\(\int \frac {\cot (c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx\) [40]

Optimal result

Integrand size = 32, antiderivative size = 62 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {(i A-B) x}{2 a}+\frac {A \log (\sin (c+d x))}{a d}+\frac {A+i B}{2 d (a+i a \tan (c+d x))} \]

[Out]

-1/2*(I*A-B)*x/a+A*ln(sin(d*x+c))/a/d+1/2*(A+I*B)/d/(a+I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3677, 3612, 3556} \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {A+i B}{2 d (a+i a \tan (c+d x))}-\frac {x (-B+i A)}{2 a}+\frac {A \log (\sin (c+d x))}{a d} \]

[In]

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

[Out]

-1/2*((I*A - B)*x)/a + (A*Log[Sin[c + d*x]])/(a*d) + (A + I*B)/(2*d*(a + I*a*Tan[c + d*x]))

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 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 {A+i B}{2 d (a+i a \tan (c+d x))}+\frac {\int \cot (c+d x) (2 a A-a (i A-B) \tan (c+d x)) \, dx}{2 a^2} \\ & = -\frac {(i A-B) x}{2 a}+\frac {A+i B}{2 d (a+i a \tan (c+d x))}+\frac {A \int \cot (c+d x) \, dx}{a} \\ & = -\frac {(i A-B) x}{2 a}+\frac {A \log (\sin (c+d x))}{a d}+\frac {A+i B}{2 d (a+i a \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.82 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.39 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {(3 A+i B) \log (i-\tan (c+d x))-4 A \log (\tan (c+d x))+(A-i B) \log (i+\tan (c+d x))+\frac {2 i (A+i B)}{-i+\tan (c+d x)}}{4 a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.37

method	result	size
risch	\(\frac {x B}{2 a}-\frac {3 i x A}{2 a}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{4 a d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} A}{4 a d}-\frac {2 i A c}{a d}+\frac {A \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a d}\)	\(85\)
derivativedivides	\(-\frac {A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}-\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}-\frac {i A}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {B}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {A \ln \left (\tan \left (d x +c \right )\right )}{a d}\)	\(111\)
default	\(-\frac {A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}-\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}-\frac {i A}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {B}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {A \ln \left (\tan \left (d x +c \right )\right )}{a d}\)	\(111\)
norman	\(\frac {\frac {i B +A}{2 a d}+\frac {\left (-i A +B \right ) x}{2 a}+\frac {\left (-i A +B \right ) \tan \left (d x +c \right )}{2 a d}+\frac {\left (-i A +B \right ) x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}}{1+\tan ^{2}\left (d x +c \right )}+\frac {A \ln \left (\tan \left (d x +c \right )\right )}{a d}-\frac {A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}\)	\(117\)

[In]

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

[Out]

1/2*x/a*B-3/2*I*x/a*A+1/4*I/a/d*exp(-2*I*(d*x+c))*B+1/4/a/d*exp(-2*I*(d*x+c))*A-2*I*A/a/d*c+A/a/d*ln(exp(2*I*(
d*x+c))-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.10 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {{\left (2 \, {\left (3 i \, A - B\right )} d x e^{\left (2 i \, d x + 2 i \, c\right )} - 4 \, A e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - A - i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \]

[In]

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

[Out]

-1/4*(2*(3*I*A - B)*d*x*e^(2*I*d*x + 2*I*c) - 4*A*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) - 1) - A - I*B)*
e^(-2*I*d*x - 2*I*c)/(a*d)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.87 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {A \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a d} + \begin {cases} \frac {\left (A + i B\right ) e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: a d e^{2 i c} \neq 0 \\x \left (- \frac {- 3 i A + B}{2 a} + \frac {\left (- 3 i A e^{2 i c} - i A + B e^{2 i c} + B\right ) e^{- 2 i c}}{2 a}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- 3 i A + B\right )}{2 a} \]

[In]

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

[Out]

A*log(exp(2*I*d*x) - exp(-2*I*c))/(a*d) + Piecewise(((A + I*B)*exp(-2*I*c)*exp(-2*I*d*x)/(4*a*d), Ne(a*d*exp(2
*I*c), 0)), (x*(-(-3*I*A + B)/(2*a) + (-3*I*A*exp(2*I*c) - I*A + B*exp(2*I*c) + B)*exp(-2*I*c)/(2*a)), True))
+ x*(-3*I*A + B)/(2*a)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.56 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {\frac {{\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a} + \frac {{\left (3 \, A + i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a} - \frac {4 \, A \log \left (\tan \left (d x + c\right )\right )}{a} - \frac {3 \, A \tan \left (d x + c\right ) + i \, B \tan \left (d x + c\right ) - 5 i \, A + 3 \, B}{a {\left (\tan \left (d x + c\right ) - i\right )}}}{4 \, d} \]

[In]

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

[Out]

-1/4*((A - I*B)*log(tan(d*x + c) + I)/a + (3*A + I*B)*log(tan(d*x + c) - I)/a - 4*A*log(tan(d*x + c))/a - (3*A
*tan(d*x + c) + I*B*tan(d*x + c) - 5*I*A + 3*B)/(a*(tan(d*x + c) - I)))/d

Mupad [B] (verification not implemented)

Time = 7.98 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.58 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {\frac {A}{2\,a}+\frac {B\,1{}\mathrm {i}}{2\,a}}{d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}+\frac {A\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (3\,A+B\,1{}\mathrm {i}\right )}{4\,a\,d} \]

[In]

int((cot(c + d*x)*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i),x)

[Out]

(A/(2*a) + (B*1i)/(2*a))/(d*(tan(c + d*x)*1i + 1)) + (A*log(tan(c + d*x)))/(a*d) + (log(tan(c + d*x) + 1i)*(A*
1i + B)*1i)/(4*a*d) - (log(tan(c + d*x) - 1i)*(3*A + B*1i))/(4*a*d)