3.126 \(\int \frac{(a+b \tan ^{-1}(c x))^3}{(d+i c d x)^4} \, dx\)

Optimal. Leaf size=360 \[ \frac{11 b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (-c x+i)}+\frac{5 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (-c x+i)^2}-\frac{b^2 \left (a+b \tan ^{-1}(c x)\right )}{18 c d^4 (-c x+i)^3}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (-c x+i)}-\frac{b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (-c x+i)^2}-\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{6 c d^4 (-c x+i)^3}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )^2}{96 c d^4}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c d^4 (1+i c x)^3}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{24 c d^4}-\frac{85 i b^3}{576 c d^4 (-c x+i)}+\frac{19 b^3}{576 c d^4 (-c x+i)^2}+\frac{i b^3}{108 c d^4 (-c x+i)^3}+\frac{85 i b^3 \tan ^{-1}(c x)}{576 c d^4} \]

[Out]

((I/108)*b^3)/(c*d^4*(I - c*x)^3) + (19*b^3)/(576*c*d^4*(I - c*x)^2) - (((85*I)/576)*b^3)/(c*d^4*(I - c*x)) +
(((85*I)/576)*b^3*ArcTan[c*x])/(c*d^4) - (b^2*(a + b*ArcTan[c*x]))/(18*c*d^4*(I - c*x)^3) + (((5*I)/48)*b^2*(a
+ b*ArcTan[c*x]))/(c*d^4*(I - c*x)^2) + (11*b^2*(a + b*ArcTan[c*x]))/(48*c*d^4*(I - c*x)) - (11*b*(a + b*ArcT
an[c*x])^2)/(96*c*d^4) - ((I/6)*b*(a + b*ArcTan[c*x])^2)/(c*d^4*(I - c*x)^3) - (b*(a + b*ArcTan[c*x])^2)/(8*c*
d^4*(I - c*x)^2) + ((I/8)*b*(a + b*ArcTan[c*x])^2)/(c*d^4*(I - c*x)) - ((I/24)*(a + b*ArcTan[c*x])^3)/(c*d^4)
+ ((I/3)*(a + b*ArcTan[c*x])^3)/(c*d^4*(1 + I*c*x)^3)

________________________________________________________________________________________

Rubi [A]  time = 0.673546, antiderivative size = 360, normalized size of antiderivative = 1., number of steps used = 42, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4864, 4862, 627, 44, 203, 4884} \[ \frac{11 b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (-c x+i)}+\frac{5 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (-c x+i)^2}-\frac{b^2 \left (a+b \tan ^{-1}(c x)\right )}{18 c d^4 (-c x+i)^3}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (-c x+i)}-\frac{b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (-c x+i)^2}-\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{6 c d^4 (-c x+i)^3}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )^2}{96 c d^4}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c d^4 (1+i c x)^3}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{24 c d^4}-\frac{85 i b^3}{576 c d^4 (-c x+i)}+\frac{19 b^3}{576 c d^4 (-c x+i)^2}+\frac{i b^3}{108 c d^4 (-c x+i)^3}+\frac{85 i b^3 \tan ^{-1}(c x)}{576 c d^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])^3/(d + I*c*d*x)^4,x]

[Out]

((I/108)*b^3)/(c*d^4*(I - c*x)^3) + (19*b^3)/(576*c*d^4*(I - c*x)^2) - (((85*I)/576)*b^3)/(c*d^4*(I - c*x)) +
(((85*I)/576)*b^3*ArcTan[c*x])/(c*d^4) - (b^2*(a + b*ArcTan[c*x]))/(18*c*d^4*(I - c*x)^3) + (((5*I)/48)*b^2*(a
+ b*ArcTan[c*x]))/(c*d^4*(I - c*x)^2) + (11*b^2*(a + b*ArcTan[c*x]))/(48*c*d^4*(I - c*x)) - (11*b*(a + b*ArcT
an[c*x])^2)/(96*c*d^4) - ((I/6)*b*(a + b*ArcTan[c*x])^2)/(c*d^4*(I - c*x)^3) - (b*(a + b*ArcTan[c*x])^2)/(8*c*
d^4*(I - c*x)^2) + ((I/8)*b*(a + b*ArcTan[c*x])^2)/(c*d^4*(I - c*x)) - ((I/24)*(a + b*ArcTan[c*x])^3)/(c*d^4)
+ ((I/3)*(a + b*ArcTan[c*x])^3)/(c*d^4*(1 + I*c*x)^3)

