\(\int (c e+d e x)^3 (a+b \arctan (c+d x))^2 \, dx\) [7]

Optimal result

Integrand size = 23, antiderivative size = 157 \[ \int (c e+d e x)^3 (a+b \arctan (c+d x))^2 \, dx=\frac {1}{2} a b e^3 x+\frac {b^2 e^3 (c+d x)^2}{12 d}+\frac {b^2 e^3 (c+d x) \arctan (c+d x)}{2 d}-\frac {b e^3 (c+d x)^3 (a+b \arctan (c+d x))}{6 d}-\frac {e^3 (a+b \arctan (c+d x))^2}{4 d}+\frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))^2}{4 d}-\frac {b^2 e^3 \log \left (1+(c+d x)^2\right )}{3 d} \]

[Out]

1/2*a*b*e^3*x+1/12*b^2*e^3*(d*x+c)^2/d+1/2*b^2*e^3*(d*x+c)*arctan(d*x+c)/d-1/6*b*e^3*(d*x+c)^3*(a+b*arctan(d*x
+c))/d-1/4*e^3*(a+b*arctan(d*x+c))^2/d+1/4*e^3*(d*x+c)^4*(a+b*arctan(d*x+c))^2/d-1/3*b^2*e^3*ln(1+(d*x+c)^2)/d

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {5151, 12, 4946, 5036, 272, 45, 4930, 266, 5004} \[ \int (c e+d e x)^3 (a+b \arctan (c+d x))^2 \, dx=\frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))^2}{4 d}-\frac {b e^3 (c+d x)^3 (a+b \arctan (c+d x))}{6 d}-\frac {e^3 (a+b \arctan (c+d x))^2}{4 d}+\frac {1}{2} a b e^3 x+\frac {b^2 e^3 (c+d x) \arctan (c+d x)}{2 d}+\frac {b^2 e^3 (c+d x)^2}{12 d}-\frac {b^2 e^3 \log \left ((c+d x)^2+1\right )}{3 d} \]

[In]

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

[Out]

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

Rule 12

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

Rule 45

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 266

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]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[b*c
*n*p, Int[x^n*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0
] && (EqQ[n, 1] || EqQ[p, 1])

Rule 4946

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] - Dist[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]

Rule 5004

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> 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]

Rule 5036

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[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]

