\(\int x^2 \cot ^{-1}(c+d \tanh (a+b x)) \, dx\) [40]

Optimal result

Integrand size = 15, antiderivative size = 355 \[ \int x^2 \cot ^{-1}(c+d \tanh (a+b x)) \, dx=\frac {1}{3} x^3 \cot ^{-1}(c+d \tanh (a+b x))-\frac {1}{6} i x^3 \log \left (1+\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}+\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}+\frac {i x \operatorname {PolyLog}\left (3,-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b^2}-\frac {i \operatorname {PolyLog}\left (4,-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{8 b^3}+\frac {i \operatorname {PolyLog}\left (4,-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{8 b^3} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.32 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.23 \[ \int x^2 \cot ^{-1}(c+d \tanh (a+b x)) \, dx=\frac {1}{3} x^3 \cot ^{-1}(c+d \tanh (a+b x))+\frac {d \left (4 b^3 x^3 \log \left (1+\frac {2 \left (1+(c+d)^2\right ) e^{2 (a+b x)}}{2+2 c^2-2 d^2-4 \sqrt {-d^2}}\right )-4 b^3 x^3 \log \left (1+\frac {\left (1+(c+d)^2\right ) e^{2 (a+b x)}}{1+c^2-d^2+2 \sqrt {-d^2}}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (2,\frac {\left (1+c^2+2 c d+d^2\right ) e^{2 (a+b x)}}{-1-c^2+d^2+2 \sqrt {-d^2}}\right )-6 b^2 x^2 \operatorname {PolyLog}\left (2,-\frac {\left (1+(c+d)^2\right ) e^{2 (a+b x)}}{1+c^2-d^2+2 \sqrt {-d^2}}\right )-6 b x \operatorname {PolyLog}\left (3,\frac {\left (1+c^2+2 c d+d^2\right ) e^{2 (a+b x)}}{-1-c^2+d^2+2 \sqrt {-d^2}}\right )+6 b x \operatorname {PolyLog}\left (3,-\frac {\left (1+(c+d)^2\right ) e^{2 (a+b x)}}{1+c^2-d^2+2 \sqrt {-d^2}}\right )+3 \operatorname {PolyLog}\left (4,-\frac {2 \left (1+(c+d)^2\right ) e^{2 (a+b x)}}{2+2 c^2-2 d^2-4 \sqrt {-d^2}}\right )-3 \operatorname {PolyLog}\left (4,-\frac {\left (1+(c+d)^2\right ) e^{2 (a+b x)}}{1+c^2-d^2+2 \sqrt {-d^2}}\right )\right )}{24 b^3 \sqrt {-d^2}} \] Input:

Integrate[x^2*ArcCot[c + d*Tanh[a + b*x]],x]
 

Output:

(x^3*ArcCot[c + d*Tanh[a + b*x]])/3 + (d*(4*b^3*x^3*Log[1 + (2*(1 + (c + d 
)^2)*E^(2*(a + b*x)))/(2 + 2*c^2 - 2*d^2 - 4*Sqrt[-d^2])] - 4*b^3*x^3*Log[ 
1 + ((1 + (c + d)^2)*E^(2*(a + b*x)))/(1 + c^2 - d^2 + 2*Sqrt[-d^2])] + 6* 
b^2*x^2*PolyLog[2, ((1 + c^2 + 2*c*d + d^2)*E^(2*(a + b*x)))/(-1 - c^2 + d 
^2 + 2*Sqrt[-d^2])] - 6*b^2*x^2*PolyLog[2, -(((1 + (c + d)^2)*E^(2*(a + b* 
x)))/(1 + c^2 - d^2 + 2*Sqrt[-d^2]))] - 6*b*x*PolyLog[3, ((1 + c^2 + 2*c*d 
 + d^2)*E^(2*(a + b*x)))/(-1 - c^2 + d^2 + 2*Sqrt[-d^2])] + 6*b*x*PolyLog[ 
3, -(((1 + (c + d)^2)*E^(2*(a + b*x)))/(1 + c^2 - d^2 + 2*Sqrt[-d^2]))] + 
3*PolyLog[4, (-2*(1 + (c + d)^2)*E^(2*(a + b*x)))/(2 + 2*c^2 - 2*d^2 - 4*S 
qrt[-d^2])] - 3*PolyLog[4, -(((1 + (c + d)^2)*E^(2*(a + b*x)))/(1 + c^2 - 
d^2 + 2*Sqrt[-d^2]))]))/(24*b^3*Sqrt[-d^2])
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.24, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5723, 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.

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

\(\Big \downarrow \) 5723

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[x^2*ArcCot[c + d*Tanh[a + b*x]],x]
 

Output:

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

Defintions of rubi rules used

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 5723

Int[ArcCot[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_ 
.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcCot[c + d*Tanh[a + b*x]]/(f*(m 
+ 1))), x] + (-Simp[I*b*((I - c - d)/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*( 
E^(2*a + 2*b*x)/(I - c + d + (I - c - d)*E^(2*a + 2*b*x))), x], x] + Simp[I 
*b*((I + c + d)/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*(E^(2*a + 2*b*x)/(I + 
c - d + (I + c + d)*E^(2*a + 2*b*x))), x], x]) /; FreeQ[{a, b, c, d, e, f}, 
 x] && IGtQ[m, 0] && NeQ[(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 = 40.82 (sec) , antiderivative size = 6916, normalized size of antiderivative = 19.48

method	result	size
risch	\(\text {Expression too large to display}\)	\(6916\)

Input:

int(x^2*arccot(c+d*tanh(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

result too large to display
 

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. 1289 vs. \(2 (263) = 526\).

