3.2.62 \(\int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx\) [162]

3.2.62.1 Optimal result

Integrand size = 21, antiderivative size = 154 \[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac {1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac {x^2 \operatorname {PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {i x \operatorname {PolyLog}\left (3,i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {\operatorname {PolyLog}\left (4,i c e^{2 i a+2 i b x}\right )}{8 b^3} \]

output

1/12*b*x^4+1/3*x^3*arccot(c+(1+I*c)*tan(b*x+a))+1/6*I*x^3*ln(1-I*c*exp(2*I 
*a+2*I*b*x))+1/4*x^2*polylog(2,I*c*exp(2*I*a+2*I*b*x))/b+1/4*I*x*polylog(3 
,I*c*exp(2*I*a+2*I*b*x))/b^2-1/8*polylog(4,I*c*exp(2*I*a+2*I*b*x))/b^3
 

3.2.62.2 Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.88 \[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\frac {1}{24} \left (8 x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))+4 i x^3 \log \left (1+\frac {i e^{-2 i (a+b x)}}{c}\right )-\frac {6 x^2 \operatorname {PolyLog}\left (2,-\frac {i e^{-2 i (a+b x)}}{c}\right )}{b}+\frac {6 i x \operatorname {PolyLog}\left (3,-\frac {i e^{-2 i (a+b x)}}{c}\right )}{b^2}+\frac {3 \operatorname {PolyLog}\left (4,-\frac {i e^{-2 i (a+b x)}}{c}\right )}{b^3}\right ) \]

input

Integrate[x^2*ArcCot[c + (1 + I*c)*Tan[a + b*x]],x]
 

output

(8*x^3*ArcCot[c + (1 + I*c)*Tan[a + b*x]] + (4*I)*x^3*Log[1 + I/(c*E^((2*I 
)*(a + b*x)))] - (6*x^2*PolyLog[2, (-I)/(c*E^((2*I)*(a + b*x)))])/b + ((6* 
I)*x*PolyLog[3, (-I)/(c*E^((2*I)*(a + b*x)))])/b^2 + (3*PolyLog[4, (-I)/(c 
*E^((2*I)*(a + b*x)))])/b^3)/24
 

3.2.62.3 Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.29, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5695, 2615, 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.

\(\displaystyle \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx\)

\(\Big \downarrow \) 5695

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

input

Int[x^2*ArcCot[c + (1 + I*c)*Tan[a + b*x]],x]
 

output

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

3.2.62.3.1 Defintions of rubi rules used

rule 2615

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

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 5695

Int[ArcCot[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_. 
), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcCot[c + d*Tan[a + b*x]]/(f*(m + 
1))), x] + Simp[I*(b/(f*(m + 1)))   Int[(e + f*x)^(m + 1)/(c + I*d + c*E^(2 
*I*a + 2*I*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && E 
qQ[(c + I*d)^2, -1]
 

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]
 

3.2.62.4 Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 2.88 (sec) , antiderivative size = 1448, normalized size of antiderivative = 9.40

method	result	size
risch	\(\text {Expression too large to display}\)	\(1448\)

input

int(x^2*arccot(c+(I*c+1)*tan(b*x+a)),x,method=_RETURNVERBOSE)
 