Rule 5151

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int e^3 x^3 (a+b \arctan (x))^2 \, dx,x,c+d x\right )}{d} \\ & = \frac {e^3 \text {Subst}\left (\int x^3 (a+b \arctan (x))^2 \, dx,x,c+d x\right )}{d} \\ & = \frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))^2}{4 d}-\frac {\left (b e^3\right ) \text {Subst}\left (\int \frac {x^4 (a+b \arctan (x))}{1+x^2} \, dx,x,c+d x\right )}{2 d} \\ & = \frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))^2}{4 d}-\frac {\left (b e^3\right ) \text {Subst}\left (\int x^2 (a+b \arctan (x)) \, dx,x,c+d x\right )}{2 d}+\frac {\left (b e^3\right ) \text {Subst}\left (\int \frac {x^2 (a+b \arctan (x))}{1+x^2} \, dx,x,c+d x\right )}{2 d} \\ & = -\frac {b e^3 (c+d x)^3 (a+b \arctan (c+d x))}{6 d}+\frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))^2}{4 d}+\frac {\left (b e^3\right ) \text {Subst}(\int (a+b \arctan (x)) \, dx,x,c+d x)}{2 d}-\frac {\left (b e^3\right ) \text {Subst}\left (\int \frac {a+b \arctan (x)}{1+x^2} \, dx,x,c+d x\right )}{2 d}+\frac {\left (b^2 e^3\right ) \text {Subst}\left (\int \frac {x^3}{1+x^2} \, dx,x,c+d x\right )}{6 d} \\ & = \frac {1}{2} a b e^3 x-\frac {b e^3 (c+d x)^3 (a+b \arctan (c+d x))}{6 d}-\frac {e^3 (a+b \arctan (c+d x))^2}{4 d}+\frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))^2}{4 d}+\frac {\left (b^2 e^3\right ) \text {Subst}\left (\int \frac {x}{1+x} \, dx,x,(c+d x)^2\right )}{12 d}+\frac {\left (b^2 e^3\right ) \text {Subst}(\int \arctan (x) \, dx,x,c+d x)}{2 d} \\ & = \frac {1}{2} a b e^3 x+\frac {b^2 e^3 (c+d x) \arctan (c+d x)}{2 d}-\frac {b e^3 (c+d x)^3 (a+b \arctan (c+d x))}{6 d}-\frac {e^3 (a+b \arctan (c+d x))^2}{4 d}+\frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))^2}{4 d}+\frac {\left (b^2 e^3\right ) \text {Subst}\left (\int \left (1+\frac {1}{-1-x}\right ) \, dx,x,(c+d x)^2\right )}{12 d}-\frac {\left (b^2 e^3\right ) \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,c+d x\right )}{2 d} \\ & = \frac {1}{2} a b e^3 x+\frac {b^2 e^3 (c+d x)^2}{12 d}+\frac {b^2 e^3 (c+d x) \arctan (c+d x)}{2 d}-\frac {b e^3 (c+d x)^3 (a+b \arctan (c+d x))}{6 d}-\frac {e^3 (a+b \arctan (c+d x))^2}{4 d}+\frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))^2}{4 d}-\frac {b^2 e^3 \log \left (1+(c+d x)^2\right )}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.38 \[ \int (c e+d e x)^3 (a+b \arctan (c+d x))^2 \, dx=\frac {e^3 \left ((c+d x) \left (b^2 (c+d x)+3 a^2 (c+d x)^3-2 a b \left (-3+c^2+2 c d x+d^2 x^2\right )\right )+2 b \left (-b \left (-3 c+c^3-3 d x+3 c^2 d x+3 c d^2 x^2+d^3 x^3\right )+3 a \left (-1+c^4+4 c^3 d x+6 c^2 d^2 x^2+4 c d^3 x^3+d^4 x^4\right )\right ) \arctan (c+d x)+3 b^2 \left (-1+c^4+4 c^3 d x+6 c^2 d^2 x^2+4 c d^3 x^3+d^4 x^4\right ) \arctan (c+d x)^2-4 b^2 \log \left (1+(c+d x)^2\right )\right )}{12 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {\frac {e^{3} a^{2} \left (d x +c \right )^{4}}{4}+e^{3} b^{2} \left (\frac {\left (d x +c \right )^{4} \arctan \left (d x +c \right )^{2}}{4}-\frac {\left (d x +c \right )^{3} \arctan \left (d x +c \right )}{6}+\frac {\left (d x +c \right ) \arctan \left (d x +c \right )}{2}-\frac {\arctan \left (d x +c \right )^{2}}{4}+\frac {\left (d x +c \right )^{2}}{12}-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{3}\right )+2 e^{3} a b \left (\frac {\left (d x +c \right )^{4} \arctan \left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3}}{12}+\frac {d x}{4}+\frac {c}{4}-\frac {\arctan \left (d x +c \right )}{4}\right )}{d}\)	\(152\)
default	\(\frac {\frac {e^{3} a^{2} \left (d x +c \right )^{4}}{4}+e^{3} b^{2} \left (\frac {\left (d x +c \right )^{4} \arctan \left (d x +c \right )^{2}}{4}-\frac {\left (d x +c \right )^{3} \arctan \left (d x +c \right )}{6}+\frac {\left (d x +c \right ) \arctan \left (d x +c \right )}{2}-\frac {\arctan \left (d x +c \right )^{2}}{4}+\frac {\left (d x +c \right )^{2}}{12}-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{3}\right )+2 e^{3} a b \left (\frac {\left (d x +c \right )^{4} \arctan \left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3}}{12}+\frac {d x}{4}+\frac {c}{4}-\frac {\arctan \left (d x +c \right )}{4}\right )}{d}\)	\(152\)
