\(\int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx\) [41]

Optimal result

Integrand size = 34, antiderivative size = 102 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {(3 A+i B) x}{2 a}-\frac {(3 A+i B) \cot (c+d x)}{2 a d}-\frac {(i A-B) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot (c+d x)}{2 d (a+i a \tan (c+d x))} \]

[Out]

-1/2*(3*A+I*B)*x/a-1/2*(3*A+I*B)*cot(d*x+c)/a/d-(I*A-B)*ln(sin(d*x+c))/a/d+1/2*(A+I*B)*cot(d*x+c)/d/(a+I*a*tan
(d*x+c))

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3677, 3610, 3612, 3556} \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {(3 A+i B) \cot (c+d x)}{2 a d}-\frac {(-B+i A) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot (c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {x (3 A+i B)}{2 a} \]

[In]

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

[Out]

-1/2*((3*A + I*B)*x)/a - ((3*A + I*B)*Cot[c + d*x])/(2*a*d) - ((I*A - B)*Log[Sin[c + d*x]])/(a*d) + ((A + I*B)
*Cot[c + d*x])/(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 3610

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

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) \cot (c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \cot ^2(c+d x) (a (3 A+i B)-2 a (i A-B) \tan (c+d x)) \, dx}{2 a^2} \\ & = -\frac {(3 A+i B) \cot (c+d x)}{2 a d}+\frac {(A+i B) \cot (c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \cot (c+d x) (-2 a (i A-B)-a (3 A+i B) \tan (c+d x)) \, dx}{2 a^2} \\ & = -\frac {(3 A+i B) x}{2 a}-\frac {(3 A+i B) \cot (c+d x)}{2 a d}+\frac {(A+i B) \cot (c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(i A-B) \int \cot (c+d x) \, dx}{a} \\ & = -\frac {(3 A+i B) x}{2 a}-\frac {(3 A+i B) \cot (c+d x)}{2 a d}-\frac {(i A-B) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot (c+d x)}{2 d (a+i a \tan (c+d x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.26 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.96 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {\frac {(A+i B) \cot ^2(c+d x)}{i+\cot (c+d x)}-(3 A+i B) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )+2 (-i A+B) (\log (\cos (c+d x))+\log (\tan (c+d x)))}{2 a d} \]

[In]

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

[Out]

(((A + I*B)*Cot[c + d*x]^2)/(I + Cot[c + d*x]) - (3*A + I*B)*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan
[c + d*x]^2] + 2*((-I)*A + B)*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]]))/(2*a*d)

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.37

method	result	size
risch	\(-\frac {3 i x B}{2 a}-\frac {5 x A}{2 a}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} B}{4 a d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} A}{4 a d}-\frac {2 i B c}{a d}-\frac {2 A c}{a d}-\frac {2 i A}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{a d}-\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A}{a d}\)	\(140\)
norman	\(\frac {-\frac {A}{a d}-\frac {\left (i B +3 A \right ) \left (\tan ^{2}\left (d x +c \right )\right )}{2 a d}-\frac {\left (i B +3 A \right ) x \tan \left (d x +c \right )}{2 a}-\frac {\left (i B +3 A \right ) x \left (\tan ^{3}\left (d x +c \right )\right )}{2 a}+\frac {\left (-i A +B \right ) \tan \left (d x +c \right )}{2 a d}}{\tan \left (d x +c \right ) \left (1+\tan ^{2}\left (d x +c \right )\right )}+\frac {\left (-i A +B \right ) \ln \left (\tan \left (d x +c \right )\right )}{a d}-\frac {\left (-i A +B \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 a d}\)	\(164\)
derivativedivides	\(-\frac {A}{2 d a \left (\tan \left (d x +c \right )-i\right )}-\frac {i B}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {i A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}-\frac {3 A \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}-\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}-\frac {i B \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}-\frac {A}{a d \tan \left (d x +c \right )}-\frac {i A \ln \left (\tan \left (d x +c \right )\right )}{a d}+\frac {B \ln \left (\tan \left (d x +c \right )\right )}{a d}\)	\(166\)
default	\(-\frac {A}{2 d a \left (\tan \left (d x +c \right )-i\right )}-\frac {i B}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {i A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}-\frac {3 A \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}-\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}-\frac {i B \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}-\frac {A}{a d \tan \left (d x +c \right )}-\frac {i A \ln \left (\tan \left (d x +c \right )\right )}{a d}+\frac {B \ln \left (\tan \left (d x +c \right )\right )}{a d}\)	\(166\)

