Integrand size = 23, antiderivative size = 271 \[ \int (c e+d e x)^2 (a+b \arctan (c+d x))^3 \, dx=a b^2 e^2 x+\frac {b^3 e^2 (c+d x) \arctan (c+d x)}{d}-\frac {b e^2 (a+b \arctan (c+d x))^2}{2 d}-\frac {b e^2 (c+d x)^2 (a+b \arctan (c+d x))^2}{2 d}-\frac {i e^2 (a+b \arctan (c+d x))^3}{3 d}+\frac {e^2 (c+d x)^3 (a+b \arctan (c+d x))^3}{3 d}-\frac {b e^2 (a+b \arctan (c+d x))^2 \log \left (\frac {2}{1+i (c+d x)}\right )}{d}-\frac {b^3 e^2 \log \left (1+(c+d x)^2\right )}{2 d}-\frac {i b^2 e^2 (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{d}-\frac {b^3 e^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i (c+d x)}\right )}{2 d} \]
a*b^2*e^2*x+b^3*e^2*(d*x+c)*arctan(d*x+c)/d-1/2*b*e^2*(a+b*arctan(d*x+c))^ 2/d-1/2*b*e^2*(d*x+c)^2*(a+b*arctan(d*x+c))^2/d-1/3*I*e^2*(a+b*arctan(d*x+ c))^3/d+1/3*e^2*(d*x+c)^3*(a+b*arctan(d*x+c))^3/d-b*e^2*(a+b*arctan(d*x+c) )^2*ln(2/(1+I*(d*x+c)))/d-1/2*b^3*e^2*ln(1+(d*x+c)^2)/d-I*b^2*e^2*(a+b*arc tan(d*x+c))*polylog(2,1-2/(1+I*(d*x+c)))/d-1/2*b^3*e^2*polylog(3,1-2/(1+I* (d*x+c)))/d
Time = 0.58 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.29 \[ \int (c e+d e x)^2 (a+b \arctan (c+d x))^3 \, dx=\frac {e^2 \left (-3 a^2 b (c+d x)^2+2 a^3 (c+d x)^3+6 a^2 b (c+d x)^3 \arctan (c+d x)+3 a^2 b \log \left (1+(c+d x)^2\right )+6 a b^2 \left (c+d x-\arctan (c+d x)-(c+d x)^2 \arctan (c+d x)+i \arctan (c+d x)^2+(c+d x)^3 \arctan (c+d x)^2-2 \arctan (c+d x) \log \left (1+e^{2 i \arctan (c+d x)}\right )+i \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c+d x)}\right )\right )+b^3 \left (6 (c+d x) \arctan (c+d x)-3 \left (1+(c+d x)^2\right ) \arctan (c+d x)^2+2 i \arctan (c+d x)^3-2 (c+d x) \arctan (c+d x)^3+2 (c+d x) \left (1+(c+d x)^2\right ) \arctan (c+d x)^3-6 \arctan (c+d x)^2 \log \left (1+e^{2 i \arctan (c+d x)}\right )+6 \log \left (\frac {1}{\sqrt {1+(c+d x)^2}}\right )+6 i \arctan (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c+d x)}\right )-3 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c+d x)}\right )\right )\right )}{6 d} \]
(e^2*(-3*a^2*b*(c + d*x)^2 + 2*a^3*(c + d*x)^3 + 6*a^2*b*(c + d*x)^3*ArcTa n[c + d*x] + 3*a^2*b*Log[1 + (c + d*x)^2] + 6*a*b^2*(c + d*x - ArcTan[c + d*x] - (c + d*x)^2*ArcTan[c + d*x] + I*ArcTan[c + d*x]^2 + (c + d*x)^3*Arc Tan[c + d*x]^2 - 2*ArcTan[c + d*x]*Log[1 + E^((2*I)*ArcTan[c + d*x])] + I* PolyLog[2, -E^((2*I)*ArcTan[c + d*x])]) + b^3*(6*(c + d*x)*ArcTan[c + d*x] - 3*(1 + (c + d*x)^2)*ArcTan[c + d*x]^2 + (2*I)*ArcTan[c + d*x]^3 - 2*(c + d*x)*ArcTan[c + d*x]^3 + 2*(c + d*x)*(1 + (c + d*x)^2)*ArcTan[c + d*x]^3 - 6*ArcTan[c + d*x]^2*Log[1 + E^((2*I)*ArcTan[c + d*x])] + 6*Log[1/Sqrt[1 + (c + d*x)^2]] + (6*I)*ArcTan[c + d*x]*PolyLog[2, -E^((2*I)*ArcTan[c + d *x])] - 3*PolyLog[3, -E^((2*I)*ArcTan[c + d*x])])))/(6*d)
Time = 1.31 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.85, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {5566, 27, 5361, 5451, 5361, 5451, 2009, 5419, 5455, 5379, 5529, 7164}
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.
