\(\int \frac {(a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx\) [115]

Optimal result

Integrand size = 22, antiderivative size = 180 \[ \int \frac {(a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\frac {i b^2}{16 c d^3 (i-c x)^2}+\frac {3 b^2}{16 c d^3 (i-c x)}-\frac {3 b^2 \arctan (c x)}{16 c d^3}-\frac {b (a+b \arctan (c x))}{4 c d^3 (i-c x)^2}+\frac {i b (a+b \arctan (c x))}{4 c d^3 (i-c x)}-\frac {i (a+b \arctan (c x))^2}{8 c d^3}+\frac {i (a+b \arctan (c x))^2}{2 c d^3 (1+i c x)^2} \]

[Out]

1/16*I*b^2/c/d^3/(I-c*x)^2+3/16*b^2/c/d^3/(I-c*x)-3/16*b^2*arctan(c*x)/c/d^3-1/4*b*(a+b*arctan(c*x))/c/d^3/(I-
c*x)^2+1/4*I*b*(a+b*arctan(c*x))/c/d^3/(I-c*x)-1/8*I*(a+b*arctan(c*x))^2/c/d^3+1/2*I*(a+b*arctan(c*x))^2/c/d^3
/(1+I*c*x)^2

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4974, 4972, 641, 46, 209, 5004} \[ \int \frac {(a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\frac {i b (a+b \arctan (c x))}{4 c d^3 (-c x+i)}-\frac {b (a+b \arctan (c x))}{4 c d^3 (-c x+i)^2}+\frac {i (a+b \arctan (c x))^2}{2 c d^3 (1+i c x)^2}-\frac {i (a+b \arctan (c x))^2}{8 c d^3}-\frac {3 b^2 \arctan (c x)}{16 c d^3}+\frac {3 b^2}{16 c d^3 (-c x+i)}+\frac {i b^2}{16 c d^3 (-c x+i)^2} \]

[In]

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

[Out]

((I/16)*b^2)/(c*d^3*(I - c*x)^2) + (3*b^2)/(16*c*d^3*(I - c*x)) - (3*b^2*ArcTan[c*x])/(16*c*d^3) - (b*(a + b*A
rcTan[c*x]))/(4*c*d^3*(I - c*x)^2) + ((I/4)*b*(a + b*ArcTan[c*x]))/(c*d^3*(I - c*x)) - ((I/8)*(a + b*ArcTan[c*
x])^2)/(c*d^3) + ((I/2)*(a + b*ArcTan[c*x])^2)/(c*d^3*(1 + I*c*x)^2)

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 209

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

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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 4972

