\(\int \cot ^4(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [254]

Optimal result

Integrand size = 31, antiderivative size = 154 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x+\frac {a \left (3 a^2 A-8 A b^2-9 a b B\right ) \cot (c+d x)}{3 d}-\frac {a^2 (5 A b+3 a B) \cot ^2(c+d x)}{6 d}-\frac {\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \log (\sin (c+d x))}{d}-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d} \] Output:

(A*a^3-3*A*a*b^2-3*B*a^2*b+B*b^3)*x+1/3*a*(3*A*a^2-8*A*b^2-9*B*a*b)*cot(d* 
x+c)/d-1/6*a^2*(5*A*b+3*B*a)*cot(d*x+c)^2/d-(3*A*a^2*b-A*b^3+B*a^3-3*B*a*b 
^2)*ln(sin(d*x+c))/d-1/3*a*A*cot(d*x+c)^3*(a+b*tan(d*x+c))^2/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.81 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.06 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {6 a \left (a^2 A-3 A b^2-3 a b B\right ) \cot (c+d x)-3 a^2 (3 A b+a B) \cot ^2(c+d x)-2 a^3 A \cot ^3(c+d x)+3 (a+i b)^3 (-i A+B) \log (i-\tan (c+d x))-6 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \log (\tan (c+d x))+3 (a-i b)^3 (i A+B) \log (i+\tan (c+d x))}{6 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05, 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, 4118, 25, 3042, 4111, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4088

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4118

\(\displaystyle \frac {1}{3} \left (\int -\cot ^2(c+d x) \left (b^2 (a A-3 b B) \tan ^2(c+d x)+3 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (3 A a^2-9 b B a-8 A b^2\right )\right )dx-\frac {a^2 (3 a B+5 A b) \cot ^2(c+d x)}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (-\int \cot ^2(c+d x) \left (b^2 (a A-3 b B) \tan ^2(c+d x)+3 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (3 A a^2-9 b B a-8 A b^2\right )\right )dx-\frac {a^2 (3 a B+5 A b) \cot ^2(c+d x)}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (-\int \frac {b^2 (a A-3 b B) \tan (c+d x)^2+3 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (3 A a^2-9 b B a-8 A b^2\right )}{\tan (c+d x)^2}dx-\frac {a^2 (3 a B+5 A b) \cot ^2(c+d x)}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4111

\(\displaystyle \frac {1}{3} \left (-\int 3 \cot (c+d x) \left (B a^3+3 A b a^2-3 b^2 B a-A b^3-\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)\right )dx+\frac {a \left (3 a^2 A-9 a b B-8 A b^2\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a B+5 A b) \cot ^2(c+d x)}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (-3 \int \cot (c+d x) \left (B a^3+3 A b a^2-3 b^2 B a-A b^3-\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)\right )dx+\frac {a \left (3 a^2 A-9 a b B-8 A b^2\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a B+5 A b) \cot ^2(c+d x)}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (-3 \int \frac {B a^3+3 A b a^2-3 b^2 B a-A b^3-\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)}{\tan (c+d x)}dx+\frac {a \left (3 a^2 A-9 a b B-8 A b^2\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a B+5 A b) \cot ^2(c+d x)}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4014

\(\displaystyle \frac {1}{3} \left (-3 \left (\left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \int \cot (c+d x)dx-x \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )\right )+\frac {a \left (3 a^2 A-9 a b B-8 A b^2\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a B+5 A b) \cot ^2(c+d x)}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (-3 \left (\left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-x \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )\right )+\frac {a \left (3 a^2 A-9 a b B-8 A b^2\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a B+5 A b) \cot ^2(c+d x)}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (-3 \left (-\left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\left (x \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )\right )\right )+\frac {a \left (3 a^2 A-9 a b B-8 A b^2\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a B+5 A b) \cot ^2(c+d x)}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {1}{3} \left (\frac {a \left (3 a^2 A-9 a b B-8 A b^2\right ) \cot (c+d x)}{d}-\frac {a^2 (3 a B+5 A b) \cot ^2(c+d x)}{2 d}-3 \left (\frac {\left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \log (-\sin (c+d x))}{d}-x \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )\right )\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\)

Input:

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

Output:

((a*(3*a^2*A - 8*A*b^2 - 9*a*b*B)*Cot[c + d*x])/d - (a^2*(5*A*b + 3*a*B)*C 
ot[c + d*x]^2)/(2*d) - 3*(-((a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*x) + ( 
(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Log[-Sin[c + d*x]])/d))/3 - (a*A*C 
ot[c + d*x]^3*(a + b*Tan[c + d*x])^2)/(3*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 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 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 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 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 = 0.24 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.12