| \(\displaystyle \int (c e+d e x)^2 (a+b \arctan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 5566 |
| \(\displaystyle \frac {\int e^2 (c+d x)^2 (a+b \arctan (c+d x))^3d(c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e^2 \int (c+d x)^2 (a+b \arctan (c+d x))^3d(c+d x)}{d}\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))^3-b \int \frac {(c+d x)^3 (a+b \arctan (c+d x))^2}{(c+d x)^2+1}d(c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 5451 |
| \(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))^3-b \left (\int (c+d x) (a+b \arctan (c+d x))^2d(c+d x)-\int \frac {(c+d x) (a+b \arctan (c+d x))^2}{(c+d x)^2+1}d(c+d x)\right )\right )}{d}\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))^3-b \left (-b \int \frac {(c+d x)^2 (a+b \arctan (c+d x))}{(c+d x)^2+1}d(c+d x)-\int \frac {(c+d x) (a+b \arctan (c+d x))^2}{(c+d x)^2+1}d(c+d x)+\frac {1}{2} (c+d x)^2 (a+b \arctan (c+d x))^2\right )\right )}{d}\) |
| \(\Big \downarrow \) 5451 |
| \(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))^3-b \left (-b \left (\int (a+b \arctan (c+d x))d(c+d x)-\int \frac {a+b \arctan (c+d x)}{(c+d x)^2+1}d(c+d x)\right )-\int \frac {(c+d x) (a+b \arctan (c+d x))^2}{(c+d x)^2+1}d(c+d x)+\frac {1}{2} (c+d x)^2 (a+b \arctan (c+d x))^2\right )\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))^3-b \left (-\int \frac {(c+d x) (a+b \arctan (c+d x))^2}{(c+d x)^2+1}d(c+d x)-b \left (-\int \frac {a+b \arctan (c+d x)}{(c+d x)^2+1}d(c+d x)+a (c+d x)+b (c+d x) \arctan (c+d x)-\frac {1}{2} b \log \left ((c+d x)^2+1\right )\right )+\frac {1}{2} (c+d x)^2 (a+b \arctan (c+d x))^2\right )\right )}{d}\) |
| \(\Big \downarrow \) 5419 |
| \(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))^3-b \left (-\int \frac {(c+d x) (a+b \arctan (c+d x))^2}{(c+d x)^2+1}d(c+d x)+\frac {1}{2} (c+d x)^2 (a+b \arctan (c+d x))^2-b \left (-\frac {(a+b \arctan (c+d x))^2}{2 b}+a (c+d x)+b (c+d x) \arctan (c+d x)-\frac {1}{2} b \log \left ((c+d x)^2+1\right )\right )\right )\right )}{d}\) |
| \(\Big \downarrow \) 5455 |
| \(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))^3-b \left (\int \frac {(a+b \arctan (c+d x))^2}{-c-d x+i}d(c+d x)+\frac {i (a+b \arctan (c+d x))^3}{3 b}+\frac {1}{2} (c+d x)^2 (a+b \arctan (c+d x))^2-b \left (-\frac {(a+b \arctan (c+d x))^2}{2 b}+a (c+d x)+b (c+d x) \arctan (c+d x)-\frac {1}{2} b \log \left ((c+d x)^2+1\right )\right )\right )\right )}{d}\) |
| \(\Big \downarrow \) 5379 |
| \(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))^3-b \left (-2 b \int \frac {(a+b \arctan (c+d x)) \log \left (\frac {2}{i (c+d x)+1}\right )}{(c+d x)^2+1}d(c+d x)+\frac {i (a+b \arctan (c+d x))^3}{3 b}+\frac {1}{2} (c+d x)^2 (a+b \arctan (c+d x))^2+\log \left (\frac {2}{1+i (c+d x)}\right ) (a+b \arctan (c+d x))^2-b \left (-\frac {(a+b \arctan (c+d x))^2}{2 b}+a (c+d x)+b (c+d x) \arctan (c+d x)-\frac {1}{2} b \log \left ((c+d x)^2+1\right )\right )\right )\right )}{d}\) |
| \(\Big \downarrow \) 5529 |
