\(\int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [63]

Optimal result

Integrand size = 22, antiderivative size = 71 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {3 i x}{4 a^2}+\frac {\log (\sin (c+d x))}{a^2 d}+\frac {3}{4 a^2 d (1+i \tan (c+d x))}+\frac {1}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

-3/4*I*x/a^2+ln(sin(d*x+c))/a^2/d+3/4/a^2/d/(1+I*tan(d*x+c))+1/4/d/(a+I*a*tan(d*x+c))^2

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3640, 3677, 3612, 3556} \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {3}{4 a^2 d (1+i \tan (c+d x))}+\frac {\log (\sin (c+d x))}{a^2 d}-\frac {3 i x}{4 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2} \]

[In]

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

[Out]

(((-3*I)/4)*x)/a^2 + Log[Sin[c + d*x]]/(a^2*d) + 3/(4*a^2*d*(1 + I*Tan[c + d*x])) + 1/(4*d*(a + I*a*Tan[c + d*
x])^2)

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3612

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c +
b*d)*(x/(a^2 + b^2)), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3640

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Dist[1/(2*a*m*(b*c - a*d))
, Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan
[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2
+ d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3677

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*
f*m*(b*c - a*d))), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\cot (c+d x) (4 a-2 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = \frac {3}{4 a^2 d (1+i \tan (c+d x))}+\frac {1}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot (c+d x) \left (8 a^2-6 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4} \\ & = -\frac {3 i x}{4 a^2}+\frac {3}{4 a^2 d (1+i \tan (c+d x))}+\frac {1}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot (c+d x) \, dx}{a^2} \\ & = -\frac {3 i x}{4 a^2}+\frac {\log (\sin (c+d x))}{a^2 d}+\frac {3}{4 a^2 d (1+i \tan (c+d x))}+\frac {1}{4 d (a+i a \tan (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {2}{(a+i a \tan (c+d x))^2}-\frac {7 \log (i-\tan (c+d x))-8 \log (\tan (c+d x))+\log (i+\tan (c+d x))+\frac {6 i}{-i+\tan (c+d x)}}{a^2}}{8 d} \]

[In]

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

[Out]

(2/(a + I*a*Tan[c + d*x])^2 - (7*Log[I - Tan[c + d*x]] - 8*Log[Tan[c + d*x]] + Log[I + Tan[c + d*x]] + (6*I)/(
-I + Tan[c + d*x]))/a^2)/(8*d)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01

method	result	size
risch	\(-\frac {7 i x}{4 a^{2}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{2 a^{2} d}+\frac {{\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}-\frac {2 i c}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\)	\(72\)
derivativedivides	\(-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}-\frac {3 i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}\)	\(90\)
default	\(-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}-\frac {3 i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}\)	\(90\)
norman	\(\frac {\frac {1}{a d}-\frac {3 i x}{4 a}-\frac {5 i \tan \left (d x +c \right )}{4 d a}-\frac {3 i \left (\tan ^{3}\left (d x +c \right )\right )}{4 d a}-\frac {3 i x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}-\frac {3 i x \left (\tan ^{4}\left (d x +c \right )\right )}{4 a}+\frac {\tan ^{2}\left (d x +c \right )}{2 a d}}{a \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}\)	\(144\)

[In]

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

[Out]

-7/4*I*x/a^2+1/2/a^2/d*exp(-2*I*(d*x+c))+1/16/a^2/d*exp(-4*I*(d*x+c))-2*I/a^2/d*c+1/a^2/d*ln(exp(2*I*(d*x+c))-
1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (-28 i \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 16 \, e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \]

[In]

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

[Out]

1/16*(-28*I*d*x*e^(4*I*d*x + 4*I*c) + 16*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) - 1) + 8*e^(2*I*d*x + 2*I
*c) + 1)*e^(-4*I*d*x - 4*I*c)/(a^2*d)

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.11 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (16 a^{2} d e^{4 i c} e^{- 2 i d x} + 2 a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{32 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (- 7 i e^{4 i c} - 4 i e^{2 i c} - i\right ) e^{- 4 i c}}{4 a^{2}} + \frac {7 i}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {7 i x}{4 a^{2}} + \frac {\log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{2} d} \]

[In]

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Piecewise(((16*a**2*d*exp(4*I*c)*exp(-2*I*d*x) + 2*a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(32*a**4*d**2)
, Ne(a**4*d**2*exp(6*I*c), 0)), (x*((-7*I*exp(4*I*c) - 4*I*exp(2*I*c) - I)*exp(-4*I*c)/(4*a**2) + 7*I/(4*a**2)
), True)) - 7*I*x/(4*a**2) + log(exp(2*I*d*x) - exp(-2*I*c))/(a**2*d)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.60 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\frac {2 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {14 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac {16 \, \log \left (\tan \left (d x + c\right )\right )}{a^{2}} - \frac {21 \, \tan \left (d x + c\right )^{2} - 54 i \, \tan \left (d x + c\right ) - 37}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

-1/16*(2*log(tan(d*x + c) + I)/a^2 + 14*log(tan(d*x + c) - I)/a^2 - 16*log(tan(d*x + c))/a^2 - (21*tan(d*x + c
)^2 - 54*I*tan(d*x + c) - 37)/(a^2*(tan(d*x + c) - I)^2))/d

Mupad [B] (verification not implemented)

Time = 4.86 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.37 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{8\,a^2\,d}-\frac {7\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{8\,a^2\,d}+\frac {\frac {3\,\mathrm {tan}\left (c+d\,x\right )}{4\,a^2}-\frac {1{}\mathrm {i}}{a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \]

[In]

int(cot(c + d*x)/(a + a*tan(c + d*x)*1i)^2,x)

[Out]

log(tan(c + d*x))/(a^2*d) - log(tan(c + d*x) + 1i)/(8*a^2*d) - (7*log(tan(c + d*x) - 1i))/(8*a^2*d) + ((3*tan(
c + d*x))/(4*a^2) - 1i/a^2)/(d*(2*tan(c + d*x) + tan(c + d*x)^2*1i - 1i))