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 4974

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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {i (a+b \arctan (c x))^2}{2 c d^3 (1+i c x)^2}-\frac {(i b) \int \left (\frac {i (a+b \arctan (c x))}{2 d^2 (-i+c x)^3}-\frac {a+b \arctan (c x)}{4 d^2 (-i+c x)^2}+\frac {a+b \arctan (c x)}{4 d^2 \left (1+c^2 x^2\right )}\right ) \, dx}{d} \\ & = \frac {i (a+b \arctan (c x))^2}{2 c d^3 (1+i c x)^2}+\frac {(i b) \int \frac {a+b \arctan (c x)}{(-i+c x)^2} \, dx}{4 d^3}-\frac {(i b) \int \frac {a+b \arctan (c x)}{1+c^2 x^2} \, dx}{4 d^3}+\frac {b \int \frac {a+b \arctan (c x)}{(-i+c x)^3} \, dx}{2 d^3} \\ & = -\frac {b (a+b \arctan (c x))}{4 c d^3 (i-c x)^2}+\frac {i b (a+b \arctan (c x))}{4 c d^3 (i-c x)}-\frac {i (a+b \arctan (c x))^2}{8 c d^3}+\frac {i (a+b \arctan (c x))^2}{2 c d^3 (1+i c x)^2}+\frac {\left (i b^2\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{4 d^3}+\frac {b^2 \int \frac {1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{4 d^3} \\ & = -\frac {b (a+b \arctan (c x))}{4 c d^3 (i-c x)^2}+\frac {i b (a+b \arctan (c x))}{4 c d^3 (i-c x)}-\frac {i (a+b \arctan (c x))^2}{8 c d^3}+\frac {i (a+b \arctan (c x))^2}{2 c d^3 (1+i c x)^2}+\frac {\left (i b^2\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{4 d^3}+\frac {b^2 \int \frac {1}{(-i+c x)^3 (i+c x)} \, dx}{4 d^3} \\ & = -\frac {b (a+b \arctan (c x))}{4 c d^3 (i-c x)^2}+\frac {i b (a+b \arctan (c x))}{4 c d^3 (i-c x)}-\frac {i (a+b \arctan (c x))^2}{8 c d^3}+\frac {i (a+b \arctan (c x))^2}{2 c d^3 (1+i c x)^2}+\frac {\left (i b^2\right ) \int \left (-\frac {i}{2 (-i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{4 d^3}+\frac {b^2 \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}{4 d^3} \\ & = \frac {i b^2}{16 c d^3 (i-c x)^2}+\frac {3 b^2}{16 c d^3 (i-c x)}-\frac {b (a+b \arctan (c x))}{4 c d^3 (i-c x)^2}+\frac {i b (a+b \arctan (c x))}{4 c d^3 (i-c x)}-\frac {i (a+b \arctan (c x))^2}{8 c d^3}+\frac {i (a+b \arctan (c x))^2}{2 c d^3 (1+i c x)^2}-\frac {b^2 \int \frac {1}{1+c^2 x^2} \, dx}{16 d^3}-\frac {b^2 \int \frac {1}{1+c^2 x^2} \, dx}{8 d^3} \\ & = \frac {i b^2}{16 c d^3 (i-c x)^2}+\frac {3 b^2}{16 c d^3 (i-c x)}-\frac {3 b^2 \arctan (c x)}{16 c d^3}-\frac {b (a+b \arctan (c x))}{4 c d^3 (i-c x)^2}+\frac {i b (a+b \arctan (c x))}{4 c d^3 (i-c x)}-\frac {i (a+b \arctan (c x))^2}{8 c d^3}+\frac {i (a+b \arctan (c x))^2}{2 c d^3 (1+i c x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.61 \[ \int \frac {(a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=-\frac {i \left (8 a^2+b^2 (-4-3 i c x)+4 a b (-2 i+c x)+b (i+c x) (b (-5-3 i c x)+4 a (-3 i+c x)) \arctan (c x)+2 b^2 \left (3-2 i c x+c^2 x^2\right ) \arctan (c x)^2\right )}{16 c d^3 (-i+c x)^2} \]

[In]

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

[Out]

((-1/16*I)*(8*a^2 + b^2*(-4 - (3*I)*c*x) + 4*a*b*(-2*I + c*x) + b*(I + c*x)*(b*(-5 - (3*I)*c*x) + 4*a*(-3*I +
c*x))*ArcTan[c*x] + 2*b^2*(3 - (2*I)*c*x + c^2*x^2)*ArcTan[c*x]^2))/(c*d^3*(-I + c*x)^2)

Maple [A] (verified)

