\(\int \cot ^5(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [247]

Optimal result

Integrand size = 31, antiderivative size = 151 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\left (2 a A b+a^2 B-b^2 B\right ) x-\frac {\left (b^2 B-a (2 A b+a B)\right ) \cot (c+d x)}{d}+\frac {\left (a^2 A-A b^2-2 a b B\right ) \cot ^2(c+d x)}{2 d}-\frac {a (2 A b+a B) \cot ^3(c+d x)}{3 d}-\frac {a^2 A \cot ^4(c+d x)}{4 d}+\frac {\left (a^2 A-A b^2-2 a b B\right ) \log (\sin (c+d x))}{d} \] Output:

(2*A*a*b+B*a^2-B*b^2)*x-(B*b^2-a*(2*A*b+B*a))*cot(d*x+c)/d+1/2*(A*a^2-A*b^ 
2-2*B*a*b)*cot(d*x+c)^2/d-1/3*a*(2*A*b+B*a)*cot(d*x+c)^3/d-1/4*a^2*A*cot(d 
*x+c)^4/d+(A*a^2-A*b^2-2*B*a*b)*ln(sin(d*x+c))/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.04 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.19 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {12 \left (2 a A b+a^2 B-b^2 B\right ) \cot (c+d x)+6 \left (a^2 A-A b^2-2 a b B\right ) \cot ^2(c+d x)-4 a (2 A b+a B) \cot ^3(c+d x)-3 a^2 A \cot ^4(c+d x)-6 \left ((a+i b)^2 (A+i B) \log (i-\tan (c+d x))+\left (-2 a^2 A+2 A b^2+4 a b B\right ) \log (\tan (c+d x))+(a-i b)^2 (A-i B) \log (i+\tan (c+d x))\right )}{12 d} \] Input:

Integrate[Cot[c + d*x]^5*(a + b*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]
 

Output:

(12*(2*a*A*b + a^2*B - b^2*B)*Cot[c + d*x] + 6*(a^2*A - A*b^2 - 2*a*b*B)*C 
ot[c + d*x]^2 - 4*a*(2*A*b + a*B)*Cot[c + d*x]^3 - 3*a^2*A*Cot[c + d*x]^4 
- 6*((a + I*b)^2*(A + I*B)*Log[I - Tan[c + d*x]] + (-2*a^2*A + 2*A*b^2 + 4 
*a*b*B)*Log[Tan[c + d*x]] + (a - I*b)^2*(A - I*B)*Log[I + Tan[c + d*x]]))/ 
(12*d)
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.01, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4087, 3042, 4111, 25, 3042, 4012, 3042, 4012, 25, 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.

\(\displaystyle \int \cot ^5(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan (c+d x)^5}dx\)

\(\Big \downarrow \) 4087

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4111

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4014

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

Input:

Int[Cot[c + d*x]^5*(a + b*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 4012

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

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 4087

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

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 
]
 

