\(\int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx\) [38]

Optimal result

Integrand size = 22, antiderivative size = 86 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=8 i a^4 x+\frac {7 a^4 \log (\cos (c+d x))}{d}+\frac {a^4 \log (\sin (c+d x))}{d}-\frac {\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {3 \left (a^4+i a^4 \tan (c+d x)\right )}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.79 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4 \log (\tan (c+d x))}{d}-\frac {8 a^4 \log (i+\tan (c+d x))}{d}-\frac {4 i a^4 \tan (c+d x)}{d}+\frac {a^4 \tan ^2(c+d x)}{2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {3042, 4039, 27, 3042, 4077, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4039

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4077

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4072

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3956

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

\(\Big \downarrow \) 4014

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

Input:

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

Output:

-1/2*(a^2 + I*a^2*Tan[c + d*x])^2/d + a*((8*I)*a^3*x + (7*a^3*Log[Cos[c + 
d*x]])/d + (a^3*Log[-Sin[c + d*x]])/d - (3*(a^3 + I*a^3*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 4039

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + 
 d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[a/(d*(m + n - 1)) 
Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + 
a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*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] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] 
 && (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 4077

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[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + 
n))), x] + Simp[1/(d*(m + n))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan 
[e + f*x])^n*Simp[a*A*d*(m + n) + B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - 
 a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*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] && 
GtQ[m, 1] &&  !LtQ[n, -1]
 

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60

method	result	size
parallelrisch	\(\frac {a^{4} \left (16 i d x +\tan \left (d x +c \right )^{2}-8 i \tan \left (d x +c \right )+2 \ln \left (\tan \left (d x +c \right )\right )-8 \ln \left (\sec \left (d x +c \right )^{2}\right )\right )}{2 d}\)	\(52\)
derivativedivides	\(\frac {a^{4} \left (-4 i \tan \left (d x +c \right )+\frac {\tan \left (d x +c \right )^{2}}{2}-4 \ln \left (1+\tan \left (d x +c \right )^{2}\right )+8 i \arctan \left (\tan \left (d x +c \right )\right )+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(58\)
default	\(\frac {a^{4} \left (-4 i \tan \left (d x +c \right )+\frac {\tan \left (d x +c \right )^{2}}{2}-4 \ln \left (1+\tan \left (d x +c \right )^{2}\right )+8 i \arctan \left (\tan \left (d x +c \right )\right )+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(58\)
norman	\(\frac {a^{4} \tan \left (d x +c \right )^{2}}{2 d}+8 i a^{4} x -\frac {4 i a^{4} \tan \left (d x +c \right )}{d}+\frac {a^{4} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {4 a^{4} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}\)	\(73\)
risch	\(-\frac {16 i a^{4} c}{d}+\frac {2 a^{4} \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}+4\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {7 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(85\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.59 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {10 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 \, a^{4} + 7 \, {\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + {\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \] Input:

integrate(cot(d*x+c)*(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 + 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) + (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 = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.22 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^{4} \left (\log {\left (e^{2 i d x} - e^{- 2 i c} \right )} + 7 \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}\right )}{d} + \frac {10 a^{4} e^{2 i c} e^{2 i d x} + 8 a^{4}}{d e^{4 i c} e^{4 i d x} + 2 d e^{2 i c} e^{2 i d x} + d} \] Input:

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

Output:

a**4*(log(exp(2*I*d*x) - exp(-2*I*c)) + 7*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 = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.78 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^{4} \tan \left (d x + c\right )^{2} + 16 i \, {\left (d x + c\right )} a^{4} - 8 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, a^{4} \log \left (\tan \left (d x + c\right )\right ) - 8 i \, a^{4} \tan \left (d x + c\right )}{2 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {8 \, a^{4} \log \left (\tan \left (d x + c\right ) + i\right )}{d} + \frac {a^{4} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} + \frac {a^{4} d \tan \left (d x + c\right )^{2} - 8 i \, a^{4} d \tan \left (d x + c\right )}{2 \, d^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.89 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d}-\frac {8\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}+\frac {a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {a^4\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.42 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^{4} \left (8 \cos \left (d x +c \right ) \sin \left (d x +c \right ) i -16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}+16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+14 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-14 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+14 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-14 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+16 \sin \left (d x +c \right )^{2} d i x -\sin \left (d x +c \right )^{2}-16 d i x \right )}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

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

Output:

(a**4*(8*cos(c + d*x)*sin(c + d*x)*i - 16*log(tan((c + d*x)/2)**2 + 1)*sin 
(c + d*x)**2 + 16*log(tan((c + d*x)/2)**2 + 1) + 14*log(tan((c + d*x)/2) - 
 1)*sin(c + d*x)**2 - 14*log(tan((c + d*x)/2) - 1) + 14*log(tan((c + d*x)/ 
2) + 1)*sin(c + d*x)**2 - 14*log(tan((c + d*x)/2) + 1) + 2*log(tan((c + d* 
x)/2))*sin(c + d*x)**2 - 2*log(tan((c + d*x)/2)) + 16*sin(c + d*x)**2*d*i* 
x - sin(c + d*x)**2 - 16*d*i*x))/(2*d*(sin(c + d*x)**2 - 1))