\(\int \frac {a+b \arctan (c x)}{(d+i c d x)^2} \, dx\) [54]

Optimal result

Integrand size = 20, antiderivative size = 69 \[ \int \frac {a+b \arctan (c x)}{(d+i c d x)^2} \, dx=\frac {i b}{2 c d^2 (i-c x)}-\frac {i b \arctan (c x)}{2 c d^2}+\frac {i (a+b \arctan (c x))}{c d^2 (1+i c x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4972, 641, 46, 209} \[ \int \frac {a+b \arctan (c x)}{(d+i c d x)^2} \, dx=\frac {i (a+b \arctan (c x))}{c d^2 (1+i c x)}-\frac {i b \arctan (c x)}{2 c d^2}+\frac {i b}{2 c d^2 (-c x+i)} \]

[In]

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

[Out]

((I/2)*b)/(c*d^2*(I - c*x)) - ((I/2)*b*ArcTan[c*x])/(c*d^2) + (I*(a + b*ArcTan[c*x]))/(c*d^2*(1 + I*c*x))

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]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.61 \[ \int \frac {a+b \arctan (c x)}{(d+i c d x)^2} \, dx=\frac {2 a-i b+(b-i b c x) \arctan (c x)}{2 c d^2 (-i+c x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {\frac {i a}{d^{2} \left (i c x +1\right )}+\frac {i b \arctan \left (c x \right )}{d^{2} \left (i c x +1\right )}-\frac {i b \arctan \left (c x \right )}{2 d^{2}}-\frac {i b}{2 d^{2} \left (c x -i\right )}}{c}\)	\(68\)
default	\(\frac {\frac {i a}{d^{2} \left (i c x +1\right )}+\frac {i b \arctan \left (c x \right )}{d^{2} \left (i c x +1\right )}-\frac {i b \arctan \left (c x \right )}{2 d^{2}}-\frac {i b}{2 d^{2} \left (c x -i\right )}}{c}\)	\(68\)
parts	\(\frac {i a}{d^{2} c \left (i c x +1\right )}+\frac {i b \arctan \left (c x \right )}{c \,d^{2} \left (i c x +1\right )}-\frac {i b \arctan \left (c x \right )}{2 c \,d^{2}}-\frac {i b}{2 c \,d^{2} \left (c x -i\right )}\)	\(76\)
risch	\(-\frac {i b \ln \left (i c x +1\right )}{2 c \,d^{2} \left (c x -i\right )}+\frac {2 i b \ln \left (-i c x +1\right )+\ln \left (-c x -i\right ) b c x -\ln \left (c x -i\right ) b c x -i \ln \left (-c x -i\right ) b +i \ln \left (c x -i\right ) b -2 i b +4 a}{4 d^{2} \left (c x -i\right ) c}\)	\(111\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.72 \[ \int \frac {a+b \arctan (c x)}{(d+i c d x)^2} \, dx=\frac {{\left (b c x + i \, b\right )} \log \left (-\frac {c x + i}{c x - i}\right ) + 4 \, a - 2 i \, b}{4 \, {\left (c^{2} d^{2} x - i \, c d^{2}\right )}} \]

[In]

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

[Out]

1/4*((b*c*x + I*b)*log(-(c*x + I)/(c*x - I)) + 4*a - 2*I*b)/(c^2*d^2*x - I*c*d^2)

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. 116 vs. \(2 (51) = 102\).

Time = 0.88 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.68 \[ \int \frac {a+b \arctan (c x)}{(d+i c d x)^2} \, dx=\frac {i b \log {\left (- i c x + 1 \right )}}{2 c^{2} d^{2} x - 2 i c d^{2}} - \frac {i b \log {\left (i c x + 1 \right )}}{2 c^{2} d^{2} x - 2 i c d^{2}} - \frac {b \left (\frac {\log {\left (b x - \frac {i b}{c} \right )}}{4} - \frac {\log {\left (b x + \frac {i b}{c} \right )}}{4}\right )}{c d^{2}} - \frac {- 2 a + i b}{2 c^{2} d^{2} x - 2 i c d^{2}} \]

[In]

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

[Out]

I*b*log(-I*c*x + 1)/(2*c**2*d**2*x - 2*I*c*d**2) - I*b*log(I*c*x + 1)/(2*c**2*d**2*x - 2*I*c*d**2) - b*(log(b*
x - I*b/c)/4 - log(b*x + I*b/c)/4)/(c*d**2) - (-2*a + I*b)/(2*c**2*d**2*x - 2*I*c*d**2)

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \arctan (c x)}{(d+i c d x)^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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