\(\int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [31]

Optimal result

Integrand size = 24, antiderivative size = 101 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=4 a^3 x+\frac {2 a^3 \cot (c+d x)}{d}-\frac {4 i a^3 \log (\sin (c+d x))}{d}-\frac {i a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac {\cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=a^3 \left (\frac {4 \cot (c+d x)}{d}-\frac {3 i \cot ^2(c+d x)}{2 d}-\frac {\cot ^3(c+d x)}{3 d}-\frac {4 i \log (\tan (c+d x))}{d}+\frac {4 i \log (i+\tan (c+d x))}{d}\right ) \] Input:

Integrate[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.14, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4031, 3042, 4028, 3042, 4025, 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.

\(\displaystyle \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4031

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4028

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4025

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4014

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

Input:

Int[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^3,x]
 

Output:

-1/3*(Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^3)/d + I*((2*I)*a*(-((a^2*Cot[ 
c + d*x])/d) + 2*(-(a^2*x) + (I*a^2*Log[-Sin[c + d*x]])/d)) - (a*Cot[c + d 
*x]^2*(a + I*a*Tan[c + d*x])^2)/(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 4025

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

rule 4028

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

rule 4031

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

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63

method	result	size
derivativedivides	\(\frac {a^{3} \left (4 \cot \left (d x +c \right )-\frac {\cot \left (d x +c \right )^{3}}{3}-\frac {3 i \cot \left (d x +c \right )^{2}}{2}+2 i \ln \left (\cot \left (d x +c \right )^{2}+1\right )-2 \pi +4 \,\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}\)	\(64\)
default	\(\frac {a^{3} \left (4 \cot \left (d x +c \right )-\frac {\cot \left (d x +c \right )^{3}}{3}-\frac {3 i \cot \left (d x +c \right )^{2}}{2}+2 i \ln \left (\cot \left (d x +c \right )^{2}+1\right )-2 \pi +4 \,\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}\)	\(64\)
parallelrisch	\(\frac {\left (-12 i \left (2 \ln \left (\tan \left (d x +c \right )\right )-\ln \left (\sec \left (d x +c \right )^{2}\right )\right )-2 \cot \left (d x +c \right )^{3}-9 i \cot \left (d x +c \right )^{2}+24 d x +24 \cot \left (d x +c \right )\right ) a^{3}}{6 d}\)	\(67\)
risch	\(-\frac {8 a^{3} c}{d}+\frac {2 i a^{3} \left (24 \,{\mathrm e}^{4 i \left (d x +c \right )}-33 \,{\mathrm e}^{2 i \left (d x +c \right )}+13\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {4 i a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(78\)
norman	\(\frac {-\frac {a^{3}}{3 d}+4 a^{3} x \tan \left (d x +c \right )^{3}+\frac {4 a^{3} \tan \left (d x +c \right )^{2}}{d}-\frac {3 i a^{3} \tan \left (d x +c \right )}{2 d}}{\tan \left (d x +c \right )^{3}}-\frac {4 i a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {2 i a^{3} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}\)	\(101\)

Input:

int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*a^3*(4*cot(d*x+c)-1/3*cot(d*x+c)^3-3/2*I*cot(d*x+c)^2+2*I*ln(cot(d*x+c 
)^2+1)-2*Pi+4*arccot(cot(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.38 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {2 \, {\left (-24 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 33 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 13 i \, a^{3} + 6 \, {\left (i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \] Input:

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

-2/3*(-24*I*a^3*e^(4*I*d*x + 4*I*c) + 33*I*a^3*e^(2*I*d*x + 2*I*c) - 13*I* 
a^3 + 6*(I*a^3*e^(6*I*d*x + 6*I*c) - 3*I*a^3*e^(4*I*d*x + 4*I*c) + 3*I*a^3 
*e^(2*I*d*x + 2*I*c) - I*a^3)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(6*I*d*x 
+ 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)
 

