3.3.62 \(\int \cot ^3(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\) [262]

3.3.62.1 Optimal result

Integrand size = 31, antiderivative size = 186 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-\left (\left (4 a^3 A b-4 a A b^3+a^4 B-6 a^2 b^2 B+b^4 B\right ) x\right )-\frac {b^3 (A b+4 a B) \log (\cos (c+d x))}{d}-\frac {a^2 \left (a^2 A-6 A b^2-4 a b B\right ) \log (\sin (c+d x))}{d}+\frac {b^2 \left (3 a A b+a^2 B+b^2 B\right ) \tan (c+d x)}{d}-\frac {a (5 A b+2 a B) \cot (c+d x) (a+b \tan (c+d x))^2}{2 d}-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d} \]

output

-(4*A*a^3*b-4*A*a*b^3+B*a^4-6*B*a^2*b^2+B*b^4)*x-b^3*(A*b+4*B*a)*ln(cos(d* 
x+c))/d-a^2*(A*a^2-6*A*b^2-4*B*a*b)*ln(sin(d*x+c))/d+b^2*(3*A*a*b+B*a^2+B* 
b^2)*tan(d*x+c)/d-1/2*a*(5*A*b+2*B*a)*cot(d*x+c)*(a+b*tan(d*x+c))^2/d-1/2* 
a*A*cot(d*x+c)^2*(a+b*tan(d*x+c))^3/d
 

3.3.62.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.70 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.75 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {-2 a^3 (4 A b+a B) \cot (c+d x)-a^4 A \cot ^2(c+d x)+(a+i b)^4 (A+i B) \log (i-\tan (c+d x))-2 a^2 \left (a^2 A-6 A b^2-4 a b B\right ) \log (\tan (c+d x))+(a-i b)^4 (A-i B) \log (i+\tan (c+d x))+2 b^4 B \tan (c+d x)}{2 d} \]

input

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

output

(-2*a^3*(4*A*b + a*B)*Cot[c + d*x] - a^4*A*Cot[c + d*x]^2 + (a + I*b)^4*(A 
 + I*B)*Log[I - Tan[c + d*x]] - 2*a^2*(a^2*A - 6*A*b^2 - 4*a*b*B)*Log[Tan[ 
c + d*x]] + (a - I*b)^4*(A - I*B)*Log[I + Tan[c + d*x]] + 2*b^4*B*Tan[c + 
d*x])/(2*d)
 

3.3.62.3 Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 4088, 3042, 4128, 27, 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.

\(\displaystyle \int \cot ^3(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4088

\(\displaystyle \frac {1}{2} \int \cot ^2(c+d x) (a+b \tan (c+d x))^2 \left (b (a A+2 b B) \tan ^2(c+d x)-2 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (5 A b+2 a B)\right )dx-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \frac {(a+b \tan (c+d x))^2 \left (b (a A+2 b B) \tan (c+d x)^2-2 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (5 A b+2 a B)\right )}{\tan (c+d x)^2}dx-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d}\)

\(\Big \downarrow \) 4128

\(\displaystyle \frac {1}{2} \left (\int -2 \cot (c+d x) (a+b \tan (c+d x)) \left (-b \left (B a^2+3 A b a+b^2 B\right ) \tan ^2(c+d x)+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (A a^2-4 b B a-6 A b^2\right )\right )dx-\frac {a (2 a B+5 A b) \cot (c+d x) (a+b \tan (c+d x))^2}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (-2 \int \cot (c+d x) (a+b \tan (c+d x)) \left (-b \left (B a^2+3 A b a+b^2 B\right ) \tan ^2(c+d x)+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (A a^2-4 b B a-6 A b^2\right )\right )dx-\frac {a (2 a B+5 A b) \cot (c+d x) (a+b \tan (c+d x))^2}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (-2 \int \frac {(a+b \tan (c+d x)) \left (-b \left (B a^2+3 A b a+b^2 B\right ) \tan (c+d x)^2+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (A a^2-4 b B a-6 A b^2\right )\right )}{\tan (c+d x)}dx-\frac {a (2 a B+5 A b) \cot (c+d x) (a+b \tan (c+d x))^2}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d}\)

