3.21 \(\int x^2 (d+i c d x)^3 (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=191 \[ -\frac{1}{6} i c^3 d^3 x^6 \left (a+b \tan ^{-1}(c x)\right )-\frac{3}{5} c^2 d^3 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac{3}{4} i c d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{30} i b c^2 d^3 x^5+\frac{7 b d^3 \log \left (c^2 x^2+1\right )}{15 c^3}+\frac{11 i b d^3 x}{12 c^2}-\frac{11 i b d^3 \tan ^{-1}(c x)}{12 c^3}+\frac{3}{20} b c d^3 x^4-\frac{7 b d^3 x^2}{15 c}-\frac{11}{36} i b d^3 x^3 \]

[Out]

(((11*I)/12)*b*d^3*x)/c^2 - (7*b*d^3*x^2)/(15*c) - ((11*I)/36)*b*d^3*x^3 + (3*b*c*d^3*x^4)/20 + (I/30)*b*c^2*d
^3*x^5 - (((11*I)/12)*b*d^3*ArcTan[c*x])/c^3 + (d^3*x^3*(a + b*ArcTan[c*x]))/3 + ((3*I)/4)*c*d^3*x^4*(a + b*Ar
cTan[c*x]) - (3*c^2*d^3*x^5*(a + b*ArcTan[c*x]))/5 - (I/6)*c^3*d^3*x^6*(a + b*ArcTan[c*x]) + (7*b*d^3*Log[1 +
c^2*x^2])/(15*c^3)

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Rubi [A]  time = 0.171228, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {43, 4872, 12, 1802, 635, 203, 260} \[ -\frac{1}{6} i c^3 d^3 x^6 \left (a+b \tan ^{-1}(c x)\right )-\frac{3}{5} c^2 d^3 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac{3}{4} i c d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{30} i b c^2 d^3 x^5+\frac{7 b d^3 \log \left (c^2 x^2+1\right )}{15 c^3}+\frac{11 i b d^3 x}{12 c^2}-\frac{11 i b d^3 \tan ^{-1}(c x)}{12 c^3}+\frac{3}{20} b c d^3 x^4-\frac{7 b d^3 x^2}{15 c}-\frac{11}{36} i b d^3 x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((11*I)/12)*b*d^3*x)/c^2 - (7*b*d^3*x^2)/(15*c) - ((11*I)/36)*b*d^3*x^3 + (3*b*c*d^3*x^4)/20 + (I/30)*b*c^2*d
^3*x^5 - (((11*I)/12)*b*d^3*ArcTan[c*x])/c^3 + (d^3*x^3*(a + b*ArcTan[c*x]))/3 + ((3*I)/4)*c*d^3*x^4*(a + b*Ar
cTan[c*x]) - (3*c^2*d^3*x^5*(a + b*ArcTan[c*x]))/5 - (I/6)*c^3*d^3*x^6*(a + b*ArcTan[c*x]) + (7*b*d^3*Log[1 +
c^2*x^2])/(15*c^3)

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 4872

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_))^(q_.), x_Symbol] :> With[{u = I
ntHide[(f*x)^m*(d + e*x)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[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]))

