3.1.20 \(\int \cot ^3(c+d x) (a+b \tan (c+d x))^3 (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\) [20]

3.1.20.1 Optimal result
3.1.20.2 Mathematica [C] (verified)
3.1.20.3 Rubi [A] (verified)
3.1.20.4 Maple [A] (verified)
3.1.20.5 Fricas [A] (verification not implemented)
3.1.20.6 Sympy [A] (verification not implemented)
3.1.20.7 Maxima [A] (verification not implemented)
3.1.20.8 Giac [A] (verification not implemented)
3.1.20.9 Mupad [B] (verification not implemented)

3.1.20.1 Optimal result

Integrand size = 40, antiderivative size = 119 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\left (\left (a^3 B-3 a b^2 B-3 a^2 b C+b^3 C\right ) x\right )-\frac {b^2 (b B+3 a C) \log (\cos (c+d x))}{d}+\frac {a^2 (3 b B+a C) \log (\sin (c+d x))}{d}+\frac {b^2 (a B+b C) \tan (c+d x)}{d}-\frac {a B \cot (c+d x) (a+b \tan (c+d x))^2}{d} \]

output
-(B*a^3-3*B*a*b^2-3*C*a^2*b+C*b^3)*x-b^2*(B*b+3*C*a)*ln(cos(d*x+c))/d+a^2* 
(3*B*b+C*a)*ln(sin(d*x+c))/d+b^2*(B*a+C*b)*tan(d*x+c)/d-a*B*cot(d*x+c)*(a+ 
b*tan(d*x+c))^2/d
 
3.1.20.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.52 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.95 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {-2 a^3 B \cot (c+d x)+i (a+i b)^3 (B+i C) \log (i-\tan (c+d x))+2 a^2 (3 b B+a C) \log (\tan (c+d x))+(i a+b)^3 (B-i C) \log (i+\tan (c+d x))+2 b^3 C \tan (c+d x)}{2 d} \]

input
Integrate[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^3*(B*Tan[c + d*x] + C*Tan[c 
+ d*x]^2),x]
 
output
(-2*a^3*B*Cot[c + d*x] + I*(a + I*b)^3*(B + I*C)*Log[I - Tan[c + d*x]] + 2 
*a^2*(3*b*B + a*C)*Log[Tan[c + d*x]] + (I*a + b)^3*(B - I*C)*Log[I + Tan[c 
 + d*x]] + 2*b^3*C*Tan[c + d*x])/(2*d)
 
3.1.20.3 Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 4115, 3042, 4088, 3042, 4120, 25, 3042, 4107, 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.

\(\displaystyle \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4088

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4120

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4107

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

input
Int[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^3*(B*Tan[c + d*x] + C*Tan[c + d*x] 
^2),x]
 
output
-((a^3*B - 3*a*b^2*B - 3*a^2*b*C + b^3*C)*x) - (b^2*(b*B + 3*a*C)*Log[Cos[ 
c + d*x]])/d + (a^2*(3*b*B + a*C)*Log[-Sin[c + d*x]])/d + (b^2*(a*B + b*C) 
*Tan[c + d*x])/d - (a*B*Cot[c + d*x]*(a + b*Tan[c + d*x])^2)/d
 

3.1.20.3.1 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 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 4107
Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2 
)/tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[B*x, x] + (Simp[A   Int[1/Tan[ 
e + f*x], x], x] + Simp[C   Int[Tan[e + f*x], x], x]) /; FreeQ[{e, f, A, B, 
 C}, x] && NeQ[A, C]
 

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]
 

rule 4120
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*Tan[e + f*x]*((c + d*Tan[e + f*x])^(n + 
 1)/(d*f*(n + 2))), x] - Simp[1/(d*(n + 2))   Int[(c + d*Tan[e + f*x])^n*Si 
mp[b*c*C - a*A*d*(n + 2) - (A*b + a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C* 
d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, 
b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && 
  !LtQ[n, -1]
 
3.1.20.4 Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99

method result size
parallelrisch \(\frac {\left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+\left (6 B \,a^{2} b +2 C \,a^{3}\right ) \ln \left (\tan \left (d x +c \right )\right )-2 B \cot \left (d x +c \right ) a^{3}+2 C \,b^{3} \tan \left (d x +c \right )-2 d x \left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right )}{2 d}\) \(118\)
derivativedivides \(\frac {C \,b^{3} \tan \left (d x +c \right )-\frac {B \,a^{3}}{\tan \left (d x +c \right )}+a^{2} \left (3 B b +C a \right ) \ln \left (\tan \left (d x +c \right )\right )+\frac {\left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-B \,a^{3}+3 B a \,b^{2}+3 C \,a^{2} b -C \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) \(124\)
default \(\frac {C \,b^{3} \tan \left (d x +c \right )-\frac {B \,a^{3}}{\tan \left (d x +c \right )}+a^{2} \left (3 B b +C a \right ) \ln \left (\tan \left (d x +c \right )\right )+\frac {\left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-B \,a^{3}+3 B a \,b^{2}+3 C \,a^{2} b -C \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) \(124\)
norman \(\frac {\left (-B \,a^{3}+3 B a \,b^{2}+3 C \,a^{2} b -C \,b^{3}\right ) x \tan \left (d x +c \right )^{2}+\frac {C \,b^{3} \tan \left (d x +c \right )^{3}}{d}-\frac {B \,a^{3} \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}+\frac {a^{2} \left (3 B b +C a \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (3 B \,a^{2} b -B \,b^{3}+C \,a^{3}-3 C a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) \(144\)
risch \(-B \,a^{3} x +3 B a \,b^{2} x +3 C \,a^{2} b x -C \,b^{3} x -\frac {2 i C \,a^{3} c}{d}+i B \,b^{3} x -i C \,a^{3} x +\frac {2 i B \,b^{3} c}{d}-\frac {2 i \left (B \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-C \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+B \,a^{3}+C \,b^{3}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {6 i C a \,b^{2} c}{d}-\frac {6 i B \,a^{2} b c}{d}+3 i C a \,b^{2} x -3 i B \,a^{2} b x -\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B \,b^{3}}{d}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) C a \,b^{2}}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C}{d}\) \(269\)