\(\Big \downarrow \) 4120

\(\displaystyle \frac {1}{2} \left (-2 \left (-\int -\cot (c+d x) \left (-\left ((A b+4 a B) \tan ^2(c+d x) b^3\right )+a^2 \left (A a^2-4 b B a-6 A b^2\right )+\left (B a^4+4 A b a^3-6 b^2 B a^2-4 A b^3 a+b^4 B\right ) \tan (c+d x)\right )dx-\frac {b^2 \left (a^2 B+3 a A b+b^2 B\right ) \tan (c+d x)}{d}\right )-\frac {a (2 a B+5 A b) \cot (c+d x) (a+b \tan (c+d x))^2}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (-2 \left (\int \cot (c+d x) \left (-\left ((A b+4 a B) \tan ^2(c+d x) b^3\right )+a^2 \left (A a^2-4 b B a-6 A b^2\right )+\left (B a^4+4 A b a^3-6 b^2 B a^2-4 A b^3 a+b^4 B\right ) \tan (c+d x)\right )dx-\frac {b^2 \left (a^2 B+3 a A b+b^2 B\right ) \tan (c+d x)}{d}\right )-\frac {a (2 a B+5 A b) \cot (c+d x) (a+b \tan (c+d x))^2}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (-2 \left (\int \frac {-\left ((A b+4 a B) \tan (c+d x)^2 b^3\right )+a^2 \left (A a^2-4 b B a-6 A b^2\right )+\left (B a^4+4 A b a^3-6 b^2 B a^2-4 A b^3 a+b^4 B\right ) \tan (c+d x)}{\tan (c+d x)}dx-\frac {b^2 \left (a^2 B+3 a A b+b^2 B\right ) \tan (c+d x)}{d}\right )-\frac {a (2 a B+5 A b) \cot (c+d x) (a+b \tan (c+d x))^2}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d}\)

\(\Big \downarrow \) 4107

\(\displaystyle \frac {1}{2} \left (-2 \left (a^2 \left (a^2 A-4 a b B-6 A b^2\right ) \int \cot (c+d x)dx-\left (b^3 (4 a B+A b) \int \tan (c+d x)dx\right )-\frac {b^2 \left (a^2 B+3 a A b+b^2 B\right ) \tan (c+d x)}{d}+x \left (a^4 B+4 a^3 A b-6 a^2 b^2 B-4 a A b^3+b^4 B\right )\right )-\frac {a (2 a B+5 A b) \cot (c+d x) (a+b \tan (c+d x))^2}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (-2 \left (a^2 \left (a^2 A-4 a b B-6 A b^2\right ) \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-\left (b^3 (4 a B+A b) \int \tan (c+d x)dx\right )-\frac {b^2 \left (a^2 B+3 a A b+b^2 B\right ) \tan (c+d x)}{d}+x \left (a^4 B+4 a^3 A b-6 a^2 b^2 B-4 a A b^3+b^4 B\right )\right )-\frac {a (2 a B+5 A b) \cot (c+d x) (a+b \tan (c+d x))^2}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (-2 \left (-a^2 \left (a^2 A-4 a b B-6 A b^2\right ) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\left (b^3 (4 a B+A b) \int \tan (c+d x)dx\right )-\frac {b^2 \left (a^2 B+3 a A b+b^2 B\right ) \tan (c+d x)}{d}+x \left (a^4 B+4 a^3 A b-6 a^2 b^2 B-4 a A b^3+b^4 B\right )\right )-\frac {a (2 a B+5 A b) \cot (c+d x) (a+b \tan (c+d x))^2}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {1}{2} \left (-2 \left (-\frac {b^2 \left (a^2 B+3 a A b+b^2 B\right ) \tan (c+d x)}{d}+\frac {a^2 \left (a^2 A-4 a b B-6 A b^2\right ) \log (-\sin (c+d x))}{d}+x \left (a^4 B+4 a^3 A b-6 a^2 b^2 B-4 a A b^3+b^4 B\right )+\frac {b^3 (4 a B+A b) \log (\cos (c+d x))}{d}\right )-\frac {a (2 a B+5 A b) \cot (c+d x) (a+b \tan (c+d x))^2}{d}\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^3}{2 d}\)

