\(\int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx\) [49]

Optimal result

Integrand size = 34, antiderivative size = 141 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {3 (3 A+i B) x}{4 a^2}-\frac {3 (3 A+i B) \cot (c+d x)}{4 a^2 d}-\frac {(2 i A-B) \log (\sin (c+d x))}{a^2 d}+\frac {(2 A+i B) \cot (c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \cot (c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3677, 3610, 3612, 3556} \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {3 (3 A+i B) \cot (c+d x)}{4 a^2 d}-\frac {(-B+2 i A) \log (\sin (c+d x))}{a^2 d}+\frac {(2 A+i B) \cot (c+d x)}{2 a^2 d (1+i \tan (c+d x))}-\frac {3 x (3 A+i B)}{4 a^2}+\frac {(A+i B) \cot (c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[In]

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

[Out]

(-3*(3*A + I*B)*x)/(4*a^2) - (3*(3*A + I*B)*Cot[c + d*x])/(4*a^2*d) - (((2*I)*A - B)*Log[Sin[c + d*x]])/(a^2*d
) + ((2*A + I*B)*Cot[c + d*x])/(2*a^2*d*(1 + I*Tan[c + d*x])) + ((A + I*B)*Cot[c + d*x])/(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 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)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\cot ^2(c+d x) (a (5 A+i B)-3 a (i A-B) \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = \frac {(2 A+i B) \cot (c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot ^2(c+d x) \left (6 a^2 (3 A+i B)-8 a^2 (2 i A-B) \tan (c+d x)\right ) \, dx}{8 a^4} \\ & = -\frac {3 (3 A+i B) \cot (c+d x)}{4 a^2 d}+\frac {(2 A+i B) \cot (c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot (c+d x) \left (-8 a^2 (2 i A-B)-6 a^2 (3 A+i B) \tan (c+d x)\right ) \, dx}{8 a^4} \\ & = -\frac {3 (3 A+i B) x}{4 a^2}-\frac {3 (3 A+i B) \cot (c+d x)}{4 a^2 d}+\frac {(2 A+i B) \cot (c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \cot (c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(2 i A-B) \int \cot (c+d x) \, dx}{a^2} \\ & = -\frac {3 (3 A+i B) x}{4 a^2}-\frac {3 (3 A+i B) \cot (c+d x)}{4 a^2 d}-\frac {(2 i A-B) \log (\sin (c+d x))}{a^2 d}+\frac {(2 A+i B) \cot (c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \cot (c+d x)}{4 d (a+i a \tan (c+d x))^2} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 2.20 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.92 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {2 (A+i B) \cot ^3(c+d x)}{(i+\cot (c+d x))^2}+\frac {4 (2 A+i B) \cot ^2(c+d x)}{i+\cot (c+d x)}-6 (3 A+i B) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )+8 (-2 i A+B) (\log (\cos (c+d x))+\log (\tan (c+d x)))}{8 a^2 d} \]

[In]

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

[Out]

((2*(A + I*B)*Cot[c + d*x]^3)/(I + Cot[c + d*x])^2 + (4*(2*A + I*B)*Cot[c + d*x]^2)/(I + Cot[c + d*x]) - 6*(3*
A + I*B)*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] + 8*((-2*I)*A + B)*(Log[Cos[c + d*x]] +
Log[Tan[c + d*x]]))/(8*a^2*d)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.26

method	result	size
risch	\(-\frac {7 i x B}{4 a^{2}}-\frac {17 x A}{4 a^{2}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} B}{2 a^{2} d}-\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} A}{4 a^{2} d}+\frac {{\mathrm e}^{-4 i \left (d x +c \right )} B}{16 a^{2} d}-\frac {i {\mathrm e}^{-4 i \left (d x +c \right )} A}{16 a^{2} d}-\frac {2 i B c}{a^{2} d}-\frac {4 A c}{a^{2} d}-\frac {2 i A}{a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{a^{2} d}-\frac {2 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A}{a^{2} d}\)	\(177\)
derivativedivides	\(\frac {i A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i B \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {i A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{a^{2} d}-\frac {9 A \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {5 A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {3 i B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {A}{a^{2} d \tan \left (d x +c \right )}-\frac {2 i A \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}+\frac {B \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}\)	\(207\)
default	\(\frac {i A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i B \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {i A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{a^{2} d}-\frac {9 A \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {5 A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {3 i B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {A}{a^{2} d \tan \left (d x +c \right )}-\frac {2 i A \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}+\frac {B \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}\)	\(207\)

[In]

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

[Out]

-7/4*I*x/a^2*B-17/4*x/a^2*A+1/2/a^2/d*exp(-2*I*(d*x+c))*B-3/4*I/a^2/d*exp(-2*I*(d*x+c))*A+1/16/a^2/d*exp(-4*I*
(d*x+c))*B-1/16*I/a^2/d*exp(-4*I*(d*x+c))*A-2*I*B/a^2/d*c-4/a^2/d*A*c-2*I*A/a^2/d/(exp(2*I*(d*x+c))-1)+1/a^2/d
*ln(exp(2*I*(d*x+c))-1)*B-2*I/a^2/d*ln(exp(2*I*(d*x+c))-1)*A

