\(\int \cot ^5(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [255]

Optimal result

Integrand size = 31, antiderivative size = 191 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x+\frac {\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot (c+d x)}{d}+\frac {a \left (2 a^2 A-5 A b^2-6 a b B\right ) \cot ^2(c+d x)}{4 d}-\frac {a^2 (3 A b+2 a B) \cot ^3(c+d x)}{6 d}+\frac {\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \log (\sin (c+d x))}{d}-\frac {a A \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.51 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.04 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {12 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot (c+d x)+6 a \left (a^2 A-3 A b^2-3 a b B\right ) \cot ^2(c+d x)-4 a^2 (3 A b+a B) \cot ^3(c+d x)-3 a^3 A \cot ^4(c+d x)-6 (a+i b)^3 (A+i B) \log (i-\tan (c+d x))+12 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \log (\tan (c+d x))-6 (a-i b)^3 (A-i B) \log (i+\tan (c+d x))}{12 d} \] Input:

Integrate[Cot[c + d*x]^5*(a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.07, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.548, Rules used = {3042, 4088, 27, 3042, 4118, 25, 3042, 4111, 27, 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.

\(\displaystyle \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan (c+d x)^5}dx\)

\(\Big \downarrow \) 4088

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4118

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4111

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4014

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

Input:

Int[Cot[c + d*x]^5*(a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]
 

Output:

((a*(2*a^2*A - 5*A*b^2 - 6*a*b*B)*Cot[c + d*x]^2)/(2*d) - (a^2*(3*A*b + 2* 
a*B)*Cot[c + d*x]^3)/(3*d) - 2*(-((3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)* 
x) - ((3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Cot[c + d*x])/d - ((a^3*A - 
3*a*A*b^2 - 3*a^2*b*B + b^3*B)*Log[-Sin[c + d*x]])/d))/2 - (a*A*Cot[c + d* 
x]^4*(a + b*Tan[c + d*x])^2)/(4*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 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]
 

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.09

method	result	size
parallelrisch	\(\frac {6 \left (-A \,a^{3}+3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+12 \left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \ln \left (\tan \left (d x +c \right )\right )-3 A \cot \left (d x +c \right )^{4} a^{3}+4 \left (-3 A \,a^{2} b -B \,a^{3}\right ) \cot \left (d x +c \right )^{3}+6 a \cot \left (d x +c \right )^{2} \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right )+12 \cot \left (d x +c \right ) \left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right )+36 x \left (A \,a^{2} b -\frac {1}{3} A \,b^{3}+\frac {1}{3} B \,a^{3}-B a \,b^{2}\right ) d}{12 d}\)	\(209\)
derivativedivides	\(\frac {\frac {\left (-A \,a^{3}+3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}}{\tan \left (d x +c \right )}+\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {A \,a^{3}}{4 \tan \left (d x +c \right )^{4}}-\frac {a^{2} \left (3 A b +B a \right )}{3 \tan \left (d x +c \right )^{3}}+\frac {a \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right )}{2 \tan \left (d x +c \right )^{2}}}{d}\)	\(212\)
default	\(\frac {\frac {\left (-A \,a^{3}+3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}}{\tan \left (d x +c \right )}+\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {A \,a^{3}}{4 \tan \left (d x +c \right )^{4}}-\frac {a^{2} \left (3 A b +B a \right )}{3 \tan \left (d x +c \right )^{3}}+\frac {a \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right )}{2 \tan \left (d x +c \right )^{2}}}{d}\)	\(212\)
norman	\(\frac {\frac {\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \tan \left (d x +c \right )^{3}}{d}+\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) x \tan \left (d x +c \right )^{4}-\frac {A \,a^{3}}{4 d}+\frac {a \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right ) \tan \left (d x +c \right )^{2}}{2 d}-\frac {a^{2} \left (3 A b +B a \right ) \tan \left (d x +c \right )}{3 d}}{\tan \left (d x +c \right )^{4}}+\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\)	\(225\)
risch	\(\frac {6 i B \,a^{2} b c}{d}-i B \,b^{3} x -\frac {2 i A \,a^{3} c}{d}-\frac {2 i \left (-9 B a \,b^{2}+12 A \,a^{2} b -3 A \,b^{3}+9 i A a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+9 i A a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+9 i B \,a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-18 i A a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-18 i B \,a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+9 i B \,a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+4 B \,a^{3}-6 i A \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+6 i A \,a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-6 i A \,a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+9 B a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-18 A \,a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+36 A \,a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-30 A \,a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-27 B a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+27 B a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 B \,a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+3 A \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+9 A \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-9 A \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-10 B \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+12 B \,a^{3} {\mathrm e}^{4 i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+3 A \,a^{2} b x -A \,b^{3} x +B \,a^{3} x -3 B a \,b^{2} x -\frac {2 i B \,b^{3} c}{d}+\frac {6 i A a \,b^{2} c}{d}-i A \,a^{3} x +3 i B \,a^{2} b x +3 i A a \,b^{2} x +\frac {A \,a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A \,b^{2}}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B \,b^{3}}{d}\)	\(577\)