parts	\(\frac {e^{3} a^{2} \left (d x +c \right )^{4}}{4 d}+\frac {e^{3} b^{2} \left (\frac {\left (d x +c \right )^{4} \arctan \left (d x +c \right )^{2}}{4}-\frac {\left (d x +c \right )^{3} \arctan \left (d x +c \right )}{6}+\frac {\left (d x +c \right ) \arctan \left (d x +c \right )}{2}-\frac {\arctan \left (d x +c \right )^{2}}{4}+\frac {\left (d x +c \right )^{2}}{12}-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{3}\right )}{d}+\frac {2 e^{3} a b \left (\frac {\left (d x +c \right )^{4} \arctan \left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3}}{12}+\frac {d x}{4}+\frac {c}{4}-\frac {\arctan \left (d x +c \right )}{4}\right )}{d}\)	\(157\)
parallelrisch	\(-\frac {18 e^{3} d \,c^{2} a^{2}+5 e^{3} d \,c^{2} b^{2}+e^{3} b^{2} d -36 x^{2} \arctan \left (d x +c \right ) a b \,c^{2} d^{3} e^{3}-24 x^{3} \arctan \left (d x +c \right ) a b c \,d^{4} e^{3}-24 x \arctan \left (d x +c \right ) a b \,c^{3} d^{2} e^{3}+6 a b c d \,e^{3}-18 a b \,c^{3} d \,e^{3}-3 x^{4} a^{2} d^{5} e^{3}+4 e^{3} b^{2} \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) d +3 e^{3} b^{2} \arctan \left (d x +c \right )^{2} d -x^{2} b^{2} d^{3} e^{3}-12 x^{3} a^{2} c \,d^{4} e^{3}+2 x^{3} a b \,d^{4} e^{3}-12 x \,a^{2} c^{3} d^{2} e^{3}-6 x a b \,d^{2} e^{3}+2 x^{3} \arctan \left (d x +c \right ) b^{2} d^{4} e^{3}-6 x \arctan \left (d x +c \right ) b^{2} d^{2} e^{3}+2 \arctan \left (d x +c \right ) b^{2} c^{3} d \,e^{3}-6 \arctan \left (d x +c \right ) b^{2} c d \,e^{3}+6 \arctan \left (d x +c \right ) a b d \,e^{3}-3 \arctan \left (d x +c \right )^{2} b^{2} c^{4} d \,e^{3}-18 x^{2} a^{2} c^{2} d^{3} e^{3}-3 d^{5} e^{3} b^{2} \arctan \left (d x +c \right )^{2} x^{4}-2 x \,b^{2} c \,d^{2} e^{3}-12 d^{4} e^{3} c \,b^{2} \arctan \left (d x +c \right )^{2} x^{3}-12 x \arctan \left (d x +c \right )^{2} b^{2} c^{3} d^{2} e^{3}+6 x a b \,c^{2} d^{2} e^{3}+6 x^{2} \arctan \left (d x +c \right ) b^{2} c \,d^{3} e^{3}-6 x^{4} \arctan \left (d x +c \right ) a b \,d^{5} e^{3}-18 x^{2} \arctan \left (d x +c \right )^{2} b^{2} c^{2} d^{3} e^{3}+6 x \arctan \left (d x +c \right ) b^{2} c^{2} d^{2} e^{3}-6 \arctan \left (d x +c \right ) a b \,c^{4} d \,e^{3}+6 x^{2} a b c \,d^{3} e^{3}+42 a^{2} c^{4} d \,e^{3}}{12 d^{2}}\)	\(594\)
risch	\(i e^{3} d^{2} a b c \,x^{3} \ln \left (1-i \left (d x +c \right )\right )+\frac {3 i e^{3} d a b \,c^{2} x^{2} \ln \left (1-i \left (d x +c \right )\right )}{2}-\frac {i e^{3} d \,b^{2} c \,x^{2} \ln \left (1-i \left (d x +c \right )\right )}{4}+i e^{3} a b \,c^{3} x \ln \left (1-i \left (d x +c \right )\right )+\frac {i e^{3} d^{3} a b \,x^{4} \ln \left (1-i \left (d x +c \right )\right )}{4}-\frac {e^{3} d^{2} b^{2} c \,x^{3} \ln \left (1-i \left (d x +c \right )\right )^{2}}{4}-\frac {3 e^{3} d \,b^{2} c^{2} x^{2} \ln \left (1-i \left (d x +c \right )\right )^{2}}{8}-\frac {i e^{3} b^{2} c^{2} x \ln \left (1-i \left (d x +c \right )\right )}{4}-\frac {i e^{3} d^{2} b^{2} x^{3} \ln \left (1-i \left (d x +c \right )\right )}{12}+\frac {e^{3} b \left (-6 i a \,d^{4} x^{4}+3 b \,d^{4} x^{4} \ln \left (1-i \left (d x +c \right )\right )-24 i a c \,d^{3} x^{3}+12 b c \,d^{3} x^{3} \ln \left (1-i \left (d x +c \right )\right )-36 i a \,c^{2} d^{2} x^{2}+2 i b \,d^{3} x^{3}+18 b \,c^{2} d^{2} x^{2} \ln \left (1-i \left (d x +c \right )\right )-24 i a \,c^{3} d x +6 i b c \,d^{2} x^{2}+12 b \,c^{3} d x \ln \left (1-i \left (d x +c \right )\right )+6 i b \,c^{2} d x +3 b \,c^{4} \ln \left (1-i \left (d x +c \right )\right )-6 i b d x -3 b \ln \left (1-i \left (d x +c \right )\right )\right ) \ln \left (1+i \left (d x +c \right )\right )}{24 d}+\frac {i e^{3} b^{2} x \ln \left (1-i \left (d x +c \right )\right )}{4}-\frac {e^{3} d^{3} b^{2} x^{4} \ln \left (1-i \left (d x +c \right )\right )^{2}}{16}-\frac {e^{3} b^{2} c^{4} \ln \left (1-i \left (d x +c \right )\right )^{2}}{16 d}-\frac {e^{3} b^{2} c^{3} x \ln \left (1-i \left (d x +c \right )\right )^{2}}{4}-\frac {e^{3} b^{2} \left (d^{4} x^{4}+4 c \,d^{3} x^{3}+6 c^{2} d^{2} x^{2}+4 c^{3} d x +c^{4}-1\right ) \ln \left (1+i \left (d x +c \right )\right )^{2}}{16 d}+\frac {a b \,e^{3} x}{2}+\frac {e^{3} a^{2} d^{3} x^{4}}{4}+e^{3} c^{3} a^{2} x +\frac {e^{3} d \,x^{2} b^{2}}{12}+\frac {e^{3} b^{2} c x}{6}+\frac {e^{3} b^{2} \ln \left (1-i \left (d x +c \right )\right )^{2}}{16 d}-\frac {e^{3} a b c d \,x^{2}}{2}-\frac {e^{3} b^{2} c^{3} \arctan \left (d x +c \right )}{6 d}+\frac {e^{3} b^{2} c \arctan \left (d x +c \right )}{2 d}-\frac {e^{3} a b \arctan \left (d x +c \right )}{2 d}-\frac {e^{3} b^{2} \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}{3 d}+e^{3} a^{2} c \,d^{2} x^{3}+\frac {3 e^{3} a^{2} c^{2} d \,x^{2}}{2}-\frac {e^{3} a b \,d^{2} x^{3}}{6}-\frac {e^{3} a b \,c^{2} x}{2}+\frac {e^{3} a b \,c^{4} \arctan \left (d x +c \right )}{2 d}\)	\(856\)