Rule 4864

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a
+ b*ArcTan[c*x])^p)/(e*(q + 1)), x] - Dist[(b*c*p)/(e*(q + 1)), Int[ExpandIntegrand[(a + b*ArcTan[c*x])^(p -
1), (d + e*x)^(q + 1)/(1 + c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] && N
eQ[q, -1]

Rule 4862

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b*
ArcTan[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{
a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 4884

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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c x)\right )^3}{(d+i c d x)^4} \, dx &=\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c d^4 (1+i c x)^3}-\frac{(i b) \int \left (\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (-i+c x)^4}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{4 d^3 (-i+c x)^3}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3 (-i+c x)^2}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}\right ) \, dx}{d}\\ &=\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c d^4 (1+i c x)^3}+\frac{(i b) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^2} \, dx}{8 d^4}-\frac{(i b) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{8 d^4}-\frac{(i b) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^4} \, dx}{2 d^4}+\frac{b \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^3} \, dx}{4 d^4}\\ &=-\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{6 c d^4 (i-c x)^3}-\frac{b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)^2}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{24 c d^4}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c d^4 (1+i c x)^3}+\frac{\left (i b^2\right ) \int \left (-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 (-i+c x)^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{4 d^4}-\frac{\left (i b^2\right ) \int \left (-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 (-i+c x)^4}+\frac{a+b \tan ^{-1}(c x)}{4 (-i+c x)^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{8 (-i+c x)^2}-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{8 \left (1+c^2 x^2\right )}\right ) \, dx}{3 d^4}+\frac{b^2 \int \left (-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 (-i+c x)^3}+\frac{a+b \tan ^{-1}(c x)}{4 (-i+c x)^2}-\frac{a+b \tan ^{-1}(c x)}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{4 d^4}\\ &=-\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{6 c d^4 (i-c x)^3}-\frac{b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)^2}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{24 c d^4}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c d^4 (1+i c x)^3}-\frac{\left (i b^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{12 d^4}-\frac{\left (i b^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{8 d^4}+\frac{b^2 \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{24 d^4}-\frac{b^2 \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{24 d^4}+\frac{b^2 \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{16 d^4}-\frac{b^2 \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{16 d^4}+\frac{b^2 \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{8 d^4}-\frac{b^2 \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{8 d^4}-\frac{b^2 \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^4} \, dx}{6 d^4}\\ &=-\frac{b^2 \left (a+b \tan ^{-1}(c x)\right )}{18 c d^4 (i-c x)^3}+\frac{5 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (i-c x)^2}+\frac{11 b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (i-c x)}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )^2}{96 c d^4}-\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{6 c d^4 (i-c x)^3}-\frac{b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)^2}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{24 c d^4}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c d^4 (1+i c x)^3}-\frac{\left (i b^3\right ) \int \frac{1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{24 d^4}-\frac{\left (i b^3\right ) \int \frac{1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{16 d^4}+\frac{b^3 \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{24 d^4}-\frac{b^3 \int \frac{1}{(-i+c x)^3 \left (1+c^2 x^2\right )} \, dx}{18 d^4}+\frac{b^3 \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{16 d^4}+\frac{b^3 \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{8 d^4}\\ &=-\frac{b^2 \left (a+b \tan ^{-1}(c x)\right )}{18 c d^4 (i-c x)^3}+\frac{5 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (i-c x)^2}+\frac{11 b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (i-c x)}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )^2}{96 c d^4}-\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{6 c d^4 (i-c x)^3}-\frac{b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)^2}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{24 c d^4}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c d^4 (1+i c x)^3}-\frac{\left (i b^3\right ) \int \frac{1}{(-i+c x)^3 (i+c x)} \, dx}{24 d^4}-\frac{\left (i b^3\right ) \int \frac{1}{(-i+c x)^3 (i+c x)} \, dx}{16 d^4}+\frac{b^3 \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{24 d^4}-\frac{b^3 \int \frac{1}{(-i+c x)^4 (i+c x)} \, dx}{18 d^4}+\frac{b^3 \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{16 d^4}+\frac{b^3 \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{8 d^4}\\ &=-\frac{b^2 \left (a+b \tan ^{-1}(c x)\right )}{18 c d^4 (i-c x)^3}+\frac{5 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (i-c x)^2}+\frac{11 b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (i-c x)}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )^2}{96 c d^4}-\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{6 c d^4 (i-c x)^3}-\frac{b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)^2}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{24 c d^4}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c d^4 (1+i c x)^3}-\frac{\left (i b^3\right ) \int \left (-\frac{i}{2 (-i+c x)^3}+\frac{1}{4 (-i+c x)^2}-\frac{1}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{24 d^4}-\frac{\left (i b^3\right ) \int \left (-\frac{i}{2 (-i+c x)^3}+\frac{1}{4 (-i+c x)^2}-\frac{1}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{16 d^4}+\frac{b^3 \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{24 d^4}-\frac{b^3 \int \left (-\frac{i}{2 (-i+c x)^4}+\frac{1}{4 (-i+c x)^3}+\frac{i}{8 (-i+c x)^2}-\frac{i}{8 \left (1+c^2 x^2\right )}\right ) \, dx}{18 d^4}+\frac{b^3 \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{16 d^4}+\frac{b^3 \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{8 d^4}\\ &=\frac{i b^3}{108 c d^4 (i-c x)^3}+\frac{19 b^3}{576 c d^4 (i-c x)^2}-\frac{85 i b^3}{576 c d^4 (i-c x)}-\frac{b^2 \left (a+b \tan ^{-1}(c x)\right )}{18 c d^4 (i-c x)^3}+\frac{5 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (i-c x)^2}+\frac{11 b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (i-c x)}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )^2}{96 c d^4}-\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{6 c d^4 (i-c x)^3}-\frac{b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)^2}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{24 c d^4}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c d^4 (1+i c x)^3}+\frac{\left (i b^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{144 d^4}+\frac{\left (i b^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{96 d^4}+\frac{\left (i b^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{64 d^4}+\frac{\left (i b^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{48 d^4}+\frac{\left (i b^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{32 d^4}+\frac{\left (i b^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{16 d^4}\\ &=\frac{i b^3}{108 c d^4 (i-c x)^3}+\frac{19 b^3}{576 c d^4 (i-c x)^2}-\frac{85 i b^3}{576 c d^4 (i-c x)}+\frac{85 i b^3 \tan ^{-1}(c x)}{576 c d^4}-\frac{b^2 \left (a+b \tan ^{-1}(c x)\right )}{18 c d^4 (i-c x)^3}+\frac{5 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (i-c x)^2}+\frac{11 b^2 \left (a+b \tan ^{-1}(c x)\right )}{48 c d^4 (i-c x)}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )^2}{96 c d^4}-\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{6 c d^4 (i-c x)^3}-\frac{b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)^2}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^4 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{24 c d^4}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c d^4 (1+i c x)^3}\\ \end{align*}