Input:

int(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

1/12*(6*(-A*a^3+3*A*a*b^2+3*B*a^2*b-B*b^3)*ln(sec(d*x+c)^2)+12*(A*a^3-3*A* 
a*b^2-3*B*a^2*b+B*b^3)*ln(tan(d*x+c))-3*A*cot(d*x+c)^4*a^3+4*(-3*A*a^2*b-B 
*a^3)*cot(d*x+c)^3+6*a*cot(d*x+c)^2*(A*a^2-3*A*b^2-3*B*a*b)+12*cot(d*x+c)* 
(3*A*a^2*b-A*b^3+B*a^3-3*B*a*b^2)+36*x*(A*a^2*b-1/3*A*b^3+1/3*B*a^3-B*a*b^ 
2)*d)/d
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.18 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {6 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{4} + 3 \, {\left (3 \, A a^{3} - 6 \, B a^{2} b - 6 \, A a b^{2} + 4 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{4} - 3 \, A a^{3} + 12 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \tan \left (d x + c\right )^{3} + 6 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} - 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{12 \, d \tan \left (d x + c\right )^{4}} \] Input:

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="f 
ricas")
 

Output:

1/12*(6*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*log(tan(d*x + c)^2/(tan(d* 
x + c)^2 + 1))*tan(d*x + c)^4 + 3*(3*A*a^3 - 6*B*a^2*b - 6*A*a*b^2 + 4*(B* 
a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*d*x)*tan(d*x + c)^4 - 3*A*a^3 + 12*(B 
*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*tan(d*x + c)^3 + 6*(A*a^3 - 3*B*a^2* 
b - 3*A*a*b^2)*tan(d*x + c)^2 - 4*(B*a^3 + 3*A*a^2*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. 400 vs. \(2 (187) = 374\).

Time = 2.68 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.09 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } A a^{3} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right )^{3} \cot ^{5}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } A a^{3} x & \text {for}\: c = - d x \\- \frac {A a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {A a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {A a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {A a^{3}}{4 d \tan ^{4}{\left (c + d x \right )}} + 3 A a^{2} b x + \frac {3 A a^{2} b}{d \tan {\left (c + d x \right )}} - \frac {A a^{2} b}{d \tan ^{3}{\left (c + d x \right )}} + \frac {3 A a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {3 A a b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {3 A a b^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - A b^{3} x - \frac {A b^{3}}{d \tan {\left (c + d x \right )}} + B a^{3} x + \frac {B a^{3}}{d \tan {\left (c + d x \right )}} - \frac {B a^{3}}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {3 B a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {3 B a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {3 B a^{2} b}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 B a b^{2} x - \frac {3 B a b^{2}}{d \tan {\left (c + d x \right )}} - \frac {B b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} & \text {otherwise} \end {cases} \] Input:

integrate(cot(d*x+c)**5*(a+b*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.13 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {12 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} {\left (d x + c\right )} - 6 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {3 \, A a^{3} - 12 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \tan \left (d x + c\right )^{3} - 6 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} + 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4}}}{12 \, d} \] Input:

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="m 
axima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.16 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} {\left (d x + c\right )}}{d} - \frac {{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} + \frac {{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {3 \, A a^{3} - 12 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \tan \left (d x + c\right )^{3} - 6 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} + 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{12 \, d \tan \left (d x + c\right )^{4}} \] Input:

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="g 
iac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.59 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.07 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,a^3-3\,B\,a^2\,b-3\,A\,a\,b^2+B\,b^3\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^4\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^3}{3}+A\,b\,a^2\right )+\frac {A\,a^3}{4}+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {A\,a^3}{2}+\frac {3\,B\,a^2\,b}{2}+\frac {3\,A\,a\,b^2}{2}\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (-B\,a^3-3\,A\,a^2\,b+3\,B\,a\,b^2+A\,b^3\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \] Input:

int(cot(c + d*x)^5*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^3,x)
 

Output:

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

Reduce [F]

\[ \int \cot ^5(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int \cot \left (d x +c \right )^{5} \left (a +\tan \left (d x +c \right ) b \right )^{3} \left (A +B \tan \left (d x +c \right )\right )d x \] Input:

int(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)
 

Output:

int(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)