Integrand size = 15, antiderivative size = 395 \[ \int x^2 \text {arctanh}(c+d \tan (a+b x)) \, dx=\frac {1}{3} x^3 \text {arctanh}(c+d \tan (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b}+\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b}+\frac {x \operatorname {PolyLog}\left (3,-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b^2}-\frac {x \operatorname {PolyLog}\left (3,-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b^2}+\frac {i \operatorname {PolyLog}\left (4,-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{8 b^3}-\frac {i \operatorname {PolyLog}\left (4,-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{8 b^3} \]
1/3*x^3*arctanh(c+d*tan(b*x+a))+1/6*x^3*ln(1+(1-c+I*d)*exp(2*I*a+2*I*b*x)/ (1-c-I*d))-1/6*x^3*ln(1+(1+c-I*d)*exp(2*I*a+2*I*b*x)/(1+c+I*d))-1/4*I*x^2* polylog(2,-(1-c+I*d)*exp(2*I*a+2*I*b*x)/(1-c-I*d))/b+1/4*I*x^2*polylog(2,- (1+c-I*d)*exp(2*I*a+2*I*b*x)/(1+c+I*d))/b+1/4*x*polylog(3,-(1-c+I*d)*exp(2 *I*a+2*I*b*x)/(1-c-I*d))/b^2-1/4*x*polylog(3,-(1+c-I*d)*exp(2*I*a+2*I*b*x) /(1+c+I*d))/b^2+1/8*I*polylog(4,-(1-c+I*d)*exp(2*I*a+2*I*b*x)/(1-c-I*d))/b ^3-1/8*I*polylog(4,-(1+c-I*d)*exp(2*I*a+2*I*b*x)/(1+c+I*d))/b^3
Time = 3.61 (sec) , antiderivative size = 349, normalized size of antiderivative = 0.88 \[ \int x^2 \text {arctanh}(c+d \tan (a+b x)) \, dx=\frac {8 b^3 x^3 \text {arctanh}(c+d \tan (a+b x))+4 b^3 x^3 \log \left (1+\frac {(-1+c+i d) e^{-2 i (a+b x)}}{-1+c-i d}\right )-4 b^3 x^3 \log \left (1+\frac {(1+c+i d) e^{-2 i (a+b x)}}{1+c-i d}\right )+6 i b^2 x^2 \operatorname {PolyLog}\left (2,\frac {(1-c-i d) e^{-2 i (a+b x)}}{-1+c-i d}\right )-6 i b^2 x^2 \operatorname {PolyLog}\left (2,\frac {(-1-c-i d) e^{-2 i (a+b x)}}{1+c-i d}\right )+6 b x \operatorname {PolyLog}\left (3,\frac {(1-c-i d) e^{-2 i (a+b x)}}{-1+c-i d}\right )-6 b x \operatorname {PolyLog}\left (3,\frac {(-1-c-i d) e^{-2 i (a+b x)}}{1+c-i d}\right )-3 i \operatorname {PolyLog}\left (4,\frac {(1-c-i d) e^{-2 i (a+b x)}}{-1+c-i d}\right )+3 i \operatorname {PolyLog}\left (4,\frac {(-1-c-i d) e^{-2 i (a+b x)}}{1+c-i d}\right )}{24 b^3} \]
(8*b^3*x^3*ArcTanh[c + d*Tan[a + b*x]] + 4*b^3*x^3*Log[1 + (-1 + c + I*d)/ ((-1 + c - I*d)*E^((2*I)*(a + b*x)))] - 4*b^3*x^3*Log[1 + (1 + c + I*d)/(( 1 + c - I*d)*E^((2*I)*(a + b*x)))] + (6*I)*b^2*x^2*PolyLog[2, (1 - c - I*d )/((-1 + c - I*d)*E^((2*I)*(a + b*x)))] - (6*I)*b^2*x^2*PolyLog[2, (-1 - c - I*d)/((1 + c - I*d)*E^((2*I)*(a + b*x)))] + 6*b*x*PolyLog[3, (1 - c - I *d)/((-1 + c - I*d)*E^((2*I)*(a + b*x)))] - 6*b*x*PolyLog[3, (-1 - c - I*d )/((1 + c - I*d)*E^((2*I)*(a + b*x)))] - (3*I)*PolyLog[4, (1 - c - I*d)/(( -1 + c - I*d)*E^((2*I)*(a + b*x)))] + (3*I)*PolyLog[4, (-1 - c - I*d)/((1 + c - I*d)*E^((2*I)*(a + b*x)))])/(24*b^3)
Time = 1.55 (sec) , antiderivative size = 522, normalized size of antiderivative = 1.32, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6821, 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.
