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

Optimal. Leaf size=271 \[ \frac{9 b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (-c x+i)}+\frac{3 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (-c x+i)^2}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (-c x+i)}-\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (-c x+i)^2}-\frac{9 b \left (a+b \tan ^{-1}(c x)\right )^2}{32 c d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^3 (1+i c x)^2}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{8 c d^3}-\frac{21 i b^3}{64 c d^3 (-c x+i)}+\frac{3 b^3}{64 c d^3 (-c x+i)^2}+\frac{21 i b^3 \tan ^{-1}(c x)}{64 c d^3} \]

[Out]

(3*b^3)/(64*c*d^3*(I - c*x)^2) - (((21*I)/64)*b^3)/(c*d^3*(I - c*x)) + (((21*I)/64)*b^3*ArcTan[c*x])/(c*d^3) +
(((3*I)/16)*b^2*(a + b*ArcTan[c*x]))/(c*d^3*(I - c*x)^2) + (9*b^2*(a + b*ArcTan[c*x]))/(16*c*d^3*(I - c*x)) -
(9*b*(a + b*ArcTan[c*x])^2)/(32*c*d^3) - (3*b*(a + b*ArcTan[c*x])^2)/(8*c*d^3*(I - c*x)^2) + (((3*I)/8)*b*(a
+ b*ArcTan[c*x])^2)/(c*d^3*(I - c*x)) - ((I/8)*(a + b*ArcTan[c*x])^3)/(c*d^3) + ((I/2)*(a + b*ArcTan[c*x])^3)/
(c*d^3*(1 + I*c*x)^2)

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Rubi [A]  time = 0.403001, antiderivative size = 271, normalized size of antiderivative = 1., number of steps used = 24, 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} \[ \frac{9 b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (-c x+i)}+\frac{3 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (-c x+i)^2}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (-c x+i)}-\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (-c x+i)^2}-\frac{9 b \left (a+b \tan ^{-1}(c x)\right )^2}{32 c d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^3 (1+i c x)^2}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{8 c d^3}-\frac{21 i b^3}{64 c d^3 (-c x+i)}+\frac{3 b^3}{64 c d^3 (-c x+i)^2}+\frac{21 i b^3 \tan ^{-1}(c x)}{64 c d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^3)/(64*c*d^3*(I - c*x)^2) - (((21*I)/64)*b^3)/(c*d^3*(I - c*x)) + (((21*I)/64)*b^3*ArcTan[c*x])/(c*d^3) +
(((3*I)/16)*b^2*(a + b*ArcTan[c*x]))/(c*d^3*(I - c*x)^2) + (9*b^2*(a + b*ArcTan[c*x]))/(16*c*d^3*(I - c*x)) -
(9*b*(a + b*ArcTan[c*x])^2)/(32*c*d^3) - (3*b*(a + b*ArcTan[c*x])^2)/(8*c*d^3*(I - c*x)^2) + (((3*I)/8)*b*(a
+ b*ArcTan[c*x])^2)/(c*d^3*(I - c*x)) - ((I/8)*(a + b*ArcTan[c*x])^3)/(c*d^3) + ((I/2)*(a + b*ArcTan[c*x])^3)/
(c*d^3*(1 + I*c*x)^2)