Time = 2.38 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.54

method	result	size
derivativedivides	\(\frac {\frac {i a^{2}}{2 d^{3} \left (i c x +1\right )^{2}}+\frac {b^{2} \left (\frac {i \arctan \left (c x \right )^{2}}{2 \left (i c x +1\right )^{2}}-i \left (\frac {i \arctan \left (c x \right ) \ln \left (c x +i\right )}{8}-\frac {i \arctan \left (c x \right ) \ln \left (c x -i\right )}{8}-\frac {i \arctan \left (c x \right )}{4 \left (c x -i\right )^{2}}+\frac {\arctan \left (c x \right )}{4 c x -4 i}-\frac {\left (\ln \left (c x +i\right )-\ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right ) \ln \left (-\frac {i \left (-c x +i\right )}{2}\right )}{16}+\frac {\ln \left (c x +i\right )^{2}}{32}-\frac {3 i \arctan \left (c x \right )}{16}-\frac {3 i}{16 \left (c x -i\right )}-\frac {1}{16 \left (c x -i\right )^{2}}-\frac {\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{16}+\frac {\ln \left (c x -i\right )^{2}}{32}\right )\right )}{d^{3}}+\frac {i a b \arctan \left (c x \right )}{d^{3} \left (i c x +1\right )^{2}}-\frac {i a b \arctan \left (c x \right )}{4 d^{3}}-\frac {a b}{4 d^{3} \left (c x -i\right )^{2}}-\frac {i a b}{4 d^{3} \left (c x -i\right )}}{c}\)	\(277\)
default	\(\frac {\frac {i a^{2}}{2 d^{3} \left (i c x +1\right )^{2}}+\frac {b^{2} \left (\frac {i \arctan \left (c x \right )^{2}}{2 \left (i c x +1\right )^{2}}-i \left (\frac {i \arctan \left (c x \right ) \ln \left (c x +i\right )}{8}-\frac {i \arctan \left (c x \right ) \ln \left (c x -i\right )}{8}-\frac {i \arctan \left (c x \right )}{4 \left (c x -i\right )^{2}}+\frac {\arctan \left (c x \right )}{4 c x -4 i}-\frac {\left (\ln \left (c x +i\right )-\ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right ) \ln \left (-\frac {i \left (-c x +i\right )}{2}\right )}{16}+\frac {\ln \left (c x +i\right )^{2}}{32}-\frac {3 i \arctan \left (c x \right )}{16}-\frac {3 i}{16 \left (c x -i\right )}-\frac {1}{16 \left (c x -i\right )^{2}}-\frac {\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{16}+\frac {\ln \left (c x -i\right )^{2}}{32}\right )\right )}{d^{3}}+\frac {i a b \arctan \left (c x \right )}{d^{3} \left (i c x +1\right )^{2}}-\frac {i a b \arctan \left (c x \right )}{4 d^{3}}-\frac {a b}{4 d^{3} \left (c x -i\right )^{2}}-\frac {i a b}{4 d^{3} \left (c x -i\right )}}{c}\)	\(277\)
parts	\(\frac {i a^{2}}{2 d^{3} \left (i c x +1\right )^{2} c}+\frac {b^{2} \left (\frac {i \arctan \left (c x \right )^{2}}{2 \left (i c x +1\right )^{2}}-i \left (\frac {i \arctan \left (c x \right ) \ln \left (c x +i\right )}{8}-\frac {i \arctan \left (c x \right ) \ln \left (c x -i\right )}{8}-\frac {i \arctan \left (c x \right )}{4 \left (c x -i\right )^{2}}+\frac {\arctan \left (c x \right )}{4 c x -4 i}-\frac {\left (\ln \left (c x +i\right )-\ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right ) \ln \left (-\frac {i \left (-c x +i\right )}{2}\right )}{16}+\frac {\ln \left (c x +i\right )^{2}}{32}-\frac {3 i \arctan \left (c x \right )}{16}-\frac {3 i}{16 \left (c x -i\right )}-\frac {1}{16 \left (c x -i\right )^{2}}-\frac {\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{16}+\frac {\ln \left (c x -i\right )^{2}}{32}\right )\right )}{d^{3} c}+\frac {i a b \arctan \left (c x \right )}{c \,d^{3} \left (i c x +1\right )^{2}}-\frac {i a b \arctan \left (c x \right )}{4 c \,d^{3}}-\frac {a b}{4 c \,d^{3} \left (c x -i\right )^{2}}-\frac {i a b}{4 c \,d^{3} \left (c x -i\right )}\)	\(291\)
risch	\(\frac {i b^{2} \left (c^{2} x^{2}-2 i c x +3\right ) \ln \left (i c x +1\right )^{2}}{32 d^{3} \left (c x -i\right )^{2} c}-\frac {\left (3 i \ln \left (-i c x +1\right ) b^{2}+i \ln \left (-i c x +1\right ) b^{2} c^{2} x^{2}+2 \ln \left (-i c x +1\right ) b^{2} c x +2 b^{2} c x -4 i b^{2}+8 a b \right ) \ln \left (i c x +1\right )}{16 d^{3} \left (c x -i\right )^{2} c}-\frac {i \left (4 i \ln \left (-i c x +1\right ) b^{2} c x -6 i b^{2} c x -b^{2} c^{2} x^{2} \ln \left (-i c x +1\right )^{2}+4 i \ln \left (\left (-3 i b c +4 a c \right ) x +4 i a +3 b \right ) a b \,c^{2} x^{2}+3 \ln \left (\left (-3 i b c +4 a c \right ) x +4 i a +3 b \right ) b^{2} c^{2} x^{2}-3 \ln \left (\left (3 i b c -4 a c \right ) x +4 i a +3 b \right ) b^{2} c^{2} x^{2}+6 i \ln \left (\left (3 i b c -4 a c \right ) x +4 i a +3 b \right ) b^{2} c x +4 i \ln \left (\left (3 i b c -4 a c \right ) x +4 i a +3 b \right ) a b -6 i \ln \left (\left (-3 i b c +4 a c \right ) x +4 i a +3 b \right ) b^{2} c x -4 i \ln \left (\left (-3 i b c +4 a c \right ) x +4 i a +3 b \right ) a b +8 \ln \left (\left (-3 i b c +4 a c \right ) x +4 i a +3 b \right ) a b c x -8 \ln \left (\left (3 i b c -4 a c \right ) x +4 i a +3 b \right ) a b c x -16 i a b +2 i b^{2} c x \ln \left (-i c x +1\right )^{2}-4 i \ln \left (\left (3 i b c -4 a c \right ) x +4 i a +3 b \right ) a b \,c^{2} x^{2}+8 a b c x -3 b^{2} \ln \left (-i c x +1\right )^{2}+16 i \ln \left (-i c x +1\right ) a b -3 \ln \left (\left (-3 i b c +4 a c \right ) x +4 i a +3 b \right ) b^{2}+3 \ln \left (\left (3 i b c -4 a c \right ) x +4 i a +3 b \right ) b^{2}+8 b^{2} \ln \left (-i c x +1\right )+16 a^{2}-8 b^{2}\right )}{32 d^{3} \left (c x -i\right )^{2} c}\)	\(624\)