[In]

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

[Out]

1/d*(1/4*e^3*a^2*(d*x+c)^4+e^3*b^2*(1/4*(d*x+c)^4*arctan(d*x+c)^2-1/6*(d*x+c)^3*arctan(d*x+c)+1/2*(d*x+c)*arct
an(d*x+c)-1/4*arctan(d*x+c)^2+1/12*(d*x+c)^2-1/3*ln(1+(d*x+c)^2))+2*e^3*a*b*(1/4*(d*x+c)^4*arctan(d*x+c)-1/12*
(d*x+c)^3+1/4*d*x+1/4*c-1/4*arctan(d*x+c)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (143) = 286\).

Time = 0.29 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.15 \[ \int (c e+d e x)^3 (a+b \arctan (c+d x))^2 \, dx=\frac {3 \, a^{2} d^{4} e^{3} x^{4} + 2 \, {\left (6 \, a^{2} c - a b\right )} d^{3} e^{3} x^{3} + {\left (18 \, a^{2} c^{2} - 6 \, a b c + b^{2}\right )} d^{2} e^{3} x^{2} + 2 \, {\left (6 \, a^{2} c^{3} - 3 \, a b c^{2} + b^{2} c + 3 \, a b\right )} d e^{3} x - 4 \, b^{2} e^{3} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) + 3 \, {\left (b^{2} d^{4} e^{3} x^{4} + 4 \, b^{2} c d^{3} e^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} e^{3} x^{2} + 4 \, b^{2} c^{3} d e^{3} x + {\left (b^{2} c^{4} - b^{2}\right )} e^{3}\right )} \arctan \left (d x + c\right )^{2} + 2 \, {\left (3 \, a b d^{4} e^{3} x^{4} + {\left (12 \, a b c - b^{2}\right )} d^{3} e^{3} x^{3} + 3 \, {\left (6 \, a b c^{2} - b^{2} c\right )} d^{2} e^{3} x^{2} + 3 \, {\left (4 \, a b c^{3} - b^{2} c^{2} + b^{2}\right )} d e^{3} x + {\left (3 \, a b c^{4} - b^{2} c^{3} + 3 \, b^{2} c - 3 \, a b\right )} e^{3}\right )} \arctan \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

