\(\int \frac {x (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx\) [114]

Optimal result

Integrand size = 23, antiderivative size = 178 \[ \int \frac {x (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=-\frac {b^2}{16 c^2 d^3 (i-c x)^2}-\frac {5 i b^2}{16 c^2 d^3 (i-c x)}+\frac {5 i b^2 \arctan (c x)}{16 c^2 d^3}-\frac {i b (a+b \arctan (c x))}{4 c^2 d^3 (i-c x)^2}+\frac {3 b (a+b \arctan (c x))}{4 c^2 d^3 (i-c x)}+\frac {(a+b \arctan (c x))^2}{8 c^2 d^3}+\frac {x^2 (a+b \arctan (c x))^2}{2 d^3 (1+i c x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {37, 4994, 4972, 641, 46, 209, 5004} \[ \int \frac {x (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\frac {3 b (a+b \arctan (c x))}{4 c^2 d^3 (-c x+i)}-\frac {i b (a+b \arctan (c x))}{4 c^2 d^3 (-c x+i)^2}+\frac {(a+b \arctan (c x))^2}{8 c^2 d^3}+\frac {x^2 (a+b \arctan (c x))^2}{2 d^3 (1+i c x)^2}+\frac {5 i b^2 \arctan (c x)}{16 c^2 d^3}-\frac {5 i b^2}{16 c^2 d^3 (-c x+i)}-\frac {b^2}{16 c^2 d^3 (-c x+i)^2} \]

[In]

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

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 4994

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_))^(q_), x_Symbol] :> With[{u
= IntHide[(f*x)^m*(d + e*x)^q, x]}, Dist[(a + b*ArcTan[c*x])^p, u, x] - Dist[b*c*p, Int[ExpandIntegrand[(a +
b*ArcTan[c*x])^(p - 1), u/(1 + c^2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && IGtQ[p, 1] && EqQ[c
^2*d^2 + e^2, 0] && IntegersQ[m, q] && NeQ[m, -1] && NeQ[q, -1] && ILtQ[m + q + 1, 0] && LtQ[m*q, 0]

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 {x^2 (a+b \arctan (c x))^2}{2 d^3 (1+i c x)^2}-(2 b c) \int \left (-\frac {i (a+b \arctan (c x))}{4 c^2 d^3 (-i+c x)^3}-\frac {3 (a+b \arctan (c x))}{8 c^2 d^3 (-i+c x)^2}-\frac {a+b \arctan (c x)}{8 c^2 d^3 \left (1+c^2 x^2\right )}\right ) \, dx \\ & = \frac {x^2 (a+b \arctan (c x))^2}{2 d^3 (1+i c x)^2}+\frac {(i b) \int \frac {a+b \arctan (c x)}{(-i+c x)^3} \, dx}{2 c d^3}+\frac {b \int \frac {a+b \arctan (c x)}{1+c^2 x^2} \, dx}{4 c d^3}+\frac {(3 b) \int \frac {a+b \arctan (c x)}{(-i+c x)^2} \, dx}{4 c d^3} \\ & = -\frac {i b (a+b \arctan (c x))}{4 c^2 d^3 (i-c x)^2}+\frac {3 b (a+b \arctan (c x))}{4 c^2 d^3 (i-c x)}+\frac {(a+b \arctan (c x))^2}{8 c^2 d^3}+\frac {x^2 (a+b \arctan (c x))^2}{2 d^3 (1+i c x)^2}+\frac {\left (i b^2\right ) \int \frac {1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{4 c d^3}+\frac {\left (3 b^2\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{4 c d^3} \\ & = -\frac {i b (a+b \arctan (c x))}{4 c^2 d^3 (i-c x)^2}+\frac {3 b (a+b \arctan (c x))}{4 c^2 d^3 (i-c x)}+\frac {(a+b \arctan (c x))^2}{8 c^2 d^3}+\frac {x^2 (a+b \arctan (c x))^2}{2 d^3 (1+i c x)^2}+\frac {\left (i b^2\right ) \int \frac {1}{(-i+c x)^3 (i+c x)} \, dx}{4 c d^3}+\frac {\left (3 b^2\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{4 c d^3} \\ & = -\frac {i b (a+b \arctan (c x))}{4 c^2 d^3 (i-c x)^2}+\frac {3 b (a+b \arctan (c x))}{4 c^2 d^3 (i-c x)}+\frac {(a+b \arctan (c x))^2}{8 c^2 d^3}+\frac {x^2 (a+b \arctan (c x))^2}{2 d^3 (1+i c x)^2}+\frac {\left (i b^2\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}{4 c d^3}+\frac {\left (3 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 c d^3} \\ & = -\frac {b^2}{16 c^2 d^3 (i-c x)^2}-\frac {5 i b^2}{16 c^2 d^3 (i-c x)}-\frac {i b (a+b \arctan (c x))}{4 c^2 d^3 (i-c x)^2}+\frac {3 b (a+b \arctan (c x))}{4 c^2 d^3 (i-c x)}+\frac {(a+b \arctan (c x))^2}{8 c^2 d^3}+\frac {x^2 (a+b \arctan (c x))^2}{2 d^3 (1+i c x)^2}-\frac {\left (i b^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{16 c d^3}+\frac {\left (3 i b^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{8 c d^3} \\ & = -\frac {b^2}{16 c^2 d^3 (i-c x)^2}-\frac {5 i b^2}{16 c^2 d^3 (i-c x)}+\frac {5 i b^2 \arctan (c x)}{16 c^2 d^3}-\frac {i b (a+b \arctan (c x))}{4 c^2 d^3 (i-c x)^2}+\frac {3 b (a+b \arctan (c x))}{4 c^2 d^3 (i-c x)}+\frac {(a+b \arctan (c x))^2}{8 c^2 d^3}+\frac {x^2 (a+b \arctan (c x))^2}{2 d^3 (1+i c x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.66 \[ \int \frac {x (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\frac {4 a b (2 i-3 c x)+b^2 (4+5 i c x)+a^2 (-8-16 i c x)+b (i+c x) (a (4 i-12 c x)+b (3+5 i c x)) \arctan (c x)-2 b^2 \left (1+2 i c x+3 c^2 x^2\right ) \arctan (c x)^2}{16 c^2 d^3 (-i+c x)^2} \]

