\(\int \cot ^4(c+d x) (a+b \tan (c+d x))^4 \, dx\) [452]

Optimal result

Integrand size = 21, antiderivative size = 117 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^4 \, dx=\left (a^4-6 a^2 b^2+b^4\right ) x+\frac {a^2 \left (3 a^2-17 b^2\right ) \cot (c+d x)}{3 d}-\frac {4 a^3 b \cot ^2(c+d x)}{3 d}-\frac {4 a b \left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d} \] Output:

(a^4-6*a^2*b^2+b^4)*x+1/3*a^2*(3*a^2-17*b^2)*cot(d*x+c)/d-4/3*a^3*b*cot(d* 
x+c)^2/d-4*a*b*(a^2-b^2)*ln(sin(d*x+c))/d-1/3*a^2*cot(d*x+c)^3*(a+b*tan(d* 
x+c))^2/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.73 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.07 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {-6 a^2 \left (a^2-6 b^2\right ) \cot (c+d x)+12 a^3 b \cot ^2(c+d x)+2 a^4 \cot ^3(c+d x)+3 i (a+i b)^4 \log (i-\tan (c+d x))+24 a b \left (a^2-b^2\right ) \log (\tan (c+d x))-3 i (a-i b)^4 \log (i+\tan (c+d x))}{6 d} \] Input:

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

Output:

-1/6*(-6*a^2*(a^2 - 6*b^2)*Cot[c + d*x] + 12*a^3*b*Cot[c + d*x]^2 + 2*a^4* 
Cot[c + d*x]^3 + (3*I)*(a + I*b)^4*Log[I - Tan[c + d*x]] + 24*a*b*(a^2 - b 
^2)*Log[Tan[c + d*x]] - (3*I)*(a - I*b)^4*Log[I + Tan[c + d*x]])/d
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 4048, 3042, 4118, 25, 3042, 4111, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4118

\(\displaystyle \frac {1}{3} \left (\int -\cot ^2(c+d x) \left (\left (3 a^2-17 b^2\right ) a^2+12 b \left (a^2-b^2\right ) \tan (c+d x) a+b^2 \left (a^2-3 b^2\right ) \tan ^2(c+d x)\right )dx-\frac {4 a^3 b \cot ^2(c+d x)}{d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (-\int \cot ^2(c+d x) \left (\left (3 a^2-17 b^2\right ) a^2+12 b \left (a^2-b^2\right ) \tan (c+d x) a+b^2 \left (a^2-3 b^2\right ) \tan ^2(c+d x)\right )dx-\frac {4 a^3 b \cot ^2(c+d x)}{d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (-\int \frac {\left (3 a^2-17 b^2\right ) a^2+12 b \left (a^2-b^2\right ) \tan (c+d x) a+b^2 \left (a^2-3 b^2\right ) \tan (c+d x)^2}{\tan (c+d x)^2}dx-\frac {4 a^3 b \cot ^2(c+d x)}{d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4111

\(\displaystyle \frac {1}{3} \left (-\int 3 \cot (c+d x) \left (4 a b \left (a^2-b^2\right )-\left (a^4-6 b^2 a^2+b^4\right ) \tan (c+d x)\right )dx-\frac {4 a^3 b \cot ^2(c+d x)}{d}+\frac {a^2 \left (3 a^2-17 b^2\right ) \cot (c+d x)}{d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (-3 \int \cot (c+d x) \left (4 a b \left (a^2-b^2\right )-\left (a^4-6 b^2 a^2+b^4\right ) \tan (c+d x)\right )dx-\frac {4 a^3 b \cot ^2(c+d x)}{d}+\frac {a^2 \left (3 a^2-17 b^2\right ) \cot (c+d x)}{d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (-3 \int \frac {4 a b \left (a^2-b^2\right )-\left (a^4-6 b^2 a^2+b^4\right ) \tan (c+d x)}{\tan (c+d x)}dx-\frac {4 a^3 b \cot ^2(c+d x)}{d}+\frac {a^2 \left (3 a^2-17 b^2\right ) \cot (c+d x)}{d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4014

\(\displaystyle \frac {1}{3} \left (-3 \left (4 a b \left (a^2-b^2\right ) \int \cot (c+d x)dx-x \left (a^4-6 a^2 b^2+b^4\right )\right )-\frac {4 a^3 b \cot ^2(c+d x)}{d}+\frac {a^2 \left (3 a^2-17 b^2\right ) \cot (c+d x)}{d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (-3 \left (4 a b \left (a^2-b^2\right ) \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-x \left (a^4-6 a^2 b^2+b^4\right )\right )-\frac {4 a^3 b \cot ^2(c+d x)}{d}+\frac {a^2 \left (3 a^2-17 b^2\right ) \cot (c+d x)}{d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (-3 \left (-4 a b \left (a^2-b^2\right ) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\left (x \left (a^4-6 a^2 b^2+b^4\right )\right )\right )-\frac {4 a^3 b \cot ^2(c+d x)}{d}+\frac {a^2 \left (3 a^2-17 b^2\right ) \cot (c+d x)}{d}\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {1}{3} \left (-\frac {4 a^3 b \cot ^2(c+d x)}{d}+\frac {a^2 \left (3 a^2-17 b^2\right ) \cot (c+d x)}{d}-3 \left (\frac {4 a b \left (a^2-b^2\right ) \log (-\sin (c+d x))}{d}-x \left (a^4-6 a^2 b^2+b^4\right )\right )\right )-\frac {a^2 \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