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.07

method	result	size
derivativedivides	\(\frac {A \,b^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+B \,b^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 A a b \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+2 B a b \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+A \,a^{2} \left (-\frac {\cot \left (d x +c \right )^{4}}{4}+\frac {\cot \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+B \,a^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\)	\(162\)
default	\(\frac {A \,b^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+B \,b^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 A a b \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+2 B a b \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+A \,a^{2} \left (-\frac {\cot \left (d x +c \right )^{4}}{4}+\frac {\cot \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+B \,a^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\)	\(162\)
parallelrisch	\(\frac {6 \left (-A \,a^{2}+A \,b^{2}+2 B a b \right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+12 \left (A \,a^{2}-A \,b^{2}-2 B a b \right ) \ln \left (\tan \left (d x +c \right )\right )-3 A \cot \left (d x +c \right )^{4} a^{2}+4 \left (-2 A a b -B \,a^{2}\right ) \cot \left (d x +c \right )^{3}+6 \cot \left (d x +c \right )^{2} \left (A \,a^{2}-A \,b^{2}-2 B a b \right )+12 \cot \left (d x +c \right ) \left (2 A a b +B \,a^{2}-B \,b^{2}\right )+24 x d \left (A a b +\frac {1}{2} B \,a^{2}-\frac {1}{2} B \,b^{2}\right )}{12 d}\)	\(170\)
norman	\(\frac {\frac {\left (2 A a b +B \,a^{2}-B \,b^{2}\right ) \tan \left (d x +c \right )^{3}}{d}+\left (2 A a b +B \,a^{2}-B \,b^{2}\right ) x \tan \left (d x +c \right )^{4}-\frac {A \,a^{2}}{4 d}+\frac {\left (A \,a^{2}-A \,b^{2}-2 B a b \right ) \tan \left (d x +c \right )^{2}}{2 d}-\frac {a \left (2 A b +B a \right ) \tan \left (d x +c \right )}{3 d}}{\tan \left (d x +c \right )^{4}}+\frac {\left (A \,a^{2}-A \,b^{2}-2 B a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (A \,a^{2}-A \,b^{2}-2 B a b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\)	\(188\)
risch	\(2 A a b x +B \,a^{2} x -B \,b^{2} x -\frac {2 i A \,a^{2} c}{d}-i A \,a^{2} x +\frac {2 i A \,b^{2} c}{d}+i A \,b^{2} x -\frac {2 i \left (6 i B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-6 i A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 i A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 A a b \,{\mathrm e}^{6 i \left (d x +c \right )}-6 B \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+3 B \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+6 i B a b \,{\mathrm e}^{6 i \left (d x +c \right )}-12 i B a b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 i A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+24 A a b \,{\mathrm e}^{4 i \left (d x +c \right )}+12 B \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-9 B \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 i A \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+3 i A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 i A \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-20 A a b \,{\mathrm e}^{2 i \left (d x +c \right )}-10 B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+9 B \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+8 A a b +4 B \,a^{2}-3 B \,b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+2 i B a b x +\frac {4 i B a b c}{d}+\frac {A \,a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A \,b^{2}}{d}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B a b}{d}\)	\(447\)

Input:

int(cot(d*x+c)^5*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

1/d*(A*b^2*(-1/2*cot(d*x+c)^2-ln(sin(d*x+c)))+B*b^2*(-cot(d*x+c)-d*x-c)+2* 
A*a*b*(-1/3*cot(d*x+c)^3+cot(d*x+c)+d*x+c)+2*B*a*b*(-1/2*cot(d*x+c)^2-ln(s 
in(d*x+c)))+A*a^2*(-1/4*cot(d*x+c)^4+1/2*cot(d*x+c)^2+ln(sin(d*x+c)))+B*a^ 
2*(-1/3*cot(d*x+c)^3+cot(d*x+c)+d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.26 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{4} + 3 \, {\left (3 \, A a^{2} - 4 \, B a b - 2 \, A b^{2} + 4 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} d x\right )} \tan \left (d x + c\right )^{4} + 12 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \tan \left (d x + c\right )^{3} - 3 \, A a^{2} + 6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \tan \left (d x + c\right )^{2} - 4 \, {\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )}{12 \, d \tan \left (d x + c\right )^{4}} \] Input:

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="f 
ricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (136) = 272\).

Time = 2.07 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.07 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } A a^{2} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right )^{2} \cot ^{5}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } A a^{2} x & \text {for}\: c = - d x \\- \frac {A a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {A a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {A a^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {A a^{2}}{4 d \tan ^{4}{\left (c + d x \right )}} + 2 A a b x + \frac {2 A a b}{d \tan {\left (c + d x \right )}} - \frac {2 A a b}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {A b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {A b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {A b^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} + B a^{2} x + \frac {B a^{2}}{d \tan {\left (c + d x \right )}} - \frac {B a^{2}}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {B a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - \frac {2 B a b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B a b}{d \tan ^{2}{\left (c + d x \right )}} - B b^{2} x - \frac {B b^{2}}{d \tan {\left (c + d x \right )}} & \text {otherwise} \end {cases} \] Input:

integrate(cot(d*x+c)**5*(a+b*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)
 

Output:

