\(\int \frac {\cot ^2(c+d x)}{a+b \tan (c+d x)} \, dx\) [464]

Optimal result

Integrand size = 21, antiderivative size = 81 \[ \int \frac {\cot ^2(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {a x}{a^2+b^2}-\frac {\cot (c+d x)}{a d}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 \left (a^2+b^2\right ) d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.19 \[ \int \frac {\cot ^2(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {\frac {\cot (c+d x)}{a}-\frac {\log (i-\cot (c+d x))}{2 (i a+b)}+\frac {\log (i+\cot (c+d x))}{2 (i a-b)}-\frac {b^3 \log (b+a \cot (c+d x))}{a^2 \left (a^2+b^2\right )}}{d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4052, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x)^2 (a+b \tan (c+d x))}dx\)

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

\(\Big \downarrow \) 4013

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

Input:

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

Output:

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

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 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 4052

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c 
+ d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + Simp[1 
/((m + 1)*(a^2 + b^2)*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + 
d*Tan[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - 
 a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x], x], x] / 
; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] 
 && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || Integ 
erQ[m]) &&  !(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 = 1.00 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16

method	result	size
derivativedivides	\(\frac {-\frac {1}{a \tan \left (d x +c \right )}-\frac {b \ln \left (\tan \left (d x +c \right )\right )}{a^{2}}+\frac {\frac {b \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}-a \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}+\frac {b^{3} \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2} \left (a^{2}+b^{2}\right )}}{d}\)	\(94\)
default	\(\frac {-\frac {1}{a \tan \left (d x +c \right )}-\frac {b \ln \left (\tan \left (d x +c \right )\right )}{a^{2}}+\frac {\frac {b \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}-a \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}+\frac {b^{3} \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2} \left (a^{2}+b^{2}\right )}}{d}\)	\(94\)
parallelrisch	\(-\frac {2 a^{3} d x +2 b^{3} \left (\ln \left (\tan \left (d x +c \right )\right )-\ln \left (a +b \tan \left (d x +c \right )\right )\right )+a^{2} b \left (2 \ln \left (\tan \left (d x +c \right )\right )-\ln \left (\sec \left (d x +c \right )^{2}\right )\right )+2 \cot \left (d x +c \right ) a \left (a^{2}+b^{2}\right )}{2 a^{2} d \left (a^{2}+b^{2}\right )}\)	\(94\)
norman	\(\frac {-\frac {1}{a d}-\frac {a x \tan \left (d x +c \right )}{a^{2}+b^{2}}}{\tan \left (d x +c \right )}+\frac {b^{3} \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2} d \left (a^{2}+b^{2}\right )}-\frac {b \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}+\frac {b \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \left (a^{2}+b^{2}\right )}\)	\(111\)
risch	\(\frac {x}{i b -a}+\frac {2 i b x}{a^{2}}+\frac {2 i b c}{a^{2} d}-\frac {2 i b^{3} x}{a^{2} \left (a^{2}+b^{2}\right )}-\frac {2 i b^{3} c}{a^{2} d \left (a^{2}+b^{2}\right )}-\frac {2 i}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{2} d \left (a^{2}+b^{2}\right )}\)	\(165\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.73 \[ \int \frac {\cot ^2(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {2 \, a^{3} d x \tan \left (d x + c\right ) - b^{3} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) + 2 \, a^{3} + 2 \, a b^{2} + {\left (a^{2} b + b^{3}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )}{2 \, {\left (a^{4} + a^{2} b^{2}\right )} d \tan \left (d x + c\right )} \] Input:

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.44 (sec) , antiderivative size = 1080, normalized size of antiderivative = 13.33 \[ \int \frac {\cot ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(cot(d*x+c)**2/(a+b*tan(d*x+c)),x)
 

Output:

Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0) & Eq(c, 0) & Eq(d, 0)), ((-x - cot(c 
 + d*x)/d)/a, Eq(b, 0)), ((log(tan(c + d*x)**2 + 1)/(2*d) - log(tan(c + d* 
x))/d - 1/(2*d*tan(c + d*x)**2))/b, Eq(a, 0)), (-3*d*x*tan(c + d*x)**2/(2* 
a*d*tan(c + d*x)**2 + 2*I*a*d*tan(c + d*x)) - 3*I*d*x*tan(c + d*x)/(2*a*d* 
tan(c + d*x)**2 + 2*I*a*d*tan(c + d*x)) - I*log(tan(c + d*x)**2 + 1)*tan(c 
 + d*x)**2/(2*a*d*tan(c + d*x)**2 + 2*I*a*d*tan(c + d*x)) + log(tan(c + d* 
x)**2 + 1)*tan(c + d*x)/(2*a*d*tan(c + d*x)**2 + 2*I*a*d*tan(c + d*x)) + 2 
*I*log(tan(c + d*x))*tan(c + d*x)**2/(2*a*d*tan(c + d*x)**2 + 2*I*a*d*tan( 
c + d*x)) - 2*log(tan(c + d*x))*tan(c + d*x)/(2*a*d*tan(c + d*x)**2 + 2*I* 
a*d*tan(c + d*x)) - 3*tan(c + d*x)/(2*a*d*tan(c + d*x)**2 + 2*I*a*d*tan(c 
+ d*x)) - 2*I/(2*a*d*tan(c + d*x)**2 + 2*I*a*d*tan(c + d*x)), Eq(b, -I*a)) 
, (-3*d*x*tan(c + d*x)**2/(2*a*d*tan(c + d*x)**2 - 2*I*a*d*tan(c + d*x)) + 
 3*I*d*x*tan(c + d*x)/(2*a*d*tan(c + d*x)**2 - 2*I*a*d*tan(c + d*x)) + I*l 
og(tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(2*a*d*tan(c + d*x)**2 - 2*I*a*d*t 
an(c + d*x)) + log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a*d*tan(c + d*x)** 
2 - 2*I*a*d*tan(c + d*x)) - 2*I*log(tan(c + d*x))*tan(c + d*x)**2/(2*a*d*t 
an(c + d*x)**2 - 2*I*a*d*tan(c + d*x)) - 2*log(tan(c + d*x))*tan(c + d*x)/ 
(2*a*d*tan(c + d*x)**2 - 2*I*a*d*tan(c + d*x)) - 3*tan(c + d*x)/(2*a*d*tan 
(c + d*x)**2 - 2*I*a*d*tan(c + d*x)) + 2*I/(2*a*d*tan(c + d*x)**2 - 2*I*a* 
d*tan(c + d*x)), Eq(b, I*a)), (zoo*x/a, Eq(c, -d*x)), (x*cot(c)**2/(a +...
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.23 \[ \int \frac {\cot ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {2 \, b^{3} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + a^{2} b^{2}} - \frac {2 \, {\left (d x + c\right )} a}{a^{2} + b^{2}} + \frac {b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac {2 \, b \log \left (\tan \left (d x + c\right )\right )}{a^{2}} - \frac {2}{a \tan \left (d x + c\right )}}{2 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.42 \[ \int \frac {\cot ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {b^{4} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b d + a^{2} b^{3} d} - \frac {{\left (d x + c\right )} a}{a^{2} d + b^{2} d} + \frac {b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, {\left (a^{2} d + b^{2} d\right )}} - \frac {b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2} d} - \frac {1}{a d \tan \left (d x + c\right )} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.19 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.33 \[ \int \frac {\cot ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (b+a\,1{}\mathrm {i}\right )}-\frac {\mathrm {cot}\left (c+d\,x\right )}{a\,d}-\frac {b\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a^2\,d}+\frac {b^3\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{a^2\,d\,\left (a^2+b^2\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (a+b\,1{}\mathrm {i}\right )} \] Input:

int(cot(c + d*x)^2/(a + b*tan(c + d*x)),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.09 \[ \int \frac {\cot ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {-\cos \left (d x +c \right ) a^{3}-\cos \left (d x +c \right ) a \,b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right ) \sin \left (d x +c \right ) b^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a^{2} b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) b^{3}-\sin \left (d x +c \right ) a^{3} d x}{\sin \left (d x +c \right ) a^{2} d \left (a^{2}+b^{2}\right )} \] Input:

int(cot(d*x+c)^2/(a+b*tan(d*x+c)),x)
 

Output:

( - cos(c + d*x)*a**3 - cos(c + d*x)*a*b**2 + log(tan((c + d*x)/2)**2 + 1) 
*sin(c + d*x)*a**2*b + log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - 
a)*sin(c + d*x)*b**3 - log(tan((c + d*x)/2))*sin(c + d*x)*a**2*b - log(tan 
((c + d*x)/2))*sin(c + d*x)*b**3 - sin(c + d*x)*a**3*d*x)/(sin(c + d*x)*a* 
*2*d*(a**2 + b**2))