Mathematica [A]  time = 0.296543, size = 269, normalized size = 0.75 \[ \frac{3 b (c x+i) \tan ^{-1}(c x) \left (-72 i a^2 \left (c^2 x^2-4 i c x-7\right )+12 a b \left (-11 c^2 x^2+32 i c x+29\right )+b^2 \left (85 i c^2 x^2+208 c x-139 i\right )\right )-72 i a^2 b \left (3 c^2 x^2-9 i c x-10\right )-576 a^3+12 a b^2 \left (-33 c^2 x^2+81 i c x+56\right )-18 i b^2 (c x+i) \tan ^{-1}(c x)^2 \left (12 a \left (c^2 x^2-4 i c x-7\right )+b \left (-11 i c^2 x^2-32 c x+29 i\right )\right )+b^3 \left (255 i c^2 x^2+567 c x-328 i\right )-72 i b^3 \left (c^3 x^3-3 i c^2 x^2-3 c x-7 i\right ) \tan ^{-1}(c x)^3}{1728 c d^4 (c x-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x])^3/(d + I*c*d*x)^4,x]

[Out]

(-576*a^3 + 12*a*b^2*(56 + (81*I)*c*x - 33*c^2*x^2) + b^3*(-328*I + 567*c*x + (255*I)*c^2*x^2) - (72*I)*a^2*b*
(-10 - (9*I)*c*x + 3*c^2*x^2) + 3*b*(I + c*x)*(12*a*b*(29 + (32*I)*c*x - 11*c^2*x^2) + b^2*(-139*I + 208*c*x +
(85*I)*c^2*x^2) - (72*I)*a^2*(-7 - (4*I)*c*x + c^2*x^2))*ArcTan[c*x] - (18*I)*b^2*(I + c*x)*(b*(29*I - 32*c*x
- (11*I)*c^2*x^2) + 12*a*(-7 - (4*I)*c*x + c^2*x^2))*ArcTan[c*x]^2 - (72*I)*b^3*(-7*I - 3*c*x - (3*I)*c^2*x^2
+ c^3*x^3)*ArcTan[c*x]^3)/(1728*c*d^4*(-I + c*x)^3)

________________________________________________________________________________________

Maple [B]  time = 0.467, size = 881, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))^3/(d+I*c*d*x)^4,x)

[Out]

139/576/c*b^3/d^4/(c*x-I)^3*arctan(c*x)-11/48/c*a*b^2/d^4*arctan(c*x)-1/8/c*a^2*b/d^4/(c*x-I)^2-41/216*I/c*b^3
/d^4/(c*x-I)^3+1/3*I/c*a^3/d^4/(1+I*c*x)^3-1/32*b^3/d^4/(c*x-I)^3*arctan(c*x)^2*x+1/18/c*a*b^2/d^4/(c*x-I)^3-1
1/48/c*a*b^2/d^4/(c*x-I)-1/8/c*a*b^2/d^4*arctan(c*x)*ln(c*x-I)-1/4/c*a*b^2/d^4*arctan(c*x)/(c*x-I)^2-1/8*I/c*a
^2*b/d^4*arctan(c*x)-1/32*I/c*a*b^2/d^4*ln(c*x+I)^2+1/6*I/c*a^2*b/d^4/(c*x-I)^3+1/8*I*b^3/d^4/(c*x-I)^3*arctan
(c*x)^3*x+23/192*I*b^3/d^4/(c*x-I)^3*arctan(c*x)*x+41/192*c*b^3/d^4/(c*x-I)^3*x^2*arctan(c*x)-1/8*c*b^3/d^4/(c
*x-I)^3*arctan(c*x)^3*x^2-11/96*c^2*b^3/d^4/(c*x-I)^3*arctan(c*x)^2*x^3+1/8/c*a*b^2/d^4*arctan(c*x)*ln(c*x+I)-
1/8*I/c*a^2*b/d^4/(c*x-I)+1/3*I/c*b^3/d^4/(1+I*c*x)^3*arctan(c*x)^3+29/96*I/c*b^3/d^4/(c*x-I)^3*arctan(c*x)^2+
5/48*I/c*a*b^2/d^4/(c*x-I)^2-1/32*I/c*a*b^2/d^4*ln(c*x-I)^2+85/576*I*c*b^3/d^4/(c*x-I)^3*x^2+1/24/c*b^3/d^4/(c
*x-I)^3*arctan(c*x)^3+7/32*I*c*b^3/d^4/(c*x-I)^3*arctan(c*x)^2*x^2+85/576*I*c^2*b^3/d^4/(c*x-I)^3*arctan(c*x)*
x^3+I/c*a^2*b/d^4/(1+I*c*x)^3*arctan(c*x)+I/c*a*b^2/d^4/(1+I*c*x)^3*arctan(c*x)^2-1/16*I/c*a*b^2/d^4*ln(-1/2*I
*(-c*x+I))*ln(-1/2*I*(c*x+I))+1/3*I/c*a*b^2/d^4*arctan(c*x)/(c*x-I)^3-1/4*I/c*a*b^2/d^4*arctan(c*x)/(c*x-I)+1/
16*I/c*a*b^2/d^4*ln(c*x-I)*ln(-1/2*I*(c*x+I))+1/16*I/c*a*b^2/d^4*ln(-1/2*I*(-c*x+I))*ln(c*x+I)-1/24*I*c^2*b^3/
d^4/(c*x-I)^3*arctan(c*x)^3*x^3+21/64*b^3/d^4/(c*x-I)^3*x