1/12*(3*a^2*d^4*e^3*x^4 + 2*(6*a^2*c - a*b)*d^3*e^3*x^3 + (18*a^2*c^2 - 6*a*b*c + b^2)*d^2*e^3*x^2 + 2*(6*a^2*
c^3 - 3*a*b*c^2 + b^2*c + 3*a*b)*d*e^3*x - 4*b^2*e^3*log(d^2*x^2 + 2*c*d*x + c^2 + 1) + 3*(b^2*d^4*e^3*x^4 + 4
*b^2*c*d^3*e^3*x^3 + 6*b^2*c^2*d^2*e^3*x^2 + 4*b^2*c^3*d*e^3*x + (b^2*c^4 - b^2)*e^3)*arctan(d*x + c)^2 + 2*(3
*a*b*d^4*e^3*x^4 + (12*a*b*c - b^2)*d^3*e^3*x^3 + 3*(6*a*b*c^2 - b^2*c)*d^2*e^3*x^2 + 3*(4*a*b*c^3 - b^2*c^2 +
b^2)*d*e^3*x + (3*a*b*c^4 - b^2*c^3 + 3*b^2*c - 3*a*b)*e^3)*arctan(d*x + c))/d

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 45.36 (sec) , antiderivative size = 583, normalized size of antiderivative = 3.71 \[ \int (c e+d e x)^3 (a+b \arctan (c+d x))^2 \, dx=\begin {cases} a^{2} c^{3} e^{3} x + \frac {3 a^{2} c^{2} d e^{3} x^{2}}{2} + a^{2} c d^{2} e^{3} x^{3} + \frac {a^{2} d^{3} e^{3} x^{4}}{4} + \frac {a b c^{4} e^{3} \operatorname {atan}{\left (c + d x \right )}}{2 d} + 2 a b c^{3} e^{3} x \operatorname {atan}{\left (c + d x \right )} + 3 a b c^{2} d e^{3} x^{2} \operatorname {atan}{\left (c + d x \right )} - \frac {a b c^{2} e^{3} x}{2} + 2 a b c d^{2} e^{3} x^{3} \operatorname {atan}{\left (c + d x \right )} - \frac {a b c d e^{3} x^{2}}{2} + \frac {a b d^{3} e^{3} x^{4} \operatorname {atan}{\left (c + d x \right )}}{2} - \frac {a b d^{2} e^{3} x^{3}}{6} + \frac {a b e^{3} x}{2} - \frac {a b e^{3} \operatorname {atan}{\left (c + d x \right )}}{2 d} + \frac {b^{2} c^{4} e^{3} \operatorname {atan}^{2}{\left (c + d x \right )}}{4 d} + b^{2} c^{3} e^{3} x \operatorname {atan}^{2}{\left (c + d x \right )} - \frac {b^{2} c^{3} e^{3} \operatorname {atan}{\left (c + d x \right )}}{6 d} + \frac {3 b^{2} c^{2} d e^{3} x^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{2} - \frac {b^{2} c^{2} e^{3} x \operatorname {atan}{\left (c + d x \right )}}{2} + b^{2} c d^{2} e^{3} x^{3} \operatorname {atan}^{2}{\left (c + d x \right )} - \frac {b^{2} c d e^{3} x^{2} \operatorname {atan}{\left (c + d x \right )}}{2} + \frac {b^{2} c e^{3} x}{6} + \frac {b^{2} c e^{3} \operatorname {atan}{\left (c + d x \right )}}{2 d} + \frac {b^{2} d^{3} e^{3} x^{4} \operatorname {atan}^{2}{\left (c + d x \right )}}{4} - \frac {b^{2} d^{2} e^{3} x^{3} \operatorname {atan}{\left (c + d x \right )}}{6} + \frac {b^{2} d e^{3} x^{2}}{12} + \frac {b^{2} e^{3} x \operatorname {atan}{\left (c + d x \right )}}{2} - \frac {2 b^{2} e^{3} \log {\left (\frac {c}{d} + x - \frac {i}{d} \right )}}{3 d} - \frac {b^{2} e^{3} \operatorname {atan}^{2}{\left (c + d x \right )}}{4 d} + \frac {2 i b^{2} e^{3} \operatorname {atan}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\c^{3} e^{3} x \left (a + b \operatorname {atan}{\left (c \right )}\right )^{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*c**3*e**3*x + 3*a**2*c**2*d*e**3*x**2/2 + a**2*c*d**2*e**3*x**3 + a**2*d**3*e**3*x**4/4 + a*b*
c**4*e**3*atan(c + d*x)/(2*d) + 2*a*b*c**3*e**3*x*atan(c + d*x) + 3*a*b*c**2*d*e**3*x**2*atan(c + d*x) - a*b*c
**2*e**3*x/2 + 2*a*b*c*d**2*e**3*x**3*atan(c + d*x) - a*b*c*d*e**3*x**2/2 + a*b*d**3*e**3*x**4*atan(c + d*x)/2
- a*b*d**2*e**3*x**3/6 + a*b*e**3*x/2 - a*b*e**3*atan(c + d*x)/(2*d) + b**2*c**4*e**3*atan(c + d*x)**2/(4*d)
+ b**2*c**3*e**3*x*atan(c + d*x)**2 - b**2*c**3*e**3*atan(c + d*x)/(6*d) + 3*b**2*c**2*d*e**3*x**2*atan(c + d*
x)**2/2 - b**2*c**2*e**3*x*atan(c + d*x)/2 + b**2*c*d**2*e**3*x**3*atan(c + d*x)**2 - b**2*c*d*e**3*x**2*atan(
c + d*x)/2 + b**2*c*e**3*x/6 + b**2*c*e**3*atan(c + d*x)/(2*d) + b**2*d**3*e**3*x**4*atan(c + d*x)**2/4 - b**2
*d**2*e**3*x**3*atan(c + d*x)/6 + b**2*d*e**3*x**2/12 + b**2*e**3*x*atan(c + d*x)/2 - 2*b**2*e**3*log(c/d + x
- I/d)/(3*d) - b**2*e**3*atan(c + d*x)**2/(4*d) + 2*I*b**2*e**3*atan(c + d*x)/(3*d), Ne(d, 0)), (c**3*e**3*x*(
a + b*atan(c))**2, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 597 vs. \(2 (143) = 286\).

