\(\int \cot ^6(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\) [265]

Optimal result

Integrand size = 31, antiderivative size = 273 \[ \int \cot ^6(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-\left (\left (a^4 A-6 a^2 A b^2+A b^4-4 a^3 b B+4 a b^3 B\right ) x\right )-\frac {\left (a^4 A-6 a^2 A b^2+A b^4-4 a^3 b B+4 a b^3 B\right ) \cot (c+d x)}{d}+\frac {a \left (40 a^2 A b-28 A b^3+10 a^3 B-55 a b^2 B\right ) \cot ^2(c+d x)}{20 d}+\frac {a^2 \left (10 a^2 A-18 A b^2-25 a b B\right ) \cot ^3(c+d x)}{30 d}+\frac {\left (4 a^3 A b-4 a A b^3+a^4 B-6 a^2 b^2 B+b^4 B\right ) \log (\sin (c+d x))}{d}-\frac {a (8 A b+5 a B) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{20 d}-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d} \] Output:

-(A*a^4-6*A*a^2*b^2+A*b^4-4*B*a^3*b+4*B*a*b^3)*x-(A*a^4-6*A*a^2*b^2+A*b^4- 
4*B*a^3*b+4*B*a*b^3)*cot(d*x+c)/d+1/20*a*(40*A*a^2*b-28*A*b^3+10*B*a^3-55* 
B*a*b^2)*cot(d*x+c)^2/d+1/30*a^2*(10*A*a^2-18*A*b^2-25*B*a*b)*cot(d*x+c)^3 
/d+(4*A*a^3*b-4*A*a*b^3+B*a^4-6*B*a^2*b^2+B*b^4)*ln(sin(d*x+c))/d-1/20*a*( 
8*A*b+5*B*a)*cot(d*x+c)^4*(a+b*tan(d*x+c))^2/d-1/5*a*A*cot(d*x+c)^5*(a+b*t 
an(d*x+c))^3/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.21 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.94 \[ \int \cot ^6(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {-60 \left (a^4 A-6 a^2 A b^2+A b^4-4 a^3 b B+4 a b^3 B\right ) \cot (c+d x)+30 a \left (4 a^2 A b-4 A b^3+a^3 B-6 a b^2 B\right ) \cot ^2(c+d x)+20 a^2 \left (a^2 A-6 A b^2-4 a b B\right ) \cot ^3(c+d x)-15 a^3 (4 A b+a B) \cot ^4(c+d x)-12 a^4 A \cot ^5(c+d x)+30 i (a+i b)^4 (A+i B) \log (i-\tan (c+d x))+60 \left (4 a^3 A b-4 a A b^3+a^4 B-6 a^2 b^2 B+b^4 B\right ) \log (\tan (c+d x))-30 (a-i b)^4 (i A+B) \log (i+\tan (c+d x))}{60 d} \] Input:

Integrate[Cot[c + d*x]^6*(a + b*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]
 

Output:

(-60*(a^4*A - 6*a^2*A*b^2 + A*b^4 - 4*a^3*b*B + 4*a*b^3*B)*Cot[c + d*x] + 
30*a*(4*a^2*A*b - 4*A*b^3 + a^3*B - 6*a*b^2*B)*Cot[c + d*x]^2 + 20*a^2*(a^ 
2*A - 6*A*b^2 - 4*a*b*B)*Cot[c + d*x]^3 - 15*a^3*(4*A*b + a*B)*Cot[c + d*x 
]^4 - 12*a^4*A*Cot[c + d*x]^5 + (30*I)*(a + I*b)^4*(A + I*B)*Log[I - Tan[c 
 + d*x]] + 60*(4*a^3*A*b - 4*a*A*b^3 + a^4*B - 6*a^2*b^2*B + b^4*B)*Log[Ta 
n[c + d*x]] - 30*(a - I*b)^4*(I*A + B)*Log[I + Tan[c + d*x]])/(60*d)
 

Rubi [A] (verified)

Time = 1.90 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.548, Rules used = {3042, 4088, 3042, 4128, 27, 3042, 4118, 3042, 4111, 27, 3042, 4012, 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.