output

1/4*I*x*polylog(3,I*exp(2*I*(b*x+a))*c)/b^2+1/12*(Pi*csgn(I/(exp(2*I*(b*x+ 
a))+1))*csgn(I*(c-I))*csgn(I*(c-I)/(exp(2*I*(b*x+a))+1))-Pi*csgn(I/(exp(2* 
I*(b*x+a))+1))*csgn(I*(exp(2*I*(b*x+a))*c+I))*csgn(I*(exp(2*I*(b*x+a))*c+I 
)/(exp(2*I*(b*x+a))+1))-Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(c-I)/(exp( 
2*I*(b*x+a))+1))^2+Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(exp(2*I*(b*x+a) 
)*c+I)/(exp(2*I*(b*x+a))+1))^2+Pi*csgn(I*exp(I*(b*x+a)))^2*csgn(I*exp(2*I* 
(b*x+a)))-2*Pi*csgn(I*exp(I*(b*x+a)))*csgn(I*exp(2*I*(b*x+a)))^2+Pi*csgn(I 
*exp(2*I*(b*x+a)))^3+Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I*(c-I)/(exp(2*I*(b* 
x+a))+1))*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))+1))-Pi*csgn(I*ex 
p(2*I*(b*x+a)))*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))+1))^2-Pi*c 
sgn(I*(exp(2*I*(b*x+a))*c+I)/(exp(2*I*(b*x+a))+1))*csgn((exp(2*I*(b*x+a))* 
c+I)/(exp(2*I*(b*x+a))+1))+Pi*csgn((exp(2*I*(b*x+a))*c+I)/(exp(2*I*(b*x+a) 
)+1))^2-Pi*csgn(I*(c-I))*csgn(I*(c-I)/(exp(2*I*(b*x+a))+1))^2+Pi*csgn(I*(e 
xp(2*I*(b*x+a))*c+I))*csgn(I*(exp(2*I*(b*x+a))*c+I)/(exp(2*I*(b*x+a))+1))^ 
2+Pi*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))+1))*csgn(exp(2*I*(b*x 
+a))*(c-I)/(exp(2*I*(b*x+a))+1))+Pi*csgn(exp(2*I*(b*x+a))*(c-I)/(exp(2*I*( 
b*x+a))+1))^2+Pi*csgn(I*(c-I)/(exp(2*I*(b*x+a))+1))^3-Pi*csgn(I*(c-I)/(exp 
(2*I*(b*x+a))+1))*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))+1))^2-Pi 
*csgn(I*(exp(2*I*(b*x+a))*c+I)/(exp(2*I*(b*x+a))+1))^3+Pi*csgn(I*(exp(2*I* 
(b*x+a))*c+I)/(exp(2*I*(b*x+a))+1))*csgn((exp(2*I*(b*x+a))*c+I)/(exp(2*...
 

3.2.62.5 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. 321 vs. \(2 (107) = 214\).

Time = 0.27 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.08 \[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\frac {b^{4} x^{4} - 2 i \, b^{3} x^{3} \log \left (\frac {{\left (c e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{c - i}\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) - a^{4} - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {4 i \, c}}{2 \, c}\right ) - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {4 i \, c}}{2 \, c}\right ) + 12 i \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 12 i \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) - 2 \, {\left (-i \, b^{3} x^{3} - i \, a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 2 \, {\left (-i \, b^{3} x^{3} - i \, a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 12 \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) - 12 \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{3}} \]

input

integrate(x^2*arccot(c+(1+I*c)*tan(b*x+a)),x, algorithm="fricas")
 

output

1/12*(b^4*x^4 - 2*I*b^3*x^3*log((c*e^(2*I*b*x + 2*I*a) + I)*e^(-2*I*b*x - 
2*I*a)/(c - I)) + 6*b^2*x^2*dilog(1/2*sqrt(4*I*c)*e^(I*b*x + I*a)) + 6*b^2 
*x^2*dilog(-1/2*sqrt(4*I*c)*e^(I*b*x + I*a)) - a^4 - 2*I*a^3*log(1/2*(2*c* 
e^(I*b*x + I*a) + I*sqrt(4*I*c))/c) - 2*I*a^3*log(1/2*(2*c*e^(I*b*x + I*a) 
 - I*sqrt(4*I*c))/c) + 12*I*b*x*polylog(3, 1/2*sqrt(4*I*c)*e^(I*b*x + I*a) 
) + 12*I*b*x*polylog(3, -1/2*sqrt(4*I*c)*e^(I*b*x + I*a)) - 2*(-I*b^3*x^3 
- I*a^3)*log(1/2*sqrt(4*I*c)*e^(I*b*x + I*a) + 1) - 2*(-I*b^3*x^3 - I*a^3) 
*log(-1/2*sqrt(4*I*c)*e^(I*b*x + I*a) + 1) - 12*polylog(4, 1/2*sqrt(4*I*c) 
*e^(I*b*x + I*a)) - 12*polylog(4, -1/2*sqrt(4*I*c)*e^(I*b*x + I*a)))/b^3
 

3.2.62.6 Sympy [F(-2)]

Exception generated. \[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\text {Exception raised: CoercionFailed} \]

input

integrate(x**2*acot(c+(1+I*c)*tan(b*x+a)),x)
 

output

Exception raised: CoercionFailed >> Cannot convert _t0**2*I + 2*c*exp(2*I* 
a) - I*exp(2*I*a) of type <class 'sympy.core.add.Add'> to QQ_I[x,b,c,_t0,e 
xp(I*a)]
 

3.2.62.7 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. 310 vs. \(2 (107) = 214\).

