\(\int x \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx\) [15]

Optimal result

Integrand size = 19, antiderivative size = 123 \[ \int x \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\frac {b x^3}{6}+\frac {1}{2} x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac {1}{4} i x^2 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac {x \operatorname {PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {i \operatorname {PolyLog}\left (3,i c e^{2 i a+2 i b x}\right )}{8 b^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.89 \[ \int x \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\frac {1}{2} x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac {i \left (2 b^2 x^2 \log \left (1+\frac {i e^{-2 i (a+b x)}}{c}\right )+2 i b x \operatorname {PolyLog}\left (2,-\frac {i e^{-2 i (a+b x)}}{c}\right )+\operatorname {PolyLog}\left (3,-\frac {i e^{-2 i (a+b x)}}{c}\right )\right )}{8 b^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.26, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {5695, 2615, 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.

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

\(\Big \downarrow \) 5695

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

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

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]
 

Maple [C] (warning: unable to verify)

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

Time = 2.25 (sec) , antiderivative size = 1413, normalized size of antiderivative = 11.49

method	result	size
risch	\(\text {Expression too large to display}\)	\(1413\)

Input:

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

Output:

1/2*I*x^2*ln(exp(I*(b*x+a)))+1/8*(Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*( 
-I+c))*csgn(I*(-I+c)/(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*(-I+c)/(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*P 
i*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*(-I+c)/(exp(2*I*(b*x+a))+1))*cs 
gn(I*exp(2*I*(b*x+a))*(-I+c)/(exp(2*I*(b*x+a))+1))-Pi*csgn(I*exp(2*I*(b*x+ 
a)))*csgn(I*exp(2*I*(b*x+a))*(-I+c)/(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*(-I+c))*csgn(I*(-I+c)/(exp(2*I*(b*x+a))+1))^2+Pi*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))^2+Pi*csg 
n(I*exp(2*I*(b*x+a))*(-I+c)/(exp(2*I*(b*x+a))+1))*csgn(exp(2*I*(b*x+a))*(- 
I+c)/(exp(2*I*(b*x+a))+1))+Pi*csgn(exp(2*I*(b*x+a))*(-I+c)/(exp(2*I*(b*x+a 
))+1))^2+Pi*csgn(I*(-I+c)/(exp(2*I*(b*x+a))+1))^3-Pi*csgn(I*(-I+c)/(exp(2* 
I*(b*x+a))+1))*csgn(I*exp(2*I*(b*x+a))*(-I+c)/(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))^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*I*...
 

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. 270 vs. \(2 (85) = 170\).

Time = 0.11 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.20 \[ \int x \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\frac {2 \, b^{3} x^{3} - 3 i \, b^{2} x^{2} \log \left (\frac {{\left (c e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{c - i}\right ) + 2 \, a^{3} + 6 \, b x {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 \, b x {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 3 i \, a^{2} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {4 i \, c}}{2 \, c}\right ) + 3 i \, a^{2} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {4 i \, c}}{2 \, c}\right ) - 3 \, {\left (-i \, b^{2} x^{2} + i \, a^{2}\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 3 \, {\left (-i \, b^{2} x^{2} + i \, a^{2}\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) + 6 i \, {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \, {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{2}} \] Input:

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

Output:

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

Sympy [F(-2)]

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

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

Output:

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

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. 218 vs. \(2 (85) = 170\).

Time = 0.05 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.77 \[ \int x \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\frac {\frac {6 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \operatorname {arccot}\left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )}{b} - \frac {{\left (-4 i \, {\left (b x + a\right )}^{3} + 12 i \, {\left (b x + a\right )}^{2} a - 6 i \, b x {\rm Li}_2\left (i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 6 \, {\left (i \, {\left (b x + a\right )}^{2} - 2 i \, {\left (b x + a\right )} a\right )} \arctan \left (c \cos \left (2 \, b x + 2 \, a\right ), c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \log \left (c^{2} \cos \left (2 \, b x + 2 \, a\right )^{2} + c^{2} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\rm Li}_{3}(i \, c e^{\left (2 i \, b x + 2 i \, a\right )})\right )} {\left (-i \, c - 1\right )}}{b {\left (c - i\right )}}}{12 \, b} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\int \mathit {acot} \left (\tan \left (b x +a \right ) c i +\tan \left (b x +a \right )+c \right ) x d x \] Input:

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

Output:

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