Time = 1.06 (sec) , antiderivative size = 597, normalized size of antiderivative = 3.80 \[ \int (c e+d e x)^3 (a+b \arctan (c+d x))^2 \, dx=\frac {1}{4} \, a^{2} d^{3} e^{3} x^{4} + a^{2} c d^{2} e^{3} x^{3} + \frac {3}{2} \, a^{2} c^{2} d e^{3} x^{2} + 3 \, {\left (x^{2} \arctan \left (d x + c\right ) - d {\left (\frac {x}{d^{2}} + \frac {{\left (c^{2} - 1\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{3}} - \frac {c \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{3}}\right )}\right )} a b c^{2} d e^{3} + {\left (2 \, x^{3} \arctan \left (d x + c\right ) - d {\left (\frac {d x^{2} - 4 \, c x}{d^{3}} - \frac {2 \, {\left (c^{3} - 3 \, c\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{4}} + \frac {{\left (3 \, c^{2} - 1\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{4}}\right )}\right )} a b c d^{2} e^{3} + \frac {1}{6} \, {\left (3 \, x^{4} \arctan \left (d x + c\right ) - d {\left (\frac {d^{2} x^{3} - 3 \, c d x^{2} + 3 \, {\left (3 \, c^{2} - 1\right )} x}{d^{4}} + \frac {3 \, {\left (c^{4} - 6 \, c^{2} + 1\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{5}} - \frac {6 \, {\left (c^{3} - c\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{5}}\right )}\right )} a b d^{3} e^{3} + a^{2} c^{3} e^{3} x + \frac {{\left (2 \, {\left (d x + c\right )} \arctan \left (d x + c\right ) - \log \left ({\left (d x + c\right )}^{2} + 1\right )\right )} a b c^{3} e^{3}}{d} + \frac {b^{2} d^{2} e^{3} x^{2} + 2 \, b^{2} c d e^{3} x - 4 \, b^{2} e^{3} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) + 3 \, {\left (b^{2} d^{4} e^{3} x^{4} + 4 \, b^{2} c d^{3} e^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} e^{3} x^{2} + 4 \, b^{2} c^{3} d e^{3} x + {\left (b^{2} c^{4} - b^{2}\right )} e^{3}\right )} \arctan \left (d x + c\right )^{2} - 2 \, {\left (b^{2} d^{3} e^{3} x^{3} + 3 \, b^{2} c d^{2} e^{3} x^{2} + 3 \, {\left (b^{2} c^{2} - b^{2}\right )} d e^{3} x + {\left (b^{2} c^{3} - 3 \, b^{2} c\right )} e^{3}\right )} \arctan \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

