\(\int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\) [273]

Optimal result

Integrand size = 31, antiderivative size = 137 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {(A b-a B) x}{a^2+b^2}+\frac {(A b-a B) \cot (c+d x)}{a^2 d}-\frac {A \cot ^2(c+d x)}{2 a d}-\frac {\left (a^2 A-A b^2+a b B\right ) \log (\sin (c+d x))}{a^3 d}-\frac {b^3 (A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{a^3 \left (a^2+b^2\right ) d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.98 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.19 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\frac {2 (A b-a B) \cot (c+d x)}{a^2}-\frac {A \cot ^2(c+d x)}{a}+\frac {(A+i B) \log (i-\tan (c+d x))}{a+i b}-\frac {2 \left (a^2 A-A b^2+a b B\right ) \log (\tan (c+d x))}{a^3}+\frac {(A-i B) \log (i+\tan (c+d x))}{a-i b}+\frac {2 b^3 (-A b+a B) \log (a+b \tan (c+d x))}{a^3 \left (a^2+b^2\right )}}{2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4092, 27, 3042, 4132, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4092

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

\(\Big \downarrow \) 4013

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

Input:

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

Output:

-1/2*(A*Cot[c + d*x]^2)/(a*d) - (-(((A*b - a*B)*Cot[c + d*x])/(a*d)) + (-( 
(a^2*(A*b - a*B)*x)/(a^2 + b^2)) + ((a^2*A - A*b^2 + a*b*B)*Log[-Sin[c + d 
*x]])/(a*d) + (b^3*(A*b - a*B)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/(a*(a 
^2 + b^2)*d))/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 = 0.29 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (a A +B b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (A b -B a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {A}{2 a \tan \left (d x +c \right )^{2}}-\frac {-A b +B a}{a^{2} \tan \left (d x +c \right )}+\frac {\left (-A \,a^{2}+A \,b^{2}-B a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{3}}-\frac {\left (A b -B a \right ) b^{3} \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) a^{3}}}{d}\)	\(152\)
default	\(\frac {\frac {\frac {\left (a A +B b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (A b -B a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {A}{2 a \tan \left (d x +c \right )^{2}}-\frac {-A b +B a}{a^{2} \tan \left (d x +c \right )}+\frac {\left (-A \,a^{2}+A \,b^{2}-B a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{3}}-\frac {\left (A b -B a \right ) b^{3} \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) a^{3}}}{d}\)	\(152\)
parallelrisch	\(\frac {\left (-2 A \,b^{4}+2 B a \,b^{3}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )+\left (A \,a^{4}+B \,a^{3} b \right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+\left (-2 A \,a^{4}+2 A \,b^{4}-2 B \,a^{3} b -2 B a \,b^{3}\right ) \ln \left (\tan \left (d x +c \right )\right )-\left (A a \cot \left (d x +c \right )^{2} \left (a^{2}+b^{2}\right )-2 \cot \left (d x +c \right ) \left (a^{2}+b^{2}\right ) \left (A b -B a \right )-2 a^{2} d x \left (A b -B a \right )\right ) a}{2 \left (a^{2}+b^{2}\right ) a^{3} d}\)	\(162\)
norman	\(\frac {\frac {\left (A b -B a \right ) \tan \left (d x +c \right )}{a^{2} d}+\frac {\left (A b -B a \right ) x \tan \left (d x +c \right )^{2}}{a^{2}+b^{2}}-\frac {A}{2 a d}}{\tan \left (d x +c \right )^{2}}+\frac {\left (a A +B b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \left (a^{2}+b^{2}\right )}-\frac {\left (A \,a^{2}-A \,b^{2}+B a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{3} d}-\frac {b^{3} \left (A b -B a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) a^{3} d}\)	\(171\)
risch	\(\frac {x B}{i b -a}+\frac {i x A}{i b -a}-\frac {2 i b^{3} B x}{a^{2} \left (a^{2}+b^{2}\right )}-\frac {2 i \left (i A a \,{\mathrm e}^{2 i \left (d x +c \right )}-A b \,{\mathrm e}^{2 i \left (d x +c \right )}+B a \,{\mathrm e}^{2 i \left (d x +c \right )}+A b -B a \right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 i b^{3} B c}{a^{2} d \left (a^{2}+b^{2}\right )}-\frac {2 i A \,b^{2} x}{a^{3}}+\frac {2 i B b x}{a^{2}}+\frac {2 i A c}{a d}+\frac {2 i B b c}{a^{2} d}+\frac {2 i b^{4} A c}{a^{3} d \left (a^{2}+b^{2}\right )}+\frac {2 i b^{4} A x}{a^{3} \left (a^{2}+b^{2}\right )}-\frac {2 i A \,b^{2} c}{a^{3} d}+\frac {2 i A x}{a}-\frac {A \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A \,b^{2}}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{a^{2} d}-\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) A}{a^{3} d \left (a^{2}+b^{2}\right )}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) B}{a^{2} d \left (a^{2}+b^{2}\right )}\)	\(415\)