Input:

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

Output:

((a^2*(3*a^2 - 17*b^2)*Cot[c + d*x])/d - (4*a^3*b*Cot[c + d*x]^2)/d - 3*(- 
((a^4 - 6*a^2*b^2 + b^4)*x) + (4*a*b*(a^2 - b^2)*Log[-Sin[c + d*x]])/d))/3 
 - (a^2*Cot[c + d*x]^3*(a + b*Tan[c + d*x])^2)/(3*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 4048

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

rule 4111

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

rule 4118

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

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {\frac {\left (4 a^{3} b -4 a \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (a^{4}-6 b^{2} a^{2}+b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {a^{4}}{3 \tan \left (d x +c \right )^{3}}+\frac {a^{2} \left (a^{2}-6 b^{2}\right )}{\tan \left (d x +c \right )}-\frac {2 a^{3} b}{\tan \left (d x +c \right )^{2}}-4 a b \left (a^{2}-b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(123\)
default	\(\frac {\frac {\left (4 a^{3} b -4 a \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (a^{4}-6 b^{2} a^{2}+b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {a^{4}}{3 \tan \left (d x +c \right )^{3}}+\frac {a^{2} \left (a^{2}-6 b^{2}\right )}{\tan \left (d x +c \right )}-\frac {2 a^{3} b}{\tan \left (d x +c \right )^{2}}-4 a b \left (a^{2}-b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(123\)
norman	\(\frac {\left (a^{4}-6 b^{2} a^{2}+b^{4}\right ) x \tan \left (d x +c \right )^{3}+\frac {a^{2} \left (a^{2}-6 b^{2}\right ) \tan \left (d x +c \right )^{2}}{d}-\frac {a^{4}}{3 d}-\frac {2 a^{3} b \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{3}}-\frac {4 a b \left (a^{2}-b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {2 a b \left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}\)	\(134\)
parallelrisch	\(\frac {-6 a^{3} b \left (2 \ln \left (\tan \left (d x +c \right )\right )-\ln \left (\sec \left (d x +c \right )^{2}\right )\right )+6 a \,b^{3} \left (2 \ln \left (\tan \left (d x +c \right )\right )-\ln \left (\sec \left (d x +c \right )^{2}\right )\right )+3 a^{4} x d -18 a^{2} b^{2} d x +3 b^{4} d x +3 \cot \left (d x +c \right ) a^{4}-18 \cot \left (d x +c \right ) b^{2} a^{2}-6 \cot \left (d x +c \right )^{2} a^{3} b -\cot \left (d x +c \right )^{3} a^{4}}{3 d}\)	\(137\)
risch	\(4 i a^{3} b x -4 i a \,b^{3} x +a^{4} x -6 a^{2} b^{2} x +b^{4} x +\frac {8 i a^{3} b c}{d}-\frac {8 i a \,b^{3} c}{d}+\frac {4 i a^{2} \left (3 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-9 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+18 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+6 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a^{2}-9 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {4 a^{3} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {4 a \,b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(218\)

Input:

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

Output:

1/d*(1/2*(4*a^3*b-4*a*b^3)*ln(1+tan(d*x+c)^2)+(a^4-6*a^2*b^2+b^4)*arctan(t 
an(d*x+c))-1/3*a^4/tan(d*x+c)^3+a^2*(a^2-6*b^2)/tan(d*x+c)-2*a^3*b/tan(d*x 
+c)^2-4*a*b*(a^2-b^2)*ln(tan(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.12 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {6 \, a^{3} b \tan \left (d x + c\right ) + 6 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} + a^{4} + 3 \, {\left (2 \, a^{3} b - {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} d x\right )} \tan \left (d x + c\right )^{3} - 3 \, {\left (a^{4} - 6 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{3 \, d \tan \left (d x + c\right )^{3}} \] Input:

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

Output:

-1/3*(6*a^3*b*tan(d*x + c) + 6*(a^3*b - a*b^3)*log(tan(d*x + c)^2/(tan(d*x 
 + c)^2 + 1))*tan(d*x + c)^3 + a^4 + 3*(2*a^3*b - (a^4 - 6*a^2*b^2 + b^4)* 
d*x)*tan(d*x + c)^3 - 3*(a^4 - 6*a^2*b^2)*tan(d*x + c)^2)/(d*tan(d*x + c)^ 
3)
 