[In]

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

[Out]

1/c*(1/2*I*a^2/d^3/(1+I*c*x)^2+b^2/d^3*(1/2*I/(1+I*c*x)^2*arctan(c*x)^2-I*(1/8*I*arctan(c*x)*ln(c*x+I)-1/8*I*a
rctan(c*x)*ln(c*x-I)-1/4*I*arctan(c*x)/(c*x-I)^2+1/4*arctan(c*x)/(c*x-I)-1/16*(ln(c*x+I)-ln(-1/2*I*(c*x+I)))*l
n(-1/2*I*(-c*x+I))+1/32*ln(c*x+I)^2-3/16*I*arctan(c*x)-3/16*I/(c*x-I)-1/16/(c*x-I)^2-1/16*ln(c*x-I)*ln(-1/2*I*
(c*x+I))+1/32*ln(c*x-I)^2))+I*a*b/d^3/(1+I*c*x)^2*arctan(c*x)-1/4*I*a*b/d^3*arctan(c*x)-1/4*a*b/d^3/(c*x-I)^2-
1/4*I*a*b/d^3/(c*x-I))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=-\frac {2 \, {\left (4 i \, a b + 3 \, b^{2}\right )} c x - {\left (i \, b^{2} c^{2} x^{2} + 2 \, b^{2} c x + 3 i \, b^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} + 16 i \, a^{2} + 16 \, a b - 8 i \, b^{2} - {\left ({\left (4 \, a b - 3 i \, b^{2}\right )} c^{2} x^{2} - 2 \, {\left (4 i \, a b + b^{2}\right )} c x + 12 \, a b - 5 i \, b^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{32 \, {\left (c^{3} d^{3} x^{2} - 2 i \, c^{2} d^{3} x - c d^{3}\right )}} \]

[In]

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

[Out]

-1/32*(2*(4*I*a*b + 3*b^2)*c*x - (I*b^2*c^2*x^2 + 2*b^2*c*x + 3*I*b^2)*log(-(c*x + I)/(c*x - I))^2 + 16*I*a^2
+ 16*a*b - 8*I*b^2 - ((4*a*b - 3*I*b^2)*c^2*x^2 - 2*(4*I*a*b + b^2)*c*x + 12*a*b - 5*I*b^2)*log(-(c*x + I)/(c*
x - I)))/(c^3*d^3*x^2 - 2*I*c^2*d^3*x - c*d^3)

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 464 vs. \(2 (144) = 288\).