Rule 12

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

Rule 1802

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 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int x^2 (d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac{1}{3} d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{3}{4} i c d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac{3}{5} c^2 d^3 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} i c^3 d^3 x^6 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac{d^3 x^3 \left (20+45 i c x-36 c^2 x^2-10 i c^3 x^3\right )}{60 \left (1+c^2 x^2\right )} \, dx\\ &=\frac{1}{3} d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{3}{4} i c d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac{3}{5} c^2 d^3 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} i c^3 d^3 x^6 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{60} \left (b c d^3\right ) \int \frac{x^3 \left (20+45 i c x-36 c^2 x^2-10 i c^3 x^3\right )}{1+c^2 x^2} \, dx\\ &=\frac{1}{3} d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{3}{4} i c d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac{3}{5} c^2 d^3 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} i c^3 d^3 x^6 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{60} \left (b c d^3\right ) \int \left (-\frac{55 i}{c^3}+\frac{56 x}{c^2}+\frac{55 i x^2}{c}-36 x^3-10 i c x^4+\frac{55 i-56 c x}{c^3 \left (1+c^2 x^2\right )}\right ) \, dx\\ &=\frac{11 i b d^3 x}{12 c^2}-\frac{7 b d^3 x^2}{15 c}-\frac{11}{36} i b d^3 x^3+\frac{3}{20} b c d^3 x^4+\frac{1}{30} i b c^2 d^3 x^5+\frac{1}{3} d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{3}{4} i c d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac{3}{5} c^2 d^3 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} i c^3 d^3 x^6 \left (a+b \tan ^{-1}(c x)\right )-\frac{\left (b d^3\right ) \int \frac{55 i-56 c x}{1+c^2 x^2} \, dx}{60 c^2}\\ &=\frac{11 i b d^3 x}{12 c^2}-\frac{7 b d^3 x^2}{15 c}-\frac{11}{36} i b d^3 x^3+\frac{3}{20} b c d^3 x^4+\frac{1}{30} i b c^2 d^3 x^5+\frac{1}{3} d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{3}{4} i c d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac{3}{5} c^2 d^3 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} i c^3 d^3 x^6 \left (a+b \tan ^{-1}(c x)\right )-\frac{\left (11 i b d^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{12 c^2}+\frac{\left (14 b d^3\right ) \int \frac{x}{1+c^2 x^2} \, dx}{15 c}\\ &=\frac{11 i b d^3 x}{12 c^2}-\frac{7 b d^3 x^2}{15 c}-\frac{11}{36} i b d^3 x^3+\frac{3}{20} b c d^3 x^4+\frac{1}{30} i b c^2 d^3 x^5-\frac{11 i b d^3 \tan ^{-1}(c x)}{12 c^3}+\frac{1}{3} d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{3}{4} i c d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac{3}{5} c^2 d^3 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} i c^3 d^3 x^6 \left (a+b \tan ^{-1}(c x)\right )+\frac{7 b d^3 \log \left (1+c^2 x^2\right )}{15 c^3}\\ \end{align*}

Mathematica [A]  time = 0.0856371, size = 234, normalized size = 1.23 \[ -\frac{1}{6} i a c^3 d^3 x^6-\frac{3}{5} a c^2 d^3 x^5+\frac{3}{4} i a c d^3 x^4+\frac{1}{3} a d^3 x^3+\frac{1}{30} i b c^2 d^3 x^5+\frac{7 b d^3 \log \left (c^2 x^2+1\right )}{15 c^3}-\frac{1}{6} i b c^3 d^3 x^6 \tan ^{-1}(c x)-\frac{3}{5} b c^2 d^3 x^5 \tan ^{-1}(c x)+\frac{11 i b d^3 x}{12 c^2}-\frac{11 i b d^3 \tan ^{-1}(c x)}{12 c^3}+\frac{3}{20} b c d^3 x^4-\frac{7 b d^3 x^2}{15 c}+\frac{3}{4} i b c d^3 x^4 \tan ^{-1}(c x)+\frac{1}{3} b d^3 x^3 \tan ^{-1}(c x)-\frac{11}{36} i b d^3 x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((11*I)/12)*b*d^3*x)/c^2 - (7*b*d^3*x^2)/(15*c) + (a*d^3*x^3)/3 - ((11*I)/36)*b*d^3*x^3 + ((3*I)/4)*a*c*d^3*x
^4 + (3*b*c*d^3*x^4)/20 - (3*a*c^2*d^3*x^5)/5 + (I/30)*b*c^2*d^3*x^5 - (I/6)*a*c^3*d^3*x^6 - (((11*I)/12)*b*d^
3*ArcTan[c*x])/c^3 + (b*d^3*x^3*ArcTan[c*x])/3 + ((3*I)/4)*b*c*d^3*x^4*ArcTan[c*x] - (3*b*c^2*d^3*x^5*ArcTan[c
*x])/5 - (I/6)*b*c^3*d^3*x^6*ArcTan[c*x] + (7*b*d^3*Log[1 + c^2*x^2])/(15*c^3)

