\(\int \cot ^5(c+d x) (a+b \tan (c+d x)) (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\) [7]

Optimal result

Integrand size = 38, antiderivative size = 87 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=(a B-b C) x+\frac {(a B-b C) \cot (c+d x)}{d}-\frac {(b B+a C) \cot ^2(c+d x)}{2 d}-\frac {a B \cot ^3(c+d x)}{3 d}-\frac {(b B+a C) \log (\sin (c+d x))}{d} \] Output:

(B*a-C*b)*x+(B*a-C*b)*cot(d*x+c)/d-1/2*(B*b+C*a)*cot(d*x+c)^2/d-1/3*a*B*co 
t(d*x+c)^3/d-(B*b+C*a)*ln(sin(d*x+c))/d
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.04 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.48 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {b B \csc ^2(c+d x)}{2 d}-\frac {a C \csc ^2(c+d x)}{2 d}-\frac {a B \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\tan ^2(c+d x)\right )}{3 d}-\frac {b C \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )}{d}-\frac {b B \log (\sin (c+d x))}{d}-\frac {a C \log (\sin (c+d x))}{d} \] Input:

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

Output:

-1/2*(b*B*Csc[c + d*x]^2)/d - (a*C*Csc[c + d*x]^2)/(2*d) - (a*B*Cot[c + d* 
x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2])/(3*d) - (b*C*Cot[c 
 + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2])/d - (b*B*Log[Sin 
[c + d*x]])/d - (a*C*Log[Sin[c + d*x]])/d
 

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 4115, 3042, 4074, 3042, 4012, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4074

\(\displaystyle \int \cot ^3(c+d x) (b B+a C-(a B-b C) \tan (c+d x))dx-\frac {a B \cot ^3(c+d x)}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {b B+a C-(a B-b C) \tan (c+d x)}{\tan (c+d x)^3}dx-\frac {a B \cot ^3(c+d x)}{3 d}\)

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4014

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

Input:

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

Output:

(a*B - b*C)*x + ((a*B - b*C)*Cot[c + d*x])/d - ((b*B + a*C)*Cot[c + d*x]^2 
)/(2*d) - (a*B*Cot[c + d*x]^3)/(3*d) - ((b*B + a*C)*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 4074

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

rule 4115

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[1/b^2   Int[(a + b*Tan[e + f*x])^(m 
+ 1)*(c + d*Tan[e + f*x])^n*(b*B - a*C + b*C*Tan[e + f*x]), x], x] /; FreeQ 
[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - 
a*b*B + a^2*C, 0]
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {B b \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+C b \left (-\cot \left (d x +c \right )-d x -c \right )+B a \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+C a \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(95\)
default	\(\frac {B b \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+C b \left (-\cot \left (d x +c \right )-d x -c \right )+B a \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+C a \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(95\)
norman	\(\frac {\frac {\left (B a -C b \right ) \tan \left (d x +c \right )^{3}}{d}+\left (B a -C b \right ) x \tan \left (d x +c \right )^{4}-\frac {\left (B b +C a \right ) \tan \left (d x +c \right )^{2}}{2 d}-\frac {B a \tan \left (d x +c \right )}{3 d}}{\tan \left (d x +c \right )^{4}}-\frac {\left (B b +C a \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {\left (B b +C a \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\)	\(125\)
parallelrisch	\(\frac {-3 B b \left (-\ln \left (\sec \left (d x +c \right )^{2}\right )+2 \ln \left (\tan \left (d x +c \right )\right )\right )-3 C a \left (-\ln \left (\sec \left (d x +c \right )^{2}\right )+2 \ln \left (\tan \left (d x +c \right )\right )\right )-2 B \cot \left (d x +c \right )^{3} a -3 B b \cot \left (d x +c \right )^{2}+6 B a d x -3 C a \cot \left (d x +c \right )^{2}-6 C b d x +6 B a \cot \left (d x +c \right )-6 C b \cot \left (d x +c \right )}{6 d}\)	\(125\)
risch	\(i B b x +i C a x +B a x -C b x +\frac {2 i B b c}{d}+\frac {2 i C a c}{d}-\frac {2 i \left (3 i B b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 i C a \,{\mathrm e}^{4 i \left (d x +c \right )}-6 B a \,{\mathrm e}^{4 i \left (d x +c \right )}+3 C b \,{\mathrm e}^{4 i \left (d x +c \right )}-3 i B b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 i C a \,{\mathrm e}^{2 i \left (d x +c \right )}+6 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-6 b \,{\mathrm e}^{2 i \left (d x +c \right )} C -4 B a +3 C b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C a}{d}\)	\(215\)