Time = 33.40 (sec) , antiderivative size = 464, normalized size of antiderivative = 2.58 \[ \int \frac {(a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=- \frac {b \left (4 a - 3 i b\right ) \log {\left (- \frac {i b \left (4 a - 3 i b\right )}{c} + x \left (4 a b - 3 i b^{2}\right ) \right )}}{32 c d^{3}} + \frac {b \left (4 a - 3 i b\right ) \log {\left (\frac {i b \left (4 a - 3 i b\right )}{c} + x \left (4 a b - 3 i b^{2}\right ) \right )}}{32 c d^{3}} + \frac {\left (- 4 a b - b^{2} c x + 2 i b^{2}\right ) \log {\left (i c x + 1 \right )}}{8 c^{3} d^{3} x^{2} - 16 i c^{2} d^{3} x - 8 c d^{3}} + \frac {\left (i b^{2} c^{2} x^{2} + 2 b^{2} c x + 3 i b^{2}\right ) \log {\left (- i c x + 1 \right )}^{2}}{32 c^{3} d^{3} x^{2} - 64 i c^{2} d^{3} x - 32 c d^{3}} + \frac {\left (i b^{2} c^{2} x^{2} + 2 b^{2} c x + 3 i b^{2}\right ) \log {\left (i c x + 1 \right )}^{2}}{32 c^{3} d^{3} x^{2} - 64 i c^{2} d^{3} x - 32 c d^{3}} + \frac {- 8 i a^{2} - 8 a b + 4 i b^{2} + x \left (- 4 i a b c - 3 b^{2} c\right )}{16 c^{3} d^{3} x^{2} - 32 i c^{2} d^{3} x - 16 c d^{3}} + \frac {\left (8 a b - i b^{2} c^{2} x^{2} \log {\left (i c x + 1 \right )} - 2 b^{2} c x \log {\left (i c x + 1 \right )} + 2 b^{2} c x - 3 i b^{2} \log {\left (i c x + 1 \right )} - 4 i b^{2}\right ) \log {\left (- i c x + 1 \right )}}{16 c^{3} d^{3} x^{2} - 32 i c^{2} d^{3} x - 16 c d^{3}} \]

[In]

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

[Out]

-b*(4*a - 3*I*b)*log(-I*b*(4*a - 3*I*b)/c + x*(4*a*b - 3*I*b**2))/(32*c*d**3) + b*(4*a - 3*I*b)*log(I*b*(4*a -
3*I*b)/c + x*(4*a*b - 3*I*b**2))/(32*c*d**3) + (-4*a*b - b**2*c*x + 2*I*b**2)*log(I*c*x + 1)/(8*c**3*d**3*x**
2 - 16*I*c**2*d**3*x - 8*c*d**3) + (I*b**2*c**2*x**2 + 2*b**2*c*x + 3*I*b**2)*log(-I*c*x + 1)**2/(32*c**3*d**3
*x**2 - 64*I*c**2*d**3*x - 32*c*d**3) + (I*b**2*c**2*x**2 + 2*b**2*c*x + 3*I*b**2)*log(I*c*x + 1)**2/(32*c**3*
d**3*x**2 - 64*I*c**2*d**3*x - 32*c*d**3) + (-8*I*a**2 - 8*a*b + 4*I*b**2 + x*(-4*I*a*b*c - 3*b**2*c))/(16*c**
3*d**3*x**2 - 32*I*c**2*d**3*x - 16*c*d**3) + (8*a*b - I*b**2*c**2*x**2*log(I*c*x + 1) - 2*b**2*c*x*log(I*c*x
+ 1) + 2*b**2*c*x - 3*I*b**2*log(I*c*x + 1) - 4*I*b**2)*log(-I*c*x + 1)/(16*c**3*d**3*x**2 - 32*I*c**2*d**3*x
- 16*c*d**3)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.76 \[ \int \frac {(a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=-\frac {{\left (4 i \, a b + 3 \, b^{2}\right )} c x - 2 \, {\left (-i \, b^{2} c^{2} x^{2} - 2 \, b^{2} c x - 3 i \, b^{2}\right )} \arctan \left (c x\right )^{2} + 8 i \, a^{2} + 8 \, a b - 4 i \, b^{2} + {\left ({\left (4 i \, a b + 3 \, b^{2}\right )} c^{2} x^{2} + 2 \, {\left (4 \, a b - i \, b^{2}\right )} c x + 12 i \, a b + 5 \, b^{2}\right )} \arctan \left (c x\right )}{16 \, {\left (c^{3} d^{3} x^{2} - 2 i \, c^{2} d^{3} x - c d^{3}\right )}} \]

[In]

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

[Out]

-1/16*((4*I*a*b + 3*b^2)*c*x - 2*(-I*b^2*c^2*x^2 - 2*b^2*c*x - 3*I*b^2)*arctan(c*x)^2 + 8*I*a^2 + 8*a*b - 4*I*
b^2 + ((4*I*a*b + 3*b^2)*c^2*x^2 + 2*(4*a*b - I*b^2)*c*x + 12*I*a*b + 5*b^2)*arctan(c*x))/(c^3*d^3*x^2 - 2*I*c
^2*d^3*x - c*d^3)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3} \,d x \]

[In]

int((a + b*atan(c*x))^2/(d + c*d*x*1i)^3,x)

[Out]

int((a + b*atan(c*x))^2/(d + c*d*x*1i)^3, x)