input
int(cot(d*x+c)^3*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x,method 
=_RETURNVERBOSE)
 
output
1/2*((-3*B*a^2*b+B*b^3-C*a^3+3*C*a*b^2)*ln(sec(d*x+c)^2)+(6*B*a^2*b+2*C*a^ 
3)*ln(tan(d*x+c))-2*B*cot(d*x+c)*a^3+2*C*b^3*tan(d*x+c)-2*d*x*(B*a^3-3*B*a 
*b^2-3*C*a^2*b+C*b^3))/d
 
3.1.20.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.22 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {2 \, C b^{3} \tan \left (d x + c\right )^{2} - 2 \, B a^{3} - 2 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} d x \tan \left (d x + c\right ) + {\left (C a^{3} + 3 \, B a^{2} b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) - {\left (3 \, C a b^{2} + B b^{3}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )} \]

input
integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, 
 algorithm="fricas")
 
output
1/2*(2*C*b^3*tan(d*x + c)^2 - 2*B*a^3 - 2*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + 
 C*b^3)*d*x*tan(d*x + c) + (C*a^3 + 3*B*a^2*b)*log(tan(d*x + c)^2/(tan(d*x 
 + c)^2 + 1))*tan(d*x + c) - (3*C*a*b^2 + B*b^3)*log(1/(tan(d*x + c)^2 + 1 
))*tan(d*x + c))/(d*tan(d*x + c))
 
3.1.20.6 Sympy [A] (verification not implemented)

Time = 1.70 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.80 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\begin {cases} \text {NaN} & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right )^{3} \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{3}{\left (c \right )} & \text {for}\: d = 0 \\\text {NaN} & \text {for}\: c = - d x \\- B a^{3} x - \frac {B a^{3}}{d \tan {\left (c + d x \right )}} - \frac {3 B a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 B a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 3 B a b^{2} x + \frac {B b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {C a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {C a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 3 C a^{2} b x + \frac {3 C a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - C b^{3} x + \frac {C b^{3} \tan {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases} \]

input
integrate(cot(d*x+c)**3*(a+b*tan(d*x+c))**3*(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))**3*(B*tan(c) + C*t 
an(c)**2)*cot(c)**3, Eq(d, 0)), (nan, Eq(c, -d*x)), (-B*a**3*x - B*a**3/(d 
*tan(c + d*x)) - 3*B*a**2*b*log(tan(c + d*x)**2 + 1)/(2*d) + 3*B*a**2*b*lo 
g(tan(c + d*x))/d + 3*B*a*b**2*x + B*b**3*log(tan(c + d*x)**2 + 1)/(2*d) - 
 C*a**3*log(tan(c + d*x)**2 + 1)/(2*d) + C*a**3*log(tan(c + d*x))/d + 3*C* 
a**2*b*x + 3*C*a*b**2*log(tan(c + d*x)**2 + 1)/(2*d) - C*b**3*x + C*b**3*t 
an(c + d*x)/d, True))
 
3.1.20.7 Maxima [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.05 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {2 \, C b^{3} \tan \left (d x + c\right ) - \frac {2 \, B a^{3}}{\tan \left (d x + c\right )} - 2 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} {\left (d x + c\right )} - {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \log \left (\tan \left (d x + c\right )\right )}{2 \, d} \]

input
integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, 
 algorithm="maxima")
 
output
1/2*(2*C*b^3*tan(d*x + c) - 2*B*a^3/tan(d*x + c) - 2*(B*a^3 - 3*C*a^2*b - 
3*B*a*b^2 + C*b^3)*(d*x + c) - (C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*log 
(tan(d*x + c)^2 + 1) + 2*(C*a^3 + 3*B*a^2*b)*log(tan(d*x + c)))/d
 
3.1.20.8 Giac [A] (verification not implemented)

Time = 1.22 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.28 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {2 \, C b^{3} \tan \left (d x + c\right ) - 2 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} {\left (d x + c\right )} - {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {2 \, {\left (C a^{3} \tan \left (d x + c\right ) + 3 \, B a^{2} b \tan \left (d x + c\right ) + B a^{3}\right )}}{\tan \left (d x + c\right )}}{2 \, d} \]

input
integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, 
 algorithm="giac")
 
output
1/2*(2*C*b^3*tan(d*x + c) - 2*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + C*b^3)*(d*x 
 + c) - (C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*log(tan(d*x + c)^2 + 1) + 
2*(C*a^3 + 3*B*a^2*b)*log(abs(tan(d*x + c))) - 2*(C*a^3*tan(d*x + c) + 3*B 
*a^2*b*tan(d*x + c) + B*a^3)/tan(d*x + c))/d
 
3.1.20.9 Mupad [B] (verification not implemented)

Time = 8.78 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.96 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (C\,a^3+3\,B\,b\,a^2\right )}{d}-\frac {B\,a^3\,\mathrm {cot}\left (c+d\,x\right )}{d}+\frac {C\,b^3\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,{\left (a-b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \]

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