\(\int x^3 \cot ^2(a+b x) \, dx\) [6]

Optimal result

Integrand size = 12, antiderivative size = 97 \[ \int x^3 \cot ^2(a+b x) \, dx=-\frac {i x^3}{b}-\frac {x^4}{4}-\frac {x^3 \cot (a+b x)}{b}+\frac {3 x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {3 i x \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^3}+\frac {3 \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^4} \] Output:

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(256\) vs. \(2(97)=194\).

Time = 0.76 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.64 \[ \int x^3 \cot ^2(a+b x) \, dx=-\frac {x^4}{4}-\frac {i e^{2 i a} \left (2 b^3 e^{-2 i a} x^3+3 i b^2 \left (1-e^{-2 i a}\right ) x^2 \log \left (1-e^{-i (a+b x)}\right )+3 i b^2 \left (1-e^{-2 i a}\right ) x^2 \log \left (1+e^{-i (a+b x)}\right )-6 b \left (1-e^{-2 i a}\right ) x \operatorname {PolyLog}\left (2,-e^{-i (a+b x)}\right )-6 b \left (1-e^{-2 i a}\right ) x \operatorname {PolyLog}\left (2,e^{-i (a+b x)}\right )+6 i \left (1-e^{-2 i a}\right ) \operatorname {PolyLog}\left (3,-e^{-i (a+b x)}\right )+6 i \left (1-e^{-2 i a}\right ) \operatorname {PolyLog}\left (3,e^{-i (a+b x)}\right )\right )}{b^4 \left (-1+e^{2 i a}\right )}+\frac {x^3 \csc (a) \csc (a+b x) \sin (b x)}{b} \] Input:

Integrate[x^3*Cot[a + b*x]^2,x]
 

Output:

-1/4*x^4 - (I*E^((2*I)*a)*((2*b^3*x^3)/E^((2*I)*a) + (3*I)*b^2*(1 - E^((-2 
*I)*a))*x^2*Log[1 - E^((-I)*(a + b*x))] + (3*I)*b^2*(1 - E^((-2*I)*a))*x^2 
*Log[1 + E^((-I)*(a + b*x))] - 6*b*(1 - E^((-2*I)*a))*x*PolyLog[2, -E^((-I 
)*(a + b*x))] - 6*b*(1 - E^((-2*I)*a))*x*PolyLog[2, E^((-I)*(a + b*x))] + 
(6*I)*(1 - E^((-2*I)*a))*PolyLog[3, -E^((-I)*(a + b*x))] + (6*I)*(1 - E^(( 
-2*I)*a))*PolyLog[3, E^((-I)*(a + b*x))]))/(b^4*(-1 + E^((2*I)*a))) + (x^3 
*Csc[a]*Csc[a + b*x]*Sin[b*x])/b
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.39, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {3042, 4203, 15, 25, 3042, 25, 4202, 2620, 3011, 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.

\(\displaystyle \int x^3 \cot ^2(a+b x) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4203

\(\displaystyle -\frac {3 \int -x^2 \cot (a+b x)dx}{b}-\int x^3dx-\frac {x^3 \cot (a+b x)}{b}\)

\(\Big \downarrow \) 15

\(\displaystyle -\frac {3 \int -x^2 \cot (a+b x)dx}{b}-\frac {x^3 \cot (a+b x)}{b}-\frac {x^4}{4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3 \int x^2 \cot (a+b x)dx}{b}-\frac {x^3 \cot (a+b x)}{b}-\frac {x^4}{4}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[x^3*Cot[a + b*x]^2,x]
 

Output:

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

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

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 4203

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symb 
ol] :> Simp[b*(c + d*x)^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] + (-Si 
mp[b*d*(m/(f*(n - 1)))   Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1), x] 
, x] - Simp[b^2   Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; Free 
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (85 ) = 170\).

Time = 0.46 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.38

method	result	size
risch	\(-\frac {x^{4}}{4}-\frac {2 i x^{3}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {6 i a^{2} x}{b^{3}}-\frac {6 i \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {3 \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {3 a^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {2 i x^{3}}{b}+\frac {6 \operatorname {polylog}\left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b^{2}}+\frac {4 i a^{3}}{b^{4}}+\frac {6 \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{4}}-\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {6 i \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}\)	\(231\)

Input:

int(x^3*cot(b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

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

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. 372 vs. \(2 (82) = 164\).

Time = 0.10 (sec) , antiderivative size = 372, normalized size of antiderivative = 3.84 \[ \int x^3 \cot ^2(a+b x) \, dx=-\frac {b^{4} x^{4} \sin \left (2 \, b x + 2 \, a\right ) + 4 \, b^{3} x^{3} \cos \left (2 \, b x + 2 \, a\right ) + 4 \, b^{3} x^{3} + 6 i \, b x {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 i \, b x {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) \sin \left (2 \, b x + 2 \, a\right ) - 3 \, {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right ) - 3 \, {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right )}{4 \, b^{4} \sin \left (2 \, b x + 2 \, a\right )} \] Input:

integrate(x^3*cot(b*x+a)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int x^3 \cot ^2(a+b x) \, dx=\int x^{3} \cot ^{2}{\left (a + b x \right )}\, dx \] Input:

integrate(x**3*cot(b*x+a)**2,x)
 

Output:

Integral(x**3*cot(a + b*x)**2, 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. 952 vs. \(2 (82) = 164\).

Time = 0.20 (sec) , antiderivative size = 952, normalized size of antiderivative = 9.81 \[ \int x^3 \cot ^2(a+b x) \, dx=\text {Too large to display} \] Input:

integrate(x^3*cot(b*x+a)^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int x^3 \cot ^2(a+b x) \, dx=\int { x^{3} \cot \left (b x + a\right )^{2} \,d x } \] Input:

integrate(x^3*cot(b*x+a)^2,x, algorithm="giac")
 

Output:

integrate(x^3*cot(b*x + a)^2, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \cot ^2(a+b x) \, dx=\int x^3\,{\mathrm {cot}\left (a+b\,x\right )}^2 \,d x \] Input:

int(x^3*cot(a + b*x)^2,x)
 

Output:

int(x^3*cot(a + b*x)^2, x)
 

Reduce [F]

\[ \int x^3 \cot ^2(a+b x) \, dx=\int \cot \left (b x +a \right )^{2} x^{3}d x \] Input:

int(x^3*cot(b*x+a)^2,x)
 

Output:

int(cot(a + b*x)**2*x**3,x)