\(\displaystyle \int \cot ^6(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x))^4 (A+B \tan (c+d x))}{\tan (c+d x)^6}dx\)

\(\Big \downarrow \) 4088

\(\displaystyle \frac {1}{5} \int \cot ^5(c+d x) (a+b \tan (c+d x))^2 \left (-b (2 a A-5 b B) \tan ^2(c+d x)-5 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (8 A b+5 a B)\right )dx-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \int \frac {(a+b \tan (c+d x))^2 \left (-b (2 a A-5 b B) \tan (c+d x)^2-5 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (8 A b+5 a B)\right )}{\tan (c+d x)^5}dx-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 4128

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int -2 \cot ^4(c+d x) (a+b \tan (c+d x)) \left (b \left (5 B a^2+12 A b a-10 b^2 B\right ) \tan ^2(c+d x)+10 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (10 A a^2-25 b B a-18 A b^2\right )\right )dx-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (-\frac {1}{2} \int \cot ^4(c+d x) (a+b \tan (c+d x)) \left (b \left (5 B a^2+12 A b a-10 b^2 B\right ) \tan ^2(c+d x)+10 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (10 A a^2-25 b B a-18 A b^2\right )\right )dx-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (-\frac {1}{2} \int \frac {(a+b \tan (c+d x)) \left (b \left (5 B a^2+12 A b a-10 b^2 B\right ) \tan (c+d x)^2+10 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (10 A a^2-25 b B a-18 A b^2\right )\right )}{\tan (c+d x)^4}dx-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 4118

\(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {a^2 \left (10 a^2 A-25 a b B-18 A b^2\right ) \cot ^3(c+d x)}{3 d}-\int \cot ^3(c+d x) \left (b^2 \left (5 B a^2+12 A b a-10 b^2 B\right ) \tan ^2(c+d x)-10 \left (A a^4-4 b B a^3-6 A b^2 a^2+4 b^3 B a+A b^4\right ) \tan (c+d x)+a \left (10 B a^3+40 A b a^2-55 b^2 B a-28 A b^3\right )\right )dx\right )-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {a^2 \left (10 a^2 A-25 a b B-18 A b^2\right ) \cot ^3(c+d x)}{3 d}-\int \frac {b^2 \left (5 B a^2+12 A b a-10 b^2 B\right ) \tan (c+d x)^2-10 \left (A a^4-4 b B a^3-6 A b^2 a^2+4 b^3 B a+A b^4\right ) \tan (c+d x)+a \left (10 B a^3+40 A b a^2-55 b^2 B a-28 A b^3\right )}{\tan (c+d x)^3}dx\right )-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 4111

\(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (-\int -10 \cot ^2(c+d x) \left (A a^4-4 b B a^3-6 A b^2 a^2+4 b^3 B a+A b^4+\left (B a^4+4 A b a^3-6 b^2 B a^2-4 A b^3 a+b^4 B\right ) \tan (c+d x)\right )dx+\frac {a^2 \left (10 a^2 A-25 a b B-18 A b^2\right ) \cot ^3(c+d x)}{3 d}+\frac {a \left (10 a^3 B+40 a^2 A b-55 a b^2 B-28 A b^3\right ) \cot ^2(c+d x)}{2 d}\right )-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (10 \int \cot ^2(c+d x) \left (A a^4-4 b B a^3-6 A b^2 a^2+4 b^3 B a+A b^4+\left (B a^4+4 A b a^3-6 b^2 B a^2-4 A b^3 a+b^4 B\right ) \tan (c+d x)\right )dx+\frac {a^2 \left (10 a^2 A-25 a b B-18 A b^2\right ) \cot ^3(c+d x)}{3 d}+\frac {a \left (10 a^3 B+40 a^2 A b-55 a b^2 B-28 A b^3\right ) \cot ^2(c+d x)}{2 d}\right )-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (10 \int \frac {A a^4-4 b B a^3-6 A b^2 a^2+4 b^3 B a+A b^4+\left (B a^4+4 A b a^3-6 b^2 B a^2-4 A b^3 a+b^4 B\right ) \tan (c+d x)}{\tan (c+d x)^2}dx+\frac {a^2 \left (10 a^2 A-25 a b B-18 A b^2\right ) \cot ^3(c+d x)}{3 d}+\frac {a \left (10 a^3 B+40 a^2 A b-55 a b^2 B-28 A b^3\right ) \cot ^2(c+d x)}{2 d}\right )-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (10 \left (\int \cot (c+d x) \left (B a^4+4 A b a^3-6 b^2 B a^2-4 A b^3 a+b^4 B-\left (A a^4-4 b B a^3-6 A b^2 a^2+4 b^3 B a+A b^4\right ) \tan (c+d x)\right )dx-\frac {\left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right ) \cot (c+d x)}{d}\right )+\frac {a^2 \left (10 a^2 A-25 a b B-18 A b^2\right ) \cot ^3(c+d x)}{3 d}+\frac {a \left (10 a^3 B+40 a^2 A b-55 a b^2 B-28 A b^3\right ) \cot ^2(c+d x)}{2 d}\right )-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (10 \left (\int \frac {B a^4+4 A b a^3-6 b^2 B a^2-4 A b^3 a+b^4 B-\left (A a^4-4 b B a^3-6 A b^2 a^2+4 b^3 B a+A b^4\right ) \tan (c+d x)}{\tan (c+d x)}dx-\frac {\left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right ) \cot (c+d x)}{d}\right )+\frac {a^2 \left (10 a^2 A-25 a b B-18 A b^2\right ) \cot ^3(c+d x)}{3 d}+\frac {a \left (10 a^3 B+40 a^2 A b-55 a b^2 B-28 A b^3\right ) \cot ^2(c+d x)}{2 d}\right )-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 4014