| \(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))^3-b \left (-2 b \left (\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{i (c+d x)+1}\right )}{(c+d x)^2+1}d(c+d x)-\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{i (c+d x)+1}\right ) (a+b \arctan (c+d x))\right )+\frac {i (a+b \arctan (c+d x))^3}{3 b}+\frac {1}{2} (c+d x)^2 (a+b \arctan (c+d x))^2+\log \left (\frac {2}{1+i (c+d x)}\right ) (a+b \arctan (c+d x))^2-b \left (-\frac {(a+b \arctan (c+d x))^2}{2 b}+a (c+d x)+b (c+d x) \arctan (c+d x)-\frac {1}{2} b \log \left ((c+d x)^2+1\right )\right )\right )\right )}{d}\) |
| \(\Big \downarrow \) 7164 |
| \(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))^3-b \left (-2 b \left (-\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{i (c+d x)+1}\right ) (a+b \arctan (c+d x))-\frac {1}{4} b \operatorname {PolyLog}\left (3,1-\frac {2}{i (c+d x)+1}\right )\right )+\frac {i (a+b \arctan (c+d x))^3}{3 b}+\frac {1}{2} (c+d x)^2 (a+b \arctan (c+d x))^2+\log \left (\frac {2}{1+i (c+d x)}\right ) (a+b \arctan (c+d x))^2-b \left (-\frac {(a+b \arctan (c+d x))^2}{2 b}+a (c+d x)+b (c+d x) \arctan (c+d x)-\frac {1}{2} b \log \left ((c+d x)^2+1\right )\right )\right )\right )}{d}\) |
(e^2*(((c + d*x)^3*(a + b*ArcTan[c + d*x])^3)/3 - b*(((c + d*x)^2*(a + b*A rcTan[c + d*x])^2)/2 + ((I/3)*(a + b*ArcTan[c + d*x])^3)/b + (a + b*ArcTan [c + d*x])^2*Log[2/(1 + I*(c + d*x))] - b*(a*(c + d*x) + b*(c + d*x)*ArcTa n[c + d*x] - (a + b*ArcTan[c + d*x])^2/(2*b) - (b*Log[1 + (c + d*x)^2])/2) - 2*b*((-1/2*I)*(a + b*ArcTan[c + d*x])*PolyLog[2, 1 - 2/(1 + I*(c + d*x) )] - (b*PolyLog[3, 1 - 2/(1 + I*(c + d*x))])/4))))/d
3.1.15.3.1 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[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] & & IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(Log[2/(1 + e*(x/d))]/e), x] + Simp[b*c*( p/e) Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 + c^2*x^2)) , x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0 ]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbo l] :> Simp[(a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e _.)*(x_)^2), x_Symbol] :> Simp[f^2/e Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x] )^p, x], x] - Simp[d*(f^2/e) Int[(f*x)^(m - 2)*((a + b*ArcTan[c*x])^p/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcTan[c*x])^(p + 1)/(b*e*(p + 1))), x] - Si mp[1/(c*d) Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]
Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2 ), x_Symbol] :> Simp[(-I)*(a + b*ArcTan[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)) , x] + Simp[b*p*(I/2) Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[2, 1 - u]/ (d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c ^2*d] && EqQ[(1 - u)^2 - (1 - 2*(I/(I - c*x)))^2, 0]
Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[(f*(x/d))^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0] && IGtQ[p, 0]
Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /; !FalseQ[w]] /; FreeQ[n, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 6.18 (sec) , antiderivative size = 1249, normalized size of antiderivative = 4.61
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1249\) |
| default | \(\text {Expression too large to display}\) | \(1249\) |
| parts | \(\text {Expression too large to display}\) | \(1257\) |
1/d*(1/3*e^2*a^3*(d*x+c)^3+e^2*b^3*(1/3*(d*x+c)^3*arctan(d*x+c)^3-1/2*(d*x +c)^2*arctan(d*x+c)^2+1/2*arctan(d*x+c)^2*ln(1+(d*x+c)^2)-arctan(d*x+c)^2* ln((1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+I*arctan(d*x+c)*polylog(2,-(1+I*(d*x +c))^2/(1+(d*x+c)^2))-1/2*polylog(3,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))+1/12*I *arctan(d*x+c)*(3*csgn(I*(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))^2*csgn(I*(1+I* (d*x+c))^2/(1+(d*x+c)^2))*Pi*arctan(d*x+c)-6*csgn(I*(1+I*(d*x+c))/(1+(d*x+ c)^2)^(1/2))*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2*Pi*arctan(d*x+c)+3*cs gn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))^3*Pi*arctan(d*x+c)-3*csgn(I*(1+I*(d*x+ c))^2/(1+(d*x+c)^2))*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(1+(1+I*(d*x+c)) ^2/(1+(d*x+c)^2))^2)^2*Pi*arctan(d*x+c)+3*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c )^2))*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2 ))^2)*csgn(I/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)*Pi*arctan(d*x+c)-3*csgn( I*(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2*csgn(I*(1+(1+I*(d*x+c))^2/(1+(d*x+c )^2))^2)*Pi*arctan(d*x+c)+6*csgn(I*(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*csgn (I*(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)^2*Pi*arctan(d*x+c)-3*csgn(I*(1+(1+ I*(d*x+c))^2/(1+(d*x+c)^2))^2)^3*Pi*arctan(d*x+c)+3*csgn(I*(1+I*(d*x+c))^2 /(1+(d*x+c)^2)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)^3*Pi*arctan(d*x+c)-3*c sgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)^2 *csgn(I/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)*Pi*arctan(d*x+c)+4*arctan(d*x +c)^2+12*I*ln(2)*arctan(d*x+c)+6*I*arctan(d*x+c)-12-12*I*(d*x+c))+ln(1+...
\[ \int (c e+d e x)^2 (a+b \arctan (c+d x))^3 \, dx=\int { {\left (d e x + c e\right )}^{2} {\left (b \arctan \left (d x + c\right ) + a\right )}^{3} \,d x } \]
integral(a^3*d^2*e^2*x^2 + 2*a^3*c*d*e^2*x + a^3*c^2*e^2 + (b^3*d^2*e^2*x^ 2 + 2*b^3*c*d*e^2*x + b^3*c^2*e^2)*arctan(d*x + c)^3 + 3*(a*b^2*d^2*e^2*x^ 2 + 2*a*b^2*c*d*e^2*x + a*b^2*c^2*e^2)*arctan(d*x + c)^2 + 3*(a^2*b*d^2*e^ 2*x^2 + 2*a^2*b*c*d*e^2*x + a^2*b*c^2*e^2)*arctan(d*x + c), x)
