\(\int \frac {c+d x}{a+b \cot (e+f x)} \, dx\) [54]

Optimal result

Integrand size = 18, antiderivative size = 126 \[ \int \frac {c+d x}{a+b \cot (e+f x)} \, dx=\frac {(c+d x)^2}{2 (a-i b) d}-\frac {b (c+d x) \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{\left (a^2+b^2\right ) f}+\frac {i b d \operatorname {PolyLog}\left (2,\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 \left (a^2+b^2\right ) f^2} \] Output:

1/2*(d*x+c)^2/(a-I*b)/d-b*(d*x+c)*ln(1-(a+I*b)*exp(2*I*(f*x+e))/(a-I*b))/( 
a^2+b^2)/f+1/2*I*b*d*polylog(2,(a+I*b)*exp(2*I*(f*x+e))/(a-I*b))/(a^2+b^2) 
/f^2
 

Mathematica [A] (warning: unable to verify)

Time = 1.46 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.44 \[ \int \frac {c+d x}{a+b \cot (e+f x)} \, dx=\frac {1}{2} b \left (\frac {2 i (c+d x)^2}{(a+i b) d \left (a \left (-1+e^{2 i e}\right )+i b \left (1+e^{2 i e}\right )\right )}-\frac {2 (c+d x) \log \left (1+\frac {(-a+i b) e^{-2 i (e+f x)}}{a+i b}\right )}{\left (a^2+b^2\right ) f}-\frac {i d \operatorname {PolyLog}\left (2,\frac {(a-i b) e^{-2 i (e+f x)}}{a+i b}\right )}{\left (a^2+b^2\right ) f^2}\right )+\frac {x (2 c+d x) \sin (e)}{2 (b \cos (e)+a \sin (e))} \] Input:

Integrate[(c + d*x)/(a + b*Cot[e + f*x]),x]
 

Output:

(b*(((2*I)*(c + d*x)^2)/((a + I*b)*d*(a*(-1 + E^((2*I)*e)) + I*b*(1 + E^(( 
2*I)*e)))) - (2*(c + d*x)*Log[1 + (-a + I*b)/((a + I*b)*E^((2*I)*(e + f*x) 
))])/((a^2 + b^2)*f) - (I*d*PolyLog[2, (a - I*b)/((a + I*b)*E^((2*I)*(e + 
f*x)))])/((a^2 + b^2)*f^2)))/2 + (x*(2*c + d*x)*Sin[e])/(2*(b*Cos[e] + a*S 
in[e]))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4214, 25, 2620, 2715, 2838}

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.

\(\displaystyle \int \frac {c+d x}{a+b \cot (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4214

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2620

\(\displaystyle 2 i b \left (\frac {i (c+d x) \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 f \left (a^2+b^2\right )}-\frac {i d \int \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )dx}{2 f \left (a^2+b^2\right )}\right )+\frac {(c+d x)^2}{2 d (a-i b)}\)

\(\Big \downarrow \) 2715

\(\displaystyle 2 i b \left (\frac {i (c+d x) \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 f \left (a^2+b^2\right )}-\frac {d \int e^{-2 i (e+f x)} \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )de^{2 i (e+f x)}}{4 f^2 \left (a^2+b^2\right )}\right )+\frac {(c+d x)^2}{2 d (a-i b)}\)

\(\Big \downarrow \) 2838

\(\displaystyle 2 i b \left (\frac {i (c+d x) \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 f \left (a^2+b^2\right )}+\frac {d \operatorname {PolyLog}\left (2,\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{4 f^2 \left (a^2+b^2\right )}\right )+\frac {(c+d x)^2}{2 d (a-i b)}\)

Input:

Int[(c + d*x)/(a + b*Cot[e + f*x]),x]
 

Output:

(c + d*x)^2/(2*(a - I*b)*d) + (2*I)*b*(((I/2)*(c + d*x)*Log[1 - ((a + I*b) 
*E^((2*I)*(e + f*x)))/(a - I*b)])/((a^2 + b^2)*f) + (d*PolyLog[2, ((a + I* 
b)*E^((2*I)*(e + f*x)))/(a - I*b)])/(4*(a^2 + b^2)*f^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4214