\(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (10 \left (\left (a^4 B+4 a^3 A b-6 a^2 b^2 B-4 a A b^3+b^4 B\right ) \int \cot (c+d x)dx-\frac {\left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right ) \cot (c+d x)}{d}-\left (x \left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right )\right )\right )+\frac {a^2 \left (10 a^2 A-25 a b B-18 A b^2\right ) \cot ^3(c+d x)}{3 d}+\frac {a \left (10 a^3 B+40 a^2 A b-55 a b^2 B-28 A b^3\right ) \cot ^2(c+d x)}{2 d}\right )-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (10 \left (\left (a^4 B+4 a^3 A b-6 a^2 b^2 B-4 a A b^3+b^4 B\right ) \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-\frac {\left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right ) \cot (c+d x)}{d}-\left (x \left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right )\right )\right )+\frac {a^2 \left (10 a^2 A-25 a b B-18 A b^2\right ) \cot ^3(c+d x)}{3 d}+\frac {a \left (10 a^3 B+40 a^2 A b-55 a b^2 B-28 A b^3\right ) \cot ^2(c+d x)}{2 d}\right )-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (10 \left (-\left (a^4 B+4 a^3 A b-6 a^2 b^2 B-4 a A b^3+b^4 B\right ) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\frac {\left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right ) \cot (c+d x)}{d}-\left (x \left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right )\right )\right )+\frac {a^2 \left (10 a^2 A-25 a b B-18 A b^2\right ) \cot ^3(c+d x)}{3 d}+\frac {a \left (10 a^3 B+40 a^2 A b-55 a b^2 B-28 A b^3\right ) \cot ^2(c+d x)}{2 d}\right )-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {a^2 \left (10 a^2 A-25 a b B-18 A b^2\right ) \cot ^3(c+d x)}{3 d}+\frac {a \left (10 a^3 B+40 a^2 A b-55 a b^2 B-28 A b^3\right ) \cot ^2(c+d x)}{2 d}+10 \left (-\frac {\left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right ) \cot (c+d x)}{d}+\frac {\left (a^4 B+4 a^3 A b-6 a^2 b^2 B-4 a A b^3+b^4 B\right ) \log (-\sin (c+d x))}{d}-\left (x \left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right )\right )\right )\right )-\frac {a (5 a B+8 A b) \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^3}{5 d}\)

Input:

Int[Cot[c + d*x]^6*(a + b*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]
 

Output:

