\(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [14]

Optimal result

Integrand size = 34, antiderivative size = 94 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-2 a^2 (i A+B) x-\frac {a^2 (3 i A+2 B) \cot (c+d x)}{2 d}-\frac {2 a^2 (A-i B) \log (\sin (c+d x))}{d}-\frac {A \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=a^2 \left (-\frac {2 i A \cot (c+d x)}{d}-\frac {B \cot (c+d x)}{d}-\frac {A \cot ^2(c+d x)}{2 d}-\frac {2 A \log (\tan (c+d x))}{d}+\frac {2 i B \log (\tan (c+d x))}{d}+\frac {2 A \log (i+\tan (c+d x))}{d}-\frac {2 i B \log (i+\tan (c+d x))}{d}\right ) \] Input:

Integrate[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]
 

Output:

a^2*(((-2*I)*A*Cot[c + d*x])/d - (B*Cot[c + d*x])/d - (A*Cot[c + d*x]^2)/( 
2*d) - (2*A*Log[Tan[c + d*x]])/d + ((2*I)*B*Log[Tan[c + d*x]])/d + (2*A*Lo 
g[I + Tan[c + d*x]])/d - ((2*I)*B*Log[I + Tan[c + d*x]])/d)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3042, 4076, 3042, 4074, 27, 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.

\(\displaystyle \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan (c+d x)^3}dx\)

\(\Big \downarrow \) 4076

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4074

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4014

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

\(\displaystyle \frac {1}{2} \left (-\frac {a^2 (2 B+3 i A) \cot (c+d x)}{d}-4 \left (\frac {a^2 (A-i B) \log (-\sin (c+d x))}{d}+a^2 x (B+i A)\right )\right )-\frac {A \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\)

Input:

Int[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]
 

Output:

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

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 4074

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

rule 4076

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^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n 
+ 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Simp[a/(d*(b*c + a*d)*(n + 1))   Int[ 
(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n 
 - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m - 1) + b 
*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] 
 && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]
 

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83

method	result	size
parallelrisch	\(-\frac {2 a^{2} \left (\frac {\left (i B -A \right ) \ln \left (\sec \left (d x +c \right )^{2}\right )}{2}+\left (-i B +A \right ) \ln \left (\tan \left (d x +c \right )\right )+\frac {A \cot \left (d x +c \right )^{2}}{4}+\cot \left (d x +c \right ) \left (i A +\frac {B}{2}\right )+\left (i A +B \right ) x d \right )}{d}\)	\(78\)
derivativedivides	\(\frac {-A \,a^{2} \ln \left (\sin \left (d x +c \right )\right )-B \,a^{2} \left (d x +c \right )+2 i A \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 i B \,a^{2} \ln \left (\sin \left (d x +c \right )\right )+A \,a^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+B \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(113\)
default	\(\frac {-A \,a^{2} \ln \left (\sin \left (d x +c \right )\right )-B \,a^{2} \left (d x +c \right )+2 i A \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 i B \,a^{2} \ln \left (\sin \left (d x +c \right )\right )+A \,a^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+B \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(113\)
risch	\(\frac {4 a^{2} B c}{d}+\frac {4 i a^{2} A c}{d}-\frac {2 i a^{2} \left (3 i A \,{\mathrm e}^{2 i \left (d x +c \right )}+B \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i A -B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {2 i a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}-\frac {2 A \,a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(120\)
norman	\(\frac {\left (-2 i A \,a^{2}-2 B \,a^{2}\right ) x \tan \left (d x +c \right )^{2}-\frac {A \,a^{2}}{2 d}-\frac {\left (2 i A \,a^{2}+B \,a^{2}\right ) \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}+\frac {\left (-i B \,a^{2}+A \,a^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}-\frac {2 \left (-i B \,a^{2}+A \,a^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(122\)

Input:

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x,method=_RETURNVER 
BOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.31 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {2 \, {\left ({\left (3 \, A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (2 \, A - i \, B\right )} a^{2} - {\left ({\left (A - i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - i \, B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \] Input:

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm= 
"fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.27 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=- \frac {2 a^{2} \left (A - i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 4 A a^{2} + 2 i B a^{2} + \left (6 A a^{2} e^{2 i c} - 2 i B a^{2} e^{2 i c}\right ) e^{2 i d x}}{d e^{4 i c} e^{4 i d x} - 2 d e^{2 i c} e^{2 i d x} + d} \] Input:

integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)
 

Output:

-2*a**2*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-4*A*a**2 + 2*I*B*a 
**2 + (6*A*a**2*exp(2*I*c) - 2*I*B*a**2*exp(2*I*c))*exp(2*I*d*x))/(d*exp(4 
*I*c)*exp(4*I*d*x) - 2*d*exp(2*I*c)*exp(2*I*d*x) + d)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {4 \, {\left (d x + c\right )} {\left (i \, A + B\right )} a^{2} - 2 \, {\left (A - i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 4 \, {\left (A - i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )\right ) - \frac {2 \, {\left (-2 i \, A - B\right )} a^{2} \tan \left (d x + c\right ) - A a^{2}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \] Input:

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm= 
"maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.98 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {2 \, {\left (A a^{2} - i \, B a^{2}\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{d} - \frac {2 \, {\left (A a^{2} - i \, B a^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {A a^{2} + 2 i \, {\left (2 \, A a^{2} - i \, B a^{2}\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )^{2}} \] Input:

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm= 
"giac")
 

Output:

2*(A*a^2 - I*B*a^2)*log(tan(d*x + c) + I)/d - 2*(A*a^2 - I*B*a^2)*log(abs( 
tan(d*x + c)))/d - 1/2*(A*a^2 + 2*I*(2*A*a^2 - I*B*a^2)*tan(d*x + c))/(d*t 
an(d*x + c)^2)
 

Mupad [B] (verification not implemented)

Time = 3.69 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {\frac {A\,a^2}{2}+\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^2+A\,a^2\,2{}\mathrm {i}\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^2}-\frac {4\,a^2\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{d} \] Input:

int(cot(c + d*x)^3*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.94 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {a^{2} \left (-8 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a i -4 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} a -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} b i -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b i -8 \sin \left (d x +c \right )^{2} a d i x +\sin \left (d x +c \right )^{2} a -8 \sin \left (d x +c \right )^{2} b d x -2 a \right )}{4 \sin \left (d x +c \right )^{2} d} \] Input:

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)
 

Output:

(a**2*( - 8*cos(c + d*x)*sin(c + d*x)*a*i - 4*cos(c + d*x)*sin(c + d*x)*b 
+ 8*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a - 8*log(tan((c + d*x)/2 
)**2 + 1)*sin(c + d*x)**2*b*i - 8*log(tan((c + d*x)/2))*sin(c + d*x)**2*a 
+ 8*log(tan((c + d*x)/2))*sin(c + d*x)**2*b*i - 8*sin(c + d*x)**2*a*d*i*x 
+ sin(c + d*x)**2*a - 8*sin(c + d*x)**2*b*d*x - 2*a))/(4*sin(c + d*x)**2*d 
)