d^4\,\left (-1056\,b\,\ln \left (c^2\,x^2+1\right )+b\,\mathrm {atan}\left (c\,x\right )\,2100{}\mathrm {i}\right )}{1260}+\frac {88\,b\,c^2\,d^4\,x^2}{105}-\frac {b\,c\,d^4\,x\,5{}\mathrm {i}}{3}}{c^3}+\frac {c\,d^4\,\left (a\,x^4\,1260{}\mathrm {i}+423\,b\,x^4+b\,x^4\,\mathrm {atan}\left (c\,x\right )\,1260{}\mathrm {i}\right )}{1260}-\frac {c^3\,d^4\,\left (a\,x^6\,840{}\mathrm {i}+30\,b\,x^6+b\,x^6\,\mathrm {atan}\left (c\,x\right )\,840{}\mathrm {i}\right )}{1260}-\frac {c^2\,d^4\,\left (1512\,a\,x^5+1512\,b\,x^5\,\mathrm {atan}\left (c\,x\right )-b\,x^5\,168{}\mathrm {i}\right )}{1260} \]
(d^4*(420*a*x^3 - b*x^3*700i + 420*b*x^3*atan(c*x)))/1260 - ((d^4*(b*atan( c*x)*2100i - 1056*b*log(c^2*x^2 + 1)))/1260 + (88*b*c^2*d^4*x^2)/105 - (b* c*d^4*x*5i)/3)/c^3 + (c^4*d^4*(180*a*x^7 + 180*b*x^7*atan(c*x)))/1260 + (c *d^4*(a*x^4*1260i + 423*b*x^4 + b*x^4*atan(c*x)*1260i))/1260 - (c^3*d^4*(a *x^6*840i + 30*b*x^6 + b*x^6*atan(c*x)*840i))/1260 - (c^2*d^4*(1512*a*x^5 - b*x^5*168i + 1512*b*x^5*atan(c*x)))/1260