\(\int x^2 \arctan (c+(i+c) \coth (a+b x)) \, dx\) [102]

Optimal result

Integrand size = 19, antiderivative size = 142 \[ \int x^2 \arctan (c+(i+c) \coth (a+b x)) \, dx=-\frac {1}{12} i b x^4+\frac {1}{3} x^3 \arctan (c+(i+c) \coth (a+b x))+\frac {1}{6} i x^3 \log \left (1-i c e^{2 a+2 b x}\right )+\frac {i x^2 \operatorname {PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{4 b}-\frac {i x \operatorname {PolyLog}\left (3,i c e^{2 a+2 b x}\right )}{4 b^2}+\frac {i \operatorname {PolyLog}\left (4,i c e^{2 a+2 b x}\right )}{8 b^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.94 \[ \int x^2 \arctan (c+(i+c) \coth (a+b x)) \, dx=\frac {8 b^3 x^3 \arctan (c+(i+c) \coth (a+b x))+4 i b^3 x^3 \log \left (1+\frac {i e^{-2 (a+b x)}}{c}\right )-6 i b^2 x^2 \operatorname {PolyLog}\left (2,-\frac {i e^{-2 (a+b x)}}{c}\right )-6 i b x \operatorname {PolyLog}\left (3,-\frac {i e^{-2 (a+b x)}}{c}\right )-3 i \operatorname {PolyLog}\left (4,-\frac {i e^{-2 (a+b x)}}{c}\right )}{24 b^3} \] Input:

Integrate[x^2*ArcTan[c + (I + c)*Coth[a + b*x]],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.18, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {5720, 25, 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.

\(\displaystyle \int x^2 \arctan (c+(c+i) \coth (a+b x)) \, dx\)

\(\Big \downarrow \) 5720

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[x^2*ArcTan[c + (I + c)*Coth[a + b*x]],x]
 

Output:

(x^3*ArcTan[c + (I + c)*Coth[a + b*x]])/3 + (b*((-1/4*I)*x^4 + I*c*((x^3*L 
og[1 - I*c*E^(2*a + 2*b*x)])/(2*b*c) - (3*(-1/2*(x^2*PolyLog[2, I*c*E^(2*a 
 + 2*b*x)])/b + ((x*PolyLog[3, I*c*E^(2*a + 2*b*x)])/(2*b) - PolyLog[4, I* 
c*E^(2*a + 2*b*x)]/(4*b^2))/b))/(2*b*c))))/3
 

Defintions of rubi rules used

rule 25

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

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 5720

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

Maple [C] (warning: unable to verify)

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

Time = 1.12 (sec) , antiderivative size = 1405, normalized size of antiderivative = 9.89

method	result	size
risch	\(\text {Expression too large to display}\)	\(1405\)

Input:

int(x^2*arctan(c+(I+c)*coth(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

1/2*I/b^3*a^2*dilog(1+I*exp(b*x+a)*(-I*c)^(1/2))+1/12*Pi*(csgn(I*(2*exp(2* 
b*x+2*a)*c+2*I))*csgn(I*(2*exp(2*b*x+2*a)*c+2*I)/(exp(2*b*x+2*a)-1))^2-csg 
n(I*(2*exp(2*b*x+2*a)*c+2*I))*csgn(I*(2*exp(2*b*x+2*a)*c+2*I)/(exp(2*b*x+2 
*a)-1))*csgn(I/(exp(2*b*x+2*a)-1))-csgn(I*(2*I*exp(2*b*x+2*a)+2*exp(2*b*x+ 
2*a)*c))*csgn(I*(2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)-1) 
)^2+csgn(I*(2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c))*csgn(I*(2*I*exp(2*b*x+ 
2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)-1))*csgn(I/(exp(2*b*x+2*a)-1))-cs 
gn(I*(2*exp(2*b*x+2*a)*c+2*I)/(exp(2*b*x+2*a)-1))^3+csgn(I*(2*exp(2*b*x+2* 
a)*c+2*I)/(exp(2*b*x+2*a)-1))^2*csgn(I/(exp(2*b*x+2*a)-1))+csgn(I*(2*exp(2 
*b*x+2*a)*c+2*I)/(exp(2*b*x+2*a)-1))*csgn((2*exp(2*b*x+2*a)*c+2*I)/(exp(2* 
b*x+2*a)-1))^2-csgn(I*(2*exp(2*b*x+2*a)*c+2*I)/(exp(2*b*x+2*a)-1))*csgn((2 
*exp(2*b*x+2*a)*c+2*I)/(exp(2*b*x+2*a)-1))+csgn(I*(2*I*exp(2*b*x+2*a)+2*ex 
p(2*b*x+2*a)*c)/(exp(2*b*x+2*a)-1))^3-csgn(I*(2*I*exp(2*b*x+2*a)+2*exp(2*b 
*x+2*a)*c)/(exp(2*b*x+2*a)-1))^2*csgn(I/(exp(2*b*x+2*a)-1))-csgn(I*(2*I*ex 
p(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)-1))*csgn((2*I*exp(2*b*x+2 
*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)-1))^2+csgn(I*(2*I*exp(2*b*x+2*a)+2 
*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)-1))*csgn((2*I*exp(2*b*x+2*a)+2*exp(2*b* 
x+2*a)*c)/(exp(2*b*x+2*a)-1))+csgn((2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c) 
/(exp(2*b*x+2*a)-1))^3-csgn((2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2 
*b*x+2*a)-1))^2+csgn((2*exp(2*b*x+2*a)*c+2*I)/(exp(2*b*x+2*a)-1))^3-csg...
 