Piecewise((zoo*A*a**2*x, Eq(c, 0) & Eq(d, 0)), (x*(A + B*tan(c))*(a + b*ta 
n(c))**2*cot(c)**5, Eq(d, 0)), (zoo*A*a**2*x, Eq(c, -d*x)), (-A*a**2*log(t 
an(c + d*x)**2 + 1)/(2*d) + A*a**2*log(tan(c + d*x))/d + A*a**2/(2*d*tan(c 
 + d*x)**2) - A*a**2/(4*d*tan(c + d*x)**4) + 2*A*a*b*x + 2*A*a*b/(d*tan(c 
+ d*x)) - 2*A*a*b/(3*d*tan(c + d*x)**3) + A*b**2*log(tan(c + d*x)**2 + 1)/ 
(2*d) - A*b**2*log(tan(c + d*x))/d - A*b**2/(2*d*tan(c + d*x)**2) + B*a**2 
*x + B*a**2/(d*tan(c + d*x)) - B*a**2/(3*d*tan(c + d*x)**3) + B*a*b*log(ta 
n(c + d*x)**2 + 1)/d - 2*B*a*b*log(tan(c + d*x))/d - B*a*b/(d*tan(c + d*x) 
**2) - B*b**2*x - B*b**2/(d*tan(c + d*x)), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.16 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {12 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} {\left (d x + c\right )} - 6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {12 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \tan \left (d x + c\right )^{3} - 3 \, A a^{2} + 6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \tan \left (d x + c\right )^{2} - 4 \, {\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4}}}{12 \, d} \] Input:

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="m 
axima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.21 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {{\left (B a^{2} + 2 \, A a b - B b^{2}\right )} {\left (d x + c\right )}}{d} - \frac {{\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} + \frac {{\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} + \frac {12 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \tan \left (d x + c\right )^{3} - 3 \, A a^{2} + 6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \tan \left (d x + c\right )^{2} - 4 \, {\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )}{12 \, d \tan \left (d x + c\right )^{4}} \] Input:

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="g 
iac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.35 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.21 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {{\mathrm {cot}\left (c+d\,x\right )}^4\,\left (\frac {A\,a^2}{4}+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {A\,a^2}{2}+B\,a\,b+\frac {A\,b^2}{2}\right )-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (B\,a^2+2\,A\,a\,b-B\,b^2\right )+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^2}{3}+\frac {2\,A\,b\,a}{3}\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-A\,a^2+2\,B\,a\,b+A\,b^2\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^2}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^2}{2\,d} \] Input:

int(cot(c + d*x)^5*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.74 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {128 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2} b -32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{3}-32 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -32 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4} a^{3}+96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4} a \,b^{2}+32 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a^{3}-96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a \,b^{2}-13 \sin \left (d x +c \right )^{4} a^{3}+96 \sin \left (d x +c \right )^{4} a^{2} b d x +24 \sin \left (d x +c \right )^{4} a \,b^{2}-32 \sin \left (d x +c \right )^{4} b^{3} d x +32 \sin \left (d x +c \right )^{2} a^{3}-48 \sin \left (d x +c \right )^{2} a \,b^{2}-8 a^{3}}{32 \sin \left (d x +c \right )^{4} d} \] Input:

int(cot(d*x+c)^5*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)
 

Output:

(128*cos(c + d*x)*sin(c + d*x)**3*a**2*b - 32*cos(c + d*x)*sin(c + d*x)**3 
*b**3 - 32*cos(c + d*x)*sin(c + d*x)*a**2*b - 32*log(tan((c + d*x)/2)**2 + 
 1)*sin(c + d*x)**4*a**3 + 96*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4 
*a*b**2 + 32*log(tan((c + d*x)/2))*sin(c + d*x)**4*a**3 - 96*log(tan((c + 
d*x)/2))*sin(c + d*x)**4*a*b**2 - 13*sin(c + d*x)**4*a**3 + 96*sin(c + d*x 
)**4*a**2*b*d*x + 24*sin(c + d*x)**4*a*b**2 - 32*sin(c + d*x)**4*b**3*d*x 
+ 32*sin(c + d*x)**2*a**3 - 48*sin(c + d*x)**2*a*b**2 - 8*a**3)/(32*sin(c 
+ d*x)**4*d)