\(\int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\) [56]

Optimal result

Integrand size = 32, antiderivative size = 131 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {(7 i A-B) x}{8 a^3}+\frac {A \log (\sin (c+d x))}{a^3 d}+\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {3 A+i B}{8 a d (a+i a \tan (c+d x))^2}+\frac {7 A+i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

-1/8*(7*I*A-B)*x/a^3+A*ln(sin(d*x+c))/a^3/d+1/6*(A+I*B)/d/(a+I*a*tan(d*x+c))^3+1/8*(3*A+I*B)/a/d/(a+I*a*tan(d*
x+c))^2+1/8*(7*A+I*B)/d/(a^3+I*a^3*tan(d*x+c))

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3677, 3612, 3556} \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {7 A+i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {x (-B+7 i A)}{8 a^3}+\frac {A \log (\sin (c+d x))}{a^3 d}+\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {3 A+i B}{8 a d (a+i a \tan (c+d x))^2} \]

[In]

Int[(Cot[c + d*x]*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

-1/8*(((7*I)*A - B)*x)/a^3 + (A*Log[Sin[c + d*x]])/(a^3*d) + (A + I*B)/(6*d*(a + I*a*Tan[c + d*x])^3) + (3*A +
I*B)/(8*a*d*(a + I*a*Tan[c + d*x])^2) + (7*A + I*B)/(8*d*(a^3 + I*a^3*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}{6 d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\cot (c+d x) (6 a A-3 a (i A-B) \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {3 A+i B}{8 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\cot (c+d x) \left (24 a^2 A-6 a^2 (3 i A-B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4} \\ & = \frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {3 A+i B}{8 a d (a+i a \tan (c+d x))^2}+\frac {7 A+i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \cot (c+d x) \left (48 a^3 A-6 a^3 (7 i A-B) \tan (c+d x)\right ) \, dx}{48 a^6} \\ & = -\frac {(7 i A-B) x}{8 a^3}+\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {3 A+i B}{8 a d (a+i a \tan (c+d x))^2}+\frac {7 A+i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {A \int \cot (c+d x) \, dx}{a^3} \\ & = -\frac {(7 i A-B) x}{8 a^3}+\frac {A \log (\sin (c+d x))}{a^3 d}+\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {3 A+i B}{8 a d (a+i a \tan (c+d x))^2}+\frac {7 A+i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.60 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.01 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {-3 (15 A+i B) \log (i-\tan (c+d x))+48 A \log (\tan (c+d x))-3 (A-i B) \log (i+\tan (c+d x))+\frac {8 i (A+i B)}{(-i+\tan (c+d x))^3}-\frac {6 (3 A+i B)}{(-i+\tan (c+d x))^2}+\frac {6 (-7 i A+B)}{-i+\tan (c+d x)}}{48 a^3 d} \]

[In]

Integrate[(Cot[c + d*x]*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(-3*(15*A + I*B)*Log[I - Tan[c + d*x]] + 48*A*Log[Tan[c + d*x]] - 3*(A - I*B)*Log[I + Tan[c + d*x]] + ((8*I)*(
A + I*B))/(-I + Tan[c + d*x])^3 - (6*(3*A + I*B))/(-I + Tan[c + d*x])^2 + (6*((-7*I)*A + B))/(-I + Tan[c + d*x
]))/(48*a^3*d)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.21

method	result	size
risch	\(\frac {x B}{8 a^{3}}-\frac {15 i x A}{8 a^{3}}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} B}{16 a^{3} d}+\frac {11 \,{\mathrm e}^{-2 i \left (d x +c \right )} A}{16 a^{3} d}+\frac {3 i {\mathrm e}^{-4 i \left (d x +c \right )} B}{32 a^{3} d}+\frac {5 \,{\mathrm e}^{-4 i \left (d x +c \right )} A}{32 a^{3} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} B}{48 a^{3} d}+\frac {{\mathrm e}^{-6 i \left (d x +c \right )} A}{48 a^{3} d}-\frac {2 i A c}{a^{3} d}+\frac {A \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{3} d}\)	\(159\)
derivativedivides	\(-\frac {A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 a^{3} d}-\frac {7 i A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {i A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {7 i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {3 A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {A \ln \left (\tan \left (d x +c \right )\right )}{a^{3} d}\)	\(193\)
default	\(-\frac {A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 a^{3} d}-\frac {7 i A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {i A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {7 i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {3 A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {A \ln \left (\tan \left (d x +c \right )\right )}{a^{3} d}\)	\(193\)

[In]

int(cot(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/8*x/a^3*B-15/8*I*x/a^3*A+3/16*I/a^3/d*exp(-2*I*(d*x+c))*B+11/16/a^3/d*exp(-2*I*(d*x+c))*A+3/32*I/a^3/d*exp(-
4*I*(d*x+c))*B+5/32/a^3/d*exp(-4*I*(d*x+c))*A+1/48*I/a^3/d*exp(-6*I*(d*x+c))*B+1/48/a^3/d*exp(-6*I*(d*x+c))*A-
2*I*A/a^3/d*c+A/a^3/d*ln(exp(2*I*(d*x+c))-1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.79 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (12 \, {\left (15 i \, A - B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} - 96 \, A e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 6 \, {\left (11 \, A + 3 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 3 \, {\left (5 \, A + 3 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, A - 2 i \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