Time = 0.21 (sec) , antiderivative size = 1289, normalized size of antiderivative = 3.63 \[ \int x^2 \cot ^{-1}(c+d \tanh (a+b x)) \, dx=\text {Too large to display} \] Input:

integrate(x^2*arccot(c+d*tanh(b*x+a)),x, algorithm="fricas")
 

Output:

1/6*(2*b^3*x^3*arctan(cosh(b*x + a)/(c*cosh(b*x + a) + d*sinh(b*x + a))) - 
 3*I*b^2*x^2*dilog(sqrt(-(c^2 - d^2 + 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 1))* 
(cosh(b*x + a) + sinh(b*x + a))) - 3*I*b^2*x^2*dilog(-sqrt(-(c^2 - d^2 + 2 
*I*d + 1)/(c^2 - 2*c*d + d^2 + 1))*(cosh(b*x + a) + sinh(b*x + a))) + 3*I* 
b^2*x^2*dilog(sqrt(-(c^2 - d^2 - 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 1))*(cosh 
(b*x + a) + sinh(b*x + a))) + 3*I*b^2*x^2*dilog(-sqrt(-(c^2 - d^2 - 2*I*d 
+ 1)/(c^2 - 2*c*d + d^2 + 1))*(cosh(b*x + a) + sinh(b*x + a))) + I*a^3*log 
(2*(c^2 + 2*c*d + d^2 + 1)*cosh(b*x + a) + 2*(c^2 + 2*c*d + d^2 + 1)*sinh( 
b*x + a) + 2*(c^2 - d^2 - 2*I*d + 1)*sqrt(-(c^2 - d^2 + 2*I*d + 1)/(c^2 - 
2*c*d + d^2 + 1))) + I*a^3*log(2*(c^2 + 2*c*d + d^2 + 1)*cosh(b*x + a) + 2 
*(c^2 + 2*c*d + d^2 + 1)*sinh(b*x + a) - 2*(c^2 - d^2 - 2*I*d + 1)*sqrt(-( 
c^2 - d^2 + 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 1))) - I*a^3*log(2*(c^2 + 2*c* 
d + d^2 + 1)*cosh(b*x + a) + 2*(c^2 + 2*c*d + d^2 + 1)*sinh(b*x + a) + 2*( 
c^2 - d^2 + 2*I*d + 1)*sqrt(-(c^2 - d^2 - 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 
1))) - I*a^3*log(2*(c^2 + 2*c*d + d^2 + 1)*cosh(b*x + a) + 2*(c^2 + 2*c*d 
+ d^2 + 1)*sinh(b*x + a) - 2*(c^2 - d^2 + 2*I*d + 1)*sqrt(-(c^2 - d^2 - 2* 
I*d + 1)/(c^2 - 2*c*d + d^2 + 1))) + 6*I*b*x*polylog(3, sqrt(-(c^2 - d^2 + 
 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 1))*(cosh(b*x + a) + sinh(b*x + a))) + 6* 
I*b*x*polylog(3, -sqrt(-(c^2 - d^2 + 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 1))*( 
cosh(b*x + a) + sinh(b*x + a))) - 6*I*b*x*polylog(3, sqrt(-(c^2 - d^2 -...
 

Sympy [F(-1)]

Timed out. \[ \int x^2 \cot ^{-1}(c+d \tanh (a+b x)) \, dx=\text {Timed out} \] Input:

integrate(x**2*acot(c+d*tanh(b*x+a)),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(x^2*arccot(c+d*tanh(b*x+a)),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/3*x^3*arctan2(e^(2*b*x + 2*a) + 1, (c*e^(2*a) + d*e^(2*a))*e^(2*b*x) + c 
 - d) + 4*b*d*integrate(1/3*x^3*e^(2*b*x + 2*a)/(c^2 - 2*c*d + d^2 + (c^2* 
e^(4*a) + 2*c*d*e^(4*a) + d^2*e^(4*a) + e^(4*a))*e^(4*b*x) + 2*(c^2*e^(2*a 
) - d^2*e^(2*a) + e^(2*a))*e^(2*b*x) + 1), x)
 

Giac [F]

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

integrate(x^2*arccot(c+d*tanh(b*x+a)),x, algorithm="giac")
 

Output:

integrate(x^2*arccot(d*tanh(b*x + a) + c), x)
 

Mupad [F(-1)]

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

int(x^2*acot(c + d*tanh(a + b*x)),x)
 

Output:

int(x^2*acot(c + d*tanh(a + b*x)), x)
 

Reduce [F]

\[ \int x^2 \cot ^{-1}(c+d \tanh (a+b x)) \, dx=\int \mathit {acot} \left (\tanh \left (b x +a \right ) d +c \right ) x^{2}d x \] Input:

int(x^2*acot(c+d*tanh(b*x+a)),x)
 

Output:

int(acot(tanh(a + b*x)*d + c)*x**2,x)