-1/5*(a*A*Cot[c + d*x]^5*(a + b*Tan[c + d*x])^3)/d + (((a*(40*a^2*A*b - 28 
*A*b^3 + 10*a^3*B - 55*a*b^2*B)*Cot[c + d*x]^2)/(2*d) + (a^2*(10*a^2*A - 1 
8*A*b^2 - 25*a*b*B)*Cot[c + d*x]^3)/(3*d) + 10*(-((a^4*A - 6*a^2*A*b^2 + A 
*b^4 - 4*a^3*b*B + 4*a*b^3*B)*x) - ((a^4*A - 6*a^2*A*b^2 + A*b^4 - 4*a^3*b 
*B + 4*a*b^3*B)*Cot[c + d*x])/d + ((4*a^3*A*b - 4*a*A*b^3 + a^4*B - 6*a^2* 
b^2*B + b^4*B)*Log[-Sin[c + d*x]])/d))/2 - (a*(8*A*b + 5*a*B)*Cot[c + d*x] 
^4*(a + b*Tan[c + d*x])^2)/(4*d))/5
 

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 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 4088

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

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]
 

rule 4128

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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(\frac {\left (-120 A \,a^{3} b +120 A a \,b^{3}-30 B \,a^{4}+180 B \,a^{2} b^{2}-30 B \,b^{4}\right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+\left (240 A \,a^{3} b -240 A a \,b^{3}+60 B \,a^{4}-360 B \,a^{2} b^{2}+60 B \,b^{4}\right ) \ln \left (\tan \left (d x +c \right )\right )-12 A \cot \left (d x +c \right )^{5} a^{4}+\left (-60 A \,a^{3} b -15 B \,a^{4}\right ) \cot \left (d x +c \right )^{4}+20 a^{2} \cot \left (d x +c \right )^{3} \left (A \,a^{2}-6 A \,b^{2}-4 B a b \right )+\left (120 A \,a^{3} b -120 A a \,b^{3}+30 B \,a^{4}-180 B \,a^{2} b^{2}\right ) \cot \left (d x +c \right )^{2}+\left (-60 A \,a^{4}+360 A \,a^{2} b^{2}-60 A \,b^{4}+240 B \,a^{3} b -240 B a \,b^{3}\right ) \cot \left (d x +c \right )-60 d x \left (A \,a^{4}-6 A \,a^{2} b^{2}+A \,b^{4}-4 B \,a^{3} b +4 B a \,b^{3}\right )}{60 d}\)	\(284\)
derivativedivides	\(\frac {\frac {\left (-4 A \,a^{3} b +4 A a \,b^{3}-B \,a^{4}+6 B \,a^{2} b^{2}-B \,b^{4}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-A \,a^{4}+6 A \,a^{2} b^{2}-A \,b^{4}+4 B \,a^{3} b -4 B a \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )+\left (4 A \,a^{3} b -4 A a \,b^{3}+B \,a^{4}-6 B \,a^{2} b^{2}+B \,b^{4}\right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {A \,a^{4}-6 A \,a^{2} b^{2}+A \,b^{4}-4 B \,a^{3} b +4 B a \,b^{3}}{\tan \left (d x +c \right )}-\frac {A \,a^{4}}{5 \tan \left (d x +c \right )^{5}}-\frac {a^{3} \left (4 A b +B a \right )}{4 \tan \left (d x +c \right )^{4}}+\frac {a \left (4 A \,a^{2} b -4 A \,b^{3}+B \,a^{3}-6 B a \,b^{2}\right )}{2 \tan \left (d x +c \right )^{2}}+\frac {a^{2} \left (A \,a^{2}-6 A \,b^{2}-4 B a b \right )}{3 \tan \left (d x +c \right )^{3}}}{d}\)	\(287\)
default	\(\frac {\frac {\left (-4 A \,a^{3} b +4 A a \,b^{3}-B \,a^{4}+6 B \,a^{2} b^{2}-B \,b^{4}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-A \,a^{4}+6 A \,a^{2} b^{2}-A \,b^{4}+4 B \,a^{3} b -4 B a \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )+\left (4 A \,a^{3} b -4 A a \,b^{3}+B \,a^{4}-6 B \,a^{2} b^{2}+B \,b^{4}\right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {A \,a^{4}-6 A \,a^{2} b^{2}+A \,b^{4}-4 B \,a^{3} b +4 B a \,b^{3}}{\tan \left (d x +c \right )}-\frac {A \,a^{4}}{5 \tan \left (d x +c \right )^{5}}-\frac {a^{3} \left (4 A b +B a \right )}{4 \tan \left (d x +c \right )^{4}}+\frac {a \left (4 A \,a^{2} b -4 A \,b^{3}+B \,a^{3}-6 B a \,b^{2}\right )}{2 \tan \left (d x +c \right )^{2}}+\frac {a^{2} \left (A \,a^{2}-6 A \,b^{2}-4 B a b \right )}{3 \tan \left (d x +c \right )^{3}}}{d}\)	\(287\)