input

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

output

-1/2*(a*A*Cot[c + d*x]^2*(a + b*Tan[c + d*x])^3)/d + (-((a*(5*A*b + 2*a*B) 
*Cot[c + d*x]*(a + b*Tan[c + d*x])^2)/d) - 2*((4*a^3*A*b - 4*a*A*b^3 + a^4 
*B - 6*a^2*b^2*B + b^4*B)*x + (b^3*(A*b + 4*a*B)*Log[Cos[c + d*x]])/d + (a 
^2*(a^2*A - 6*A*b^2 - 4*a*b*B)*Log[-Sin[c + d*x]])/d - (b^2*(3*a*A*b + a^2 
*B + b^2*B)*Tan[c + d*x])/d))/2
 

3.3.62.3.1 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 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]
 

rule 4128

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[(A*d^2 + c*(c*C - B*d))*(a + b*Tan[e + 
 f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Sim 
p[1/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e 
 + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c*m + a*d* 
(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b 
*(d*(B*c - A*d)*(m + n + 1) - C*(c^2*m - d^2*(n + 1)))*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 
[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

3.3.62.4 Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.90

method	result	size
parallelrisch	\(\frac {\left (A \,a^{4}-6 A \,a^{2} b^{2}+A \,b^{4}-4 B \,a^{3} b +4 B a \,b^{3}\right ) \ln \left (\sec ^{2}\left (d x +c \right )\right )+\left (-2 A \,a^{4}+12 A \,a^{2} b^{2}+8 B \,a^{3} b \right ) \ln \left (\tan \left (d x +c \right )\right )-A \left (\cot ^{2}\left (d x +c \right )\right ) a^{4}+\left (-8 A \,a^{3} b -2 B \,a^{4}\right ) \cot \left (d x +c \right )+2 B \,b^{4} \tan \left (d x +c \right )-8 d \left (A \,a^{3} b -A a \,b^{3}+\frac {1}{4} B \,a^{4}-\frac {3}{2} B \,a^{2} b^{2}+\frac {1}{4} B \,b^{4}\right ) x}{2 d}\)	\(168\)
derivativedivides	\(\frac {B \,b^{4} \tan \left (d x +c \right )+\frac {\left (A \,a^{4}-6 A \,a^{2} b^{2}+A \,b^{4}-4 B \,a^{3} b +4 B a \,b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-4 A \,a^{3} b +4 A a \,b^{3}-B \,a^{4}+6 B \,a^{2} b^{2}-B \,b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {A \,a^{4}}{2 \tan \left (d x +c \right )^{2}}-\frac {a^{3} \left (4 A b +B a \right )}{\tan \left (d x +c \right )}-a^{2} \left (A \,a^{2}-6 A \,b^{2}-4 B a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(172\)
default	\(\frac {B \,b^{4} \tan \left (d x +c \right )+\frac {\left (A \,a^{4}-6 A \,a^{2} b^{2}+A \,b^{4}-4 B \,a^{3} b +4 B a \,b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-4 A \,a^{3} b +4 A a \,b^{3}-B \,a^{4}+6 B \,a^{2} b^{2}-B \,b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {A \,a^{4}}{2 \tan \left (d x +c \right )^{2}}-\frac {a^{3} \left (4 A b +B a \right )}{\tan \left (d x +c \right )}-a^{2} \left (A \,a^{2}-6 A \,b^{2}-4 B a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(172\)
norman	\(\frac {\left (-4 A \,a^{3} b +4 A a \,b^{3}-B \,a^{4}+6 B \,a^{2} b^{2}-B \,b^{4}\right ) x \left (\tan ^{2}\left (d x +c \right )\right )+\frac {B \,b^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{d}-\frac {A \,a^{4}}{2 d}-\frac {a^{3} \left (4 A b +B a \right ) \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}+\frac {\left (A \,a^{4}-6 A \,a^{2} b^{2}+A \,b^{4}-4 B \,a^{3} b +4 B a \,b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{2} \left (A \,a^{2}-6 A \,b^{2}-4 B a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(187\)
risch	\(\frac {2 i A \,b^{4} c}{d}+i A \,a^{4} x +4 i B a \,b^{3} x -\frac {A \,a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A \,b^{4}}{d}+\frac {6 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A \,b^{2}}{d}+\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}-4 A \,a^{3} b x +4 A a \,b^{3} x +6 B \,a^{2} b^{2} x +\frac {2 i \left (-i A \,a^{4} {\mathrm e}^{4 i \left (d x +c \right )}-4 A \,a^{3} b \,{\mathrm e}^{4 i \left (d x +c \right )}-B \,a^{4} {\mathrm e}^{4 i \left (d x +c \right )}+B \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-i A \,a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-2 B \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+4 A \,a^{3} b +B \,a^{4}+B \,b^{4}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {2 i a^{4} A c}{d}-B \,a^{4} x -B \,b^{4} x -\frac {12 i A \,a^{2} b^{2} c}{d}+\frac {8 i B a \,b^{3} c}{d}+i A \,b^{4} x -\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B a \,b^{3}}{d}-6 i A \,a^{2} b^{2} x -4 i B \,a^{3} b x -\frac {8 i B \,a^{3} b c}{d}\)	\(399\)

