\(\int x^3 (d+i c d x)^2 (a+b \arctan (c x)) \, dx\) [10]

Optimal result

Integrand size = 23, antiderivative size = 166 \[ \int x^3 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=\frac {5 b d^2 x}{12 c^3}+\frac {i b d^2 x^2}{5 c^2}-\frac {5 b d^2 x^3}{36 c}-\frac {1}{10} i b d^2 x^4+\frac {1}{30} b c d^2 x^5-\frac {5 b d^2 \arctan (c x)}{12 c^4}+\frac {1}{4} d^2 x^4 (a+b \arctan (c x))+\frac {2}{5} i c d^2 x^5 (a+b \arctan (c x))-\frac {1}{6} c^2 d^2 x^6 (a+b \arctan (c x))-\frac {i b d^2 \log \left (1+c^2 x^2\right )}{5 c^4} \] Output:

5/12*b*d^2*x/c^3+1/5*I*b*d^2*x^2/c^2-5/36*b*d^2*x^3/c-1/10*I*b*d^2*x^4+1/3 
0*b*c*d^2*x^5-5/12*b*d^2*arctan(c*x)/c^4+1/4*d^2*x^4*(a+b*arctan(c*x))+2/5 
*I*c*d^2*x^5*(a+b*arctan(c*x))-1/6*c^2*d^2*x^6*(a+b*arctan(c*x))-1/5*I*b*d 
^2*ln(c^2*x^2+1)/c^4
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.75 \[ \int x^3 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=\frac {d^2 \left (3 a c^4 x^4 \left (15+24 i c x-10 c^2 x^2\right )+b c x \left (75+36 i c x-25 c^2 x^2-18 i c^3 x^3+6 c^4 x^4\right )+3 b \left (-25+15 c^4 x^4+24 i c^5 x^5-10 c^6 x^6\right ) \arctan (c x)-36 i b \log \left (1+c^2 x^2\right )\right )}{180 c^4} \] Input:

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

Output:

(d^2*(3*a*c^4*x^4*(15 + (24*I)*c*x - 10*c^2*x^2) + b*c*x*(75 + (36*I)*c*x 
- 25*c^2*x^2 - (18*I)*c^3*x^3 + 6*c^4*x^4) + 3*b*(-25 + 15*c^4*x^4 + (24*I 
)*c^5*x^5 - 10*c^6*x^6)*ArcTan[c*x] - (36*I)*b*Log[1 + c^2*x^2]))/(180*c^4 
)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {5407, 27, 2333, 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.

\(\displaystyle \int x^3 (d+i c d x)^2 (a+b \arctan (c x)) \, dx\)

\(\Big \downarrow \) 5407

\(\displaystyle -b c \int \frac {d^2 x^4 \left (-10 c^2 x^2+24 i c x+15\right )}{60 \left (c^2 x^2+1\right )}dx-\frac {1}{6} c^2 d^2 x^6 (a+b \arctan (c x))+\frac {2}{5} i c d^2 x^5 (a+b \arctan (c x))+\frac {1}{4} d^2 x^4 (a+b \arctan (c x))\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{60} b c d^2 \int \frac {x^4 \left (-10 c^2 x^2+24 i c x+15\right )}{c^2 x^2+1}dx-\frac {1}{6} c^2 d^2 x^6 (a+b \arctan (c x))+\frac {2}{5} i c d^2 x^5 (a+b \arctan (c x))+\frac {1}{4} d^2 x^4 (a+b \arctan (c x))\)