________________________________________________________________________________________

Maxima [A]  time = 1.86376, size = 439, normalized size = 1.22 \begin{align*} -\frac{{\left (216 i \, a^{2} b + 396 \, a b^{2} - 255 i \, b^{3}\right )} c^{2} x^{2} +{\left (72 i \, b^{3} c^{3} x^{3} + 216 \, b^{3} c^{2} x^{2} - 216 i \, b^{3} c x + 504 \, b^{3}\right )} \arctan \left (c x\right )^{3} + 576 \, a^{3} - 720 i \, a^{2} b - 672 \, a b^{2} + 328 i \, b^{3} + 81 \,{\left (8 \, a^{2} b - 12 i \, a b^{2} - 7 \, b^{3}\right )} c x -{\left (18 \,{\left (-12 i \, a b^{2} - 11 \, b^{3}\right )} c^{3} x^{3} -{\left (648 \, a b^{2} - 378 i \, b^{3}\right )} c^{2} x^{2} - 1512 \, a b^{2} + 522 i \, b^{3} + 54 \,{\left (12 i \, a b^{2} - b^{3}\right )} c x\right )} \arctan \left (c x\right )^{2} +{\left ({\left (216 i \, a^{2} b + 396 \, a b^{2} - 255 i \, b^{3}\right )} c^{3} x^{3} + 9 \,{\left (72 \, a^{2} b - 84 i \, a b^{2} - 41 \, b^{3}\right )} c^{2} x^{2} + 1512 \, a^{2} b - 1044 i \, a b^{2} - 417 \, b^{3} +{\left (-648 i \, a^{2} b + 108 \, a b^{2} - 207 i \, b^{3}\right )} c x\right )} \arctan \left (c x\right )}{1728 \, c^{4} d^{4} x^{3} - 5184 i \, c^{3} d^{4} x^{2} - 5184 \, c^{2} d^{4} x + 1728 i \, c d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^3/(d+I*c*d*x)^4,x, algorithm="maxima")

[Out]

-((216*I*a^2*b + 396*a*b^2 - 255*I*b^3)*c^2*x^2 + (72*I*b^3*c^3*x^3 + 216*b^3*c^2*x^2 - 216*I*b^3*c*x + 504*b^
3)*arctan(c*x)^3 + 576*a^3 - 720*I*a^2*b - 672*a*b^2 + 328*I*b^3 + 81*(8*a^2*b - 12*I*a*b^2 - 7*b^3)*c*x - (18
*(-12*I*a*b^2 - 11*b^3)*c^3*x^3 - (648*a*b^2 - 378*I*b^3)*c^2*x^2 - 1512*a*b^2 + 522*I*b^3 + 54*(12*I*a*b^2 -
b^3)*c*x)*arctan(c*x)^2 + ((216*I*a^2*b + 396*a*b^2 - 255*I*b^3)*c^3*x^3 + 9*(72*a^2*b - 84*I*a*b^2 - 41*b^3)*
c^2*x^2 + 1512*a^2*b - 1044*I*a*b^2 - 417*b^3 + (-648*I*a^2*b + 108*a*b^2 - 207*I*b^3)*c*x)*arctan(c*x))/(1728
*c^4*d^4*x^3 - 5184*I*c^3*d^4*x^2 - 5184*c^2*d^4*x + 1728*I*c*d^4)