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Maple [A]  time = 0.027, size = 197, normalized size = 1. \begin{align*} -{\frac{i}{6}}{c}^{3}{d}^{3}a{x}^{6}-{\frac{3\,{c}^{2}{d}^{3}a{x}^{5}}{5}}+{\frac{3\,i}{4}}c{d}^{3}a{x}^{4}+{\frac{{d}^{3}a{x}^{3}}{3}}-{\frac{i}{6}}{c}^{3}{d}^{3}b\arctan \left ( cx \right ){x}^{6}-{\frac{3\,{c}^{2}{d}^{3}b\arctan \left ( cx \right ){x}^{5}}{5}}+{\frac{3\,i}{4}}c{d}^{3}b\arctan \left ( cx \right ){x}^{4}+{\frac{{d}^{3}b\arctan \left ( cx \right ){x}^{3}}{3}}+{\frac{{\frac{11\,i}{12}}b{d}^{3}x}{{c}^{2}}}+{\frac{i}{30}}b{c}^{2}{d}^{3}{x}^{5}+{\frac{3\,bc{d}^{3}{x}^{4}}{20}}-{\frac{11\,i}{36}}b{d}^{3}{x}^{3}-{\frac{7\,{d}^{3}b{x}^{2}}{15\,c}}+{\frac{7\,{d}^{3}b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{15\,{c}^{3}}}-{\frac{{\frac{11\,i}{12}}b{d}^{3}\arctan \left ( cx \right ) }{{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*I*c^3*d^3*a*x^6-3/5*c^2*d^3*a*x^5+3/4*I*c*d^3*a*x^4+1/3*d^3*a*x^3-1/6*I*c^3*d^3*b*arctan(c*x)*x^6-3/5*c^2
*d^3*b*arctan(c*x)*x^5+3/4*I*c*d^3*b*arctan(c*x)*x^4+1/3*d^3*b*arctan(c*x)*x^3+11/12*I*b*d^3*x/c^2+1/30*I*b*c^
2*d^3*x^5+3/20*b*c*d^3*x^4-11/36*I*b*d^3*x^3-7/15*b*d^3*x^2/c+7/15*b*d^3*ln(c^2*x^2+1)/c^3-11/12*I*b*d^3*arcta
n(c*x)/c^3

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Maxima [A]  time = 1.4773, size = 327, normalized size = 1.71 \begin{align*} -\frac{1}{6} i \, a c^{3} d^{3} x^{6} - \frac{3}{5} \, a c^{2} d^{3} x^{5} + \frac{3}{4} i \, a c d^{3} x^{4} - \frac{1}{90} 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^{3} - \frac{3}{20} \,{\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^{3} + \frac{1}{3} \, a d^{3} x^{3} + \frac{1}{4} 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^{3} + \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^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*I*a*c^3*d^3*x^6 - 3/5*a*c^2*d^3*x^5 + 3/4*I*a*c*d^3*x^4 - 1/90*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^3 - 3/20*(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^3 + 1/3*a*d^3*x^3 + 1/4*I*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arcta
n(c*x)/c^5))*b*c*d^3 + 1/6*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*d^3

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Fricas [A]  time = 2.79283, size = 447, normalized size = 2.34 \begin{align*} \frac{-60 i \, a c^{6} d^{3} x^{6} - 12 \,{\left (18 \, a - i \, b\right )} c^{5} d^{3} x^{5} +{\left (270 i \, a + 54 \, b\right )} c^{4} d^{3} x^{4} + 10 \,{\left (12 \, a - 11 i \, b\right )} c^{3} d^{3} x^{3} - 168 \, b c^{2} d^{3} x^{2} + 330 i \, b c d^{3} x + 333 \, b d^{3} \log \left (\frac{c x + i}{c}\right ) + 3 \, b d^{3} \log \left (\frac{c x - i}{c}\right ) +{\left (30 \, b c^{6} d^{3} x^{6} - 108 i \, b c^{5} d^{3} x^{5} - 135 \, b c^{4} d^{3} x^{4} + 60 i \, b c^{3} d^{3} x^{3}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{360 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/360*(-60*I*a*c^6*d^3*x^6 - 12*(18*a - I*b)*c^5*d^3*x^5 + (270*I*a + 54*b)*c^4*d^3*x^4 + 10*(12*a - 11*I*b)*c
^3*d^3*x^3 - 168*b*c^2*d^3*x^2 + 330*I*b*c*d^3*x + 333*b*d^3*log((c*x + I)/c) + 3*b*d^3*log((c*x - I)/c) + (30
*b*c^6*d^3*x^6 - 108*I*b*c^5*d^3*x^5 - 135*b*c^4*d^3*x^4 + 60*I*b*c^3*d^3*x^3)*log(-(c*x + I)/(c*x - I)))/c^3