\(\Big \downarrow \) 2333

\(\displaystyle -\frac {1}{60} b c d^2 \int \left (-10 x^4+\frac {24 i x^3}{c}+\frac {25 x^2}{c^2}-\frac {24 i x}{c^3}+\frac {24 i c x+25}{c^4 \left (c^2 x^2+1\right )}-\frac {25}{c^4}\right )dx-\frac {1}{6} c^2 d^2 x^6 (a+b \arctan (c x))+\frac {2}{5} i c d^2 x^5 (a+b \arctan (c x))+\frac {1}{4} d^2 x^4 (a+b \arctan (c x))\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {1}{6} c^2 d^2 x^6 (a+b \arctan (c x))+\frac {2}{5} i c d^2 x^5 (a+b \arctan (c x))+\frac {1}{4} d^2 x^4 (a+b \arctan (c x))-\frac {1}{60} b c d^2 \left (\frac {25 \arctan (c x)}{c^5}-\frac {25 x}{c^4}-\frac {12 i x^2}{c^3}+\frac {25 x^3}{3 c^2}+\frac {12 i \log \left (c^2 x^2+1\right )}{c^5}+\frac {6 i x^4}{c}-2 x^5\right )\)

Input:

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

Output:

(d^2*x^4*(a + b*ArcTan[c*x]))/4 + ((2*I)/5)*c*d^2*x^5*(a + b*ArcTan[c*x]) 
- (c^2*d^2*x^6*(a + b*ArcTan[c*x]))/6 - (b*c*d^2*((-25*x)/c^4 - ((12*I)*x^ 
2)/c^3 + (25*x^3)/(3*c^2) + ((6*I)*x^4)/c - 2*x^5 + (25*ArcTan[c*x])/c^5 + 
 ((12*I)*Log[1 + c^2*x^2])/c^5))/60
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2333

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] 
&& PolyQ[Pq, x] && IGtQ[p, -2]
 

rule 5407

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]) 
)
 

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.79

method	result	size
parts	\(d^{2} a \left (-\frac {1}{6} c^{2} x^{6}+\frac {2}{5} i c \,x^{5}+\frac {1}{4} x^{4}\right )+\frac {d^{2} b \left (-\frac {c^{6} x^{6} \arctan \left (c x \right )}{6}+\frac {2 i \arctan \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}+\frac {5 c x}{12}+\frac {c^{5} x^{5}}{30}-\frac {i c^{4} x^{4}}{10}-\frac {5 c^{3} x^{3}}{36}+\frac {i c^{2} x^{2}}{5}-\frac {i \ln \left (c^{2} x^{2}+1\right )}{5}-\frac {5 \arctan \left (c x \right )}{12}\right )}{c^{4}}\)	\(131\)
derivativedivides	\(\frac {d^{2} a \left (-\frac {1}{6} c^{6} x^{6}+\frac {2}{5} i c^{5} x^{5}+\frac {1}{4} c^{4} x^{4}\right )+d^{2} b \left (-\frac {c^{6} x^{6} \arctan \left (c x \right )}{6}+\frac {2 i \arctan \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}+\frac {5 c x}{12}+\frac {c^{5} x^{5}}{30}-\frac {i c^{4} x^{4}}{10}-\frac {5 c^{3} x^{3}}{36}+\frac {i c^{2} x^{2}}{5}-\frac {i \ln \left (c^{2} x^{2}+1\right )}{5}-\frac {5 \arctan \left (c x \right )}{12}\right )}{c^{4}}\)	\(137\)
default	\(\frac {d^{2} a \left (-\frac {1}{6} c^{6} x^{6}+\frac {2}{5} i c^{5} x^{5}+\frac {1}{4} c^{4} x^{4}\right )+d^{2} b \left (-\frac {c^{6} x^{6} \arctan \left (c x \right )}{6}+\frac {2 i \arctan \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}+\frac {5 c x}{12}+\frac {c^{5} x^{5}}{30}-\frac {i c^{4} x^{4}}{10}-\frac {5 c^{3} x^{3}}{36}+\frac {i c^{2} x^{2}}{5}-\frac {i \ln \left (c^{2} x^{2}+1\right )}{5}-\frac {5 \arctan \left (c x \right )}{12}\right )}{c^{4}}\)	\(137\)
parallelrisch	\(-\frac {30 c^{6} d^{2} b \arctan \left (c x \right ) x^{6}-72 i c^{5} b \,d^{2} \arctan \left (c x \right ) x^{5}+30 a \,c^{6} d^{2} x^{6}-72 i x^{5} a \,c^{5} d^{2}-6 b \,c^{5} d^{2} x^{5}+18 i x^{4} b \,c^{4} d^{2}-45 d^{2} b \arctan \left (c x \right ) x^{4} c^{4}-45 a \,c^{4} d^{2} x^{4}+25 b \,c^{3} d^{2} x^{3}-36 i x^{2} b \,c^{2} d^{2}+36 i b \,d^{2} \ln \left (c^{2} x^{2}+1\right )-75 b c \,d^{2} x +75 b \,d^{2} \arctan \left (c x \right )}{180 c^{4}}\)	\(178\)
risch	\(\frac {i d^{2} b \left (10 c^{2} x^{6}-24 i c \,x^{5}-15 x^{4}\right ) \ln \left (i c x +1\right )}{120}-\frac {d^{2} c^{2} a \,x^{6}}{6}-\frac {i d^{2} c^{2} x^{6} b \ln \left (-i c x +1\right )}{12}-\frac {d^{2} c b \,x^{5} \ln \left (-i c x +1\right )}{5}+\frac {b c \,d^{2} x^{5}}{30}+\frac {2 i d^{2} c \,x^{5} a}{5}+\frac {d^{2} a \,x^{4}}{4}+\frac {i d^{2} x^{4} b \ln \left (-i c x +1\right )}{8}-\frac {i b \,d^{2} x^{4}}{10}-\frac {5 b \,d^{2} x^{3}}{36 c}+\frac {i b \,d^{2} x^{2}}{5 c^{2}}+\frac {5 b \,d^{2} x}{12 c^{3}}-\frac {5 b \,d^{2} \arctan \left (c x \right )}{12 c^{4}}-\frac {i d^{2} b \ln \left (625 c^{2} x^{2}+625\right )}{5 c^{4}}\)	\(216\)