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)^3} \, dx &=\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^3 (1+i c x)^2}-\frac{(3 i b) \int \left (\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 d^2 (-i+c x)^3}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{4 d^2 (-i+c x)^2}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{4 d^2 \left (1+c^2 x^2\right )}\right ) \, dx}{2 d}\\ &=\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^3 (1+i c x)^2}+\frac{(3 i b) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^2} \, dx}{8 d^3}-\frac{(3 i b) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{8 d^3}+\frac{(3 b) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^3} \, dx}{4 d^3}\\ &=-\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)^2}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{8 c d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^3 (1+i c x)^2}+\frac{\left (3 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^3}+\frac{\left (3 b^2\right ) \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^3}\\ &=-\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)^2}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{8 c d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^3 (1+i c x)^2}-\frac{\left (3 i b^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{8 d^3}+\frac{\left (3 b^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{16 d^3}-\frac{\left (3 b^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{16 d^3}+\frac{\left (3 b^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{8 d^3}-\frac{\left (3 b^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{8 d^3}\\ &=\frac{3 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (i-c x)^2}+\frac{9 b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (i-c x)}-\frac{9 b \left (a+b \tan ^{-1}(c x)\right )^2}{32 c d^3}-\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)^2}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{8 c d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^3 (1+i c x)^2}-\frac{\left (3 i b^3\right ) \int \frac{1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{16 d^3}+\frac{\left (3 b^3\right ) \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{16 d^3}+\frac{\left (3 b^3\right ) \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{8 d^3}\\ &=\frac{3 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (i-c x)^2}+\frac{9 b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (i-c x)}-\frac{9 b \left (a+b \tan ^{-1}(c x)\right )^2}{32 c d^3}-\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)^2}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{8 c d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^3 (1+i c x)^2}-\frac{\left (3 i b^3\right ) \int \frac{1}{(-i+c x)^3 (i+c x)} \, dx}{16 d^3}+\frac{\left (3 b^3\right ) \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{16 d^3}+\frac{\left (3 b^3\right ) \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{8 d^3}\\ &=\frac{3 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (i-c x)^2}+\frac{9 b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (i-c x)}-\frac{9 b \left (a+b \tan ^{-1}(c x)\right )^2}{32 c d^3}-\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)^2}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{8 c d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^3 (1+i c x)^2}-\frac{\left (3 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^3}+\frac{\left (3 b^3\right ) \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{16 d^3}+\frac{\left (3 b^3\right ) \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{8 d^3}\\ &=\frac{3 b^3}{64 c d^3 (i-c x)^2}-\frac{21 i b^3}{64 c d^3 (i-c x)}+\frac{3 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (i-c x)^2}+\frac{9 b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (i-c x)}-\frac{9 b \left (a+b \tan ^{-1}(c x)\right )^2}{32 c d^3}-\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)^2}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{8 c d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^3 (1+i c x)^2}+\frac{\left (3 i b^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{64 d^3}+\frac{\left (3 i b^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{32 d^3}+\frac{\left (3 i b^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{16 d^3}\\ &=\frac{3 b^3}{64 c d^3 (i-c x)^2}-\frac{21 i b^3}{64 c d^3 (i-c x)}+\frac{21 i b^3 \tan ^{-1}(c x)}{64 c d^3}+\frac{3 i b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (i-c x)^2}+\frac{9 b^2 \left (a+b \tan ^{-1}(c x)\right )}{16 c d^3 (i-c x)}-\frac{9 b \left (a+b \tan ^{-1}(c x)\right )^2}{32 c d^3}-\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)^2}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{8 c d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^3 (1+i c x)^2}\\ \end{align*}

Mathematica [A]  time = 0.273388, size = 183, normalized size = 0.68 \[ -\frac{i \left (3 b (c x+i) \tan ^{-1}(c x) \left (8 a^2 (c x-3 i)+4 a b (-5-3 i c x)+b^2 (-7 c x+9 i)\right )+24 a^2 b (c x-2 i)+32 a^3+12 a b^2 (-4-3 i c x)+6 b^2 (c x+i) \tan ^{-1}(c x)^2 (4 a (c x-3 i)+b (-5-3 i c x))+8 b^3 \left (c^2 x^2-2 i c x+3\right ) \tan ^{-1}(c x)^3+3 b^3 (-7 c x+8 i)\right )}{64 c d^3 (c x-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/64)*(32*a^3 + 3*b^3*(8*I - 7*c*x) + 12*a*b^2*(-4 - (3*I)*c*x) + 24*a^2*b*(-2*I + c*x) + 3*b*(I + c*x)*(b^
2*(9*I - 7*c*x) + 4*a*b*(-5 - (3*I)*c*x) + 8*a^2*(-3*I + c*x))*ArcTan[c*x] + 6*b^2*(I + c*x)*(b*(-5 - (3*I)*c*
x) + 4*a*(-3*I + c*x))*ArcTan[c*x]^2 + 8*b^3*(3 - (2*I)*c*x + c^2*x^2)*ArcTan[c*x]^3))/(c*d^3*(-I + c*x)^2)