________________________________________________________________________________________

Fricas [A]  time = 2.36914, size = 900, normalized size = 2.5 \begin{align*} \frac{{\left (-432 i \, a^{2} b - 792 \, a b^{2} + 510 i \, b^{3}\right )} c^{2} x^{2} -{\left (18 \, b^{3} c^{3} x^{3} - 54 i \, b^{3} c^{2} x^{2} - 54 \, b^{3} c x - 126 i \, b^{3}\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{3} - 1152 \, a^{3} + 1440 i \, a^{2} b + 1344 \, a b^{2} - 656 i \, b^{3} -{\left (1296 \, a^{2} b - 1944 i \, a b^{2} - 1134 \, b^{3}\right )} c x +{\left ({\left (108 i \, a b^{2} + 99 \, b^{3}\right )} c^{3} x^{3} + 27 \,{\left (12 \, a b^{2} - 7 i \, b^{3}\right )} c^{2} x^{2} + 756 \, a b^{2} - 261 i \, b^{3} +{\left (-324 i \, a b^{2} + 27 \, b^{3}\right )} c x\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2} +{\left ({\left (216 \, a^{2} b - 396 i \, a b^{2} - 255 \, b^{3}\right )} c^{3} x^{3} +{\left (-648 i \, a^{2} b - 756 \, a b^{2} + 369 i \, b^{3}\right )} c^{2} x^{2} - 1512 i \, a^{2} b - 1044 \, a b^{2} + 417 i \, b^{3} -{\left (648 \, a^{2} b + 108 i \, a b^{2} + 207 \, b^{3}\right )} c x\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{3456 \, c^{4} d^{4} x^{3} - 10368 i \, c^{3} d^{4} x^{2} - 10368 \, c^{2} d^{4} x + 3456 i \, c d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^3/(d+I*c*d*x)^4,x, algorithm="fricas")

[Out]

((-432*I*a^2*b - 792*a*b^2 + 510*I*b^3)*c^2*x^2 - (18*b^3*c^3*x^3 - 54*I*b^3*c^2*x^2 - 54*b^3*c*x - 126*I*b^3)
*log(-(c*x + I)/(c*x - I))^3 - 1152*a^3 + 1440*I*a^2*b + 1344*a*b^2 - 656*I*b^3 - (1296*a^2*b - 1944*I*a*b^2 -
1134*b^3)*c*x + ((108*I*a*b^2 + 99*b^3)*c^3*x^3 + 27*(12*a*b^2 - 7*I*b^3)*c^2*x^2 + 756*a*b^2 - 261*I*b^3 + (
-324*I*a*b^2 + 27*b^3)*c*x)*log(-(c*x + I)/(c*x - I))^2 + ((216*a^2*b - 396*I*a*b^2 - 255*b^3)*c^3*x^3 + (-648
*I*a^2*b - 756*a*b^2 + 369*I*b^3)*c^2*x^2 - 1512*I*a^2*b - 1044*a*b^2 + 417*I*b^3 - (648*a^2*b + 108*I*a*b^2 +
207*b^3)*c*x)*log(-(c*x + I)/(c*x - I)))/(3456*c^4*d^4*x^3 - 10368*I*c^3*d^4*x^2 - 10368*c^2*d^4*x + 3456*I*c
*d^4)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))**3/(d+I*c*d*x)**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}^{3}}{{\left (i \, c d x + d\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^3/(d+I*c*d*x)^4,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x) + a)^3/(I*c*d*x + d)^4, x)