[In]

integrate(cot(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/96*(12*(15*I*A - B)*d*x*e^(6*I*d*x + 6*I*c) - 96*A*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) - 1) - 6*(11
*A + 3*I*B)*e^(4*I*d*x + 4*I*c) - 3*(5*A + 3*I*B)*e^(2*I*d*x + 2*I*c) - 2*A - 2*I*B)*e^(-6*I*d*x - 6*I*c)/(a^3
*d)

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.23 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {A \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{3} d} + \begin {cases} \frac {\left (\left (512 A a^{6} d^{2} e^{6 i c} + 512 i B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (3840 A a^{6} d^{2} e^{8 i c} + 2304 i B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (16896 A a^{6} d^{2} e^{10 i c} + 4608 i B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {- 15 i A + B}{8 a^{3}} + \frac {\left (- 15 i A e^{6 i c} - 11 i A e^{4 i c} - 5 i A e^{2 i c} - i A + B e^{6 i c} + 3 B e^{4 i c} + 3 B e^{2 i c} + B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- 15 i A + B\right )}{8 a^{3}} \]

[In]

integrate(cot(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**3,x)

[Out]

A*log(exp(2*I*d*x) - exp(-2*I*c))/(a**3*d) + Piecewise((((512*A*a**6*d**2*exp(6*I*c) + 512*I*B*a**6*d**2*exp(6
*I*c))*exp(-6*I*d*x) + (3840*A*a**6*d**2*exp(8*I*c) + 2304*I*B*a**6*d**2*exp(8*I*c))*exp(-4*I*d*x) + (16896*A*
a**6*d**2*exp(10*I*c) + 4608*I*B*a**6*d**2*exp(10*I*c))*exp(-2*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(a**9
*d**3*exp(12*I*c), 0)), (x*(-(-15*I*A + B)/(8*a**3) + (-15*I*A*exp(6*I*c) - 11*I*A*exp(4*I*c) - 5*I*A*exp(2*I*
c) - I*A + B*exp(6*I*c) + 3*B*exp(4*I*c) + 3*B*exp(2*I*c) + B)*exp(-6*I*c)/(8*a**3)), True)) + x*(-15*I*A + B)
/(8*a**3)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(cot(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.95 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.09 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {6 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} + \frac {6 \, {\left (15 \, A + i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac {96 \, A \log \left (\tan \left (d x + c\right )\right )}{a^{3}} - \frac {165 \, A \tan \left (d x + c\right )^{3} + 11 i \, B \tan \left (d x + c\right )^{3} - 579 i \, A \tan \left (d x + c\right )^{2} + 45 \, B \tan \left (d x + c\right )^{2} - 699 \, A \tan \left (d x + c\right ) - 69 i \, B \tan \left (d x + c\right ) + 301 i \, A - 51 \, B}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]

[In]

integrate(cot(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/96*(6*(A - I*B)*log(tan(d*x + c) + I)/a^3 + 6*(15*A + I*B)*log(tan(d*x + c) - I)/a^3 - 96*A*log(tan(d*x + c
))/a^3 - (165*A*tan(d*x + c)^3 + 11*I*B*tan(d*x + c)^3 - 579*I*A*tan(d*x + c)^2 + 45*B*tan(d*x + c)^2 - 699*A*
tan(d*x + c) - 69*I*B*tan(d*x + c) + 301*I*A - 51*B)/(a^3*(tan(d*x + c) - I)^3))/d

Mupad [B] (verification not implemented)

Time = 8.30 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.25 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {17\,A}{12\,a^3}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {7\,A}{8\,a^3}+\frac {B\,1{}\mathrm {i}}{8\,a^3}\right )+\frac {B\,5{}\mathrm {i}}{12\,a^3}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {3\,B}{8\,a^3}+\frac {A\,17{}\mathrm {i}}{8\,a^3}\right )}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {A\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^3\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (15\,A+B\,1{}\mathrm {i}\right )}{16\,a^3\,d} \]

[In]

int((cot(c + d*x)*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

((17*A)/(12*a^3) - tan(c + d*x)^2*((7*A)/(8*a^3) + (B*1i)/(8*a^3)) + (B*5i)/(12*a^3) + tan(c + d*x)*((A*17i)/(
8*a^3) - (3*B)/(8*a^3)))/(d*(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) + (A*log(tan(c + d*x
)))/(a^3*d) + (log(tan(c + d*x) + 1i)*(A*1i + B)*1i)/(16*a^3*d) - (log(tan(c + d*x) - 1i)*(15*A + B*1i))/(16*a
^3*d)