Input:

int(cot(d*x+c)^5*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x,method=_ 
RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.39 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {3 \, {\left (C a + B b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} - 3 \, {\left (2 \, {\left (B a - C b\right )} d x - C a - B b\right )} \tan \left (d x + c\right )^{3} - 6 \, {\left (B a - C b\right )} \tan \left (d x + c\right )^{2} + 2 \, B a + 3 \, {\left (C a + B b\right )} \tan \left (d x + c\right )}{6 \, d \tan \left (d x + c\right )^{3}} \] Input:

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, a 
lgorithm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (75) = 150\).

Time = 1.95 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.99 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\begin {cases} \text {NaN} & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right ) \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{5}{\left (c \right )} & \text {for}\: d = 0 \\\text {NaN} & \text {for}\: c = - d x \\B a x + \frac {B a}{d \tan {\left (c + d x \right )}} - \frac {B a}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {B b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B b}{2 d \tan ^{2}{\left (c + d x \right )}} + \frac {C a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {C a \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {C a}{2 d \tan ^{2}{\left (c + d x \right )}} - C b x - \frac {C b}{d \tan {\left (c + d x \right )}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((nan, Eq(c, 0) & Eq(d, 0)), (x*(a + b*tan(c))*(B*tan(c) + C*tan( 
c)**2)*cot(c)**5, Eq(d, 0)), (nan, Eq(c, -d*x)), (B*a*x + B*a/(d*tan(c + d 
*x)) - B*a/(3*d*tan(c + d*x)**3) + B*b*log(tan(c + d*x)**2 + 1)/(2*d) - B* 
b*log(tan(c + d*x))/d - B*b/(2*d*tan(c + d*x)**2) + C*a*log(tan(c + d*x)** 
2 + 1)/(2*d) - C*a*log(tan(c + d*x))/d - C*a/(2*d*tan(c + d*x)**2) - C*b*x 
 - C*b/(d*tan(c + d*x)), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.20 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {6 \, {\left (B a - C b\right )} {\left (d x + c\right )} + 3 \, {\left (C a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, {\left (C a + B b\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {6 \, {\left (B a - C b\right )} \tan \left (d x + c\right )^{2} - 2 \, B a - 3 \, {\left (C a + B b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{3}}}{6 \, d} \] Input:

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, a 
lgorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.75 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.29 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {{\left (B a - C b\right )} {\left (d x + c\right )}}{d} + \frac {{\left (C a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} - \frac {{\left (C a + B b\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} + \frac {6 \, {\left (B a - C b\right )} \tan \left (d x + c\right )^{2} - 2 \, B a - 3 \, {\left (C a + B b\right )} \tan \left (d x + c\right )}{6 \, d \tan \left (d x + c\right )^{3}} \] Input:

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, a 
lgorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 5.24 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.46 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (\left (C\,b-B\,a\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+\left (\frac {B\,b}{2}+\frac {C\,a}{2}\right )\,\mathrm {tan}\left (c+d\,x\right )+\frac {B\,a}{3}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,b+C\,a\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,\left (b+a\,1{}\mathrm {i}\right )}{2\,d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.67 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a b -12 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b c -4 \cos \left (d x +c \right ) a b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} a c +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} b^{2}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a c -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} b^{2}+12 \sin \left (d x +c \right )^{3} a b d x +3 \sin \left (d x +c \right )^{3} a c +3 \sin \left (d x +c \right )^{3} b^{2}-12 \sin \left (d x +c \right )^{3} b c d x -6 \sin \left (d x +c \right ) a c -6 \sin \left (d x +c \right ) b^{2}}{12 \sin \left (d x +c \right )^{3} d} \] Input:

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

Output:

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