3.1.14 \(\int \cot ^4(c+d x) (a+b \tan (c+d x))^2 (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\) [14]

3.1.14.1 Optimal result

Integrand size = 40, antiderivative size = 88 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\left (b^2 C-a (2 b B+a C)\right ) x-\frac {a (2 b B+a C) \cot (c+d x)}{d}-\frac {a^2 B \cot ^2(c+d x)}{2 d}-\frac {\left (a^2 B-b^2 B-2 a b C\right ) \log (\sin (c+d x))}{d} \]

output

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

3.1.14.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.40 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {-2 a (2 b B+a C) \cot (c+d x)-a^2 B \cot ^2(c+d x)+(a+i b)^2 (B+i C) \log (i-\tan (c+d x))-2 \left (a^2 B-b^2 B-2 a b C\right ) \log (\tan (c+d x))+(a-i b)^2 (B-i C) \log (i+\tan (c+d x))}{2 d} \]

input

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

output

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

3.1.14.3 Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 4115, 3042, 4087, 3042, 4111, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4087

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4111

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4014

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

input

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

output

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

3.1.14.3.1 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 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 
]
 

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]
 

3.1.14.4 Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.22

method	result	size
derivativedivides	\(\frac {B \,b^{2} \ln \left (\sin \left (d x +c \right )\right )+C \,b^{2} \left (d x +c \right )+2 B a b \left (-\cot \left (d x +c \right )-d x -c \right )+2 C a b \ln \left (\sin \left (d x +c \right )\right )+B \,a^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+C \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(107\)
default	\(\frac {B \,b^{2} \ln \left (\sin \left (d x +c \right )\right )+C \,b^{2} \left (d x +c \right )+2 B a b \left (-\cot \left (d x +c \right )-d x -c \right )+2 C a b \ln \left (\sin \left (d x +c \right )\right )+B \,a^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+C \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(107\)
parallelrisch	\(\frac {\left (B \,a^{2}-B \,b^{2}-2 C a b \right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+\left (-2 B \,a^{2}+2 B \,b^{2}+4 C a b \right ) \ln \left (\tan \left (d x +c \right )\right )-B \,a^{2} \cot \left (d x +c \right )^{2}+\left (-4 B a b -2 C \,a^{2}\right ) \cot \left (d x +c \right )-4 d x \left (B a b +\frac {1}{2} C \,a^{2}-\frac {1}{2} C \,b^{2}\right )}{2 d}\)	\(114\)
norman	\(\frac {\left (-2 B a b -C \,a^{2}+C \,b^{2}\right ) x \tan \left (d x +c \right )^{3}-\frac {B \,a^{2} \tan \left (d x +c \right )}{2 d}-\frac {a \left (2 B b +C a \right ) \tan \left (d x +c \right )^{2}}{d}}{\tan \left (d x +c \right )^{3}}-\frac {\left (B \,a^{2}-B \,b^{2}-2 C a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {\left (B \,a^{2}-B \,b^{2}-2 C a b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\)	\(138\)
risch	\(-i B \,b^{2} x -\frac {2 i B \,b^{2} c}{d}+\frac {2 i B \,a^{2} c}{d}-2 B a b x -C \,a^{2} x +C \,b^{2} x -\frac {4 i C a b c}{d}-\frac {2 i a \left (2 B b \,{\mathrm e}^{2 i \left (d x +c \right )}+C a \,{\mathrm e}^{2 i \left (d x +c \right )}+i B a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B b -C a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-2 i C a b x +i B \,a^{2} x -\frac {B \,a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B \,b^{2}}{d}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C a b}{d}\)	\(205\)

input

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

output

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

3.1.14.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.39 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {{\left (B a^{2} - 2 \, C a b - B b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + B a^{2} + {\left (B a^{2} + 2 \, {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} d x\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (C a^{2} + 2 \, B a b\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )^{2}} \]

input

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

output

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

3.1.14.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (78) = 156\).

Time = 1.76 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.34 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\begin {cases} \text {NaN} & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right )^{2} \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{4}{\left (c \right )} & \text {for}\: d = 0 \\\text {NaN} & \text {for}\: c = - d x \\\frac {B a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B a^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - 2 B a b x - \frac {2 B a b}{d \tan {\left (c + d x \right )}} - \frac {B b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - C a^{2} x - \frac {C a^{2}}{d \tan {\left (c + d x \right )}} - \frac {C a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {2 C a b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + C b^{2} x & \text {otherwise} \end {cases} \]

input

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

3.1.14.7 Maxima [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.36 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} {\left (d x + c\right )} - {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {B a^{2} + 2 \, {\left (C a^{2} + 2 \, B a b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

input

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

output

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

3.1.14.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (86) = 172\).

Time = 0.87 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.69 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} {\left (d x + c\right )} - 8 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) + 8 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

input

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

output

-1/8*(B*a^2*tan(1/2*d*x + 1/2*c)^2 - 4*C*a^2*tan(1/2*d*x + 1/2*c) - 8*B*a* 
b*tan(1/2*d*x + 1/2*c) + 8*(C*a^2 + 2*B*a*b - C*b^2)*(d*x + c) - 8*(B*a^2 
- 2*C*a*b - B*b^2)*log(tan(1/2*d*x + 1/2*c)^2 + 1) + 8*(B*a^2 - 2*C*a*b - 
B*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) - (12*B*a^2*tan(1/2*d*x + 1/2*c)^2 - 
 24*C*a*b*tan(1/2*d*x + 1/2*c)^2 - 12*B*b^2*tan(1/2*d*x + 1/2*c)^2 - 4*C*a 
^2*tan(1/2*d*x + 1/2*c) - 8*B*a*b*tan(1/2*d*x + 1/2*c) - B*a^2)/tan(1/2*d* 
x + 1/2*c)^2)/d
 

3.1.14.9 Mupad [B] (verification not implemented)

Time = 8.70 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.44 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B\,a^2+2\,C\,a\,b+B\,b^2\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\frac {B\,a^2}{2}+\mathrm {tan}\left (c+d\,x\right )\,\left (C\,a^2+2\,B\,b\,a\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^2}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^2}{2\,d} \]

input

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

output

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