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + Pi*(k_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(d*(m + 1)*(a + I*b)), x] + Simp 
[2*I*b   Int[(c + d*x)^m*E^(2*I*k*Pi)*(E^Simp[2*I*(e + f*x), x]/((a + I*b)^ 
2 + (a^2 + b^2)*E^(2*I*k*Pi)*E^Simp[2*I*(e + f*x), x])), x], x] /; FreeQ[{a 
, b, c, d, e, f}, x] && IntegerQ[4*k] && NeQ[a^2 + b^2, 0] && IGtQ[m, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 444 vs. \(2 (112 ) = 224\).

Time = 0.53 (sec) , antiderivative size = 445, normalized size of antiderivative = 3.53

method	result	size
risch	\(\frac {d \,x^{2}}{2 i b +2 a}+\frac {c x}{i b +a}+\frac {b c \ln \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+i b \,{\mathrm e}^{2 i \left (f x +e \right )}-a +i b \right )}{f \left (i b +a \right ) \left (i b -a \right )}-\frac {2 b c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f \left (i b +a \right ) \left (i b -a \right )}-\frac {b d \ln \left (1-\frac {\left (i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b +a}\right ) x}{f \left (i b +a \right ) \left (-i b +a \right )}+\frac {i b d \,x^{2}}{\left (i b +a \right ) \left (-i b +a \right )}-\frac {b d \ln \left (1-\frac {\left (i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b +a}\right ) e}{f^{2} \left (i b +a \right ) \left (-i b +a \right )}+\frac {2 i b d e x}{f \left (i b +a \right ) \left (-i b +a \right )}+\frac {i b d \,e^{2}}{f^{2} \left (i b +a \right ) \left (-i b +a \right )}+\frac {i b d \operatorname {polylog}\left (2, \frac {\left (i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b +a}\right )}{2 f^{2} \left (i b +a \right ) \left (-i b +a \right )}-\frac {b d e \ln \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+i b \,{\mathrm e}^{2 i \left (f x +e \right )}-a +i b \right )}{f^{2} \left (i b +a \right ) \left (i b -a \right )}+\frac {2 b d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2} \left (i b +a \right ) \left (i b -a \right )}\)	\(445\)

Input:

int((d*x+c)/(a+b*cot(f*x+e)),x,method=_RETURNVERBOSE)
 

Output:

1/2/(a+I*b)*d*x^2+1/(a+I*b)*c*x+1/f*b/(a+I*b)*c/(I*b-a)*ln(a*exp(2*I*(f*x+ 
e))+I*b*exp(2*I*(f*x+e))-a+I*b)-2/f*b/(a+I*b)*c/(I*b-a)*ln(exp(I*(f*x+e))) 
-1/f*b/(a+I*b)/(a-I*b)*d*ln(1-(a+I*b)*exp(2*I*(f*x+e))/(a-I*b))*x+I*b/(a+I 
*b)/(a-I*b)*d*x^2-1/f^2*b/(a+I*b)/(a-I*b)*d*ln(1-(a+I*b)*exp(2*I*(f*x+e))/ 
(a-I*b))*e+2*I/f*b/(a+I*b)/(a-I*b)*d*e*x+I/f^2*b/(a+I*b)/(a-I*b)*d*e^2+1/2 
*I/f^2*b/(a+I*b)/(a-I*b)*d*polylog(2,(a+I*b)*exp(2*I*(f*x+e))/(a-I*b))-1/f 
^2*b/(a+I*b)*d*e/(I*b-a)*ln(a*exp(2*I*(f*x+e))+I*b*exp(2*I*(f*x+e))-a+I*b) 
+2/f^2*b/(a+I*b)*d*e/(I*b-a)*ln(exp(I*(f*x+e)))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 475 vs. \(2 (105) = 210\).