input

int(cot(d*x+c)^3*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x,method=_RETURNVERBO 
SE)
 

output

1/2*((A*a^4-6*A*a^2*b^2+A*b^4-4*B*a^3*b+4*B*a*b^3)*ln(sec(d*x+c)^2)+(-2*A* 
a^4+12*A*a^2*b^2+8*B*a^3*b)*ln(tan(d*x+c))-A*cot(d*x+c)^2*a^4+(-8*A*a^3*b- 
2*B*a^4)*cot(d*x+c)+2*B*b^4*tan(d*x+c)-8*d*(A*a^3*b-A*a*b^3+1/4*B*a^4-3/2* 
B*a^2*b^2+1/4*B*b^4)*x)/d
 

3.3.62.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.07 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {2 \, B b^{4} \tan \left (d x + c\right )^{3} - A a^{4} - {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} - {\left (4 \, B a b^{3} + A b^{4}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} - {\left (A a^{4} + 2 \, {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} d x\right )} \tan \left (d x + c\right )^{2} - 2 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )^{2}} \]

input

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="f 
ricas")
 

output

1/2*(2*B*b^4*tan(d*x + c)^3 - A*a^4 - (A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2)*lo 
g(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^2 - (4*B*a*b^3 + A*b^4 
)*log(1/(tan(d*x + c)^2 + 1))*tan(d*x + c)^2 - (A*a^4 + 2*(B*a^4 + 4*A*a^3 
*b - 6*B*a^2*b^2 - 4*A*a*b^3 + B*b^4)*d*x)*tan(d*x + c)^2 - 2*(B*a^4 + 4*A 
*a^3*b)*tan(d*x + c))/(d*tan(d*x + c)^2)
 

3.3.62.6 Sympy [A] (verification not implemented)

Time = 1.79 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.66 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } A a^{4} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right )^{4} \cot ^{3}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } A a^{4} x & \text {for}\: c = - d x \\\frac {A a^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {A a^{4} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {A a^{4}}{2 d \tan ^{2}{\left (c + d x \right )}} - 4 A a^{3} b x - \frac {4 A a^{3} b}{d \tan {\left (c + d x \right )}} - \frac {3 A a^{2} b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {6 A a^{2} b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 4 A a b^{3} x + \frac {A b^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - B a^{4} x - \frac {B a^{4}}{d \tan {\left (c + d x \right )}} - \frac {2 B a^{3} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {4 B a^{3} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 6 B a^{2} b^{2} x + \frac {2 B a b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - B b^{4} x + \frac {B b^{4} \tan {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases} \]

input

integrate(cot(d*x+c)**3*(a+b*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)
 

