\(\int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [73]

Optimal result

Integrand size = 22, antiderivative size = 98 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {7 i x}{8 a^3}+\frac {\log (\sin (c+d x))}{a^3 d}+\frac {1}{6 d (a+i a \tan (c+d x))^3}+\frac {3}{8 a d (a+i a \tan (c+d x))^2}+\frac {7}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.10 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {-\frac {45 \log (i-\tan (c+d x))}{a^3}+\frac {48 \log (\tan (c+d x))}{a^3}-\frac {3 \log (i+\tan (c+d x))}{a^3}-\frac {18}{a^3 (-i+\tan (c+d x))^2}+\frac {8}{(a+i a \tan (c+d x))^3}+\frac {42}{a^2 (a+i a \tan (c+d x))}}{48 d} \] Input:

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

Output:

((-45*Log[I - Tan[c + d*x]])/a^3 + (48*Log[Tan[c + d*x]])/a^3 - (3*Log[I + 
 Tan[c + d*x]])/a^3 - 18/(a^3*(-I + Tan[c + d*x])^2) + 8/(a + I*a*Tan[c + 
d*x])^3 + 42/(a^2*(a + I*a*Tan[c + d*x])))/(48*d)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.22, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {3042, 4042, 27, 3042, 4079, 27, 3042, 4079, 3042, 4014, 3042, 25, 3956}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x) (a+i a \tan (c+d x))^3}dx\)

\(\Big \downarrow \) 4042

\(\displaystyle \frac {\int \frac {3 \cot (c+d x) (2 a-i a \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\cot (c+d x) (2 a-i a \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{2 a^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {2 a-i a \tan (c+d x)}{\tan (c+d x) (i \tan (c+d x) a+a)^2}dx}{2 a^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int \frac {2 \cot (c+d x) \left (4 a^2-3 i a^2 \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{4 a^2}+\frac {3 a}{4 d (a+i a \tan (c+d x))^2}}{2 a^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\cot (c+d x) \left (4 a^2-3 i a^2 \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{2 a^2}+\frac {3 a}{4 d (a+i a \tan (c+d x))^2}}{2 a^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {4 a^2-3 i a^2 \tan (c+d x)}{\tan (c+d x) (i \tan (c+d x) a+a)}dx}{2 a^2}+\frac {3 a}{4 d (a+i a \tan (c+d x))^2}}{2 a^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\frac {\int \cot (c+d x) \left (8 a^3-7 i a^3 \tan (c+d x)\right )dx}{2 a^2}+\frac {7 a^2}{2 d (a+i a \tan (c+d x))}}{2 a^2}+\frac {3 a}{4 d (a+i a \tan (c+d x))^2}}{2 a^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {8 a^3-7 i a^3 \tan (c+d x)}{\tan (c+d x)}dx}{2 a^2}+\frac {7 a^2}{2 d (a+i a \tan (c+d x))}}{2 a^2}+\frac {3 a}{4 d (a+i a \tan (c+d x))^2}}{2 a^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 4014

\(\displaystyle \frac {\frac {\frac {8 a^3 \int \cot (c+d x)dx-7 i a^3 x}{2 a^2}+\frac {7 a^2}{2 d (a+i a \tan (c+d x))}}{2 a^2}+\frac {3 a}{4 d (a+i a \tan (c+d x))^2}}{2 a^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {8 a^3 \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-7 i a^3 x}{2 a^2}+\frac {7 a^2}{2 d (a+i a \tan (c+d x))}}{2 a^2}+\frac {3 a}{4 d (a+i a \tan (c+d x))^2}}{2 a^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {-8 a^3 \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-7 i a^3 x}{2 a^2}+\frac {7 a^2}{2 d (a+i a \tan (c+d x))}}{2 a^2}+\frac {3 a}{4 d (a+i a \tan (c+d x))^2}}{2 a^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {\frac {\frac {7 a^2}{2 d (a+i a \tan (c+d x))}+\frac {\frac {8 a^3 \log (-\sin (c+d x))}{d}-7 i a^3 x}{2 a^2}}{2 a^2}+\frac {3 a}{4 d (a+i a \tan (c+d x))^2}}{2 a^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3}\)

Input:

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

Output:

