\(\int \cot ^4(c+d x) (a+b \tan (c+d x))^3 (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\) [21]

Optimal result

Integrand size = 40, antiderivative size = 127 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\left (\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) x\right )-\frac {a^2 (2 b B+a C) \cot (c+d x)}{d}-\frac {b^3 C \log (\cos (c+d x))}{d}-\frac {a \left (a^2 B-3 b^2 B-3 a b C\right ) \log (\sin (c+d x))}{d}-\frac {a B \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.99 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {-2 a^2 (3 b B+a C) \cot (c+d x)-a^3 B \cot ^2(c+d x)+(a+i b)^3 (B+i C) \log (i-\tan (c+d x))-2 a \left (a^2 B-3 b^2 B-3 a b C\right ) \log (\tan (c+d x))+(a-i b)^3 (B-i C) \log (i+\tan (c+d x))}{2 d} \] Input:

Integrate[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^3*(B*Tan[c + d*x] + C*Tan[c 
+ d*x]^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.325, Rules used = {3042, 4115, 3042, 4088, 27, 3042, 4118, 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.

\(\displaystyle \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4088

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4118

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4107

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

Input:

Int[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^3*(B*Tan[c + d*x] + C*Tan[c + d*x] 
^2),x]
 

Output:

-((3*a^2*b*B - b^3*B + a^3*C - 3*a*b^2*C)*x) - (a^2*(2*b*B + a*C)*Cot[c + 
d*x])/d - (b^3*C*Log[Cos[c + d*x]])/d - (a*(a^2*B - 3*b^2*B - 3*a*b*C)*Log 
[-Sin[c + d*x]])/d - (a*B*Cot[c + d*x]^2*(a + b*Tan[c + d*x])^2)/(2*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 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 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 = 0.21 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.07

method	result	size
parallelrisch	\(\frac {\left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+\left (-2 B \,a^{3}+6 B a \,b^{2}+6 C \,a^{2} b \right ) \ln \left (\tan \left (d x +c \right )\right )-B \cot \left (d x +c \right )^{2} a^{3}+\left (-6 B \,a^{2} b -2 C \,a^{3}\right ) \cot \left (d x +c \right )-6 x \left (B \,a^{2} b -\frac {1}{3} B \,b^{3}+\frac {1}{3} C \,a^{3}-C a \,b^{2}\right ) d}{2 d}\)	\(136\)
derivativedivides	\(\frac {\frac {\left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {B \,a^{3}}{2 \tan \left (d x +c \right )^{2}}-\frac {a^{2} \left (3 B b +C a \right )}{\tan \left (d x +c \right )}-a \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(140\)
default	\(\frac {\frac {\left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {B \,a^{3}}{2 \tan \left (d x +c \right )^{2}}-\frac {a^{2} \left (3 B b +C a \right )}{\tan \left (d x +c \right )}-a \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(140\)
norman	\(\frac {\left (-3 B \,a^{2} b +B \,b^{3}-C \,a^{3}+3 C a \,b^{2}\right ) x \tan \left (d x +c \right )^{3}-\frac {B \,a^{3} \tan \left (d x +c \right )}{2 d}-\frac {a^{2} \left (3 B b +C a \right ) \tan \left (d x +c \right )^{2}}{d}}{\tan \left (d x +c \right )^{3}}+\frac {\left (B \,a^{3}-3 B a \,b^{2}-3 C \,a^{2} b +C \,b^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}-\frac {a \left (B \,a^{2}-3 B \,b^{2}-3 C a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(158\)
risch	\(i B \,a^{3} x -\frac {2 i a^{2} \left (3 B b \,{\mathrm e}^{2 i \left (d x +c \right )}+C a \,{\mathrm e}^{2 i \left (d x +c \right )}+i B a \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B b -C a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-3 i B a \,b^{2} x -3 i C \,a^{2} b x -3 B \,a^{2} b x +B \,b^{3} x -C \,a^{3} x +3 C a \,b^{2} x +\frac {2 i B \,a^{3} c}{d}+\frac {2 i C \,b^{3} c}{d}+i C \,b^{3} x -\frac {6 i C \,a^{2} b c}{d}-\frac {6 i B a \,b^{2} c}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) C \,b^{3}}{d}-\frac {B \,a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B \,b^{2}}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C b}{d}\)	\(267\)

Input:

int(cot(d*x+c)^4*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x,method 
=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.28 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {C b^{3} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + B a^{3} + {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + {\left (B a^{3} + 2 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )^{2}} \] Input:

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, 
 algorithm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (121) = 242\).

Time = 2.46 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.99 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\begin {cases} \text {NaN} & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right )^{3} \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{4}{\left (c \right )} & \text {for}\: d = 0 \\\text {NaN} & \text {for}\: c = - d x \\\frac {B a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 B a^{2} b x - \frac {3 B a^{2} b}{d \tan {\left (c + d x \right )}} - \frac {3 B a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 B a b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + B b^{3} x - C a^{3} x - \frac {C a^{3}}{d \tan {\left (c + d x \right )}} - \frac {3 C a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 C a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 3 C a b^{2} x + \frac {C b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \text {otherwise} \end {cases} \] Input:

integrate(cot(d*x+c)**4*(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)**4, Eq(d, 0)), (nan, Eq(c, -d*x)), (B*a**3*log(tan(c + d* 
x)**2 + 1)/(2*d) - B*a**3*log(tan(c + d*x))/d - B*a**3/(2*d*tan(c + d*x)** 
2) - 3*B*a**2*b*x - 3*B*a**2*b/(d*tan(c + d*x)) - 3*B*a*b**2*log(tan(c + d 
*x)**2 + 1)/(2*d) + 3*B*a*b**2*log(tan(c + d*x))/d + B*b**3*x - C*a**3*x - 
 C*a**3/(d*tan(c + d*x)) - 3*C*a**2*b*log(tan(c + d*x)**2 + 1)/(2*d) + 3*C 
*a**2*b*log(tan(c + d*x))/d + 3*C*a*b**2*x + C*b**3*log(tan(c + d*x)**2 + 
1)/(2*d), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.12 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} {\left (d x + c\right )} - {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {B a^{3} + 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2}}}{2 \, d} \] Input:

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

Giac [A] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.19 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {{\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} {\left (d x + c\right )}}{d} + \frac {{\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} - \frac {{\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {B a^{3} + 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )^{2}} \] Input:

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, 
 algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 5.52 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.06 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B\,a^3+3\,C\,a^2\,b+3\,B\,a\,b^2\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (C\,a^3+3\,B\,b\,a^2\right )+\frac {B\,a^3}{2}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \] Input:

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

Reduce [F]

\[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\int \cot \left (d x +c \right )^{4} \left (a +\tan \left (d x +c \right ) b \right )^{3} \left (B \tan \left (d x +c \right )+C \tan \left (d x +c \right )^{2}\right )d x \] Input:

int(cot(d*x+c)^4*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x)
 

Output:

int(cot(d*x+c)^4*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x)