[In]

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

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (156 ) = 312\).

Time = 2.33 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.77

method	result	size
derivativedivides	\(\frac {\frac {a^{2} \left (\frac {1}{2 \left (c x -i\right )^{2}}-\frac {i}{c x -i}\right )}{d^{3}}+\frac {b^{2} \left (\frac {\arctan \left (c x \right )^{2}}{2 \left (c x -i\right )^{2}}-\frac {i \arctan \left (c x \right )^{2}}{c x -i}-\frac {3 i \arctan \left (c x \right ) \ln \left (c x +i\right )}{8}+\frac {3 i \arctan \left (c x \right ) \ln \left (c x -i\right )}{8}-\frac {i \arctan \left (c x \right )}{4 \left (c x -i\right )^{2}}-\frac {3 \arctan \left (c x \right )}{4 \left (c x -i\right )}+\frac {5 i \arctan \left (c x \right )}{16}+\frac {5 i}{16 \left (c x -i\right )}-\frac {1}{16 \left (c x -i\right )^{2}}+\frac {3 \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{16}-\frac {3 \ln \left (c x -i\right )^{2}}{32}+\frac {3 \left (\ln \left (c x +i\right )-\ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right ) \ln \left (-\frac {i \left (-c x +i\right )}{2}\right )}{16}-\frac {3 \ln \left (c x +i\right )^{2}}{32}\right )}{d^{3}}+\frac {a b \arctan \left (c x \right )}{d^{3} \left (c x -i\right )^{2}}-\frac {2 i a b \arctan \left (c x \right )}{d^{3} \left (c x -i\right )}-\frac {3 a b \arctan \left (c x \right )}{4 d^{3}}-\frac {i a b}{4 d^{3} \left (c x -i\right )^{2}}-\frac {3 a b}{4 d^{3} \left (c x -i\right )}}{c^{2}}\)	\(315\)
default	\(\frac {\frac {a^{2} \left (\frac {1}{2 \left (c x -i\right )^{2}}-\frac {i}{c x -i}\right )}{d^{3}}+\frac {b^{2} \left (\frac {\arctan \left (c x \right )^{2}}{2 \left (c x -i\right )^{2}}-\frac {i \arctan \left (c x \right )^{2}}{c x -i}-\frac {3 i \arctan \left (c x \right ) \ln \left (c x +i\right )}{8}+\frac {3 i \arctan \left (c x \right ) \ln \left (c x -i\right )}{8}-\frac {i \arctan \left (c x \right )}{4 \left (c x -i\right )^{2}}-\frac {3 \arctan \left (c x \right )}{4 \left (c x -i\right )}+\frac {5 i \arctan \left (c x \right )}{16}+\frac {5 i}{16 \left (c x -i\right )}-\frac {1}{16 \left (c x -i\right )^{2}}+\frac {3 \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{16}-\frac {3 \ln \left (c x -i\right )^{2}}{32}+\frac {3 \left (\ln \left (c x +i\right )-\ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right ) \ln \left (-\frac {i \left (-c x +i\right )}{2}\right )}{16}-\frac {3 \ln \left (c x +i\right )^{2}}{32}\right )}{d^{3}}+\frac {a b \arctan \left (c x \right )}{d^{3} \left (c x -i\right )^{2}}-\frac {2 i a b \arctan \left (c x \right )}{d^{3} \left (c x -i\right )}-\frac {3 a b \arctan \left (c x \right )}{4 d^{3}}-\frac {i a b}{4 d^{3} \left (c x -i\right )^{2}}-\frac {3 a b}{4 d^{3} \left (c x -i\right )}}{c^{2}}\)	\(315\)