Input:

int(x^3*(d+I*c*d*x)^2*(a+b*arctan(c*x)),x,method=_RETURNVERBOSE)
 

Output:

d^2*a*(-1/6*c^2*x^6+2/5*I*c*x^5+1/4*x^4)+d^2*b/c^4*(-1/6*c^6*x^6*arctan(c* 
x)+2/5*I*arctan(c*x)*c^5*x^5+1/4*c^4*x^4*arctan(c*x)+5/12*c*x+1/30*c^5*x^5 
-1/10*I*c^4*x^4-5/36*c^3*x^3+1/5*I*c^2*x^2-1/5*I*ln(c^2*x^2+1)-5/12*arctan 
(c*x))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.04 \[ \int x^3 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=-\frac {60 \, a c^{6} d^{2} x^{6} + 12 \, {\left (-12 i \, a - b\right )} c^{5} d^{2} x^{5} - 18 \, {\left (5 \, a - 2 i \, b\right )} c^{4} d^{2} x^{4} + 50 \, b c^{3} d^{2} x^{3} - 72 i \, b c^{2} d^{2} x^{2} - 150 \, b c d^{2} x + 147 i \, b d^{2} \log \left (\frac {c x + i}{c}\right ) - 3 i \, b d^{2} \log \left (\frac {c x - i}{c}\right ) + 3 \, {\left (10 i \, b c^{6} d^{2} x^{6} + 24 \, b c^{5} d^{2} x^{5} - 15 i \, b c^{4} d^{2} x^{4}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{360 \, c^{4}} \] Input:

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

Output:

-1/360*(60*a*c^6*d^2*x^6 + 12*(-12*I*a - b)*c^5*d^2*x^5 - 18*(5*a - 2*I*b) 
*c^4*d^2*x^4 + 50*b*c^3*d^2*x^3 - 72*I*b*c^2*d^2*x^2 - 150*b*c*d^2*x + 147 
*I*b*d^2*log((c*x + I)/c) - 3*I*b*d^2*log((c*x - I)/c) + 3*(10*I*b*c^6*d^2 
*x^6 + 24*b*c^5*d^2*x^5 - 15*I*b*c^4*d^2*x^4)*log(-(c*x + I)/(c*x - I)))/c 
^4
 

Sympy [A] (verification not implemented)