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. 293 vs. \(2 (105) = 210\).

Time = 0.10 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.06 \[ \int x^2 \arctan (c+(i+c) \coth (a+b x)) \, dx=\frac {-i \, b^{4} x^{4} + 2 i \, b^{3} x^{3} \log \left (-\frac {{\left (c + i\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c e^{\left (2 \, b x + 2 \, a\right )} + i}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) + i \, a^{4} - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (b x + a\right )} + i \, \sqrt {4 i \, c}}{2 \, c}\right ) - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (b x + a\right )} - i \, \sqrt {4 i \, c}}{2 \, c}\right ) - 12 i \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) - 12 i \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) - 2 \, {\left (-i \, b^{3} x^{3} - i \, a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )} + 1\right ) - 2 \, {\left (-i \, b^{3} x^{3} - i \, a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )} + 1\right ) + 12 i \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) + 12 i \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right )}{12 \, b^{3}} \] Input:

integrate(x^2*arctan(c+(I+c)*coth(b*x+a)),x, algorithm="fricas")
 

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int x^2 \arctan (c+(i+c) \coth (a+b x)) \, dx=\text {Exception raised: CoercionFailed} \] Input:

integrate(x**2*atan(c+(I+c)*coth(b*x+a)),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 1.72 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.91 \[ \int x^2 \arctan (c+(i+c) \coth (a+b x)) \, dx=\frac {1}{3} \, x^{3} \arctan \left ({\left (c + i\right )} \coth \left (b x + a\right ) + c\right ) + \frac {4}{9} \, {\left (\frac {3 \, x^{4}}{4 i \, c - 4} - \frac {4 \, b^{3} x^{3} \log \left (-i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (i \, c e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}(i \, c e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}(i \, c e^{\left (2 \, b x + 2 \, a\right )})}{-2 \, b^{4} {\left (-i \, c + 1\right )}}\right )} b {\left (c + i\right )} \] Input:

integrate(x^2*arctan(c+(I+c)*coth(b*x+a)),x, algorithm="maxima")
 

Output:

1/3*x^3*arctan((c + I)*coth(b*x + a) + c) + 4/9*(3*x^4/(4*I*c - 4) - (4*b^ 
3*x^3*log(-I*c*e^(2*b*x + 2*a) + 1) + 6*b^2*x^2*dilog(I*c*e^(2*b*x + 2*a)) 
 - 6*b*x*polylog(3, I*c*e^(2*b*x + 2*a)) + 3*polylog(4, I*c*e^(2*b*x + 2*a 
)))/(b^4*(2*I*c - 2)))*b*(c + I)
 

Giac [F]

\[ \int x^2 \arctan (c+(i+c) \coth (a+b x)) \, dx=\int { x^{2} \arctan \left ({\left (c + i\right )} \coth \left (b x + a\right ) + c\right ) \,d x } \] Input:

integrate(x^2*arctan(c+(I+c)*coth(b*x+a)),x, algorithm="giac")
 

Output:

integrate(x^2*arctan((c + I)*coth(b*x + a) + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \arctan (c+(i+c) \coth (a+b x)) \, dx=\int x^2\,\mathrm {atan}\left (c+\mathrm {coth}\left (a+b\,x\right )\,\left (c+1{}\mathrm {i}\right )\right ) \,d x \] Input:

int(x^2*atan(c + coth(a + b*x)*(c + 1i)),x)
 

Output:

int(x^2*atan(c + coth(a + b*x)*(c + 1i)), x)
 

Reduce [F]

\[ \int x^2 \arctan (c+(i+c) \coth (a+b x)) \, dx=\int \mathit {atan} \left (\coth \left (b x +a \right ) c +\coth \left (b x +a \right ) i +c \right ) x^{2}d x \] Input:

int(x^2*atan(c+(I+c)*coth(b*x+a)),x)
 

Output:

int(atan(coth(a + b*x)*c + coth(a + b*x)*i + c)*x**2,x)