parts	\(\frac {a^{2} \left (\frac {i}{c^{2} \left (-c x +i\right )}+\frac {1}{2 \left (-c x +i\right )^{2} c^{2}}\right )}{d^{3}}+\frac {b^{2} \left (\frac {\arctan \left (c x \right )^{2}}{2 \left (c x -i\right )^{2}}-\frac {i \arctan \left (c x \right )^{2}}{c x -i}-\frac {3 i \arctan \left (c x \right ) \ln \left (c x +i\right )}{8}+\frac {3 i \arctan \left (c x \right ) \ln \left (c x -i\right )}{8}-\frac {i \arctan \left (c x \right )}{4 \left (c x -i\right )^{2}}-\frac {3 \arctan \left (c x \right )}{4 \left (c x -i\right )}+\frac {5 i \arctan \left (c x \right )}{16}+\frac {5 i}{16 \left (c x -i\right )}-\frac {1}{16 \left (c x -i\right )^{2}}+\frac {3 \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{16}-\frac {3 \ln \left (c x -i\right )^{2}}{32}+\frac {3 \left (\ln \left (c x +i\right )-\ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right ) \ln \left (-\frac {i \left (-c x +i\right )}{2}\right )}{16}-\frac {3 \ln \left (c x +i\right )^{2}}{32}\right )}{d^{3} c^{2}}-\frac {2 i a b \arctan \left (c x \right )}{d^{3} c^{2} \left (c x -i\right )}+\frac {a b \arctan \left (c x \right )}{d^{3} c^{2} \left (c x -i\right )^{2}}-\frac {3 a b \arctan \left (c x \right )}{4 d^{3} c^{2}}-\frac {i a b}{4 d^{3} c^{2} \left (c x -i\right )^{2}}-\frac {3 a b}{4 d^{3} c^{2} \left (c x -i\right )}\)	\(337\)
risch	\(\frac {\left (3 b^{2} c^{2} x^{2}+2 i b^{2} c x +b^{2}\right ) \ln \left (i c x +1\right )^{2}}{32 c^{2} d^{3} \left (c x -i\right )^{2}}-\frac {\left (2 i \ln \left (-i c x +1\right ) b^{2} c x +b^{2} \ln \left (-i c x +1\right )+3 b^{2} \ln \left (-i c x +1\right ) c^{2} x^{2}-6 i b^{2} c x +16 a b c x -8 i a b -4 b^{2}\right ) \ln \left (i c x +1\right )}{16 c^{2} d^{3} \left (c x -i\right )^{2}}-\frac {i \left (32 c x \,a^{2}-16 a b -10 b^{2} c x -24 i a b c x -16 i a^{2}-2 \ln \left (-i c x +1\right )^{2} b^{2} c x +12 \ln \left (-i c x +1\right ) b^{2} c x +10 \ln \left (\left (12 i a c +5 b c \right ) x -5 i b +12 a \right ) b^{2} c x -10 \ln \left (\left (-12 i a c -5 b c \right ) x -5 i b +12 a \right ) b^{2} c x +32 i \ln \left (-i c x +1\right ) a b c x +24 i \ln \left (\left (12 i a c +5 b c \right ) x -5 i b +12 a \right ) a b c x -24 i \ln \left (\left (-12 i a c -5 b c \right ) x -5 i b +12 a \right ) a b c x +8 i b^{2}+16 \ln \left (-i c x +1\right ) a b +i \ln \left (-i c x +1\right )^{2} b^{2}+12 \ln \left (\left (12 i a c +5 b c \right ) x -5 i b +12 a \right ) a b -12 \ln \left (\left (-12 i a c -5 b c \right ) x -5 i b +12 a \right ) a b -8 i \ln \left (-i c x +1\right ) b^{2}-5 i \ln \left (\left (12 i a c +5 b c \right ) x -5 i b +12 a \right ) b^{2}+5 i \ln \left (\left (-12 i a c -5 b c \right ) x -5 i b +12 a \right ) b^{2}-12 \ln \left (\left (12 i a c +5 b c \right ) x -5 i b +12 a \right ) a b \,c^{2} x^{2}+12 \ln \left (\left (-12 i a c -5 b c \right ) x -5 i b +12 a \right ) a b \,c^{2} x^{2}+3 i b^{2} c^{2} x^{2} \ln \left (-i c x +1\right )^{2}+5 i \ln \left (\left (12 i a c +5 b c \right ) x -5 i b +12 a \right ) b^{2} c^{2} x^{2}-5 i \ln \left (\left (-12 i a c -5 b c \right ) x -5 i b +12 a \right ) b^{2} c^{2} x^{2}\right )}{32 c^{2} d^{3} \left (c x -i\right )^{2}}\)	\(657\)