norman	\(\frac {\left (-A \,a^{4}+6 A \,a^{2} b^{2}-A \,b^{4}+4 B \,a^{3} b -4 B a \,b^{3}\right ) x \tan \left (d x +c \right )^{5}-\frac {A \,a^{4}}{5 d}-\frac {\left (A \,a^{4}-6 A \,a^{2} b^{2}+A \,b^{4}-4 B \,a^{3} b +4 B a \,b^{3}\right ) \tan \left (d x +c \right )^{4}}{d}+\frac {a \left (4 A \,a^{2} b -4 A \,b^{3}+B \,a^{3}-6 B a \,b^{2}\right ) \tan \left (d x +c \right )^{3}}{2 d}+\frac {a^{2} \left (A \,a^{2}-6 A \,b^{2}-4 B a b \right ) \tan \left (d x +c \right )^{2}}{3 d}-\frac {a^{3} \left (4 A b +B a \right ) \tan \left (d x +c \right )}{4 d}}{\tan \left (d x +c \right )^{5}}+\frac {\left (4 A \,a^{3} b -4 A a \,b^{3}+B \,a^{4}-6 B \,a^{2} b^{2}+B \,b^{4}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (4 A \,a^{3} b -4 A a \,b^{3}+B \,a^{4}-6 B \,a^{2} b^{2}+B \,b^{4}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\)	\(304\)
risch	\(-\frac {8 i A \,a^{3} b c}{d}-\frac {2 i B \,b^{4} c}{d}-i B \,a^{4} x +\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B \,b^{4}}{d}+\frac {a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}+\frac {12 i B \,a^{2} b^{2} c}{d}+6 i B \,a^{2} b^{2} x +4 i A a \,b^{3} x -\frac {2 i B \,a^{4} c}{d}+\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A b}{d}-A \,a^{4} x +6 A \,a^{2} b^{2} x +4 B \,a^{3} b x -4 B a \,b^{3} x -\frac {4 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A \,b^{3}}{d}-\frac {6 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B \,b^{2}}{d}-4 i A \,a^{3} b x -A \,b^{4} x -\frac {2 i \left (60 B a \,b^{3}-120 A \,a^{2} b^{2}-80 B \,a^{3} b +15 A \,b^{4}+23 A \,a^{4}+140 A \,a^{4} {\mathrm e}^{4 i \left (d x +c \right )}+90 A \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-70 A \,a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+60 i B \,a^{4} {\mathrm e}^{6 i \left (d x +c \right )}+280 B \,a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}-240 B a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+60 B a \,b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+540 A \,a^{2} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+360 B \,a^{3} b \,{\mathrm e}^{6 i \left (d x +c \right )}-240 B a \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+420 A \,a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-660 A \,a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-440 B \,a^{3} b \,{\mathrm e}^{4 i \left (d x +c \right )}-180 A \,a^{2} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-120 B \,a^{3} b \,{\mathrm e}^{8 i \left (d x +c \right )}-30 i B \,a^{4} {\mathrm e}^{8 i \left (d x +c \right )}-60 i B \,a^{4} {\mathrm e}^{4 i \left (d x +c \right )}+30 i B \,a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+360 B a \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+45 A \,a^{4} {\mathrm e}^{8 i \left (d x +c \right )}+15 A \,b^{4} {\mathrm e}^{8 i \left (d x +c \right )}-90 A \,a^{4} {\mathrm e}^{6 i \left (d x +c \right )}-60 A \,b^{4} {\mathrm e}^{6 i \left (d x +c \right )}-60 A \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-270 i B \,a^{2} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-60 i A a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-90 i B \,a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-120 i A \,a^{3} b \,{\mathrm e}^{8 i \left (d x +c \right )}+60 i A a \,b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+90 i B \,a^{2} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-240 i A \,a^{3} b \,{\mathrm e}^{4 i \left (d x +c \right )}+180 i A a \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+270 i B \,a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+120 i A \,a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}+240 i A \,a^{3} b \,{\mathrm e}^{6 i \left (d x +c \right )}-180 i A a \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-i B \,b^{4} x +\frac {8 i A a \,b^{3} c}{d}\)	\(906\)