| \(\displaystyle \int x^2 \text {arctanh}(d \tan (a+b x)+c) \, dx\) |
| \(\Big \downarrow \) 6821 |
| \(\displaystyle -\frac {1}{3} b (i c+d+i) \int \frac {e^{2 i a+2 i b x} x^3}{c+(c-i d+1) e^{2 i a+2 i b x}+i d+1}dx+\frac {1}{3} b (-d+i (1-c)) \int \frac {e^{2 i a+2 i b x} x^3}{-c+(-c+i d+1) e^{2 i a+2 i b x}-i d+1}dx+\frac {1}{3} x^3 \text {arctanh}(d \tan (a+b x)+c)\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {1}{3} b (-d+i (1-c)) \left (\frac {x^3 \log \left (1+\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b (-d+i (1-c))}-\frac {3 \int x^2 \log \left (\frac {e^{2 i a+2 i b x} (-c+i d+1)}{-c-i d+1}+1\right )dx}{2 b (-d+i (1-c))}\right )-\frac {1}{3} b (i c+d+i) \left (\frac {x^3 \log \left (1+\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 (b d+i (b c+b))}-\frac {3 \int x^2 \log \left (\frac {e^{2 i a+2 i b x} (c-i d+1)}{c+i d+1}+1\right )dx}{2 (b d+i (b c+b))}\right )+\frac {1}{3} x^3 \text {arctanh}(d \tan (a+b x)+c)\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {1}{3} b (-d+i (1-c)) \left (\frac {x^3 \log \left (1+\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b (-d+i (1-c))}-\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )dx}{b}\right )}{2 b (-d+i (1-c))}\right )-\frac {1}{3} b (i c+d+i) \left (\frac {x^3 \log \left (1+\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 (b d+i (b c+b))}-\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )dx}{b}\right )}{2 (b d+i (b c+b))}\right )+\frac {1}{3} x^3 \text {arctanh}(d \tan (a+b x)+c)\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {1}{3} b (-d+i (1-c)) \left (\frac {x^3 \log \left (1+\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b (-d+i (1-c))}-\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b}-\frac {i \left (\frac {i \int \operatorname {PolyLog}\left (3,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )dx}{2 b}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b}\right )}{b}\right )}{2 b (-d+i (1-c))}\right )-\frac {1}{3} b (i c+d+i) \left (\frac {x^3 \log \left (1+\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 (b d+i (b c+b))}-\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 b}-\frac {i \left (\frac {i \int \operatorname {PolyLog}\left (3,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )dx}{2 b}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 b}\right )}{b}\right )}{2 (b d+i (b c+b))}\right )+\frac {1}{3} x^3 \text {arctanh}(d \tan (a+b x)+c)\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {1}{3} b (-d+i (1-c)) \left (\frac {x^3 \log \left (1+\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b (-d+i (1-c))}-\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b}-\frac {i \left (\frac {\int e^{-2 i a-2 i b x} \operatorname {PolyLog}\left (3,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )de^{2 i a+2 i b x}}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b}\right )}{b}\right )}{2 b (-d+i (1-c))}\right )-\frac {1}{3} b (i c+d+i) \left (\frac {x^3 \log \left (1+\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 (b d+i (b c+b))}-\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 b}-\frac {i \left (\frac {\int e^{-2 i a-2 i b x} \operatorname {PolyLog}\left (3,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )de^{2 i a+2 i b