\(\int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\) [274]

Optimal result

Integrand size = 31, antiderivative size = 169 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {(a A+b B) x}{a^2+b^2}+\frac {\left (a^2 A-A b^2+a b B\right ) \cot (c+d x)}{a^3 d}+\frac {(A b-a B) \cot ^2(c+d x)}{2 a^2 d}-\frac {A \cot ^3(c+d x)}{3 a d}+\frac {\left (a^2-b^2\right ) (A b-a B) \log (\sin (c+d x))}{a^4 d}+\frac {b^4 (A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{a^4 \left (a^2+b^2\right ) d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.94 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.15 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\frac {6 \left (a^2 A-A b^2+a b B\right ) \cot (c+d x)}{a^3}+\frac {3 (A b-a B) \cot ^2(c+d x)}{a^2}-\frac {2 A \cot ^3(c+d x)}{a}+\frac {3 (-i A+B) \log (i-\tan (c+d x))}{a+i b}+\frac {6 (a-b) (a+b) (A b-a B) \log (\tan (c+d x))}{a^4}+\frac {3 (i A+B) \log (i+\tan (c+d x))}{a-i b}+\frac {6 b^4 (A b-a B) \log (a+b \tan (c+d x))}{a^4 \left (a^2+b^2\right )}}{6 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.47 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.14, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4092, 27, 3042, 4132, 27, 3042, 4132, 3042, 4134, 3042, 25, 3956, 4013}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4092

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

\(\Big \downarrow \) 4013

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

Input:

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

Output:

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

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 4013

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)* 
(x_)]), x_Symbol] :> Simp[(c/(b*f))*Log[RemoveContent[a*Cos[e + f*x] + b*Si 
n[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] && EqQ[a*c + b*d, 0]
 

rule 4092

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*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1) 
/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^ 
2 + b^2))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b* 
B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2 
)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n 
+ 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && 
 NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] 
&& (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] 
 || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4132

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*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[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[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4134

Int[((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^ 
2)/(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) 
*(x_)])), x_Symbol] :> Simp[(a*(A*c - c*C + B*d) + b*(B*c - A*d + C*d))*(x/ 
((a^2 + b^2)*(c^2 + d^2))), x] + (Simp[(A*b^2 - a*b*B + a^2*C)/((b*c - a*d) 
*(a^2 + b^2))   Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] - Sim 
p[(c^2*C - B*c*d + A*d^2)/((b*c - a*d)*(c^2 + d^2))   Int[(d - c*Tan[e + f* 
x])/(c + d*Tan[e + f*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]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (-A b +B a \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (a A +B b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {-A \,a^{2}+A \,b^{2}-B a b}{a^{3} \tan \left (d x +c \right )}+\frac {\left (A \,a^{2} b -A \,b^{3}-B \,a^{3}+B a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{4}}-\frac {A}{3 a \tan \left (d x +c \right )^{3}}-\frac {-A b +B a}{2 a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (A b -B a \right ) b^{4} \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) a^{4}}}{d}\)	\(189\)
