\(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx\) [40]

Optimal result

Integrand size = 24, antiderivative size = 103 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=-8 i a^4 x-\frac {a^4 \log (\cos (c+d x))}{d}-\frac {7 a^4 \log (\sin (c+d x))}{d}-\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {3 i \cot (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d} \] Output:

-8*I*a^4*x-a^4*ln(cos(d*x+c))/d-7*a^4*ln(sin(d*x+c))/d-1/2*cot(d*x+c)^2*(a 
^2+I*a^2*tan(d*x+c))^2/d-3*I*cot(d*x+c)*(a^4+I*a^4*tan(d*x+c))/d
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.59 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=a^4 \left (-\frac {4 i \cot (c+d x)}{d}-\frac {\cot ^2(c+d x)}{2 d}-\frac {7 \log (\tan (c+d x))}{d}+\frac {8 \log (i+\tan (c+d x))}{d}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3042, 4036, 27, 3042, 4076, 25, 3042, 4072, 3042, 3956, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4036

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4076

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4072

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3956

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

\(\Big \downarrow \) 4014

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

\(\displaystyle -\frac {7 a^4 \log (-\sin (c+d x))}{d}-\frac {a^4 \log (\cos (c+d x))}{d}-\frac {3 i \cot (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d}-8 i a^4 x-\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}\)

Input:

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

Output:

(-8*I)*a^4*x - (a^4*Log[Cos[c + d*x]])/d - (7*a^4*Log[-Sin[c + d*x]])/d - 
(Cot[c + d*x]^2*(a^2 + I*a^2*Tan[c + d*x])^2)/(2*d) - ((3*I)*Cot[c + d*x]* 
(a^4 + I*a^4*Tan[c + d*x]))/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 4036

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x] 
)^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] + Si 
mp[a/(d*(b*c + a*d)*(n + 1))   Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[ 
e + f*x])^(n + 1)*Simp[b*(b*c*(m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) 
 + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], 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] && GtQ[m, 1] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4072

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_ 
.)*(x_)]))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*(d/ 
b)   Int[Tan[e + f*x], x], x] + Simp[1/b   Int[Simp[A*b*c + (A*b*d + B*(b*c 
 - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d 
, e, f, A, B}, x] && NeQ[b*c - a*d, 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 = 1.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.50

method	result	size
parallelrisch	\(-\frac {a^{4} \left (16 i d x +14 \ln \left (\tan \left (d x +c \right )\right )-8 \ln \left (\sec \left (d x +c \right )^{2}\right )+8 i \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}\right )}{2 d}\)	\(52\)
derivativedivides	\(\frac {a^{4} \left (-4 i \cot \left (d x +c \right )-\frac {\cot \left (d x +c \right )^{2}}{2}+4 \ln \left (\cot \left (d x +c \right )^{2}+1\right )+8 i \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )-\ln \left (\cot \left (d x +c \right )\right )\right )}{d}\)	\(66\)
default	\(\frac {a^{4} \left (-4 i \cot \left (d x +c \right )-\frac {\cot \left (d x +c \right )^{2}}{2}+4 \ln \left (\cot \left (d x +c \right )^{2}+1\right )+8 i \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )-\ln \left (\cot \left (d x +c \right )\right )\right )}{d}\)	\(66\)
norman	\(\frac {-\frac {a^{4}}{2 d}-\frac {4 i a^{4} \tan \left (d x +c \right )}{d}-8 i a^{4} x \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{2}}-\frac {7 a^{4} \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {4 a^{4} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}\)	\(84\)
risch	\(\frac {16 i a^{4} c}{d}+\frac {2 a^{4} \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}-4\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {7 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(86\)

Input:

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

Output:

-1/2*a^4*(16*I*d*x+14*ln(tan(d*x+c))-8*ln(sec(d*x+c)^2)+8*I*cot(d*x+c)+cot 
(d*x+c)^2)/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.34 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {10 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 8 \, a^{4} - {\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 7 \, {\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \] Input:

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

Output:

(10*a^4*e^(2*I*d*x + 2*I*c) - 8*a^4 - (a^4*e^(4*I*d*x + 4*I*c) - 2*a^4*e^( 
2*I*d*x + 2*I*c) + a^4)*log(e^(2*I*d*x + 2*I*c) + 1) - 7*(a^4*e^(4*I*d*x + 
 4*I*c) - 2*a^4*e^(2*I*d*x + 2*I*c) + a^4)*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 = 0.54 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.04 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^{4} \left (- 7 \log {\left (e^{2 i d x} - e^{- 2 i c} \right )} - \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}\right )}{d} + \frac {10 a^{4} e^{2 i c} e^{2 i d x} - 8 a^{4}}{d e^{4 i c} e^{4 i d x} - 2 d e^{2 i c} e^{2 i d x} + d} \] Input:

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

Output:

a**4*(-7*log(exp(2*I*d*x) - exp(-2*I*c)) - log(exp(2*I*d*x) + exp(-2*I*c)) 
)/d + (10*a**4*exp(2*I*c)*exp(2*I*d*x) - 8*a**4)/(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 = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.66 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {16 i \, {\left (d x + c\right )} a^{4} - 8 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 14 \, a^{4} \log \left (\tan \left (d x + c\right )\right ) + \frac {8 i \, a^{4} \tan \left (d x + c\right ) + a^{4}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \] Input:

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

Output:

-1/2*(16*I*(d*x + c)*a^4 - 8*a^4*log(tan(d*x + c)^2 + 1) + 14*a^4*log(tan( 
d*x + c)) + (8*I*a^4*tan(d*x + c) + a^4)/tan(d*x + c)^2)/d
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.60 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {8 \, a^{4} \log \left (\tan \left (d x + c\right ) + i\right )}{d} - \frac {7 \, a^{4} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {8 i \, a^{4} \tan \left (d x + c\right ) + a^{4}}{2 \, d \tan \left (d x + c\right )^{2}} \] Input:

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

Output:

8*a^4*log(tan(d*x + c) + I)/d - 7*a^4*log(abs(tan(d*x + c)))/d - 1/2*(8*I* 
a^4*tan(d*x + c) + a^4)/(d*tan(d*x + c)^2)
 

Mupad [B] (verification not implemented)

Time = 0.90 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.63 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {8\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}-\frac {a^4\,{\mathrm {cot}\left (c+d\,x\right )}^2}{2\,d}-\frac {7\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {a^4\,\mathrm {cot}\left (c+d\,x\right )\,4{}\mathrm {i}}{d} \] Input:

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

Output:

(8*a^4*log(tan(c + d*x) + 1i))/d - (a^4*cot(c + d*x)*4i)/d - (a^4*cot(c + 
d*x)^2)/(2*d) - (7*a^4*log(tan(c + d*x)))/d
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.38 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^{4} \left (-16 \cos \left (d x +c \right ) \sin \left (d x +c \right ) i +32 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-28 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-32 \sin \left (d x +c \right )^{2} d i x +\sin \left (d x +c \right )^{2}-2\right )}{4 \sin \left (d x +c \right )^{2} d} \] Input:

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

Output:

(a**4*( - 16*cos(c + d*x)*sin(c + d*x)*i + 32*log(tan((c + d*x)/2)**2 + 1) 
*sin(c + d*x)**2 - 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 4*log(tan 
((c + d*x)/2) + 1)*sin(c + d*x)**2 - 28*log(tan((c + d*x)/2))*sin(c + d*x) 
**2 - 32*sin(c + d*x)**2*d*i*x + sin(c + d*x)**2 - 2))/(4*sin(c + d*x)**2* 
d)