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Maple [B]  time = 0.395, size = 711, normalized size = 2.6 \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)^3,x)

[Out]

3/8/c*b^3/d^3/(c*x-I)^2+1/2*I/c*a^3/d^3/(1+I*c*x)^2+21/64*I*b^3/d^3/(c*x-I)^2*x-15/32/c*b^3/d^3/(c*x-I)^2*arct
an(c*x)^2+3/32*b^3/d^3/(c*x-I)^2*arctan(c*x)*x+3/2*I/c*a*b^2/d^3/(1+I*c*x)^2*arctan(c*x)^2-3/4*I/c*a*b^2/d^3*a
rctan(c*x)/(c*x-I)+3/16*I/c*a*b^2/d^3*ln(-1/2*I*(-c*x+I))*ln(c*x+I)-3/16*I/c*a*b^2/d^3*ln(-1/2*I*(-c*x+I))*ln(
-1/2*I*(c*x+I))+3/16*I/c*a*b^2/d^3*ln(c*x-I)*ln(-1/2*I*(c*x+I))+3/2*I/c*a^2*b/d^3/(1+I*c*x)^2*arctan(c*x)-1/8*
I*c*b^3/d^3/(c*x-I)^2*arctan(c*x)^3*x^2+21/64*I*c*b^3/d^3/(c*x-I)^2*x^2*arctan(c*x)-3/8/c*a^2*b/d^3/(c*x-I)^2-
9/16/c*a*b^2/d^3/(c*x-I)-9/16/c*a*b^2/d^3*arctan(c*x)-1/4*b^3/d^3/(c*x-I)^2*arctan(c*x)^3*x-3/32*I/c*a*b^2/d^3
*ln(c*x-I)^2-3/8*I/c*a^2*b/d^3*arctan(c*x)-3/8*I/c*a^2*b/d^3/(c*x-I)+1/2*I/c*b^3/d^3/(1+I*c*x)^2*arctan(c*x)^3
-3/8/c*a*b^2/d^3*arctan(c*x)*ln(c*x-I)-3/4/c*a*b^2/d^3/(c*x-I)^2*arctan(c*x)+3/8/c*a*b^2/d^3*arctan(c*x)*ln(c*
x+I)-9/32*c*b^3/d^3/(c*x-I)^2*arctan(c*x)^2*x^2+3/16*I*b^3/d^3/(c*x-I)^2*arctan(c*x)^2*x+1/8*I/c*b^3/d^3/(c*x-
I)^2*arctan(c*x)^3+27/64*I/c*b^3/d^3/(c*x-I)^2*arctan(c*x)-3/32*I/c*a*b^2/d^3*ln(c*x+I)^2+3/16*I/c*a*b^2/d^3/(
c*x-I)^2

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Maxima [A]  time = 1.48003, size = 313, normalized size = 1.15 \begin{align*} -\frac{{\left (8 i \, b^{3} c^{2} x^{2} + 16 \, b^{3} c x + 24 i \, b^{3}\right )} \arctan \left (c x\right )^{3} + 32 i \, a^{3} + 48 \, a^{2} b - 48 i \, a b^{2} - 24 \, b^{3} +{\left (24 i \, a^{2} b + 36 \, a b^{2} - 21 i \, b^{3}\right )} c x -{\left (6 \,{\left (-4 i \, a b^{2} - 3 \, b^{3}\right )} c^{2} x^{2} - 72 i \, a b^{2} - 30 \, b^{3} -{\left (48 \, a b^{2} - 12 i \, b^{3}\right )} c x\right )} \arctan \left (c x\right )^{2} +{\left ({\left (24 i \, a^{2} b + 36 \, a b^{2} - 21 i \, b^{3}\right )} c^{2} x^{2} + 72 i \, a^{2} b + 60 \, a b^{2} - 27 i \, b^{3} + 6 \,{\left (8 \, a^{2} b - 4 i \, a b^{2} - b^{3}\right )} c x\right )} \arctan \left (c x\right )}{64 \, c^{3} d^{3} x^{2} - 128 i \, c^{2} d^{3} x - 64 \, c d^{3}} \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)^3,x, algorithm="maxima")