Input:

int(cot(d*x+c)^6*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

1/60*((-120*A*a^3*b+120*A*a*b^3-30*B*a^4+180*B*a^2*b^2-30*B*b^4)*ln(sec(d* 
x+c)^2)+(240*A*a^3*b-240*A*a*b^3+60*B*a^4-360*B*a^2*b^2+60*B*b^4)*ln(tan(d 
*x+c))-12*A*cot(d*x+c)^5*a^4+(-60*A*a^3*b-15*B*a^4)*cot(d*x+c)^4+20*a^2*co 
t(d*x+c)^3*(A*a^2-6*A*b^2-4*B*a*b)+(120*A*a^3*b-120*A*a*b^3+30*B*a^4-180*B 
*a^2*b^2)*cot(d*x+c)^2+(-60*A*a^4+360*A*a^2*b^2-60*A*b^4+240*B*a^3*b-240*B 
*a*b^3)*cot(d*x+c)-60*d*x*(A*a^4-6*A*a^2*b^2+A*b^4-4*B*a^3*b+4*B*a*b^3))/d
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.10 \[ \int \cot ^6(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {30 \, {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{5} + 15 \, {\left (3 \, B a^{4} + 12 \, A a^{3} b - 12 \, B a^{2} b^{2} - 8 \, A a b^{3} - 4 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} d x\right )} \tan \left (d x + c\right )^{5} - 12 \, A a^{4} - 60 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \tan \left (d x + c\right )^{4} + 30 \, {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} \tan \left (d x + c\right )^{3} + 20 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2} - 15 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \tan \left (d x + c\right )}{60 \, d \tan \left (d x + c\right )^{5}} \] Input:

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="f 
ricas")
 

Output:

1/60*(30*(B*a^4 + 4*A*a^3*b - 6*B*a^2*b^2 - 4*A*a*b^3 + B*b^4)*log(tan(d*x 
 + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^5 + 15*(3*B*a^4 + 12*A*a^3*b - 
12*B*a^2*b^2 - 8*A*a*b^3 - 4*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2 + 4*B*a*b^3 
+ A*b^4)*d*x)*tan(d*x + c)^5 - 12*A*a^4 - 60*(A*a^4 - 4*B*a^3*b - 6*A*a^2* 
b^2 + 4*B*a*b^3 + A*b^4)*tan(d*x + c)^4 + 30*(B*a^4 + 4*A*a^3*b - 6*B*a^2* 
b^2 - 4*A*a*b^3)*tan(d*x + c)^3 + 20*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2)*tan 
(d*x + c)^2 - 15*(B*a^4 + 4*A*a^3*b)*tan(d*x + c))/(d*tan(d*x + c)^5)
 

Sympy [A] (verification not implemented)

