\(\int x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x)) \, dx\) [192]

Optimal result

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

[Out]

1/12*I*b*x^4+1/3*x^3*arccot(c+(I+c)*tanh(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*e
xp(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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {5304, 2215, 2221, 2611, 6744, 2320, 6724} \[ \int x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x)) \, dx=-\frac {i \operatorname {PolyLog}\left (4,-i c e^{2 a+2 b x}\right )}{8 b^3}+\frac {i x \operatorname {PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{4 b^2}-\frac {i x^2 \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{4 b}-\frac {1}{6} i x^3 \log \left (1+i c e^{2 a+2 b x}\right )+\frac {1}{3} x^3 \cot ^{-1}(c+(c+i) \tanh (a+b x))+\frac {1}{12} i b x^4 \]

[In]

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

[Out]

(I/12)*b*x^4 + (x^3*ArcCot[c + (I + c)*Tanh[a + b*x]])/3 - (I/6)*x^3*Log[1 + I*c*E^(2*a + 2*b*x)] - ((I/4)*x^2
*PolyLog[2, (-I)*c*E^(2*a + 2*b*x)])/b + ((I/4)*x*PolyLog[3, (-I)*c*E^(2*a + 2*b*x)])/b^2 - ((I/8)*PolyLog[4,
(-I)*c*E^(2*a + 2*b*x)])/b^3

Rule 2215

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] - Dist[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 2221

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]
- Dist[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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{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 2611

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] + Dist[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 5304

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] + Dist[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 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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 6744

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] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x^3 \cot ^{-1}(c+(i+c) \tanh (a+b x))+\frac {1}{3} b \int \frac {x^3}{-i+c e^{2 a+2 b x}} \, dx \\ & = \frac {1}{12} i b x^4+\frac {1}{3} x^3 \cot ^{-1}(c+(i+c) \tanh (a+b x))-\frac {1}{3} (i b c) \int \frac {e^{2 a+2 b x} x^3}{-i+c e^{2 a+2 b x}} \, dx \\ & = \frac {1}{12} i b x^4+\frac {1}{3} x^3 \cot ^{-1}(c+(i+c) \tanh (a+b x))-\frac {1}{6} i x^3 \log \left (1+i c e^{2 a+2 b x}\right )+\frac {1}{2} i \int x^2 \log \left (1+i c e^{2 a+2 b x}\right ) \, dx \\ & = \frac {1}{12} i b x^4+\frac {1}{3} x^3 \cot ^{-1}(c+(i+c) \tanh (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 \int x \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right ) \, dx}{2 b} \\ & = \frac {1}{12} i b x^4+\frac {1}{3} x^3 \cot ^{-1}(c+(i+c) \tanh (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 \int \operatorname {PolyLog}\left (3,-i c e^{2 a+2 b x}\right ) \, dx}{4 b^2} \\ & = \frac {1}{12} i b x^4+\frac {1}{3} x^3 \cot ^{-1}(c+(i+c) \tanh (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 \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,-i c x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3} \\ & = \frac {1}{12} i b x^4+\frac {1}{3} x^3 \cot ^{-1}(c+(i+c) \tanh (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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.94 \[ \int x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x)) \, dx=\frac {8 b^3 x^3 \cot ^{-1}(c+(i+c) \tanh (a+b x))-4 i b^3 x^3 \log \left (1-\frac {i e^{-2 (a+b x)}}{c}\right )+6 i b^2 x^2 \operatorname {PolyLog}\left (2,\frac {i e^{-2 (a+b x)}}{c}\right )+6 i b x \operatorname {PolyLog}\left (3,\frac {i e^{-2 (a+b x)}}{c}\right )+3 i \operatorname {PolyLog}\left (4,\frac {i e^{-2 (a+b x)}}{c}\right )}{24 b^3} \]

[In]

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

[Out]

(8*b^3*x^3*ArcCot[c + (I + c)*Tanh[a + b*x]] - (4*I)*b^3*x^3*Log[1 - I/(c*E^(2*(a + b*x)))] + (6*I)*b^2*x^2*Po
lyLog[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)

Maple [C] (warning: unable to verify)

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

Time = 2.53 (sec) , antiderivative size = 1405, normalized size of antiderivative = 9.89

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

[In]