[Out]

-((8*I*b^3*c^2*x^2 + 16*b^3*c*x + 24*I*b^3)*arctan(c*x)^3 + 32*I*a^3 + 48*a^2*b - 48*I*a*b^2 - 24*b^3 + (24*I*
a^2*b + 36*a*b^2 - 21*I*b^3)*c*x - (6*(-4*I*a*b^2 - 3*b^3)*c^2*x^2 - 72*I*a*b^2 - 30*b^3 - (48*a*b^2 - 12*I*b^
3)*c*x)*arctan(c*x)^2 + ((24*I*a^2*b + 36*a*b^2 - 21*I*b^3)*c^2*x^2 + 72*I*a^2*b + 60*a*b^2 - 27*I*b^3 + 6*(8*
a^2*b - 4*I*a*b^2 - b^3)*c*x)*arctan(c*x))/(64*c^3*d^3*x^2 - 128*I*c^2*d^3*x - 64*c*d^3)

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Fricas [A]  time = 1.9108, size = 618, normalized size = 2.28 \begin{align*} -\frac{{\left (2 \, b^{3} c^{2} x^{2} - 4 i \, b^{3} c x + 6 \, b^{3}\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{3} + 64 i \, a^{3} + 96 \, a^{2} b - 96 i \, a b^{2} - 48 \, b^{3} -{\left (-48 i \, a^{2} b - 72 \, a b^{2} + 42 i \, b^{3}\right )} c x -{\left ({\left (12 i \, a b^{2} + 9 \, b^{3}\right )} c^{2} x^{2} + 36 i \, a b^{2} + 15 \, b^{3} + 6 \,{\left (4 \, a b^{2} - i \, b^{3}\right )} c x\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2} -{\left ({\left (24 \, a^{2} b - 36 i \, a b^{2} - 21 \, b^{3}\right )} c^{2} x^{2} + 72 \, a^{2} b - 60 i \, a b^{2} - 27 \, b^{3} +{\left (-48 i \, a^{2} b - 24 \, a b^{2} + 6 i \, b^{3}\right )} c x\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{128 \, c^{3} d^{3} x^{2} - 256 i \, c^{2} d^{3} x - 128 \, c d^{3}} \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)^3,x, algorithm="fricas")

[Out]

-((2*b^3*c^2*x^2 - 4*I*b^3*c*x + 6*b^3)*log(-(c*x + I)/(c*x - I))^3 + 64*I*a^3 + 96*a^2*b - 96*I*a*b^2 - 48*b^
3 - (-48*I*a^2*b - 72*a*b^2 + 42*I*b^3)*c*x - ((12*I*a*b^2 + 9*b^3)*c^2*x^2 + 36*I*a*b^2 + 15*b^3 + 6*(4*a*b^2
- I*b^3)*c*x)*log(-(c*x + I)/(c*x - I))^2 - ((24*a^2*b - 36*I*a*b^2 - 21*b^3)*c^2*x^2 + 72*a^2*b - 60*I*a*b^2
- 27*b^3 + (-48*I*a^2*b - 24*a*b^2 + 6*I*b^3)*c*x)*log(-(c*x + I)/(c*x - I)))/(128*c^3*d^3*x^2 - 256*I*c^2*d^
3*x - 128*c*d^3)

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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)**3,x)

[Out]

Timed out

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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 )}^{3}}\,{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)^3,x, algorithm="giac")

[Out]

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