1/4*a^2*d^3*e^3*x^4 + a^2*c*d^2*e^3*x^3 + 3/2*a^2*c^2*d*e^3*x^2 + 3*(x^2*arctan(d*x + c) - d*(x/d^2 + (c^2 - 1
)*arctan((d^2*x + c*d)/d)/d^3 - c*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^3))*a*b*c^2*d*e^3 + (2*x^3*arctan(d*x + c
) - d*((d*x^2 - 4*c*x)/d^3 - 2*(c^3 - 3*c)*arctan((d^2*x + c*d)/d)/d^4 + (3*c^2 - 1)*log(d^2*x^2 + 2*c*d*x + c
^2 + 1)/d^4))*a*b*c*d^2*e^3 + 1/6*(3*x^4*arctan(d*x + c) - d*((d^2*x^3 - 3*c*d*x^2 + 3*(3*c^2 - 1)*x)/d^4 + 3*
(c^4 - 6*c^2 + 1)*arctan((d^2*x + c*d)/d)/d^5 - 6*(c^3 - c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^5))*a*b*d^3*e^3
+ a^2*c^3*e^3*x + (2*(d*x + c)*arctan(d*x + c) - log((d*x + c)^2 + 1))*a*b*c^3*e^3/d + 1/12*(b^2*d^2*e^3*x^2
+ 2*b^2*c*d*e^3*x - 4*b^2*e^3*log(d^2*x^2 + 2*c*d*x + c^2 + 1) + 3*(b^2*d^4*e^3*x^4 + 4*b^2*c*d^3*e^3*x^3 + 6*
b^2*c^2*d^2*e^3*x^2 + 4*b^2*c^3*d*e^3*x + (b^2*c^4 - b^2)*e^3)*arctan(d*x + c)^2 - 2*(b^2*d^3*e^3*x^3 + 3*b^2*
c*d^2*e^3*x^2 + 3*(b^2*c^2 - b^2)*d*e^3*x + (b^2*c^3 - 3*b^2*c)*e^3)*arctan(d*x + c))/d

Giac [F]

