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

Optimal. Leaf size=145 \[ \frac {i \text {Li}_4\left (i c e^{2 a+2 b x}\right )}{8 b^3}-\frac {i x \text {Li}_3\left (i c e^{2 a+2 b x}\right )}{4 b^2}+\frac {i x^2 \text {Li}_2\left (i c e^{2 a+2 b x}\right )}{4 b}+\frac {1}{6} i x^3 \log \left (1-i c e^{2 a+2 b x}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(-c+i) \tanh (a+b x))-\frac {1}{12} i b x^4 \]

[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]  time = 0.23, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {5196, 2184, 2190, 2531, 6609, 2282, 6589} \[ -\frac {i x \text {PolyLog}\left (3,i c e^{2 a+2 b x}\right )}{4 b^2}+\frac {i \text {PolyLog}\left (4,i c e^{2 a+2 b x}\right )}{8 b^3}+\frac {i x^2 \text {PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{4 b}+\frac {1}{6} i x^3 \log \left (1-i c e^{2 a+2 b x}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(-c+i) \tanh (a+b x))-\frac {1}{12} i b x^4 \]

Antiderivative was successfully verified.

[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 2184

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2282

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 2531

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 5196

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 6589

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 6609

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*} \int x^2 \cot ^{-1}(c-(i-c) \tanh (a+b x)) \, dx &=\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 \text {Li}_2\left (i c e^{2 a+2 b x}\right )}{4 b}-\frac {i \int x \text {Li}_2\left (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 \text {Li}_2\left (i c e^{2 a+2 b x}\right )}{4 b}-\frac {i x \text {Li}_3\left (i c e^{2 a+2 b x}\right )}{4 b^2}+\frac {i \int \text {Li}_3\left (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 \text {Li}_2\left (i c e^{2 a+2 b x}\right )}{4 b}-\frac {i x \text {Li}_3\left (i c e^{2 a+2 b x}\right )}{4 b^2}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_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 \text {Li}_2\left (i c e^{2 a+2 b x}\right )}{4 b}-\frac {i x \text {Li}_3\left (i c e^{2 a+2 b x}\right )}{4 b^2}+\frac {i \text {Li}_4\left (i c e^{2 a+2 b x}\right )}{8 b^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.23, size = 128, normalized size = 0.88 \[ \frac {i \left (4 b^3 x^3 \log \left (1+\frac {i e^{-2 (a+b x)}}{c}\right )-6 b^2 x^2 \text {Li}_2\left (-\frac {i e^{-2 (a+b x)}}{c}\right )-6 b x \text {Li}_3\left (-\frac {i e^{-2 (a+b x)}}{c}\right )-3 \text {Li}_4\left (-\frac {i e^{-2 (a+b x)}}{c}\right )\right )}{24 b^3}+\frac {1}{3} x^3 \cot ^{-1}(c+(c-i) \tanh (a+b x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [C]  time = 0.65, size = 290, normalized size = 2.00 \[ \frac {-i \, b^{4} x^{4} + 2 i \, b^{3} x^{3} \log \left (\frac {{\left (c - i\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c e^{\left (2 \, b x + 2 \, a\right )} + i}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) + i \, a^{4} - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (b x + a\right )} + i \, \sqrt {4 i \, c}}{2 \, c}\right ) - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (b x + a\right )} - i \, \sqrt {4 i \, c}}{2 \, c}\right ) - 12 i \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) - 12 i \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) + {\left (2 i \, b^{3} x^{3} + 2 i \, a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )} + 1\right ) + {\left (2 i \, b^{3} x^{3} + 2 i \, a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )} + 1\right ) + 12 i \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) + 12 i \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right )}{12 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 - I)*e^(2*b*x + 2*a)/(c*e^(2*b*x + 2*a) + 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*sqr
t(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 + 2*I*a^3)*log(1/2*sq
rt(4*I*c)*e^(b*x + a) + 1) + (2*I*b^3*x^3 + 2*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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {arccot}\left ({\left (c - i\right )} \tanh \left (b x + a\right ) + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [C]  time = 6.24, size = 1570, normalized size = 10.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.00, size = 129, normalized size = 0.89 \[ \frac {1}{3} \, x^{3} \operatorname {arccot}\left ({\left (c - i\right )} \tanh \left (b x + a\right ) + c\right ) + \frac {4}{9} \, {\left (\frac {3 \, x^{4}}{4 i \, c + 4} - \frac {4 \, b^{3} x^{3} \log \left (-i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (i \, c e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}(i \, c e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}(i \, c e^{\left (2 \, b x + 2 \, a\right )})}{-2 \, b^{4} {\left (-i \, c - 1\right )}}\right )} b {\left (c - i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {acot}\left (c+\mathrm {tanh}\left (a+b\,x\right )\,\left (c-\mathrm {i}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: CoercionFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: CoercionFailed

________________________________________________________________________________________