default	\(\frac {\frac {\frac {\left (-A b +B a \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (a A +B b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {-A \,a^{2}+A \,b^{2}-B a b}{a^{3} \tan \left (d x +c \right )}+\frac {\left (A \,a^{2} b -A \,b^{3}-B \,a^{3}+B a \,b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{4}}-\frac {A}{3 a \tan \left (d x +c \right )^{3}}-\frac {-A b +B a}{2 a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (A b -B a \right ) b^{4} \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) a^{4}}}{d}\)	\(189\)
parallelrisch	\(\frac {\left (6 A \,b^{5}-6 B a \,b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )+\left (-3 A \,a^{4} b +3 B \,a^{5}\right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+\left (6 A \,a^{4} b -6 A \,b^{5}-6 B \,a^{5}+6 B a \,b^{4}\right ) \ln \left (\tan \left (d x +c \right )\right )+6 \left (-\frac {A \,a^{2} \cot \left (d x +c \right )^{3} \left (a^{2}+b^{2}\right )}{3}+\frac {\cot \left (d x +c \right )^{2} a \left (a^{2}+b^{2}\right ) \left (A b -B a \right )}{2}+\cot \left (d x +c \right ) \left (a^{2}+b^{2}\right ) \left (A \,a^{2}-A \,b^{2}+B a b \right )+a^{3} d x \left (a A +B b \right )\right ) a}{6 \left (a^{2}+b^{2}\right ) a^{4} d}\)	\(198\)
norman	\(\frac {\frac {\left (a A +B b \right ) x \tan \left (d x +c \right )^{3}}{a^{2}+b^{2}}+\frac {\left (A \,a^{2}-A \,b^{2}+B a b \right ) \tan \left (d x +c \right )^{2}}{a^{3} d}-\frac {A}{3 a d}+\frac {\left (A b -B a \right ) \tan \left (d x +c \right )}{2 a^{2} d}}{\tan \left (d x +c \right )^{3}}+\frac {\left (A b -B a \right ) \left (a^{2}-b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{4} d}+\frac {b^{4} \left (A b -B a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) a^{4} d}-\frac {\left (A b -B a \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \left (a^{2}+b^{2}\right )}\)	\(202\)
risch	\(-\frac {2 i b^{5} A c}{\left (a^{2}+b^{2}\right ) a^{4} d}-\frac {x A}{i b -a}-\frac {2 i B \,b^{2} c}{a^{3} d}+\frac {2 i A \,b^{3} c}{a^{4} d}+\frac {2 i b^{4} B c}{\left (a^{2}+b^{2}\right ) a^{3} d}-\frac {2 i A b x}{a^{2}}+\frac {2 i x B}{a}+\frac {2 i A \,b^{3} x}{a^{4}}-\frac {2 i A b c}{a^{2} d}+\frac {i x B}{i b -a}+\frac {2 i b^{4} B x}{\left (a^{2}+b^{2}\right ) a^{3}}-\frac {2 i \left (-3 i A a b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 i B \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-3 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 i A a b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 i B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+6 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+6 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 A \,a^{2}+3 A \,b^{2}-3 B a b \right )}{3 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {2 i b^{5} A x}{\left (a^{2}+b^{2}\right ) a^{4}}+\frac {2 i B c}{a d}-\frac {2 i B \,b^{2} x}{a^{3}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A b}{a^{2} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A \,b^{3}}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{a d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B \,b^{2}}{a^{3} d}+\frac {b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) A}{\left (a^{2}+b^{2}\right ) a^{4} d}-\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) B}{\left (a^{2}+b^{2}\right ) a^{3} d}\)	\(585\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.73 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {2 \, A a^{5} + 2 \, A a^{3} b^{2} + 3 \, {\left (B a^{5} - A a^{4} b - B a b^{4} + A b^{5}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} + 3 \, {\left (B a b^{4} - A b^{5}\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} + 3 \, {\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A a^{2} b^{3} - 2 \, {\left (A a^{5} + B a^{4} b\right )} d x\right )} \tan \left (d x + c\right )^{3} - 6 \, {\left (A a^{5} + B a^{4} b + B a^{2} b^{3} - A a b^{4}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A a^{2} b^{3}\right )} \tan \left (d x + c\right )}{6 \, {\left (a^{6} + a^{4} b^{2}\right )} d \tan \left (d x + c\right )^{3}} \] Input:

integrate(cot(d*x+c)^4*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="fri 
cas")
 

Output:

-1/6*(2*A*a^5 + 2*A*a^3*b^2 + 3*(B*a^5 - A*a^4*b - B*a*b^4 + A*b^5)*log(ta 
n(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^3 + 3*(B*a*b^4 - A*b^5)*lo 
g((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1))*ta 
n(d*x + c)^3 + 3*(B*a^5 - A*a^4*b + B*a^3*b^2 - A*a^2*b^3 - 2*(A*a^5 + B*a 
^4*b)*d*x)*tan(d*x + c)^3 - 6*(A*a^5 + B*a^4*b + B*a^2*b^3 - A*a*b^4)*tan( 
d*x + c)^2 + 3*(B*a^5 - A*a^4*b + B*a^3*b^2 - A*a^2*b^3)*tan(d*x + c))/((a 
^6 + a^4*b^2)*d*tan(d*x + c)^3)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.67 (sec) , antiderivative size = 3016, normalized size of antiderivative = 17.85 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((zoo*A*x, Eq(a, 0) & Eq(b, 0) & Eq(c, 0) & Eq(d, 0)), ((A*x + A/ 
(d*tan(c + d*x)) - A/(3*d*tan(c + d*x)**3) + B*log(tan(c + d*x)**2 + 1)/(2 
*d) - B*log(tan(c + d*x))/d - B/(2*d*tan(c + d*x)**2))/a, Eq(b, 0)), ((-A* 
log(tan(c + d*x)**2 + 1)/(2*d) + A*log(tan(c + d*x))/d + A/(2*d*tan(c + d* 
x)**2) - A/(4*d*tan(c + d*x)**4) + B*x + B/(d*tan(c + d*x)) - B/(3*d*tan(c 
 + d*x)**3))/b, Eq(a, 0)), (15*A*d*x*tan(c + d*x)**4/(6*a*d*tan(c + d*x)** 
4 + 6*I*a*d*tan(c + d*x)**3) + 15*I*A*d*x*tan(c + d*x)**3/(6*a*d*tan(c + d 
*x)**4 + 6*I*a*d*tan(c + d*x)**3) + 6*I*A*log(tan(c + d*x)**2 + 1)*tan(c + 
 d*x)**4/(6*a*d*tan(c + d*x)**4 + 6*I*a*d*tan(c + d*x)**3) - 6*A*log(tan(c 
 + d*x)**2 + 1)*tan(c + d*x)**3/(6*a*d*tan(c + d*x)**4 + 6*I*a*d*tan(c + d 
*x)**3) - 12*I*A*log(tan(c + d*x))*tan(c + d*x)**4/(6*a*d*tan(c + d*x)**4 
+ 6*I*a*d*tan(c + d*x)**3) + 12*A*log(tan(c + d*x))*tan(c + d*x)**3/(6*a*d 
*tan(c + d*x)**4 + 6*I*a*d*tan(c + d*x)**3) + 15*A*tan(c + d*x)**3/(6*a*d* 
tan(c + d*x)**4 + 6*I*a*d*tan(c + d*x)**3) + 9*I*A*tan(c + d*x)**2/(6*a*d* 
tan(c + d*x)**4 + 6*I*a*d*tan(c + d*x)**3) + A*tan(c + d*x)/(6*a*d*tan(c + 
 d*x)**4 + 6*I*a*d*tan(c + d*x)**3) - 2*I*A/(6*a*d*tan(c + d*x)**4 + 6*I*a 
*d*tan(c + d*x)**3) - 9*I*B*d*x*tan(c + d*x)**4/(6*a*d*tan(c + d*x)**4 + 6 
*I*a*d*tan(c + d*x)**3) + 9*B*d*x*tan(c + d*x)**3/(6*a*d*tan(c + d*x)**4 + 
 6*I*a*d*tan(c + d*x)**3) + 6*B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**4/( 
6*a*d*tan(c + d*x)**4 + 6*I*a*d*tan(c + d*x)**3) + 6*I*B*log(tan(c + d*...
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.18 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\frac {6 \, {\left (A a + B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} - \frac {6 \, {\left (B a b^{4} - A b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + a^{4} b^{2}} + \frac {3 \, {\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac {6 \, {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{4}} - \frac {2 \, A a^{2} - 6 \, {\left (A a^{2} + B a b - A b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (B a^{2} - A a b\right )} \tan \left (d x + c\right )}{a^{3} \tan \left (d x + c\right )^{3}}}{6 \, d} \] Input:

integrate(cot(d*x+c)^4*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="max 
ima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.30 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {{\left (A a + B b\right )} {\left (d x + c\right )}}{a^{2} d + b^{2} d} + \frac {{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, {\left (a^{2} d + b^{2} d\right )}} - \frac {{\left (B a b^{5} - A b^{6}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b d + a^{4} b^{3} d} - \frac {{\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{4} d} - \frac {2 \, A a^{3} - 6 \, {\left (A a^{3} + B a^{2} b - A a b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (B a^{3} - A a^{2} b\right )} \tan \left (d x + c\right )}{6 \, a^{4} d \tan \left (d x + c\right )^{3}} \] Input:

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

Output:

(A*a + B*b)*(d*x + c)/(a^2*d + b^2*d) + 1/2*(B*a - A*b)*log(tan(d*x + c)^2 
 + 1)/(a^2*d + b^2*d) - (B*a*b^5 - A*b^6)*log(abs(b*tan(d*x + c) + a))/(a^ 
6*b*d + a^4*b^3*d) - (B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*log(abs(tan(d*x + 
 c)))/(a^4*d) - 1/6*(2*A*a^3 - 6*(A*a^3 + B*a^2*b - A*a*b^2)*tan(d*x + c)^ 
2 + 3*(B*a^3 - A*a^2*b)*tan(d*x + c))/(a^4*d*tan(d*x + c)^3)
 

Mupad [B] (verification not implemented)

Time = 5.76 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.23 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (A\,a^2+B\,a\,b-A\,b^2\right )}{a^3}-\frac {A}{3\,a}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,b-B\,a\right )}{2\,a^2}\right )}{d}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,b^5-B\,a\,b^4\right )}{d\,\left (a^6+a^4\,b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^3-A\,a^2\,b-B\,a\,b^2+A\,b^3\right )}{a^4\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (b+a\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (a+b\,1{}\mathrm {i}\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.17 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {-\cot \left (d x +c \right )^{3}+3 \cot \left (d x +c \right )+3 d x}{3 d} \] Input:

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

Output:

( - cot(c + d*x)**3 + 3*cot(c + d*x) + 3*d*x)/(3*d)