\(\int x^3 \cot (a+b x) \, dx\) [1]

Optimal result

Integrand size = 10, antiderivative size = 101 \[ \int x^3 \cot (a+b x) \, dx=-\frac {i x^4}{4}+\frac {x^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {3 i x^2 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}+\frac {3 x \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 i \operatorname {PolyLog}\left (4,e^{2 i (a+b x)}\right )}{4 b^4} \] Output:

-1/4*I*x^4+x^3*ln(1-exp(2*I*(b*x+a)))/b-3/2*I*x^2*polylog(2,exp(2*I*(b*x+a 
)))/b^2+3/2*x*polylog(3,exp(2*I*(b*x+a)))/b^3+3/4*I*polylog(4,exp(2*I*(b*x 
+a)))/b^4
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.82 \[ \int x^3 \cot (a+b x) \, dx=\frac {i b^4 x^4+4 b^3 x^3 \log \left (1-e^{-i (a+b x)}\right )+4 b^3 x^3 \log \left (1+e^{-i (a+b x)}\right )+12 i b^2 x^2 \operatorname {PolyLog}\left (2,-e^{-i (a+b x)}\right )+12 i b^2 x^2 \operatorname {PolyLog}\left (2,e^{-i (a+b x)}\right )+24 b x \operatorname {PolyLog}\left (3,-e^{-i (a+b x)}\right )+24 b x \operatorname {PolyLog}\left (3,e^{-i (a+b x)}\right )-24 i \operatorname {PolyLog}\left (4,-e^{-i (a+b x)}\right )-24 i \operatorname {PolyLog}\left (4,e^{-i (a+b x)}\right )}{4 b^4} \] Input:

Integrate[x^3*Cot[a + b*x],x]
 

Output:

(I*b^4*x^4 + 4*b^3*x^3*Log[1 - E^((-I)*(a + b*x))] + 4*b^3*x^3*Log[1 + E^( 
(-I)*(a + b*x))] + (12*I)*b^2*x^2*PolyLog[2, -E^((-I)*(a + b*x))] + (12*I) 
*b^2*x^2*PolyLog[2, E^((-I)*(a + b*x))] + 24*b*x*PolyLog[3, -E^((-I)*(a + 
b*x))] + 24*b*x*PolyLog[3, E^((-I)*(a + b*x))] - (24*I)*PolyLog[4, -E^((-I 
)*(a + b*x))] - (24*I)*PolyLog[4, E^((-I)*(a + b*x))])/(4*b^4)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.48, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 25, 4202, 2620, 3011, 7163, 2720, 7143}

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.

\(\displaystyle \int x^3 \cot (a+b x) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4202

\(\displaystyle 2 i \int \frac {e^{i (2 a+2 b x+\pi )} x^3}{1+e^{i (2 a+2 b x+\pi )}}dx-\frac {i x^4}{4}\)

\(\Big \downarrow \) 2620

\(\displaystyle 2 i \left (\frac {3 i \int x^2 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )dx}{2 b}-\frac {i x^3 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )-\frac {i x^4}{4}\)

\(\Big \downarrow \) 3011

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )dx}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )-\frac {i x^4}{4}\)

\(\Big \downarrow \) 7163

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {i \left (\frac {i \int \operatorname {PolyLog}\left (3,-e^{i (2 a+2 b x+\pi )}\right )dx}{2 b}-\frac {i x \operatorname {PolyLog}\left (3,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )-\frac {i x^4}{4}\)

\(\Big \downarrow \) 2720

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {i \left (\frac {\int e^{-i (2 a+2 b x+\pi )} \operatorname {PolyLog}\left (3,-e^{i (2 a+2 b x+\pi )}\right )de^{i (2 a+2 b x+\pi )}}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )-\frac {i x^4}{4}\)

\(\Big \downarrow \) 7143

\(\displaystyle 2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {i \left (\frac {\operatorname {PolyLog}\left (4,-e^{i (2 a+2 b x+\pi )}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )-\frac {i x^4}{4}\)