method	result	size
parallelrisch	\(\frac {3 \left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+6 \left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )-2 A \cot \left (d x +c \right )^{3} a^{3}+3 \left (-3 A \,a^{2} b -B \,a^{3}\right ) \cot \left (d x +c \right )^{2}+6 a \cot \left (d x +c \right ) \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right )+6 d x \left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right )}{6 d}\)	\(172\)
derivativedivides	\(\frac {\frac {\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )+\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {A \,a^{3}}{3 \tan \left (d x +c \right )^{3}}-\frac {a^{2} \left (3 A b +B a \right )}{2 \tan \left (d x +c \right )^{2}}+\frac {a \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right )}{\tan \left (d x +c \right )}}{d}\)	\(174\)
default	\(\frac {\frac {\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )+\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {A \,a^{3}}{3 \tan \left (d x +c \right )^{3}}-\frac {a^{2} \left (3 A b +B a \right )}{2 \tan \left (d x +c \right )^{2}}+\frac {a \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right )}{\tan \left (d x +c \right )}}{d}\)	\(174\)
norman	\(\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) x \tan \left (d x +c \right )^{3}+\frac {a \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right ) \tan \left (d x +c \right )^{2}}{d}-\frac {A \,a^{3}}{3 d}-\frac {a^{2} \left (3 A b +B a \right ) \tan \left (d x +c \right )}{2 d}}{\tan \left (d x +c \right )^{3}}-\frac {\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\)	\(188\)
risch	\(A \,a^{3} x -3 A a \,b^{2} x -3 B \,a^{2} b x +B \,b^{3} x -\frac {2 i A \,b^{3} c}{d}+3 i A \,a^{2} b x +\frac {2 i B \,a^{3} c}{d}-3 i B a \,b^{2} x -\frac {6 i B a \,b^{2} c}{d}+i B \,a^{3} x -i A \,b^{3} x +\frac {6 i A \,a^{2} b c}{d}-\frac {2 i a \left (-6 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}+9 i A a b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 i B \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-18 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-18 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-9 i A a b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 i B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-4 A \,a^{2}+9 A \,b^{2}+9 B a b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A b}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A \,b^{3}}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B a \,b^{2}}{d}\)	\(383\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.18 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {3 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} + 2 \, A a^{3} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b - 2 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{3} - 6 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{6 \, d \tan \left (d x + c\right )^{3}} \] Input:

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

Output:

-1/6*(3*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*log(tan(d*x + c)^2/(tan(d* 
x + c)^2 + 1))*tan(d*x + c)^3 + 2*A*a^3 + 3*(B*a^3 + 3*A*a^2*b - 2*(A*a^3 
- 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*d*x)*tan(d*x + c)^3 - 6*(A*a^3 - 3*B*a^2* 
b - 3*A*a*b^2)*tan(d*x + c)^2 + 3*(B*a^3 + 3*A*a^2*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. 332 vs. \(2 (150) = 300\).

Time = 2.06 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.16 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } A a^{3} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right )^{3} \cot ^{4}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } A a^{3} x & \text {for}\: c = - d x \\A a^{3} x + \frac {A a^{3}}{d \tan {\left (c + d x \right )}} - \frac {A a^{3}}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {3 A a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {3 A a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {3 A a^{2} b}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 A a b^{2} x - \frac {3 A a b^{2}}{d \tan {\left (c + d x \right )}} - \frac {A b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {A b^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {B a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 B a^{2} b x - \frac {3 B a^{2} b}{d \tan {\left (c + d x \right )}} - \frac {3 B a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 B a b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + B b^{3} x & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.17 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {6 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} {\left (d x + c\right )} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {2 \, A a^{3} - 6 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{3}}}{6 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} {\left (d x + c\right )}}{d} + \frac {{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} - \frac {{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {2 \, A a^{3} - 6 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{6 \, d \tan \left (d x + c\right )^{3}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 3.55 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.10 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B\,a^3-3\,A\,a^2\,b+3\,B\,a\,b^2+A\,b^3\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^3}{2}+\frac {3\,A\,b\,a^2}{2}\right )+\frac {A\,a^3}{3}+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-A\,a^3+3\,B\,a^2\,b+3\,A\,a\,b^2\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,{\left (a-b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 19.12 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.58 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{4}-18 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}-\cos \left (d x +c \right ) a^{4}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} a^{3} b -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} a \,b^{3}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a^{3} b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a \,b^{3}+3 \sin \left (d x +c \right )^{3} a^{4} d x +3 \sin \left (d x +c \right )^{3} a^{3} b -18 \sin \left (d x +c \right )^{3} a^{2} b^{2} d x +3 \sin \left (d x +c \right )^{3} b^{4} d x -6 \sin \left (d x +c \right ) a^{3} b}{3 \sin \left (d x +c \right )^{3} d} \] Input:

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

Output:

(4*cos(c + d*x)*sin(c + d*x)**2*a**4 - 18*cos(c + d*x)*sin(c + d*x)**2*a** 
2*b**2 - cos(c + d*x)*a**4 + 12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)* 
*3*a**3*b - 12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**3*a*b**3 - 12*lo 
g(tan((c + d*x)/2))*sin(c + d*x)**3*a**3*b + 12*log(tan((c + d*x)/2))*sin( 
c + d*x)**3*a*b**3 + 3*sin(c + d*x)**3*a**4*d*x + 3*sin(c + d*x)**3*a**3*b 
 - 18*sin(c + d*x)**3*a**2*b**2*d*x + 3*sin(c + d*x)**3*b**4*d*x - 6*sin(c 
 + d*x)*a**3*b)/(3*sin(c + d*x)**3*d)