\[ \int (c e+d e x)^2 (a+b \arctan (c+d x))^3 \, dx=e^{2} \left (\int a^{3} c^{2}\, dx + \int a^{3} d^{2} x^{2}\, dx + \int b^{3} c^{2} \operatorname {atan}^{3}{\left (c + d x \right )}\, dx + \int 3 a b^{2} c^{2} \operatorname {atan}^{2}{\left (c + d x \right )}\, dx + \int 3 a^{2} b c^{2} \operatorname {atan}{\left (c + d x \right )}\, dx + \int 2 a^{3} c d x\, dx + \int b^{3} d^{2} x^{2} \operatorname {atan}^{3}{\left (c + d x \right )}\, dx + \int 3 a b^{2} d^{2} x^{2} \operatorname {atan}^{2}{\left (c + d x \right )}\, dx + \int 3 a^{2} b d^{2} x^{2} \operatorname {atan}{\left (c + d x \right )}\, dx + \int 2 b^{3} c d x \operatorname {atan}^{3}{\left (c + d x \right )}\, dx + \int 6 a b^{2} c d x \operatorname {atan}^{2}{\left (c + d x \right )}\, dx + \int 6 a^{2} b c d x \operatorname {atan}{\left (c + d x \right )}\, dx\right ) \]
e**2*(Integral(a**3*c**2, x) + Integral(a**3*d**2*x**2, x) + Integral(b**3 *c**2*atan(c + d*x)**3, x) + Integral(3*a*b**2*c**2*atan(c + d*x)**2, x) + Integral(3*a**2*b*c**2*atan(c + d*x), x) + Integral(2*a**3*c*d*x, x) + In tegral(b**3*d**2*x**2*atan(c + d*x)**3, x) + Integral(3*a*b**2*d**2*x**2*a tan(c + d*x)**2, x) + Integral(3*a**2*b*d**2*x**2*atan(c + d*x), x) + Inte gral(2*b**3*c*d*x*atan(c + d*x)**3, x) + Integral(6*a*b**2*c*d*x*atan(c + d*x)**2, x) + Integral(6*a**2*b*c*d*x*atan(c + d*x), x))
\[ \int (c e+d e x)^2 (a+b \arctan (c+d x))^3 \, dx=\int { {\left (d e x + c e\right )}^{2} {\left (b \arctan \left (d x + c\right ) + a\right )}^{3} \,d x } \]
7/8*b^3*c^4*e^2*arctan(d*x + c)^3*arctan((d^2*x + c*d)/d)/d + 3*a*b^2*c^4* e^2*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)/d - (3*arctan(d*x + c)*arcta n((d^2*x + c*d)/d)^2/d - arctan((d^2*x + c*d)/d)^3/d)*a*b^2*c^4*e^2 - 7/32 *(6*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)^2/d - 4*arctan(d*x + c)*arct an((d^2*x + c*d)/d)^3/d + arctan((d^2*x + c*d)/d)^4/d)*b^3*c^4*e^2 + 1/3*a ^3*d^2*e^2*x^3 + 7/8*b^3*c^2*e^2*arctan(d*x + c)^3*arctan((d^2*x + c*d)/d) /d + 28*b^3*d^4*e^2*integrate(1/32*x^4*arctan(d*x + c)^3/(d^2*x^2 + 2*c*d* x + c^2 + 1), x) + 3*b^3*d^4*e^2*integrate(1/32*x^4*arctan(d*x + c)*log(d^ 2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 96*a*b^2* d^4*e^2*integrate(1/32*x^4*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1) , x) + 112*b^3*c*d^3*e^2*integrate(1/32*x^3*arctan(d*x + c)^3/(d^2*x^2 + 2 *c*d*x + c^2 + 1), x) + 4*b^3*d^4*e^2*integrate(1/32*x^4*arctan(d*x + c)*l og(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 12*b^3 *c*d^3*e^2*integrate(1/32*x^3*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 384*a*b^2*c*d^3*e^2*integrate(1 /32*x^3*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 168*b^3*c^2* d^2*e^2*integrate(1/32*x^2*arctan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1) , x) + 16*b^3*c*d^3*e^2*integrate(1/32*x^3*arctan(d*x + c)*log(d^2*x^2 + 2 *c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 18*b^3*c^2*d^2*e^2*i ntegrate(1/32*x^2*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d...
\[ \int (c e+d e x)^2 (a+b \arctan (c+d x))^3 \, dx=\int { {\left (d e x + c e\right )}^{2} {\left (b \arctan \left (d x + c\right ) + a\right )}^{3} \,d x } \]
Timed out. \[ \int (c e+d e x)^2 (a+b \arctan (c+d x))^3 \, dx=\int {\left (c\,e+d\,e\,x\right )}^2\,{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3 \,d x \]