Input:

Int[x^3*Cot[a + b*x],x]
 

Output:

(-1/4*I)*x^4 + (2*I)*(((-1/2*I)*x^3*Log[1 + E^(I*(2*a + Pi + 2*b*x))])/b + 
 (((3*I)/2)*(((I/2)*x^2*PolyLog[2, -E^(I*(2*a + Pi + 2*b*x))])/b - (I*(((- 
1/2*I)*x*PolyLog[3, -E^(I*(2*a + Pi + 2*b*x))])/b + PolyLog[4, -E^(I*(2*a 
+ Pi + 2*b*x))]/(4*b^2)))/b))/b)
 

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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

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

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^m*(E^(2*I*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[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. 239 vs. \(2 (82 ) = 164\).

Time = 0.38 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.38

method	result	size
risch	\(-\frac {i x^{4}}{4}-\frac {3 i \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {3 i \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {2 i a^{3} x}{b^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{3}}{b}+\frac {a^{3} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {\ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{3}}{b}+\frac {6 \operatorname {polylog}\left (3, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {6 \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{4}}+\frac {2 a^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {3 i a^{4}}{2 b^{4}}+\frac {6 i \operatorname {polylog}\left (4, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {6 i \operatorname {polylog}\left (4, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}\)	\(240\)

Input:

int(x^3*cot(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

-1/4*I*x^4-3*I/b^2*polylog(2,exp(I*(b*x+a)))*x^2-3*I/b^2*polylog(2,-exp(I* 
(b*x+a)))*x^2-2*I/b^3*a^3*x+1/b*ln(exp(I*(b*x+a))+1)*x^3+1/b^4*a^3*ln(1-ex 
p(I*(b*x+a)))+1/b*ln(1-exp(I*(b*x+a)))*x^3+6/b^3*polylog(3,exp(I*(b*x+a))) 
*x+6/b^3*polylog(3,-exp(I*(b*x+a)))*x-1/b^4*a^3*ln(exp(I*(b*x+a))-1)+2/b^4 
*a^3*ln(exp(I*(b*x+a)))-3/2*I/b^4*a^4+6*I/b^4*polylog(4,exp(I*(b*x+a)))+6* 
I/b^4*polylog(4,-exp(I*(b*x+a)))
 

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. 306 vs. \(2 (78) = 156\).

Time = 0.09 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.03 \[ \int x^3 \cot (a+b x) \, dx=\frac {-6 i \, b^{2} x^{2} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) - 4 \, a^{3} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) - 4 \, a^{3} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 6 \, b x {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 6 \, b x {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 4 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 4 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 i \, {\rm polylog}\left (4, \cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) - 3 i \, {\rm polylog}\left (4, \cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right )}{8 \, b^{4}} \] Input:

integrate(x^3*cot(b*x+a),x, algorithm="fricas")
 

Output:

1/8*(-6*I*b^2*x^2*dilog(cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a)) + 6*I*b^2*x 
^2*dilog(cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a)) - 4*a^3*log(-1/2*cos(2*b*x 
 + 2*a) + 1/2*I*sin(2*b*x + 2*a) + 1/2) - 4*a^3*log(-1/2*cos(2*b*x + 2*a) 
- 1/2*I*sin(2*b*x + 2*a) + 1/2) + 6*b*x*polylog(3, cos(2*b*x + 2*a) + I*si 
n(2*b*x + 2*a)) + 6*b*x*polylog(3, cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a)) 
+ 4*(b^3*x^3 + a^3)*log(-cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + 1) + 4*(b 
^3*x^3 + a^3)*log(-cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + 1) + 3*I*polylo 
g(4, cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a)) - 3*I*polylog(4, cos(2*b*x + 2 
*a) - I*sin(2*b*x + 2*a)))/b^4
 

Sympy [F]

\[ \int x^3 \cot (a+b x) \, dx=\int x^{3} \cot {\left (a + b x \right )}\, dx \] Input:

integrate(x**3*cot(b*x+a),x)
 

Output:

Integral(x**3*cot(a + b*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. 391 vs. \(2 (78) = 156\).