\[ \int (c e+d e x)^3 (a+b \arctan (c+d x))^2 \, dx=\int { {\left (d e x + c e\right )}^{3} {\left (b \arctan \left (d x + c\right ) + a\right )}^{2} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 3.68 (sec) , antiderivative size = 633, normalized size of antiderivative = 4.03 \[ \int (c e+d e x)^3 (a+b \arctan (c+d x))^2 \, dx=x\,\left (\frac {c\,e^3\,\left (20\,a^2\,c^2+6\,a^2-6\,a\,b\,c+b^2\right )}{2}+\frac {\left (6\,c^2+6\right )\,\left (2\,a^2\,c\,d^2\,e^3+\frac {a\,d^2\,e^3\,\left (b-10\,a\,c\right )}{2}\right )}{6\,d^2}-\frac {2\,c\,\left (\frac {2\,c\,\left (2\,a^2\,c\,d^2\,e^3+\frac {a\,d^2\,e^3\,\left (b-10\,a\,c\right )}{2}\right )}{d}+\frac {d\,e^3\,\left (60\,a^2\,c^2+6\,a^2-12\,a\,b\,c+b^2\right )}{6}-\frac {a^2\,d\,e^3\,\left (6\,c^2+6\right )}{6}\right )}{d}\right )+x^2\,\left (\frac {c\,\left (2\,a^2\,c\,d^2\,e^3+\frac {a\,d^2\,e^3\,\left (b-10\,a\,c\right )}{2}\right )}{d}+\frac {d\,e^3\,\left (60\,a^2\,c^2+6\,a^2-12\,a\,b\,c+b^2\right )}{12}-\frac {a^2\,d\,e^3\,\left (6\,c^2+6\right )}{12}\right )-x^3\,\left (\frac {2\,a^2\,c\,d^2\,e^3}{3}+\frac {a\,d^2\,e^3\,\left (b-10\,a\,c\right )}{6}\right )+{\mathrm {atan}\left (c+d\,x\right )}^2\,\left (b^2\,c^3\,e^3\,x-\frac {b^2\,e^3-b^2\,c^4\,e^3}{4\,d}+\frac {b^2\,d^3\,e^3\,x^4}{4}+\frac {3\,b^2\,c^2\,d\,e^3\,x^2}{2}+b^2\,c\,d^2\,e^3\,x^3\right )-d^2\,\mathrm {atan}\left (c+d\,x\right )\,\left (x^3\,\left (\frac {b^2\,e^3}{6}-2\,a\,b\,c\,e^3\right )-\frac {x\,\left (-b^2\,c^2\,e^3+b^2\,e^3+4\,a\,b\,c^3\,e^3\right )}{2\,d^2}+\frac {x^2\,\left (b^2\,c\,e^3-6\,a\,b\,c^2\,e^3\right )}{2\,d}-\frac {a\,b\,d\,e^3\,x^4}{2}\right )+\frac {a^2\,d^3\,e^3\,x^4}{4}-\frac {b^2\,e^3\,\ln \left (c^2+2\,c\,d\,x+d^2\,x^2+1\right )}{3\,d}+\frac {b\,e^3\,\mathrm {atan}\left (\frac {\frac {b\,c\,e^3\,\left (-3\,a\,c^4+b\,c^3-3\,b\,c+3\,a\right )}{6}+\frac {b\,d\,e^3\,x\,\left (-3\,a\,c^4+b\,c^3-3\,b\,c+3\,a\right )}{6}}{-\frac {b^2\,c^3\,e^3}{6}+\frac {b^2\,c\,e^3}{2}+\frac {a\,b\,c^4\,e^3}{2}-\frac {a\,b\,e^3}{2}}\right )\,\left (-3\,a\,c^4+b\,c^3-3\,b\,c+3\,a\right )}{6\,d} \]

[In]

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

[Out]

x*((c*e^3*(6*a^2 + b^2 + 20*a^2*c^2 - 6*a*b*c))/2 + ((6*c^2 + 6)*(2*a^2*c*d^2*e^3 + (a*d^2*e^3*(b - 10*a*c))/2
))/(6*d^2) - (2*c*((2*c*(2*a^2*c*d^2*e^3 + (a*d^2*e^3*(b - 10*a*c))/2))/d + (d*e^3*(6*a^2 + b^2 + 60*a^2*c^2 -
12*a*b*c))/6 - (a^2*d*e^3*(6*c^2 + 6))/6))/d) + x^2*((c*(2*a^2*c*d^2*e^3 + (a*d^2*e^3*(b - 10*a*c))/2))/d + (
d*e^3*(6*a^2 + b^2 + 60*a^2*c^2 - 12*a*b*c))/12 - (a^2*d*e^3*(6*c^2 + 6))/12) - x^3*((2*a^2*c*d^2*e^3)/3 + (a*
d^2*e^3*(b - 10*a*c))/6) + atan(c + d*x)^2*(b^2*c^3*e^3*x - (b^2*e^3 - b^2*c^4*e^3)/(4*d) + (b^2*d^3*e^3*x^4)/
4 + (3*b^2*c^2*d*e^3*x^2)/2 + b^2*c*d^2*e^3*x^3) - d^2*atan(c + d*x)*(x^3*((b^2*e^3)/6 - 2*a*b*c*e^3) - (x*(b^
2*e^3 - b^2*c^2*e^3 + 4*a*b*c^3*e^3))/(2*d^2) + (x^2*(b^2*c*e^3 - 6*a*b*c^2*e^3))/(2*d) - (a*b*d*e^3*x^4)/2) +
(a^2*d^3*e^3*x^4)/4 - (b^2*e^3*log(c^2 + d^2*x^2 + 2*c*d*x + 1))/(3*d) + (b*e^3*atan(((b*c*e^3*(3*a - 3*b*c -
3*a*c^4 + b*c^3))/6 + (b*d*e^3*x*(3*a - 3*b*c - 3*a*c^4 + b*c^3))/6)/((b^2*c*e^3)/2 - (b^2*c^3*e^3)/6 - (a*b*
e^3)/2 + (a*b*c^4*e^3)/2))*(3*a - 3*b*c - 3*a*c^4 + b*c^3))/(6*d)