Input:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.71 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {A a^{4} + A a^{2} b^{2} + {\left (A a^{4} + B a^{3} b + B a b^{3} - A b^{4}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} - {\left (B a b^{3} - A b^{4}\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + {\left (A a^{4} + A a^{2} b^{2} + 2 \, {\left (B a^{4} - A a^{3} b\right )} d x\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (B a^{4} - A a^{3} b + B a^{2} b^{2} - A a b^{3}\right )} \tan \left (d x + c\right )}{2 \, {\left (a^{5} + a^{3} b^{2}\right )} d \tan \left (d x + c\right )^{2}} \] Input:

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="fri 
cas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.24 (sec) , antiderivative size = 2599, normalized size of antiderivative = 18.97 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(cot(d*x+c)**3*(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*log(ta 
n(c + d*x)**2 + 1)/(2*d) - A*log(tan(c + d*x))/d - A/(2*d*tan(c + d*x)**2) 
 - B*x - B/(d*tan(c + d*x)))/a, Eq(b, 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))/b, Eq(a, 0)), (-3*I*A*d*x*tan(c + d*x)** 
3/(2*a*d*tan(c + d*x)**3 + 2*I*a*d*tan(c + d*x)**2) + 3*A*d*x*tan(c + d*x) 
**2/(2*a*d*tan(c + d*x)**3 + 2*I*a*d*tan(c + d*x)**2) + 2*A*log(tan(c + d* 
x)**2 + 1)*tan(c + d*x)**3/(2*a*d*tan(c + d*x)**3 + 2*I*a*d*tan(c + d*x)** 
2) + 2*I*A*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(2*a*d*tan(c + d*x)**3 
 + 2*I*a*d*tan(c + d*x)**2) - 4*A*log(tan(c + d*x))*tan(c + d*x)**3/(2*a*d 
*tan(c + d*x)**3 + 2*I*a*d*tan(c + d*x)**2) - 4*I*A*log(tan(c + d*x))*tan( 
c + d*x)**2/(2*a*d*tan(c + d*x)**3 + 2*I*a*d*tan(c + d*x)**2) - 3*I*A*tan( 
c + d*x)**2/(2*a*d*tan(c + d*x)**3 + 2*I*a*d*tan(c + d*x)**2) + A*tan(c + 
d*x)/(2*a*d*tan(c + d*x)**3 + 2*I*a*d*tan(c + d*x)**2) - I*A/(2*a*d*tan(c 
+ d*x)**3 + 2*I*a*d*tan(c + d*x)**2) - 3*B*d*x*tan(c + d*x)**3/(2*a*d*tan( 
c + d*x)**3 + 2*I*a*d*tan(c + d*x)**2) - 3*I*B*d*x*tan(c + d*x)**2/(2*a*d* 
tan(c + d*x)**3 + 2*I*a*d*tan(c + d*x)**2) - I*B*log(tan(c + d*x)**2 + 1)* 
tan(c + d*x)**3/(2*a*d*tan(c + d*x)**3 + 2*I*a*d*tan(c + d*x)**2) + B*log( 
tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(2*a*d*tan(c + d*x)**3 + 2*I*a*d*tan( 
c + d*x)**2) + 2*I*B*log(tan(c + d*x))*tan(c + d*x)**3/(2*a*d*tan(c + d...
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.15 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {\frac {2 \, {\left (B a - A b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} - \frac {2 \, {\left (B a b^{3} - A b^{4}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{5} + a^{3} b^{2}} - \frac {{\left (A a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {2 \, {\left (A a^{2} + B a b - A b^{2}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{3}} + \frac {A a + 2 \, {\left (B a - A b\right )} \tan \left (d x + c\right )}{a^{2} \tan \left (d x + c\right )^{2}}}{2 \, d} \] Input:

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="max 
ima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.30 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {{\left (B a - A b\right )} {\left (d x + c\right )}}{a^{2} d + b^{2} d} + \frac {{\left (A a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, {\left (a^{2} d + b^{2} d\right )}} + \frac {{\left (B a b^{4} - A b^{5}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{5} b d + a^{3} b^{3} d} - \frac {{\left (A a^{2} + B a b - A b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{3} d} - \frac {A a^{2} + 2 \, {\left (B a^{2} - A a b\right )} \tan \left (d x + c\right )}{2 \, a^{3} d \tan \left (d x + c\right )^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 5.49 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.28 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\frac {A}{2\,a}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,b-B\,a\right )}{a^2}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (-b+a\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,a^2+B\,a\,b-A\,b^2\right )}{a^3\,d}-\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,b^4-B\,a\,b^3\right )}{d\,\left (a^5+a^3\,b^2\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (a-b\,1{}\mathrm {i}\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.49 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+\sin \left (d x +c \right )^{2}-2}{4 \sin \left (d x +c \right )^{2} d} \] Input:

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

Output:

(4*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 - 4*log(tan((c + d*x)/2))* 
sin(c + d*x)**2 + sin(c + d*x)**2 - 2)/(4*sin(c + d*x)**2*d)