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.08 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {4 \, {\left (17 \, A + 7 i \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} - 4 \, {\left ({\left (17 \, A + 7 i \, B\right )} d x - 11 i \, A + 2 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (11 i \, A - 7 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 16 \, {\left ({\left (2 i \, A - B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-2 i \, A + B\right )} e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - i \, A + B}{16 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]

[In]

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

[Out]

-1/16*(4*(17*A + 7*I*B)*d*x*e^(6*I*d*x + 6*I*c) - 4*((17*A + 7*I*B)*d*x - 11*I*A + 2*B)*e^(4*I*d*x + 4*I*c) -
(11*I*A - 7*B)*e^(2*I*d*x + 2*I*c) + 16*((2*I*A - B)*e^(6*I*d*x + 6*I*c) + (-2*I*A + B)*e^(4*I*d*x + 4*I*c))*l
og(e^(2*I*d*x + 2*I*c) - 1) - I*A + B)/(a^2*d*e^(6*I*d*x + 6*I*c) - a^2*d*e^(4*I*d*x + 4*I*c))

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.89 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=- \frac {2 i A}{a^{2} d e^{2 i c} e^{2 i d x} - a^{2} d} + \begin {cases} \frac {\left (\left (- 4 i A a^{2} d e^{2 i c} + 4 B a^{2} d e^{2 i c}\right ) e^{- 4 i d x} + \left (- 48 i A a^{2} d e^{4 i c} + 32 B a^{2} d e^{4 i c}\right ) e^{- 2 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (- \frac {- 17 A - 7 i B}{4 a^{2}} + \frac {\left (- 17 A e^{4 i c} - 6 A e^{2 i c} - A - 7 i B e^{4 i c} - 4 i B e^{2 i c} - i B\right ) e^{- 4 i c}}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- 17 A - 7 i B\right )}{4 a^{2}} - \frac {i \left (2 A + i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{2} d} \]

[In]

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

[Out]

-2*I*A/(a**2*d*exp(2*I*c)*exp(2*I*d*x) - a**2*d) + Piecewise((((-4*I*A*a**2*d*exp(2*I*c) + 4*B*a**2*d*exp(2*I*
c))*exp(-4*I*d*x) + (-48*I*A*a**2*d*exp(4*I*c) + 32*B*a**2*d*exp(4*I*c))*exp(-2*I*d*x))*exp(-6*I*c)/(64*a**4*d
**2), Ne(a**4*d**2*exp(6*I*c), 0)), (x*(-(-17*A - 7*I*B)/(4*a**2) + (-17*A*exp(4*I*c) - 6*A*exp(2*I*c) - A - 7
*I*B*exp(4*I*c) - 4*I*B*exp(2*I*c) - I*B)*exp(-4*I*c)/(4*a**2)), True)) + x*(-17*A - 7*I*B)/(4*a**2) - I*(2*A
+ I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/(a**2*d)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.82 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.14 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} - \frac {2 \, {\left (-17 i \, A + 7 \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac {16 \, {\left (2 i \, A - B\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{2}} - \frac {16 \, {\left (-2 i \, A \tan \left (d x + c\right ) + B \tan \left (d x + c\right ) + A\right )}}{a^{2} \tan \left (d x + c\right )} - \frac {51 i \, A \tan \left (d x + c\right )^{2} - 21 \, B \tan \left (d x + c\right )^{2} + 122 \, A \tan \left (d x + c\right ) + 54 i \, B \tan \left (d x + c\right ) - 75 i \, A + 37 \, B}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

1/16*(2*(-I*A - B)*log(tan(d*x + c) + I)/a^2 - 2*(-17*I*A + 7*B)*log(tan(d*x + c) - I)/a^2 - 16*(2*I*A - B)*lo
g(tan(d*x + c))/a^2 - 16*(-2*I*A*tan(d*x + c) + B*tan(d*x + c) + A)/(a^2*tan(d*x + c)) - (51*I*A*tan(d*x + c)^
2 - 21*B*tan(d*x + c)^2 + 122*A*tan(d*x + c) + 54*I*B*tan(d*x + c) - 75*I*A + 37*B)/(a^2*(tan(d*x + c) - I)^2)
)/d

Mupad [B] (verification not implemented)

Time = 8.02 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.16 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {3\,B}{4\,a^2}+\frac {A\,9{}\mathrm {i}}{4\,a^2}\right )-\frac {A\,1{}\mathrm {i}}{a^2}+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {7\,A}{2\,a^2}+\frac {B\,1{}\mathrm {i}}{a^2}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}+2\,{\mathrm {tan}\left (c+d\,x\right )}^2-\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B+A\,2{}\mathrm {i}\right )}{a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{8\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-7\,B+A\,17{}\mathrm {i}\right )}{8\,a^2\,d} \]

[In]

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

[Out]

(log(tan(c + d*x) - 1i)*(A*17i - 7*B))/(8*a^2*d) - (log(tan(c + d*x))*(A*2i - B))/(a^2*d) - (log(tan(c + d*x)
+ 1i)*(A*1i + B))/(8*a^2*d) - (tan(c + d*x)^2*((A*9i)/(4*a^2) - (3*B)/(4*a^2)) - (A*1i)/a^2 + tan(c + d*x)*((7
*A)/(2*a^2) + (B*1i)/a^2))/(d*(2*tan(c + d*x)^2 - tan(c + d*x)*1i + tan(c + d*x)^3*1i))