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

[Out]

-1/12*I/b^3*c/(I+c)*a^4+1/12*Pi*(csgn(I/(exp(2*b*x+2*a)+1))*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))-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))-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))^2+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
)/(exp(2*b*x+2*a)+1))^2-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
+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*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))*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))*cs
gn((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*exp(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))*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*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))^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)/(ex
p(2*b*x+2*a)+1))^3+csgn((2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)+1))^2)*x^3-1/2*I/b^3*a^3*ln(1+I*exp(b*x+a)*(I
*c)^(1/2))-1/2*I/b^2*a^2*ln(1+I*exp(b*x+a)*(I*c)^(1/2))*x+1/6*I/b^3*a^3*ln(-exp(2*b*x+2*a)*c+I)-1/2*I/b^2*a^2*
ln(1-I*exp(b*x+a)*(I*c)^(1/2))*x+1/12*I*b*c/(I+c)*x^4+1/2*I/b^2*ln(1+I*c*exp(2*b*x+2*a))*a^2*x+1/6*I*x^3*ln(2*
exp(2*b*x+2*a)*c-2*I)+1/3*I/b^3*ln(1+I*c*exp(2*b*x+2*a))*a^3+1/4*I/b^3*polylog(2,-I*c*exp(2*b*x+2*a))*a^2-1/4*
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^3-1/6*I*x^3*ln(2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)-1/6*I*x^3*ln(1+I*c*exp(2*b*x+2*a))-1/2*I/b^3*
a^3*ln(1-I*exp(b*x+a)*(I*c)^(1/2))-1/12*b/(I+c)*x^4-1/2*I/b^3*a^2*dilog(1+I*exp(b*x+a)*(I*c)^(1/2))-1/2*I/b^3*
a^2*dilog(1-I*exp(b*x+a)*(I*c)^(1/2))+1/12/b^3/(I+c)*a^4

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

Time = 0.27 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.06 \[ \int x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x)) \, dx=\frac {i \, b^{4} x^{4} + 2 i \, b^{3} x^{3} \log \left (\frac {{\left (c e^{\left (2 \, b x + 2 \, a\right )} - i\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{c + i}\right ) - 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) - 6 i \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) - i \, a^{4} + 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (b x + a\right )} + i \, \sqrt {-4 i \, c}}{2 \, c}\right ) + 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (b x + a\right )} - i \, \sqrt {-4 i \, c}}{2 \, c}\right ) + 12 i \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) + 12 i \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) - 2 \, {\left (i \, b^{3} x^{3} + i \, a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )} + 1\right ) - 2 \, {\left (i \, b^{3} x^{3} + i \, a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )} + 1\right ) - 12 i \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) - 12 i \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right )}{12 \, b^{3}} \]

[In]

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

[Out]

1/12*(I*b^4*x^4 + 2*I*b^3*x^3*log((c*e^(2*b*x + 2*a) - I)*e^(-2*b*x - 2*a)/(c + I)) - 6*I*b^2*x^2*dilog(1/2*sq
rt(-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*polylog(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. \[ \int x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x)) \, dx=\text {Exception raised: CoercionFailed} \]

[In]

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

[Out]

Exception raised: CoercionFailed >> Cannot convert 2*_t0**4*c**2*exp(4*a) + _t0**4*I*c*exp(4*a) - 3*_t0**2*I*c
*exp(2*a) + _t0**2*exp(2*a) - 1 of type <class 'sympy.core.add.Add'> to QQ_I[x,b,c,_t0,exp(a)]

Maxima [A] (verification not implemented)

none

Time = 1.27 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.91 \[ \int x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x)) \, dx=\frac {1}{3} \, x^{3} \operatorname {arccot}\left ({\left (c + i\right )} \tanh \left (b x + a\right ) + c\right ) - \frac {4}{9} \, {\left (\frac {3 \, x^{4}}{4 i \, c - 4} - \frac {4 \, b^{3} x^{3} \log \left (i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-i \, c e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-i \, c e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}(-i \, c e^{\left (2 \, b x + 2 \, a\right )})}{-2 \, b^{4} {\left (-i \, c + 1\right )}}\right )} b {\left (c + i\right )} \]

[In]

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

[Out]

1/3*x^3*arccot((c + I)*tanh(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]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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