3.89 \(\int x^2 \tan ^{-1}(c-(i-c) \tanh (a+b x)) \, dx\)
Optimal. Leaf size=145 \[ \frac{i x \text{PolyLog}\left (3,i c e^{2 a+2 b x}\right )}{4 b^2}-\frac{i \text{PolyLog}\left (4,i c e^{2 a+2 b x}\right )}{8 b^3}-\frac{i x^2 \text{PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{4 b}-\frac{1}{6} i x^3 \log \left (1-i c e^{2 a+2 b x}\right )+\frac{1}{3} x^3 \tan ^{-1}(c-(-c+i) \tanh (a+b x))+\frac{1}{12} i b x^4 \]
[Out]
(I/12)*b*x^4 + (x^3*ArcTan[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
________________________________________________________________________________________
Rubi [A] time = 0.220334, antiderivative size = 145, normalized size of antiderivative = 1.,
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 =
{5195, 2184, 2190, 2531, 6609, 2282, 6589} \[ \frac{i x \text{PolyLog}\left (3,i c e^{2 a+2 b x}\right )}{4 b^2}-\frac{i \text{PolyLog}\left (4,i c e^{2 a+2 b x}\right )}{8 b^3}-\frac{i x^2 \text{PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{4 b}-\frac{1}{6} i x^3 \log \left (1-i c e^{2 a+2 b x}\right )+\frac{1}{3} x^3 \tan ^{-1}(c-(-c+i) \tanh (a+b x))+\frac{1}{12} i b x^4 \]
Antiderivative was successfully verified.
[In]
Int[x^2*ArcTan[c - (I - c)*Tanh[a + b*x]],x]
[Out]
(I/12)*b*x^4 + (x^3*ArcTan[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 5195
Int[ArcTan[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*ArcTan[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 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 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 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]
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 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]
Rubi steps
\begin{align*} \int x^2 \tan ^{-1}(c-(i-c) \tanh (a+b x)) \, dx &=\frac{1}{3} x^3 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 = 5.10254, size = 128, normalized size = 0.88 \[ \frac{1}{3} x^3 \tan ^{-1}(c+(c-i) \tanh (a+b x))-\frac{i \left (-6 b^2 x^2 \text{PolyLog}\left (2,-\frac{i e^{-2 (a+b x)}}{c}\right )-6 b x \text{PolyLog}\left (3,-\frac{i e^{-2 (a+b x)}}{c}\right )-3 \text{PolyLog}\left (4,-\frac{i e^{-2 (a+b x)}}{c}\right )+4 b^3 x^3 \log \left (1+\frac{i e^{-2 (a+b x)}}{c}\right )\right )}{24 b^3} \]
Antiderivative was successfully verified.
[In]
Integrate[x^2*ArcTan[c - (I - c)*Tanh[a + b*x]],x]
[Out]
(x^3*ArcTan[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
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Maple [C] time = 10.954, size = 1570, normalized size = 10.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*arctan(c-(I-c)*tanh(b*x+a)),x)
[Out]
-1/6*Pi*x^3+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))^2+1/2*I/b^2*ln(1-I*c
*exp(2*b*x+2*a))*x*a^2-1/2*I/b^2*a^2*ln(1-I*exp(b*x+a)*(-I*c)^(1/2))*x-1/8*I*polylog(4,I*c*exp(2*b*x+2*a))/b^3
+1/4*I*x*polylog(3,I*c*exp(2*b*x+2*a))/b^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
/6*I*x^3*ln(-2*exp(2*b*x+2*a)*c-2*I)-1/6*I*x^3*ln(2*I*exp(2*b*x+2*a)-2*exp(2*b*x+2*a)*c)+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+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/b^3/(I-c)*a^4-1/12*b/(I-c)*x^4-1/6*I*x^3*ln(1-I*c*exp(2*b*x+2*a))+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^2-1/2*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((2*exp(2*b*x+2*a)*c+2*I)/(exp(2*b*x+2*a)+1))^3+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*c/(I-c)*a^4+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))+1/6*I/b^3*a^3*ln(exp(2
*b*x+2*a)*c+I)-1/12*I*c*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)/(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))*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))*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))*c
sgn((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))+1/3/b^3*a^3/(I-c)*ln(ex
p(b*x+a))-1/3/b^2/(I-c)*x*a^3+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/(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))-1/4*I*x^2*polylog(2,
I*c*exp(2*b*x+2*a))/b-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+1/3*I/b^3*c*a^3/(I-c)*ln(exp(b*x+a))-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))^2-1/3*I/b^2*c/(I-c)*x*a^3-1/2*I/
b^2*a^2*ln(1+I*exp(b*x+a)*(-I*c)^(1/2))*x
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Maxima [A] time = 5.8979, size = 174, normalized size = 1.2 \begin{align*} \frac{1}{3} \, x^{3} \arctan \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 )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*arctan(c-(I-c)*tanh(b*x+a)),x, algorithm="maxima")
[Out]
1/3*x^3*arctan((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)
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Fricas [C] time = 2.01688, size = 857, normalized size = 5.91 \begin{align*} \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 ) +{\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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*arctan(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*s
qrt(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*sq
rt(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*
sqrt(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
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{i b \int \frac{x^{3}}{i c e^{2 a} e^{2 b x} - 1}\, dx}{3} + \frac{i x^{3} \log{\left (- i c + \frac{i c}{e^{2 a} e^{2 b x} + 1} - \frac{i c e^{a} e^{b x}}{e^{a} e^{b x} + e^{- a} e^{- b x}} + 1 + \frac{1}{e^{2 a} e^{2 b x} + 1} - \frac{e^{a} e^{b x}}{e^{a} e^{b x} + e^{- a} e^{- b x}} \right )}}{6} - \frac{i x^{3} \log{\left (i c - \frac{i c}{e^{2 a} e^{2 b x} + 1} + \frac{i c e^{a} e^{b x}}{e^{a} e^{b x} + e^{- a} e^{- b x}} + 1 - \frac{1}{e^{2 a} e^{2 b x} + 1} + \frac{e^{a} e^{b x}}{e^{a} e^{b x} + e^{- a} e^{- b x}} \right )}}{6} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2*atan(c-(I-c)*tanh(b*x+a)),x)
[Out]
-I*b*Integral(x**3/(I*c*exp(2*a)*exp(2*b*x) - 1), x)/3 + I*x**3*log(-I*c + I*c/(exp(2*a)*exp(2*b*x) + 1) - I*c
*exp(a)*exp(b*x)/(exp(a)*exp(b*x) + exp(-a)*exp(-b*x)) + 1 + 1/(exp(2*a)*exp(2*b*x) + 1) - exp(a)*exp(b*x)/(ex
p(a)*exp(b*x) + exp(-a)*exp(-b*x)))/6 - I*x**3*log(I*c - I*c/(exp(2*a)*exp(2*b*x) + 1) + I*c*exp(a)*exp(b*x)/(
exp(a)*exp(b*x) + exp(-a)*exp(-b*x)) + 1 - 1/(exp(2*a)*exp(2*b*x) + 1) + exp(a)*exp(b*x)/(exp(a)*exp(b*x) + ex
p(-a)*exp(-b*x)))/6
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \arctan \left ({\left (c - i\right )} \tanh \left (b x + a\right ) + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*arctan(c-(I-c)*tanh(b*x+a)),x, algorithm="giac")
[Out]
integrate(x^2*arctan((c - I)*tanh(b*x + a) + c), x)