Time = 5.62 (sec) , antiderivative size = 546, normalized size of antiderivative = 2.00 \[ \int \cot ^6(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } A a^{4} 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 )^{4} \cot ^{6}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } A a^{4} x & \text {for}\: c = - d x \\- A a^{4} x - \frac {A a^{4}}{d \tan {\left (c + d x \right )}} + \frac {A a^{4}}{3 d \tan ^{3}{\left (c + d x \right )}} - \frac {A a^{4}}{5 d \tan ^{5}{\left (c + d x \right )}} - \frac {2 A a^{3} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {4 A a^{3} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {2 A a^{3} b}{d \tan ^{2}{\left (c + d x \right )}} - \frac {A a^{3} b}{d \tan ^{4}{\left (c + d x \right )}} + 6 A a^{2} b^{2} x + \frac {6 A a^{2} b^{2}}{d \tan {\left (c + d x \right )}} - \frac {2 A a^{2} b^{2}}{d \tan ^{3}{\left (c + d x \right )}} + \frac {2 A a b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - \frac {4 A a b^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {2 A a b^{3}}{d \tan ^{2}{\left (c + d x \right )}} - A b^{4} x - \frac {A b^{4}}{d \tan {\left (c + d x \right )}} - \frac {B a^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B a^{4} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {B a^{4}}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {B a^{4}}{4 d \tan ^{4}{\left (c + d x \right )}} + 4 B a^{3} b x + \frac {4 B a^{3} b}{d \tan {\left (c + d x \right )}} - \frac {4 B a^{3} b}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {3 B a^{2} b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - \frac {6 B a^{2} b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {3 B a^{2} b^{2}}{d \tan ^{2}{\left (c + d x \right )}} - 4 B a b^{3} x - \frac {4 B a b^{3}}{d \tan {\left (c + d x \right )}} - \frac {B b^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b^{4} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} & \text {otherwise} \end {cases} \] Input:

integrate(cot(d*x+c)**6*(a+b*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)
 

Output:

Piecewise((zoo*A*a**4*x, Eq(c, 0) & Eq(d, 0)), (x*(A + B*tan(c))*(a + b*ta 
n(c))**4*cot(c)**6, Eq(d, 0)), (zoo*A*a**4*x, Eq(c, -d*x)), (-A*a**4*x - A 
*a**4/(d*tan(c + d*x)) + A*a**4/(3*d*tan(c + d*x)**3) - A*a**4/(5*d*tan(c 
+ d*x)**5) - 2*A*a**3*b*log(tan(c + d*x)**2 + 1)/d + 4*A*a**3*b*log(tan(c 
+ d*x))/d + 2*A*a**3*b/(d*tan(c + d*x)**2) - A*a**3*b/(d*tan(c + d*x)**4) 
+ 6*A*a**2*b**2*x + 6*A*a**2*b**2/(d*tan(c + d*x)) - 2*A*a**2*b**2/(d*tan( 
c + d*x)**3) + 2*A*a*b**3*log(tan(c + d*x)**2 + 1)/d - 4*A*a*b**3*log(tan( 
c + d*x))/d - 2*A*a*b**3/(d*tan(c + d*x)**2) - A*b**4*x - A*b**4/(d*tan(c 
+ d*x)) - B*a**4*log(tan(c + d*x)**2 + 1)/(2*d) + B*a**4*log(tan(c + d*x)) 
/d + B*a**4/(2*d*tan(c + d*x)**2) - B*a**4/(4*d*tan(c + d*x)**4) + 4*B*a** 
3*b*x + 4*B*a**3*b/(d*tan(c + d*x)) - 4*B*a**3*b/(3*d*tan(c + d*x)**3) + 3 
*B*a**2*b**2*log(tan(c + d*x)**2 + 1)/d - 6*B*a**2*b**2*log(tan(c + d*x))/ 
d - 3*B*a**2*b**2/(d*tan(c + d*x)**2) - 4*B*a*b**3*x - 4*B*a*b**3/(d*tan(c 
 + d*x)) - B*b**4*log(tan(c + d*x)**2 + 1)/(2*d) + B*b**4*log(tan(c + d*x) 
)/d, True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.06 \[ \int \cot ^6(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-\frac {60 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} {\left (d x + c\right )} + 30 \, {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 60 \, {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {12 \, A a^{4} + 60 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \tan \left (d x + c\right )^{4} - 30 \, {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} \tan \left (d x + c\right )^{3} - 20 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2} + 15 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{5}}}{60 \, d} \] Input:

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="m 
axima")
 

Output:

-1/60*(60*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2 + 4*B*a*b^3 + A*b^4)*(d*x + c) 
+ 30*(B*a^4 + 4*A*a^3*b - 6*B*a^2*b^2 - 4*A*a*b^3 + B*b^4)*log(tan(d*x + c 
)^2 + 1) - 60*(B*a^4 + 4*A*a^3*b - 6*B*a^2*b^2 - 4*A*a*b^3 + B*b^4)*log(ta 
n(d*x + c)) + (12*A*a^4 + 60*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2 + 4*B*a*b^3 
+ A*b^4)*tan(d*x + c)^4 - 30*(B*a^4 + 4*A*a^3*b - 6*B*a^2*b^2 - 4*A*a*b^3) 
*tan(d*x + c)^3 - 20*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2)*tan(d*x + c)^2 + 15 
*(B*a^4 + 4*A*a^3*b)*tan(d*x + c))/tan(d*x + c)^5)/d
 

Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.09 \[ \int \cot ^6(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-\frac {{\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} {\left (d x + c\right )}}{d} - \frac {{\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} + \frac {{\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {12 \, A a^{4} + 60 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \tan \left (d x + c\right )^{4} - 30 \, {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} \tan \left (d x + c\right )^{3} - 20 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2} + 15 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \tan \left (d x + c\right )}{60 \, d \tan \left (d x + c\right )^{5}} \] Input:

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="g 
iac")
 

Output:

-(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2 + 4*B*a*b^3 + A*b^4)*(d*x + c)/d - 1/2*( 
B*a^4 + 4*A*a^3*b - 6*B*a^2*b^2 - 4*A*a*b^3 + B*b^4)*log(tan(d*x + c)^2 + 
1)/d + (B*a^4 + 4*A*a^3*b - 6*B*a^2*b^2 - 4*A*a*b^3 + B*b^4)*log(abs(tan(d 
*x + c)))/d - 1/60*(12*A*a^4 + 60*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2 + 4*B*a 
*b^3 + A*b^4)*tan(d*x + c)^4 - 30*(B*a^4 + 4*A*a^3*b - 6*B*a^2*b^2 - 4*A*a 
*b^3)*tan(d*x + c)^3 - 20*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2)*tan(d*x + c)^2 
 + 15*(B*a^4 + 4*A*a^3*b)*tan(d*x + c))/(d*tan(d*x + c)^5)
 

Mupad [B] (verification not implemented)

Time = 3.77 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.96 \[ \int \cot ^6(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^4+4\,A\,a^3\,b-6\,B\,a^2\,b^2-4\,A\,a\,b^3+B\,b^4\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^5\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^4}{4}+A\,b\,a^3\right )+\frac {A\,a^4}{5}-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {B\,a^4}{2}+2\,A\,a^3\,b-3\,B\,a^2\,b^2-2\,A\,a\,b^3\right )+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {A\,a^4}{3}+\frac {4\,B\,a^3\,b}{3}+2\,A\,a^2\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (A\,a^4-4\,B\,a^3\,b-6\,A\,a^2\,b^2+4\,B\,a\,b^3+A\,b^4\right )\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^4}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^4}{2\,d} \] Input:

int(cot(c + d*x)^6*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^4,x)
 

Output:

(log(tan(c + d*x))*(B*a^4 + B*b^4 - 6*B*a^2*b^2 - 4*A*a*b^3 + 4*A*a^3*b))/ 
d - (cot(c + d*x)^5*(tan(c + d*x)*((B*a^4)/4 + A*a^3*b) + (A*a^4)/5 - tan( 
c + d*x)^3*((B*a^4)/2 - 3*B*a^2*b^2 - 2*A*a*b^3 + 2*A*a^3*b) + tan(c + d*x 
)^2*(2*A*a^2*b^2 - (A*a^4)/3 + (4*B*a^3*b)/3) + tan(c + d*x)^4*(A*a^4 + A* 
b^4 - 6*A*a^2*b^2 + 4*B*a*b^3 - 4*B*a^3*b)))/d + (log(tan(c + d*x) - 1i)*( 
A*1i - B)*(a*1i - b)^4)/(2*d) - (log(tan(c + d*x) + 1i)*(A*1i + B)*(a*1i + 
 b)^4)/(2*d)
 

Reduce [F]

\[ \int \cot ^6(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\int \cot \left (d x +c \right )^{6} \left (a +\tan \left (d x +c \right ) b \right )^{4} \left (A +B \tan \left (d x +c \right )\right )d x \] Input:

int(cot(d*x+c)^6*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)
 

Output:

int(cot(d*x+c)^6*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)