[In]

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

[Out]

1/c^2*(a^2/d^3*(1/2/(c*x-I)^2-I/(c*x-I))+b^2/d^3*(1/2*arctan(c*x)^2/(c*x-I)^2-I*arctan(c*x)^2/(c*x-I)-3/8*I*ar
ctan(c*x)*ln(c*x+I)+3/8*I*arctan(c*x)*ln(c*x-I)-1/4*I*arctan(c*x)/(c*x-I)^2-3/4*arctan(c*x)/(c*x-I)+5/16*I*arc
tan(c*x)+5/16*I/(c*x-I)-1/16/(c*x-I)^2+3/16*ln(c*x-I)*ln(-1/2*I*(c*x+I))-3/32*ln(c*x-I)^2+3/16*(ln(c*x+I)-ln(-
1/2*I*(c*x+I)))*ln(-1/2*I*(-c*x+I))-3/32*ln(c*x+I)^2)+a*b/d^3*arctan(c*x)/(c*x-I)^2-2*I*a*b/d^3*arctan(c*x)/(c
*x-I)-3/4*a*b/d^3*arctan(c*x)-1/4*I*a*b/d^3/(c*x-I)^2-3/4*a*b/d^3/(c*x-I))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.93 \[ \int \frac {x (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=-\frac {2 \, {\left (16 i \, a^{2} + 12 \, a b - 5 i \, b^{2}\right )} c x - {\left (3 \, b^{2} c^{2} x^{2} + 2 i \, b^{2} c x + b^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} + 16 \, a^{2} - 16 i \, a b - 8 \, b^{2} - {\left ({\left (-12 i \, a b - 5 \, b^{2}\right )} c^{2} x^{2} + 2 \, {\left (4 \, a b - i \, b^{2}\right )} c x - 4 i \, a b - 3 \, b^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{32 \, {\left (c^{4} d^{3} x^{2} - 2 i \, c^{3} d^{3} x - c^{2} d^{3}\right )}} \]

[In]

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

[Out]

-1/32*(2*(16*I*a^2 + 12*a*b - 5*I*b^2)*c*x - (3*b^2*c^2*x^2 + 2*I*b^2*c*x + b^2)*log(-(c*x + I)/(c*x - I))^2 +
16*a^2 - 16*I*a*b - 8*b^2 - ((-12*I*a*b - 5*b^2)*c^2*x^2 + 2*(4*a*b - I*b^2)*c*x - 4*I*a*b - 3*b^2)*log(-(c*x
+ I)/(c*x - I)))/(c^4*d^3*x^2 - 2*I*c^3*d^3*x - c^2*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. 502 vs. \(2 (156) = 312\).