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.35 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=- \frac {4 i a^{3} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {48 i a^{3} e^{4 i c} e^{4 i d x} - 66 i a^{3} e^{2 i c} e^{2 i d x} + 26 i a^{3}}{3 d e^{6 i c} e^{6 i d x} - 9 d e^{4 i c} e^{4 i d x} + 9 d e^{2 i c} e^{2 i d x} - 3 d} \] Input:

integrate(cot(d*x+c)**4*(a+I*a*tan(d*x+c))**3,x)
 

Output:

-4*I*a**3*log(exp(2*I*d*x) - exp(-2*I*c))/d + (48*I*a**3*exp(4*I*c)*exp(4* 
I*d*x) - 66*I*a**3*exp(2*I*c)*exp(2*I*d*x) + 26*I*a**3)/(3*d*exp(6*I*c)*ex 
p(6*I*d*x) - 9*d*exp(4*I*c)*exp(4*I*d*x) + 9*d*exp(2*I*c)*exp(2*I*d*x) - 3 
*d)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {24 \, {\left (d x + c\right )} a^{3} + 12 i \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 24 i \, a^{3} \log \left (\tan \left (d x + c\right )\right ) + \frac {24 \, a^{3} \tan \left (d x + c\right )^{2} - 9 i \, a^{3} \tan \left (d x + c\right ) - 2 \, a^{3}}{\tan \left (d x + c\right )^{3}}}{6 \, d} \] Input:

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

1/6*(24*(d*x + c)*a^3 + 12*I*a^3*log(tan(d*x + c)^2 + 1) - 24*I*a^3*log(ta 
n(d*x + c)) + (24*a^3*tan(d*x + c)^2 - 9*I*a^3*tan(d*x + c) - 2*a^3)/tan(d 
*x + c)^3)/d
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {4 i \, a^{3} \log \left (\tan \left (d x + c\right ) + i\right )}{d} - \frac {4 i \, a^{3} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} + \frac {24 \, a^{3} \tan \left (d x + c\right )^{2} - 9 i \, a^{3} \tan \left (d x + c\right ) - 2 \, a^{3}}{6 \, d \tan \left (d x + c\right )^{3}} \] Input:

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

4*I*a^3*log(tan(d*x + c) + I)/d - 4*I*a^3*log(abs(tan(d*x + c)))/d + 1/6*( 
24*a^3*tan(d*x + c)^2 - 9*I*a^3*tan(d*x + c) - 2*a^3)/(d*tan(d*x + c)^3)
 

Mupad [B] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {4\,a^3\,\mathrm {cot}\left (c+d\,x\right )}{d}+\frac {8\,a^3\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}-\frac {a^3\,{\mathrm {cot}\left (c+d\,x\right )}^3}{3\,d}-\frac {a^3\,{\mathrm {cot}\left (c+d\,x\right )}^2\,3{}\mathrm {i}}{2\,d} \] Input:

int(cot(c + d*x)^4*(a + a*tan(c + d*x)*1i)^3,x)
 

Output:

(4*a^3*cot(c + d*x))/d + (8*a^3*atan(2*tan(c + d*x) + 1i))/d - (a^3*cot(c 
+ d*x)^2*3i)/(2*d) - (a^3*cot(c + d*x)^3)/(3*d)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.18 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^{3} \left (52 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )+48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} i -48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} i +48 \sin \left (d x +c \right )^{3} d x +9 \sin \left (d x +c \right )^{3} i -18 \sin \left (d x +c \right ) i \right )}{12 \sin \left (d x +c \right )^{3} d} \] Input:

int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x)
 

Output:

(a**3*(52*cos(c + d*x)*sin(c + d*x)**2 - 4*cos(c + d*x) + 48*log(tan((c + 
d*x)/2)**2 + 1)*sin(c + d*x)**3*i - 48*log(tan((c + d*x)/2))*sin(c + d*x)* 
*3*i + 48*sin(c + d*x)**3*d*x + 9*sin(c + d*x)**3*i - 18*sin(c + d*x)*i))/ 
(12*sin(c + d*x)**3*d)