Time = 2.30 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.63 \[ \int x^3 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=- \frac {a c^{2} d^{2} x^{6}}{6} - \frac {5 b d^{2} x^{3}}{36 c} + \frac {i b d^{2} x^{2}}{5 c^{2}} + \frac {5 b d^{2} x}{12 c^{3}} - \frac {b d^{2} \left (- \frac {i \log {\left (291 b c d^{2} x - 291 i b d^{2} \right )}}{120} + \frac {71 i \log {\left (291 b c d^{2} x + 291 i b d^{2} \right )}}{210}\right )}{c^{4}} - x^{5} \left (- \frac {2 i a c d^{2}}{5} - \frac {b c d^{2}}{30}\right ) - x^{4} \left (- \frac {a d^{2}}{4} + \frac {i b d^{2}}{10}\right ) + \left (\frac {i b c^{2} d^{2} x^{6}}{12} + \frac {b c d^{2} x^{5}}{5} - \frac {i b d^{2} x^{4}}{8}\right ) \log {\left (i c x + 1 \right )} + \frac {\left (- 70 i b c^{6} d^{2} x^{6} - 168 b c^{5} d^{2} x^{5} + 105 i b c^{4} d^{2} x^{4} - 59 i b d^{2}\right ) \log {\left (- i c x + 1 \right )}}{840 c^{4}} \] Input:

integrate(x**3*(d+I*c*d*x)**2*(a+b*atan(c*x)),x)
 

Output:

-a*c**2*d**2*x**6/6 - 5*b*d**2*x**3/(36*c) + I*b*d**2*x**2/(5*c**2) + 5*b* 
d**2*x/(12*c**3) - b*d**2*(-I*log(291*b*c*d**2*x - 291*I*b*d**2)/120 + 71* 
I*log(291*b*c*d**2*x + 291*I*b*d**2)/210)/c**4 - x**5*(-2*I*a*c*d**2/5 - b 
*c*d**2/30) - x**4*(-a*d**2/4 + I*b*d**2/10) + (I*b*c**2*d**2*x**6/12 + b* 
c*d**2*x**5/5 - I*b*d**2*x**4/8)*log(I*c*x + 1) + (-70*I*b*c**6*d**2*x**6 
- 168*b*c**5*d**2*x**5 + 105*I*b*c**4*d**2*x**4 - 59*I*b*d**2)*log(-I*c*x 
+ 1)/(840*c**4)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.11 \[ \int x^3 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=-\frac {1}{6} \, a c^{2} d^{2} x^{6} + \frac {2}{5} i \, a c d^{2} x^{5} + \frac {1}{4} \, a d^{2} x^{4} - \frac {1}{90} \, {\left (15 \, x^{6} \arctan \left (c x\right ) - c {\left (\frac {3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac {15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} b c^{2} d^{2} + \frac {1}{10} i \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c d^{2} + \frac {1}{12} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b d^{2} \] Input:

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

Output:

-1/6*a*c^2*d^2*x^6 + 2/5*I*a*c*d^2*x^5 + 1/4*a*d^2*x^4 - 1/90*(15*x^6*arct 
an(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 
^2*d^2 + 1/10*I*(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*d^2 + 1/12*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 
+ 3*arctan(c*x)/c^5))*b*d^2
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.04 \[ \int x^3 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=-\frac {60 \, b c^{6} d^{2} x^{6} \arctan \left (c x\right ) + 60 \, a c^{6} d^{2} x^{6} - 144 i \, b c^{5} d^{2} x^{5} \arctan \left (c x\right ) - 144 i \, a c^{5} d^{2} x^{5} - 12 \, b c^{5} d^{2} x^{5} - 90 \, b c^{4} d^{2} x^{4} \arctan \left (c x\right ) - 90 \, a c^{4} d^{2} x^{4} + 36 i \, b c^{4} d^{2} x^{4} + 50 \, b c^{3} d^{2} x^{3} - 72 i \, b c^{2} d^{2} x^{2} - 150 \, b c d^{2} x - 3 i \, b d^{2} \log \left (i \, c x + 1\right ) + 147 i \, b d^{2} \log \left (-i \, c x + 1\right )}{360 \, c^{4}} \] Input:

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

Output:

-1/360*(60*b*c^6*d^2*x^6*arctan(c*x) + 60*a*c^6*d^2*x^6 - 144*I*b*c^5*d^2* 
x^5*arctan(c*x) - 144*I*a*c^5*d^2*x^5 - 12*b*c^5*d^2*x^5 - 90*b*c^4*d^2*x^ 
4*arctan(c*x) - 90*a*c^4*d^2*x^4 + 36*I*b*c^4*d^2*x^4 + 50*b*c^3*d^2*x^3 - 
 72*I*b*c^2*d^2*x^2 - 150*b*c*d^2*x - 3*I*b*d^2*log(I*c*x + 1) + 147*I*b*d 
^2*log(-I*c*x + 1))/c^4
 

Mupad [B] (verification not implemented)

Time = 1.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.92 \[ \int x^3 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=-\frac {\frac {d^2\,\left (75\,b\,\mathrm {atan}\left (c\,x\right )+b\,\ln \left (c^2\,x^2+1\right )\,36{}\mathrm {i}\right )}{180}+\frac {5\,b\,c^3\,d^2\,x^3}{36}-\frac {5\,b\,c\,d^2\,x}{12}-\frac {b\,c^2\,d^2\,x^2\,1{}\mathrm {i}}{5}}{c^4}+\frac {d^2\,\left (45\,a\,x^4+45\,b\,x^4\,\mathrm {atan}\left (c\,x\right )-b\,x^4\,18{}\mathrm {i}\right )}{180}-\frac {c^2\,d^2\,\left (30\,a\,x^6+30\,b\,x^6\,\mathrm {atan}\left (c\,x\right )\right )}{180}+\frac {c\,d^2\,\left (a\,x^5\,72{}\mathrm {i}+6\,b\,x^5+b\,x^5\,\mathrm {atan}\left (c\,x\right )\,72{}\mathrm {i}\right )}{180} \] Input:

int(x^3*(a + b*atan(c*x))*(d + c*d*x*1i)^2,x)
 

Output:

(d^2*(45*a*x^4 - b*x^4*18i + 45*b*x^4*atan(c*x)))/180 - ((d^2*(75*b*atan(c 
*x) + b*log(c^2*x^2 + 1)*36i))/180 - (b*c^2*d^2*x^2*1i)/5 + (5*b*c^3*d^2*x 
^3)/36 - (5*b*c*d^2*x)/12)/c^4 - (c^2*d^2*(30*a*x^6 + 30*b*x^6*atan(c*x))) 
/180 + (c*d^2*(a*x^5*72i + 6*b*x^5 + b*x^5*atan(c*x)*72i))/180
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.85 \[ \int x^3 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=\frac {d^{2} \left (-30 \mathit {atan} \left (c x \right ) b \,c^{6} x^{6}+72 \mathit {atan} \left (c x \right ) b \,c^{5} i \,x^{5}+45 \mathit {atan} \left (c x \right ) b \,c^{4} x^{4}-75 \mathit {atan} \left (c x \right ) b -36 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b i -30 a \,c^{6} x^{6}+72 a \,c^{5} i \,x^{5}+45 a \,c^{4} x^{4}+6 b \,c^{5} x^{5}-18 b \,c^{4} i \,x^{4}-25 b \,c^{3} x^{3}+36 b \,c^{2} i \,x^{2}+75 b c x \right )}{180 c^{4}} \] Input:

int(x^3*(d+I*c*d*x)^2*(a+b*atan(c*x)),x)
 

Output:

(d**2*( - 30*atan(c*x)*b*c**6*x**6 + 72*atan(c*x)*b*c**5*i*x**5 + 45*atan( 
c*x)*b*c**4*x**4 - 75*atan(c*x)*b - 36*log(c**2*x**2 + 1)*b*i - 30*a*c**6* 
x**6 + 72*a*c**5*i*x**5 + 45*a*c**4*x**4 + 6*b*c**5*x**5 - 18*b*c**4*i*x** 
4 - 25*b*c**3*x**3 + 36*b*c**2*i*x**2 + 75*b*c*x))/(180*c**4)