Time = 0.10 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.77 \[ \int \frac {c+d x}{a+b \cot (e+f x)} \, dx=\frac {2 \, a d f^{2} x^{2} + 4 \, a c f^{2} x + i \, b d {\rm Li}_2\left (-\frac {a^{2} + b^{2} - {\left (a^{2} + 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (-i \, a^{2} + 2 \, a b + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}} + 1\right ) - i \, b d {\rm Li}_2\left (-\frac {a^{2} + b^{2} - {\left (a^{2} - 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (i \, a^{2} + 2 \, a b - i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}} + 1\right ) + 2 \, {\left (b d e - b c f\right )} \log \left (\frac {1}{2} \, a^{2} + i \, a b - \frac {1}{2} \, b^{2} - \frac {1}{2} \, {\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + \frac {1}{2} \, {\left (i \, a^{2} + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )\right ) + 2 \, {\left (b d e - b c f\right )} \log \left (-\frac {1}{2} \, a^{2} + i \, a b + \frac {1}{2} \, b^{2} + \frac {1}{2} \, {\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + \frac {1}{2} \, {\left (i \, a^{2} + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )\right ) - 2 \, {\left (b d f x + b d e\right )} \log \left (\frac {a^{2} + b^{2} - {\left (a^{2} + 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (-i \, a^{2} + 2 \, a b + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) - 2 \, {\left (b d f x + b d e\right )} \log \left (\frac {a^{2} + b^{2} - {\left (a^{2} - 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (i \, a^{2} + 2 \, a b - i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right )}{4 \, {\left (a^{2} + b^{2}\right )} f^{2}} \] Input:

integrate((d*x+c)/(a+b*cot(f*x+e)),x, algorithm="fricas")
 

Output:

1/4*(2*a*d*f^2*x^2 + 4*a*c*f^2*x + I*b*d*dilog(-(a^2 + b^2 - (a^2 + 2*I*a* 
b - b^2)*cos(2*f*x + 2*e) + (-I*a^2 + 2*a*b + I*b^2)*sin(2*f*x + 2*e))/(a^ 
2 + b^2) + 1) - I*b*d*dilog(-(a^2 + b^2 - (a^2 - 2*I*a*b - b^2)*cos(2*f*x 
+ 2*e) + (I*a^2 + 2*a*b - I*b^2)*sin(2*f*x + 2*e))/(a^2 + b^2) + 1) + 2*(b 
*d*e - b*c*f)*log(1/2*a^2 + I*a*b - 1/2*b^2 - 1/2*(a^2 + b^2)*cos(2*f*x + 
2*e) + 1/2*(I*a^2 + I*b^2)*sin(2*f*x + 2*e)) + 2*(b*d*e - b*c*f)*log(-1/2* 
a^2 + I*a*b + 1/2*b^2 + 1/2*(a^2 + b^2)*cos(2*f*x + 2*e) + 1/2*(I*a^2 + I* 
b^2)*sin(2*f*x + 2*e)) - 2*(b*d*f*x + b*d*e)*log((a^2 + b^2 - (a^2 + 2*I*a 
*b - b^2)*cos(2*f*x + 2*e) + (-I*a^2 + 2*a*b + I*b^2)*sin(2*f*x + 2*e))/(a 
^2 + b^2)) - 2*(b*d*f*x + b*d*e)*log((a^2 + b^2 - (a^2 - 2*I*a*b - b^2)*co 
s(2*f*x + 2*e) + (I*a^2 + 2*a*b - I*b^2)*sin(2*f*x + 2*e))/(a^2 + b^2)))/( 
(a^2 + b^2)*f^2)
 

Sympy [F]

\[ \int \frac {c+d x}{a+b \cot (e+f x)} \, dx=\int \frac {c + d x}{a + b \cot {\left (e + f x \right )}}\, dx \] Input:

integrate((d*x+c)/(a+b*cot(f*x+e)),x)
 

Output:

Integral((c + d*x)/(a + b*cot(e + f*x)), x)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 406 vs. \(2 (105) = 210\).