output

Piecewise((zoo*A*a**4*x, Eq(c, 0) & Eq(d, 0)), (x*(A + B*tan(c))*(a + b*ta 
n(c))**4*cot(c)**3, Eq(d, 0)), (zoo*A*a**4*x, Eq(c, -d*x)), (A*a**4*log(ta 
n(c + d*x)**2 + 1)/(2*d) - A*a**4*log(tan(c + d*x))/d - A*a**4/(2*d*tan(c 
+ d*x)**2) - 4*A*a**3*b*x - 4*A*a**3*b/(d*tan(c + d*x)) - 3*A*a**2*b**2*lo 
g(tan(c + d*x)**2 + 1)/d + 6*A*a**2*b**2*log(tan(c + d*x))/d + 4*A*a*b**3* 
x + A*b**4*log(tan(c + d*x)**2 + 1)/(2*d) - B*a**4*x - B*a**4/(d*tan(c + d 
*x)) - 2*B*a**3*b*log(tan(c + d*x)**2 + 1)/d + 4*B*a**3*b*log(tan(c + d*x) 
)/d + 6*B*a**2*b**2*x + 2*B*a*b**3*log(tan(c + d*x)**2 + 1)/d - B*b**4*x + 
 B*b**4*tan(c + d*x)/d, True))
 

3.3.62.7 Maxima [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.93 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {2 \, B b^{4} \tan \left (d x + c\right ) - 2 \, {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} {\left (d x + c\right )} + {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {A a^{4} + 2 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

input

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="m 
axima")
 

output

1/2*(2*B*b^4*tan(d*x + c) - 2*(B*a^4 + 4*A*a^3*b - 6*B*a^2*b^2 - 4*A*a*b^3 
 + B*b^4)*(d*x + c) + (A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2 + 4*B*a*b^3 + A*b^4 
)*log(tan(d*x + c)^2 + 1) - 2*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2)*log(tan(d* 
x + c)) - (A*a^4 + 2*(B*a^4 + 4*A*a^3*b)*tan(d*x + c))/tan(d*x + c)^2)/d
 

3.3.62.8 Giac [A] (verification not implemented)

Time = 1.39 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.20 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {2 \, B b^{4} \tan \left (d x + c\right ) - 2 \, {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} {\left (d x + c\right )} + {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) + \frac {3 \, A a^{4} \tan \left (d x + c\right )^{2} - 12 \, B a^{3} b \tan \left (d x + c\right )^{2} - 18 \, A a^{2} b^{2} \tan \left (d x + c\right )^{2} - 2 \, B a^{4} \tan \left (d x + c\right ) - 8 \, A a^{3} b \tan \left (d x + c\right ) - A a^{4}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

input

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="g 
iac")
 

output

1/2*(2*B*b^4*tan(d*x + c) - 2*(B*a^4 + 4*A*a^3*b - 6*B*a^2*b^2 - 4*A*a*b^3 
 + B*b^4)*(d*x + c) + (A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2 + 4*B*a*b^3 + A*b^4 
)*log(tan(d*x + c)^2 + 1) - 2*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2)*log(abs(ta 
n(d*x + c))) + (3*A*a^4*tan(d*x + c)^2 - 12*B*a^3*b*tan(d*x + c)^2 - 18*A* 
a^2*b^2*tan(d*x + c)^2 - 2*B*a^4*tan(d*x + c) - 8*A*a^3*b*tan(d*x + c) - A 
*a^4)/tan(d*x + c)^2)/d
 

3.3.62.9 Mupad [B] (verification not implemented)

Time = 7.40 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.80 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-A\,a^4+4\,B\,a^3\,b+6\,A\,a^2\,b^2\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^4+4\,A\,b\,a^3\right )+\frac {A\,a^4}{2}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^4}{2\,d}+\frac {B\,b^4\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^4}{2\,d} \]

input

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

output

(log(tan(c + d*x))*(6*A*a^2*b^2 - A*a^4 + 4*B*a^3*b))/d - (cot(c + d*x)^2* 
(tan(c + d*x)*(B*a^4 + 4*A*a^3*b) + (A*a^4)/2))/d + (log(tan(c + d*x) + 1i 
)*(A - B*1i)*(a*1i + b)^4)/(2*d) + (B*b^4*tan(c + d*x))/d + (log(tan(c + d 
*x) - 1i)*(A + B*1i)*(a*1i - b)^4)/(2*d)