Time = 42.26 (sec) , antiderivative size = 502, normalized size of antiderivative = 2.82 \[ \int \frac {x (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\frac {i b \left (12 a - 5 i b\right ) \log {\left (- \frac {i b \left (12 a - 5 i b\right )}{c} + x \left (12 a b - 5 i b^{2}\right ) \right )}}{32 c^{2} d^{3}} - \frac {i b \left (12 a - 5 i b\right ) \log {\left (\frac {i b \left (12 a - 5 i b\right )}{c} + x \left (12 a b - 5 i b^{2}\right ) \right )}}{32 c^{2} d^{3}} + \frac {\left (3 b^{2} c^{2} x^{2} + 2 i b^{2} c x + b^{2}\right ) \log {\left (- i c x + 1 \right )}^{2}}{32 c^{4} d^{3} x^{2} - 64 i c^{3} d^{3} x - 32 c^{2} d^{3}} + \frac {\left (3 b^{2} c^{2} x^{2} + 2 i b^{2} c x + b^{2}\right ) \log {\left (i c x + 1 \right )}^{2}}{32 c^{4} d^{3} x^{2} - 64 i c^{3} d^{3} x - 32 c^{2} d^{3}} + \frac {- 8 a^{2} + 8 i a b + 4 b^{2} + x \left (- 16 i a^{2} c - 12 a b c + 5 i b^{2} c\right )}{16 c^{4} d^{3} x^{2} - 32 i c^{3} d^{3} x - 16 c^{2} d^{3}} + \frac {\left (16 a b c x - 8 i a b - 3 b^{2} c^{2} x^{2} \log {\left (i c x + 1 \right )} - 2 i b^{2} c x \log {\left (i c x + 1 \right )} - 6 i b^{2} c x - b^{2} \log {\left (i c x + 1 \right )} - 4 b^{2}\right ) \log {\left (- i c x + 1 \right )}}{16 c^{4} d^{3} x^{2} - 32 i c^{3} d^{3} x - 16 c^{2} d^{3}} + \frac {\left (- 8 a b c x + 4 i a b + 3 i b^{2} c x + 2 b^{2}\right ) \log {\left (i c x + 1 \right )}}{8 c^{4} d^{3} x^{2} - 16 i c^{3} d^{3} x - 8 c^{2} d^{3}} \]

[In]

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

[Out]

I*b*(12*a - 5*I*b)*log(-I*b*(12*a - 5*I*b)/c + x*(12*a*b - 5*I*b**2))/(32*c**2*d**3) - I*b*(12*a - 5*I*b)*log(
I*b*(12*a - 5*I*b)/c + x*(12*a*b - 5*I*b**2))/(32*c**2*d**3) + (3*b**2*c**2*x**2 + 2*I*b**2*c*x + b**2)*log(-I
*c*x + 1)**2/(32*c**4*d**3*x**2 - 64*I*c**3*d**3*x - 32*c**2*d**3) + (3*b**2*c**2*x**2 + 2*I*b**2*c*x + b**2)*
log(I*c*x + 1)**2/(32*c**4*d**3*x**2 - 64*I*c**3*d**3*x - 32*c**2*d**3) + (-8*a**2 + 8*I*a*b + 4*b**2 + x*(-16
*I*a**2*c - 12*a*b*c + 5*I*b**2*c))/(16*c**4*d**3*x**2 - 32*I*c**3*d**3*x - 16*c**2*d**3) + (16*a*b*c*x - 8*I*
a*b - 3*b**2*c**2*x**2*log(I*c*x + 1) - 2*I*b**2*c*x*log(I*c*x + 1) - 6*I*b**2*c*x - b**2*log(I*c*x + 1) - 4*b
**2)*log(-I*c*x + 1)/(16*c**4*d**3*x**2 - 32*I*c**3*d**3*x - 16*c**2*d**3) + (-8*a*b*c*x + 4*I*a*b + 3*I*b**2*
c*x + 2*b**2)*log(I*c*x + 1)/(8*c**4*d**3*x**2 - 16*I*c**3*d**3*x - 8*c**2*d**3)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.79 \[ \int \frac {x (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=-\frac {{\left (16 i \, a^{2} + 12 \, a b - 5 i \, b^{2}\right )} c x + 2 \, {\left (3 \, b^{2} c^{2} x^{2} + 2 i \, b^{2} c x + b^{2}\right )} \arctan \left (c x\right )^{2} + 8 \, a^{2} - 8 i \, a b - 4 \, b^{2} + {\left ({\left (12 \, a b - 5 i \, b^{2}\right )} c^{2} x^{2} - 2 \, {\left (-4 i \, a b - b^{2}\right )} c x + 4 \, a b - 3 i \, b^{2}\right )} \arctan \left (c x\right )}{16 \, {\left (c^{4} d^{3} x^{2} - 2 i \, c^{3} d^{3} x - c^{2} d^{3}\right )}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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