Time = 0.29 (sec) , antiderivative size = 406, normalized size of antiderivative = 3.22 \[ \int \frac {c+d x}{a+b \cot (e+f x)} \, dx=\frac {{\left (a + i \, b\right )} d f^{2} x^{2} + 2 \, {\left (a + i \, b\right )} c f^{2} x - 2 i \, b d f x \arctan \left (-\frac {2 \, a b \cos \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} - b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, f x + 2 \, e\right ) + a^{2} + b^{2} - {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) - b d f x \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} - 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) - 2 i \, b c f \arctan \left (b \cos \left (2 \, f x + 2 \, e\right ) + a \sin \left (2 \, f x + 2 \, e\right ) + b, a \cos \left (2 \, f x + 2 \, e\right ) - b \sin \left (2 \, f x + 2 \, e\right ) - a\right ) - b c f \log \left ({\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} - 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right ) + i \, b d {\rm Li}_2\left (\frac {{\left (i \, a - b\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}{i \, a + b}\right )}{2 \, {\left (a^{2} + b^{2}\right )} f^{2}} \] Input:

integrate((d*x+c)/(a+b*cot(f*x+e)),x, algorithm="maxima")
 

Output:

1/2*((a + I*b)*d*f^2*x^2 + 2*(a + I*b)*c*f^2*x - 2*I*b*d*f*x*arctan2(-(2*a 
*b*cos(2*f*x + 2*e) + (a^2 - b^2)*sin(2*f*x + 2*e))/(a^2 + b^2), (2*a*b*si 
n(2*f*x + 2*e) + a^2 + b^2 - (a^2 - b^2)*cos(2*f*x + 2*e))/(a^2 + b^2)) - 
b*d*f*x*log(((a^2 + b^2)*cos(2*f*x + 2*e)^2 + 4*a*b*sin(2*f*x + 2*e) + (a^ 
2 + b^2)*sin(2*f*x + 2*e)^2 + a^2 + b^2 - 2*(a^2 - b^2)*cos(2*f*x + 2*e))/ 
(a^2 + b^2)) - 2*I*b*c*f*arctan2(b*cos(2*f*x + 2*e) + a*sin(2*f*x + 2*e) + 
 b, a*cos(2*f*x + 2*e) - b*sin(2*f*x + 2*e) - a) - b*c*f*log((a^2 + b^2)*c 
os(2*f*x + 2*e)^2 + 4*a*b*sin(2*f*x + 2*e) + (a^2 + b^2)*sin(2*f*x + 2*e)^ 
2 + a^2 + b^2 - 2*(a^2 - b^2)*cos(2*f*x + 2*e)) + I*b*d*dilog((I*a - b)*e^ 
(2*I*f*x + 2*I*e)/(I*a + b)))/((a^2 + b^2)*f^2)
 

Giac [F]

\[ \int \frac {c+d x}{a+b \cot (e+f x)} \, dx=\int { \frac {d x + c}{b \cot \left (f x + e\right ) + a} \,d x } \] Input:

integrate((d*x+c)/(a+b*cot(f*x+e)),x, algorithm="giac")
 

Output:

integrate((d*x + c)/(b*cot(f*x + e) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x}{a+b \cot (e+f x)} \, dx=\int \frac {c+d\,x}{a+b\,\mathrm {cot}\left (e+f\,x\right )} \,d x \] Input:

int((c + d*x)/(a + b*cot(e + f*x)),x)
                                                                                    
                                                                                    
 

Output:

int((c + d*x)/(a + b*cot(e + f*x)), x)
 

Reduce [F]

\[ \int \frac {c+d x}{a+b \cot (e+f x)} \, dx=\frac {\left (\int \frac {x}{\cot \left (f x +e \right ) b +a}d x \right ) a^{2} d f +\left (\int \frac {x}{\cot \left (f x +e \right ) b +a}d x \right ) b^{2} d f +\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) b c -\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b -2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a -b \right ) b c +a c f x}{f \left (a^{2}+b^{2}\right )} \] Input:

int((d*x+c)/(a+b*cot(f*x+e)),x)
 

Output:

(int(x/(cot(e + f*x)*b + a),x)*a**2*d*f + int(x/(cot(e + f*x)*b + a),x)*b* 
*2*d*f + log(tan((e + f*x)/2)**2 + 1)*b*c - log(tan((e + f*x)/2)**2*b - 2* 
tan((e + f*x)/2)*a - b)*b*c + a*c*f*x)/(f*(a**2 + b**2))