1/(6*d*(a + I*a*Tan[c + d*x])^3) + ((3*a)/(4*d*(a + I*a*Tan[c + d*x])^2) + 
 (((-7*I)*a^3*x + (8*a^3*Log[-Sin[c + d*x]])/d)/(2*a^2) + (7*a^2)/(2*d*(a 
+ I*a*Tan[c + d*x])))/(2*a^2))/(2*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3956

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

rule 4014

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] + Simp[(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] && N 
eQ[a*c + b*d, 0]
 

rule 4042

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e 
 + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   In 
t[(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 4079

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(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] + Simp[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[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] /; Free 
Q[{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]
 

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {15 i x}{8 a^{3}}+\frac {11 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}+\frac {5 \,{\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}+\frac {{\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}-\frac {2 i c}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{3} d}\)	\(89\)
derivativedivides	\(\frac {i}{6 d \,a^{3} \left (i+\cot \left (d x +c \right )\right )^{3}}-\frac {17 i}{8 d \,a^{3} \left (i+\cot \left (d x +c \right )\right )}-\frac {7}{8 d \,a^{3} \left (i+\cot \left (d x +c \right )\right )^{2}}-\frac {\ln \left (\cot \left (d x +c \right )^{2}+1\right )}{2 d \,a^{3}}+\frac {7 i \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{8 d \,a^{3}}\)	\(102\)
default	\(\frac {i}{6 d \,a^{3} \left (i+\cot \left (d x +c \right )\right )^{3}}-\frac {17 i}{8 d \,a^{3} \left (i+\cot \left (d x +c \right )\right )}-\frac {7}{8 d \,a^{3} \left (i+\cot \left (d x +c \right )\right )^{2}}-\frac {\ln \left (\cot \left (d x +c \right )^{2}+1\right )}{2 d \,a^{3}}+\frac {7 i \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{8 d \,a^{3}}\)	\(102\)
norman	\(\frac {-\frac {7 i x}{8 a}+\frac {17}{12 a d}+\frac {\tan \left (d x +c \right )^{4}}{2 a d}+\frac {5 \tan \left (d x +c \right )^{2}}{4 a d}-\frac {17 i \tan \left (d x +c \right )}{8 d a}-\frac {7 i \tan \left (d x +c \right )^{3}}{3 d a}-\frac {7 i \tan \left (d x +c \right )^{5}}{8 a d}-\frac {21 i x \tan \left (d x +c \right )^{2}}{8 a}-\frac {21 i x \tan \left (d x +c \right )^{4}}{8 a}-\frac {7 i x \tan \left (d x +c \right )^{6}}{8 a}}{a^{2} \left (1+\tan \left (d x +c \right )^{2}\right )^{3}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{3} d}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \,a^{3}}\)	\(193\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.79 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (-180 i \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 96 \, e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 66 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 15 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \] Input:

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

Output:

1/96*(-180*I*d*x*e^(6*I*d*x + 6*I*c) + 96*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d 
*x + 2*I*c) - 1) + 66*e^(4*I*d*x + 4*I*c) + 15*e^(2*I*d*x + 2*I*c) + 2)*e^ 
(-6*I*d*x - 6*I*c)/(a^3*d)
 

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.91 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (16896 a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 3840 a^{6} d^{2} e^{8 i c} e^{- 4 i d x} + 512 a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (- 15 i e^{6 i c} - 11 i e^{4 i c} - 5 i e^{2 i c} - i\right ) e^{- 6 i c}}{8 a^{3}} + \frac {15 i}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {15 i x}{8 a^{3}} + \frac {\log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{3} d} \] Input:

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

Output:

Piecewise(((16896*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) + 3840*a**6*d**2*exp 
(8*I*c)*exp(-4*I*d*x) + 512*a**6*d**2*exp(6*I*c)*exp(-6*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*exp(6*I*c) 
 - 11*I*exp(4*I*c) - 5*I*exp(2*I*c) - I)*exp(-6*I*c)/(8*a**3) + 15*I/(8*a* 
*3)), True)) - 15*I*x/(8*a**3) + log(exp(2*I*d*x) - exp(-2*I*c))/(a**3*d)
 