Time = 0.11 (sec) , antiderivative size = 391, normalized size of antiderivative = 3.87 \[ \int x^3 \cot (a+b x) \, dx=-\frac {i \, {\left (b x + a\right )}^{4} - 4 i \, {\left (b x + a\right )}^{3} a + 6 i \, {\left (b x + a\right )}^{2} a^{2} + 4 \, a^{3} \log \left (\sin \left (b x + a\right )\right ) - 24 \, b x {\rm Li}_{3}(-e^{\left (i \, b x + i \, a\right )}) - 24 \, b x {\rm Li}_{3}(e^{\left (i \, b x + i \, a\right )}) + 4 \, {\left (-i \, {\left (b x + a\right )}^{3} + 3 i \, {\left (b x + a\right )}^{2} a - 3 i \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) + 4 \, {\left (i \, {\left (b x + a\right )}^{3} - 3 i \, {\left (b x + a\right )}^{2} a + 3 i \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + 12 \, {\left (i \, {\left (b x + a\right )}^{2} - 2 i \, {\left (b x + a\right )} a + i \, a^{2}\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + 12 \, {\left (i \, {\left (b x + a\right )}^{2} - 2 i \, {\left (b x + a\right )} a + i \, a^{2}\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) - 2 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) - 2 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) - 24 i \, {\rm Li}_{4}(-e^{\left (i \, b x + i \, a\right )}) - 24 i \, {\rm Li}_{4}(e^{\left (i \, b x + i \, a\right )})}{4 \, b^{4}} \] Input:

integrate(x^3*cot(b*x+a),x, algorithm="maxima")
 

Output:

-1/4*(I*(b*x + a)^4 - 4*I*(b*x + a)^3*a + 6*I*(b*x + a)^2*a^2 + 4*a^3*log( 
sin(b*x + a)) - 24*b*x*polylog(3, -e^(I*b*x + I*a)) - 24*b*x*polylog(3, e^ 
(I*b*x + I*a)) + 4*(-I*(b*x + a)^3 + 3*I*(b*x + a)^2*a - 3*I*(b*x + a)*a^2 
)*arctan2(sin(b*x + a), cos(b*x + a) + 1) + 4*(I*(b*x + a)^3 - 3*I*(b*x + 
a)^2*a + 3*I*(b*x + a)*a^2)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + 12* 
(I*(b*x + a)^2 - 2*I*(b*x + a)*a + I*a^2)*dilog(-e^(I*b*x + I*a)) + 12*(I* 
(b*x + a)^2 - 2*I*(b*x + a)*a + I*a^2)*dilog(e^(I*b*x + I*a)) - 2*((b*x + 
a)^3 - 3*(b*x + a)^2*a + 3*(b*x + a)*a^2)*log(cos(b*x + a)^2 + sin(b*x + a 
)^2 + 2*cos(b*x + a) + 1) - 2*((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(b*x + a) 
*a^2)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) - 24*I*pol 
ylog(4, -e^(I*b*x + I*a)) - 24*I*polylog(4, e^(I*b*x + I*a)))/b^4
 

Giac [F]

\[ \int x^3 \cot (a+b x) \, dx=\int { x^{3} \cot \left (b x + a\right ) \,d x } \] Input:

integrate(x^3*cot(b*x+a),x, algorithm="giac")
 

Output:

integrate(x^3*cot(b*x + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \cot (a+b x) \, dx=\int x^3\,\mathrm {cot}\left (a+b\,x\right ) \,d x \] Input:

int(x^3*cot(a + b*x),x)
 

Output:

int(x^3*cot(a + b*x), x)
 

Reduce [F]

\[ \int x^3 \cot (a+b x) \, dx=\int \cot \left (b x +a \right ) x^{3}d x \] Input:

int(x^3*cot(b*x+a),x)
 

Output:

int(cot(a + b*x)*x**3,x)