Time = 0.21 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.01 \[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\frac {\frac {4 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2}\right )} \operatorname {arccot}\left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )}{b^{2}} - \frac {{\left (-3 i \, {\left (b x + a\right )}^{4} + 12 i \, {\left (b x + a\right )}^{3} a - 18 i \, {\left (b x + a\right )}^{2} a^{2} - 2 \, {\left (4 i \, {\left (b x + a\right )}^{3} - 9 i \, {\left (b x + a\right )}^{2} a + 9 i \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (c \cos \left (2 \, b x + 2 \, a\right ), c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) - 3 \, {\left (4 i \, {\left (b x + a\right )}^{2} - 6 i \, {\left (b x + a\right )} a + 3 i \, a^{2}\right )} {\rm Li}_2\left (i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + {\left (4 \, {\left (b x + a\right )}^{3} - 9 \, {\left (b x + a\right )}^{2} a + 9 \, {\left (b x + a\right )} a^{2}\right )} \log \left (c^{2} \cos \left (2 \, b x + 2 \, a\right )^{2} + c^{2} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left (4 \, b x + a\right )} {\rm Li}_{3}(i \, c e^{\left (2 i \, b x + 2 i \, a\right )}) + 6 i \, {\rm Li}_{4}(i \, c e^{\left (2 i \, b x + 2 i \, a\right )})\right )} {\left (-i \, c - 1\right )}}{b^{2} {\left (c - i\right )}}}{12 \, b} \]

input

integrate(x^2*arccot(c+(1+I*c)*tan(b*x+a)),x, algorithm="maxima")
 

output

1/12*(4*((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(b*x + a)*a^2)*arccot((I*c + 1) 
*tan(b*x + a) + c)/b^2 - (-3*I*(b*x + a)^4 + 12*I*(b*x + a)^3*a - 18*I*(b* 
x + a)^2*a^2 - 2*(4*I*(b*x + a)^3 - 9*I*(b*x + a)^2*a + 9*I*(b*x + a)*a^2) 
*arctan2(c*cos(2*b*x + 2*a), c*sin(2*b*x + 2*a) + 1) - 3*(4*I*(b*x + a)^2 
- 6*I*(b*x + a)*a + 3*I*a^2)*dilog(I*c*e^(2*I*b*x + 2*I*a)) + (4*(b*x + a) 
^3 - 9*(b*x + a)^2*a + 9*(b*x + a)*a^2)*log(c^2*cos(2*b*x + 2*a)^2 + c^2*s 
in(2*b*x + 2*a)^2 + 2*c*sin(2*b*x + 2*a) + 1) + 3*(4*b*x + a)*polylog(3, I 
*c*e^(2*I*b*x + 2*I*a)) + 6*I*polylog(4, I*c*e^(2*I*b*x + 2*I*a)))*(-I*c - 
 1)/(b^2*(c - I)))/b
 

3.2.62.8 Giac [F]

\[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\int { x^{2} \operatorname {arccot}\left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right ) \,d x } \]

input

integrate(x^2*arccot(c+(1+I*c)*tan(b*x+a)),x, algorithm="giac")
 

output

integrate(x^2*arccot((I*c + 1)*tan(b*x + a) + c), x)
 

3.2.62.9 Mupad [F(-1)]

Timed out. \[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\int x^2\,\mathrm {acot}\left (c+\mathrm {tan}\left (a+b\,x\right )\,\left (1+c\,1{}\mathrm {i}\right )\right ) \,d x \]

input

int(x^2*acot(c + tan(a + b*x)*(c*1i + 1)),x)
 

output

int(x^2*acot(c + tan(a + b*x)*(c*1i + 1)), x)