Maxima [F(-2)]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.90 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\log \left (\tan \left (d x + c\right ) + i\right )}{16 \, a^{3} d} - \frac {15 \, \log \left (\tan \left (d x + c\right ) - i\right )}{16 \, a^{3} d} + \frac {\log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{3} d} - \frac {i \, {\left (21 \, \tan \left (d x + c\right )^{2} - 51 i \, \tan \left (d x + c\right ) - 34\right )}}{24 \, a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} \] Input:

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

Output:

-1/16*log(tan(d*x + c) + I)/(a^3*d) - 15/16*log(tan(d*x + c) - I)/(a^3*d) 
+ log(abs(tan(d*x + c)))/(a^3*d) - 1/24*I*(21*tan(d*x + c)^2 - 51*I*tan(d* 
x + c) - 34)/(a^3*d*(tan(d*x + c) - I)^3)
 

Mupad [B] (verification not implemented)

Time = 1.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.22 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {17}{12\,a^3}-\frac {7\,{\mathrm {tan}\left (c+d\,x\right )}^2}{8\,a^3}+\frac {\mathrm {tan}\left (c+d\,x\right )\,17{}\mathrm {i}}{8\,a^3}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}-\frac {15\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{16\,a^3\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a^3\,d} \] Input:

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

Output:

((tan(c + d*x)*17i)/(8*a^3) + 17/(12*a^3) - (7*tan(c + d*x)^2)/(8*a^3))/(d 
*(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) - (15*log(t 
an(c + d*x) - 1i))/(16*a^3*d) - log(tan(c + d*x) + 1i)/(16*a^3*d) + log(ta 
n(c + d*x))/(a^3*d)
 

Reduce [F]

\[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(cot(d*x+c)/(a+I*a*tan(d*x+c))^3,x)
 

Output:

(40*int(tan((c + d*x)/2)**4/(tan((c + d*x)/2)**6 - 6*tan((c + d*x)/2)**5*i 
 - 15*tan((c + d*x)/2)**4 + 20*tan((c + d*x)/2)**3*i + 15*tan((c + d*x)/2) 
**2 - 6*tan((c + d*x)/2)*i - 1),x)*d*i + 132*int(tan((c + d*x)/2)**3/(tan( 
(c + d*x)/2)**6 - 6*tan((c + d*x)/2)**5*i - 15*tan((c + d*x)/2)**4 + 20*ta 
n((c + d*x)/2)**3*i + 15*tan((c + d*x)/2)**2 - 6*tan((c + d*x)/2)*i - 1),x 
)*d - 208*int(tan((c + d*x)/2)**2/(tan((c + d*x)/2)**6 - 6*tan((c + d*x)/2 
)**5*i - 15*tan((c + d*x)/2)**4 + 20*tan((c + d*x)/2)**3*i + 15*tan((c + d 
*x)/2)**2 - 6*tan((c + d*x)/2)*i - 1),x)*d*i - 168*int(tan((c + d*x)/2)/(t 
an((c + d*x)/2)**6 - 6*tan((c + d*x)/2)**5*i - 15*tan((c + d*x)/2)**4 + 20 
*tan((c + d*x)/2)**3*i + 15*tan((c + d*x)/2)**2 - 6*tan((c + d*x)/2)*i - 1 
),x)*d + 12*int(1/(tan((c + d*x)/2)**7 - 6*tan((c + d*x)/2)**6*i - 15*tan( 
(c + d*x)/2)**5 + 20*tan((c + d*x)/2)**4*i + 15*tan((c + d*x)/2)**3 - 6*ta 
n((c + d*x)/2)**2*i - tan((c + d*x)/2)),x)*d + 72*int(1/(tan((c + d*x)/2)* 
*6 - 6*tan((c + d*x)/2)**5*i - 15*tan((c + d*x)/2)**4 + 20*tan((c + d*x)/2 
)**3*i + 15*tan((c + d*x)/2)**2 - 6*tan((c + d*x)/2)*i - 1),x)*d*i - log(t 
an((c + d*x)/2)**6 - 6*tan((c + d*x)/2)**5*i - 15*tan((c + d*x)/2)**4 + 20 
*tan((c + d*x)/2)**3*i + 15*tan((c + d*x)/2)**2 - 6*tan((c + d*x)/2)*i - 1 
) - 10*log(tan((c + d*x)/2)**2 + 1) + 25*log(tan((c + d*x)/2)))/(a**3*d)