Sympy [A] (verification not implemented)

Time = 1.94 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.58 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^4 \, dx=\begin {cases} \tilde {\infty } a^{4} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right )^{4} \cot ^{4}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } a^{4} x & \text {for}\: c = - d x \\a^{4} x + \frac {a^{4}}{d \tan {\left (c + d x \right )}} - \frac {a^{4}}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {2 a^{3} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - \frac {4 a^{3} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {2 a^{3} b}{d \tan ^{2}{\left (c + d x \right )}} - 6 a^{2} b^{2} x - \frac {6 a^{2} b^{2}}{d \tan {\left (c + d x \right )}} - \frac {2 a b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {4 a b^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + b^{4} x & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((zoo*a**4*x, Eq(c, 0) & Eq(d, 0)), (x*(a + b*tan(c))**4*cot(c)** 
4, Eq(d, 0)), (zoo*a**4*x, Eq(c, -d*x)), (a**4*x + a**4/(d*tan(c + d*x)) - 
 a**4/(3*d*tan(c + d*x)**3) + 2*a**3*b*log(tan(c + d*x)**2 + 1)/d - 4*a**3 
*b*log(tan(c + d*x))/d - 2*a**3*b/(d*tan(c + d*x)**2) - 6*a**2*b**2*x - 6* 
a**2*b**2/(d*tan(c + d*x)) - 2*a*b**3*log(tan(c + d*x)**2 + 1)/d + 4*a*b** 
3*log(tan(c + d*x))/d + b**4*x, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.04 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {3 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} {\left (d x + c\right )} + 6 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 12 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {6 \, a^{3} b \tan \left (d x + c\right ) + a^{4} - 3 \, {\left (a^{4} - 6 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \] Input:

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

Output:

1/3*(3*(a^4 - 6*a^2*b^2 + b^4)*(d*x + c) + 6*(a^3*b - a*b^3)*log(tan(d*x + 
 c)^2 + 1) - 12*(a^3*b - a*b^3)*log(tan(d*x + c)) - (6*a^3*b*tan(d*x + c) 
+ a^4 - 3*(a^4 - 6*a^2*b^2)*tan(d*x + c)^2)/tan(d*x + c)^3)/d
 

Giac [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.10 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} {\left (d x + c\right )}}{d} + \frac {2 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{d} - \frac {4 \, {\left (a^{3} b - a b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {6 \, a^{3} b \tan \left (d x + c\right ) + a^{4} - 3 \, {\left (a^{4} - 6 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{3 \, d \tan \left (d x + c\right )^{3}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.18 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (\frac {a^4}{3}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^4-6\,a^2\,b^2\right )+2\,a^3\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {4\,a\,b\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a^2-b^2\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,{\left (a-b\,1{}\mathrm {i}\right )}^4\,1{}\mathrm {i}}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^4\,1{}\mathrm {i}}{2\,d} \] Input:

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

Output:

(log(tan(c + d*x) + 1i)*(a - b*1i)^4*1i)/(2*d) - (cot(c + d*x)^3*(a^4/3 - 
tan(c + d*x)^2*(a^4 - 6*a^2*b^2) + 2*a^3*b*tan(c + d*x)))/d - (log(tan(c + 
 d*x) - 1i)*(a*1i - b)^4*1i)/(2*d) - (4*a*b*log(tan(c + d*x))*(a^2 - b^2)) 
/d
 

Reduce [B] (verification not implemented)

Time = 20.13 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.09 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{4}-18 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}-\cos \left (d x +c \right ) a^{4}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} a^{3} b -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} a \,b^{3}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a^{3} b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a \,b^{3}+3 \sin \left (d x +c \right )^{3} a^{4} d x +3 \sin \left (d x +c \right )^{3} a^{3} b -18 \sin \left (d x +c \right )^{3} a^{2} b^{2} d x +3 \sin \left (d x +c \right )^{3} b^{4} d x -6 \sin \left (d x +c \right ) a^{3} b}{3 \sin \left (d x +c \right )^{3} d} \] Input:

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

Output:

(4*cos(c + d*x)*sin(c + d*x)**2*a**4 - 18*cos(c + d*x)*sin(c + d*x)**2*a** 
2*b**2 - cos(c + d*x)*a**4 + 12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)* 
*3*a**3*b - 12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**3*a*b**3 - 12*lo 
g(tan((c + d*x)/2))*sin(c + d*x)**3*a**3*b + 12*log(tan((c + d*x)/2))*sin( 
c + d*x)**3*a*b**3 + 3*sin(c + d*x)**3*a**4*d*x + 3*sin(c + d*x)**3*a**3*b 
 - 18*sin(c + d*x)**3*a**2*b**2*d*x + 3*sin(c + d*x)**3*b**4*d*x - 6*sin(c 
 + d*x)*a**3*b)/(3*sin(c + d*x)**3*d)