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Sympy [A]  time = 3.31909, size = 275, normalized size = 1.44 \begin{align*} - \frac{i a c^{3} d^{3} x^{6}}{6} - \frac{7 b d^{3} x^{2}}{15 c} + \frac{11 i b d^{3} x}{12 c^{2}} - \frac{b d^{3} \left (- \frac{\log{\left (x - \frac{i}{c} \right )}}{120} - \frac{37 \log{\left (x + \frac{i}{c} \right )}}{40}\right )}{c^{3}} - x^{5} \left (\frac{3 a c^{2} d^{3}}{5} - \frac{i b c^{2} d^{3}}{30}\right ) - x^{4} \left (- \frac{3 i a c d^{3}}{4} - \frac{3 b c d^{3}}{20}\right ) - x^{3} \left (- \frac{a d^{3}}{3} + \frac{11 i b d^{3}}{36}\right ) + \left (- \frac{b c^{3} d^{3} x^{6}}{12} + \frac{3 i b c^{2} d^{3} x^{5}}{10} + \frac{3 b c d^{3} x^{4}}{8} - \frac{i b d^{3} x^{3}}{6}\right ) \log{\left (i c x + 1 \right )} + \left (\frac{b c^{3} d^{3} x^{6}}{12} - \frac{3 i b c^{2} d^{3} x^{5}}{10} - \frac{3 b c d^{3} x^{4}}{8} + \frac{i b d^{3} x^{3}}{6}\right ) \log{\left (- i c x + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a*c**3*d**3*x**6/6 - 7*b*d**3*x**2/(15*c) + 11*I*b*d**3*x/(12*c**2) - b*d**3*(-log(x - I/c)/120 - 37*log(x
+ I/c)/40)/c**3 - x**5*(3*a*c**2*d**3/5 - I*b*c**2*d**3/30) - x**4*(-3*I*a*c*d**3/4 - 3*b*c*d**3/20) - x**3*(-
a*d**3/3 + 11*I*b*d**3/36) + (-b*c**3*d**3*x**6/12 + 3*I*b*c**2*d**3*x**5/10 + 3*b*c*d**3*x**4/8 - I*b*d**3*x*
*3/6)*log(I*c*x + 1) + (b*c**3*d**3*x**6/12 - 3*I*b*c**2*d**3*x**5/10 - 3*b*c*d**3*x**4/8 + I*b*d**3*x**3/6)*l
og(-I*c*x + 1)

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Giac [A]  time = 1.19851, size = 279, normalized size = 1.46 \begin{align*} -\frac{60 \, b c^{6} d^{3} i x^{6} \arctan \left (c x\right ) + 60 \, a c^{6} d^{3} i x^{6} - 12 \, b c^{5} d^{3} i x^{5} + 216 \, b c^{5} d^{3} x^{5} \arctan \left (c x\right ) + 216 \, a c^{5} d^{3} x^{5} - 270 \, b c^{4} d^{3} i x^{4} \arctan \left (c x\right ) - 270 \, a c^{4} d^{3} i x^{4} - 54 \, b c^{4} d^{3} x^{4} + 110 \, b c^{3} d^{3} i x^{3} - 120 \, b c^{3} d^{3} x^{3} \arctan \left (c x\right ) - 120 \, a c^{3} d^{3} x^{3} + 168 \, b c^{2} d^{3} x^{2} - 330 \, b c d^{3} i x - 333 \, b d^{3} \log \left (c x + i\right ) - 3 \, b d^{3} \log \left (c x - i\right )}{360 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/360*(60*b*c^6*d^3*i*x^6*arctan(c*x) + 60*a*c^6*d^3*i*x^6 - 12*b*c^5*d^3*i*x^5 + 216*b*c^5*d^3*x^5*arctan(c*
x) + 216*a*c^5*d^3*x^5 - 270*b*c^4*d^3*i*x^4*arctan(c*x) - 270*a*c^4*d^3*i*x^4 - 54*b*c^4*d^3*x^4 + 110*b*c^3*
d^3*i*x^3 - 120*b*c^3*d^3*x^3*arctan(c*x) - 120*a*c^3*d^3*x^3 + 168*b*c^2*d^3*x^2 - 330*b*c*d^3*i*x - 333*b*d^
3*log(c*x + i) - 3*b*d^3*log(c*x - i))/c^3