[In]

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

[Out]

-3/2*I*x/a*B-5/2*x/a*A+1/4/a/d*exp(-2*I*(d*x+c))*B-1/4*I/a/d*exp(-2*I*(d*x+c))*A-2*I/a/d*B*c-2/a/d*A*c-2*I*A/a
/d/(exp(2*I*(d*x+c))-1)+1/a/d*ln(exp(2*I*(d*x+c))-1)*B-I/a/d*ln(exp(2*I*(d*x+c))-1)*A

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.26 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {2 \, {\left (5 \, A + 3 i \, B\right )} d x e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (2 \, {\left (5 \, A + 3 i \, B\right )} d x - 9 i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, {\left ({\left (i \, A - B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - i \, A + B}{4 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

[In]

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

[Out]

-1/4*(2*(5*A + 3*I*B)*d*x*e^(4*I*d*x + 4*I*c) - (2*(5*A + 3*I*B)*d*x - 9*I*A + B)*e^(2*I*d*x + 2*I*c) + 4*((I*
A - B)*e^(4*I*d*x + 4*I*c) + (-I*A + B)*e^(2*I*d*x + 2*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - I*A + B)/(a*d*e^(4
*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.55 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=- \frac {2 i A}{a d e^{2 i c} e^{2 i d x} - a d} + \begin {cases} \frac {\left (- i A + B\right ) e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: a d e^{2 i c} \neq 0 \\x \left (- \frac {- 5 A - 3 i B}{2 a} + \frac {\left (- 5 A e^{2 i c} - A - 3 i B e^{2 i c} - i B\right ) e^{- 2 i c}}{2 a}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- 5 A - 3 i B\right )}{2 a} - \frac {i \left (A + i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a d} \]

[In]

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

[Out]

-2*I*A/(a*d*exp(2*I*c)*exp(2*I*d*x) - a*d) + Piecewise(((-I*A + B)*exp(-2*I*c)*exp(-2*I*d*x)/(4*a*d), Ne(a*d*e
xp(2*I*c), 0)), (x*(-(-5*A - 3*I*B)/(2*a) + (-5*A*exp(2*I*c) - A - 3*I*B*exp(2*I*c) - I*B)*exp(-2*I*c)/(2*a)),
True)) + x*(-5*A - 3*I*B)/(2*a) - I*(A + I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/(a*d)

Maxima [F(-2)]

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

[In]

integrate(cot(d*x+c)^2*(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 = 0.55 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.30 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {\frac {2 \, {\left (i \, A + B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a} + \frac {2 \, {\left (-5 i \, A + 3 \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac {8 \, {\left (i \, A - B\right )} \log \left (\tan \left (d x + c\right )\right )}{a} + \frac {A \tan \left (d x + c\right )^{2} - i \, B \tan \left (d x + c\right )^{2} - 13 i \, A \tan \left (d x + c\right ) + 3 \, B \tan \left (d x + c\right ) - 8 \, A}{{\left (-i \, \tan \left (d x + c\right )^{2} - \tan \left (d x + c\right )\right )} a}}{8 \, d} \]

[In]

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

[Out]

-1/8*(2*(I*A + B)*log(tan(d*x + c) + I)/a + 2*(-5*I*A + 3*B)*log(tan(d*x + c) - I)/a + 8*(I*A - B)*log(tan(d*x
+ c))/a + (A*tan(d*x + c)^2 - I*B*tan(d*x + c)^2 - 13*I*A*tan(d*x + c) + 3*B*tan(d*x + c) - 8*A)/((-I*tan(d*x
+ c)^2 - tan(d*x + c))*a))/d

Mupad [B] (verification not implemented)

Time = 7.72 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.24 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {\frac {A}{a}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{2\,a}+\frac {A\,3{}\mathrm {i}}{2\,a}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{4\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-3\,B+A\,5{}\mathrm {i}\right )}{4\,a\,d} \]

[In]

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

[Out]

(log(tan(c + d*x) - 1i)*(A*5i - 3*B))/(4*a*d) - (log(tan(c + d*x))*(A*1i - B))/(a*d) - (log(tan(c + d*x) + 1i)
*(A*1i + B))/(4*a*d) - (A/a + tan(c + d*x)*((A*3i)/(2*a) - B/(2*a)))/(d*(tan(c + d*x) + tan(c + d*x)^2*1i))