\(\int \cot ^4(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [246]

Optimal result

Integrand size = 31, antiderivative size = 118 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\left (a^2 A-A b^2-2 a b B\right ) x+\frac {\left (a^2 A-A b^2-2 a b B\right ) \cot (c+d x)}{d}-\frac {a (2 A b+a B) \cot ^2(c+d x)}{2 d}-\frac {a^2 A \cot ^3(c+d x)}{3 d}+\frac {\left (b^2 B-a (2 A b+a B)\right ) \log (\sin (c+d x))}{d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.82 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.29 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {6 \left (a^2 A-A b^2-2 a b B\right ) \cot (c+d x)-3 a (2 A b+a B) \cot ^2(c+d x)-2 a^2 A \cot ^3(c+d x)+3 (a+i b)^2 (-i A+B) \log (i-\tan (c+d x))-6 \left (2 a A b+a^2 B-b^2 B\right ) \log (\tan (c+d x))+3 (a-i b)^2 (i A+B) \log (i+\tan (c+d x))}{6 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4087, 3042, 4111, 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.

\(\displaystyle \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4087

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4111

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4014

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

Input:

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

Output:

(a^2*A - A*b^2 - 2*a*b*B)*x + ((a^2*A - A*b^2 - 2*a*b*B)*Cot[c + d*x])/d - 
 (a*(2*A*b + a*B)*Cot[c + d*x]^2)/(2*d) - (a^2*A*Cot[c + d*x]^3)/(3*d) - ( 
(2*a*A*b + a^2*B - b^2*B)*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 4087

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

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 
]
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {A \,b^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+B \,b^{2} \ln \left (\sin \left (d x +c \right )\right )+2 A a b \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+2 B a b \left (-\cot \left (d x +c \right )-d x -c \right )+A \,a^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+B \,a^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(136\)
default	\(\frac {A \,b^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+B \,b^{2} \ln \left (\sin \left (d x +c \right )\right )+2 A a b \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+2 B a b \left (-\cot \left (d x +c \right )-d x -c \right )+A \,a^{2} \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+B \,a^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(136\)
parallelrisch	\(\frac {3 \left (2 A a b +B \,a^{2}-B \,b^{2}\right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+6 \left (-2 A a b -B \,a^{2}+B \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )-2 A \cot \left (d x +c \right )^{3} a^{2}+3 \left (-2 A a b -B \,a^{2}\right ) \cot \left (d x +c \right )^{2}+6 \cot \left (d x +c \right ) \left (A \,a^{2}-A \,b^{2}-2 B a b \right )+6 d x \left (A \,a^{2}-A \,b^{2}-2 B a b \right )}{6 d}\)	\(143\)
norman	\(\frac {\frac {\left (A \,a^{2}-A \,b^{2}-2 B a b \right ) \tan \left (d x +c \right )^{2}}{d}+\left (A \,a^{2}-A \,b^{2}-2 B a b \right ) x \tan \left (d x +c \right )^{3}-\frac {A \,a^{2}}{3 d}-\frac {a \left (2 A b +B a \right ) \tan \left (d x +c \right )}{2 d}}{\tan \left (d x +c \right )^{3}}-\frac {\left (2 A a b +B \,a^{2}-B \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {\left (2 A a b +B \,a^{2}-B \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\)	\(159\)
risch	\(2 i A a b x +i B \,a^{2} x -\frac {2 i \left (6 i A a b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 i B \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 i A a b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 i B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+6 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 A \,a^{2}+3 A \,b^{2}+6 B a b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+A \,a^{2} x -A \,b^{2} x -2 B a b x -i B \,b^{2} x +\frac {4 i A a b c}{d}-\frac {2 i B \,b^{2} c}{d}+\frac {2 i B \,a^{2} c}{d}-\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B \,b^{2}}{d}\)	\(324\)

Input:

int(cot(d*x+c)^4*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.33 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {3 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} + 3 \, {\left (B a^{2} + 2 \, A a b - 2 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} d x\right )} \tan \left (d x + c\right )^{3} + 2 \, A a^{2} - 6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )}{6 \, d \tan \left (d x + c\right )^{3}} \] Input:

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="f 
ricas")
 

Output:

-1/6*(3*(B*a^2 + 2*A*a*b - B*b^2)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1)) 
*tan(d*x + c)^3 + 3*(B*a^2 + 2*A*a*b - 2*(A*a^2 - 2*B*a*b - A*b^2)*d*x)*ta 
n(d*x + c)^3 + 2*A*a^2 - 6*(A*a^2 - 2*B*a*b - A*b^2)*tan(d*x + c)^2 + 3*(B 
*a^2 + 2*A*a*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. 260 vs. \(2 (107) = 214\).

Time = 1.12 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.20 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } A a^{2} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right )^{2} \cot ^{4}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } A a^{2} x & \text {for}\: c = - d x \\A a^{2} x + \frac {A a^{2}}{d \tan {\left (c + d x \right )}} - \frac {A a^{2}}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {A a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - \frac {2 A a b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {A a b}{d \tan ^{2}{\left (c + d x \right )}} - A b^{2} x - \frac {A b^{2}}{d \tan {\left (c + d x \right )}} + \frac {B a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B a^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - 2 B a b x - \frac {2 B a b}{d \tan {\left (c + d x \right )}} - \frac {B b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} & \text {otherwise} \end {cases} \] Input:

integrate(cot(d*x+c)**4*(a+b*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.26 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )} + 3 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {2 \, A a^{2} - 6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{3}}}{6 \, d} \] Input:

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="m 
axima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.32 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {{\left (A a^{2} - 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )}}{d} + \frac {{\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} - \frac {{\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {2 \, A a^{2} - 6 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )}{6 \, d \tan \left (d x + c\right )^{3}} \] Input:

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="g 
iac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.37 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.32 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (\frac {A\,a^2}{3}+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-A\,a^2+2\,B\,a\,b+A\,b^2\right )+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^2}{2}+A\,b\,a\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^2+2\,A\,a\,b-B\,b^2\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^2}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^2}{2\,d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.89 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}-36 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}-4 \cos \left (d x +c \right ) a^{3}+36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} a^{2} b -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} b^{3}-36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a^{2} b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} b^{3}+12 \sin \left (d x +c \right )^{3} a^{3} d x +9 \sin \left (d x +c \right )^{3} a^{2} b -36 \sin \left (d x +c \right )^{3} a \,b^{2} d x -18 \sin \left (d x +c \right ) a^{2} b}{12 \sin \left (d x +c \right )^{3} d} \] Input:

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

Output:

(16*cos(c + d*x)*sin(c + d*x)**2*a**3 - 36*cos(c + d*x)*sin(c + d*x)**2*a* 
b**2 - 4*cos(c + d*x)*a**3 + 36*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)* 
*3*a**2*b - 12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**3*b**3 - 36*log( 
tan((c + d*x)/2))*sin(c + d*x)**3*a**2*b + 12*log(tan((c + d*x)/2))*sin(c 
+ d*x)**3*b**3 + 12*sin(c + d*x)**3*a**3*d*x + 9*sin(c + d*x)**3*a**2*b - 
36*sin(c + d*x)**3*a*b**2*d*x - 18*sin(c + d*x)*a**2*b)/(12*sin(c + d*x)** 
3*d)