x}}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 b}\right )}{b}\right )}{2 (b d+i (b c+b))}\right )+\frac {1}{3} x^3 \text {arctanh}(d \tan (a+b x)+c)\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {1}{3} x^3 \text {arctanh}(d \tan (a+b x)+c)+\frac {1}{3} b (-d+i (1-c)) \left (\frac {x^3 \log \left (1+\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b (-d+i (1-c))}-\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b}-\frac {i \left (\frac {\operatorname {PolyLog}\left (4,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{2 b}\right )}{b}\right )}{2 b (-d+i (1-c))}\right )-\frac {1}{3} b (i c+d+i) \left (\frac {x^3 \log \left (1+\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 (b d+i (b c+b))}-\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 b}-\frac {i \left (\frac {\operatorname {PolyLog}\left (4,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{2 b}\right )}{b}\right )}{2 (b d+i (b c+b))}\right )\) |
(x^3*ArcTanh[c + d*Tan[a + b*x]])/3 + (b*(I*(1 - c) - d)*((x^3*Log[1 + ((1 - c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - c - I*d)])/(2*b*(I*(1 - c) - d)) - (3*(((I/2)*x^2*PolyLog[2, -(((1 - c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - c - I*d))])/b - (I*(((-1/2*I)*x*PolyLog[3, -(((1 - c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - c - I*d))])/b + PolyLog[4, -(((1 - c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - c - I*d))]/(4*b^2)))/b))/(2*b*(I*(1 - c) - d))))/3 - (b* (I + I*c + d)*((x^3*Log[1 + ((1 + c - I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c + I*d)])/(2*(I*(b + b*c) + b*d)) - (3*(((I/2)*x^2*PolyLog[2, -(((1 + c - I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c + I*d))])/b - (I*(((-1/2*I)*x*PolyLog [3, -(((1 + c - I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c + I*d))])/b + PolyLog [4, -(((1 + c - I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c + I*d))]/(4*b^2)))/b) )/(2*(I*(b + b*c) + b*d))))/3
3.4.17.3.1 Defintions of rubi rules used
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]
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]]]
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]
Int[ArcTanh[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_ .), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[c + d*Tan[a + b*x]]/(f*(m + 1))), x] + (-Simp[I*b*((1 + c - I*d)/(f*(m + 1))) Int[(e + f*x)^(m + 1) *(E^(2*I*a + 2*I*b*x)/(1 + c + I*d + (1 + c - I*d)*E^(2*I*a + 2*I*b*x))), x ], x] + Simp[I*b*((1 - c + I*d)/(f*(m + 1))) Int[(e + f*x)^(m + 1)*(E^(2* I*a + 2*I*b*x)/(1 - c - I*d + (1 - c + I*d)*E^(2*I*a + 2*I*b*x))), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && NeQ[(c + I*d)^2, 1]
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 21.30 (sec) , antiderivative size = 6921, normalized size of antiderivative = 17.52
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2165 vs. \(2 (279) = 558\).
Time = 0.34 (sec) , antiderivative size = 2165, normalized size of antiderivative = 5.48 \[ \int x^2 \text {arctanh}(c+d \tan (a+b x)) \, dx=\text {Too large to display} \]
1/48*(8*b^3*x^3*log(-(d*tan(b*x + a) + c + 1)/(d*tan(b*x + a) + c - 1)) - 6*I*b^2*x^2*dilog(2*((I*(c + 1)*d - d^2)*tan(b*x + a)^2 - c^2 - I*(c + 1)* d + (I*c^2 - 2*(c + 1)*d - I*d^2 + 2*I*c + I)*tan(b*x + a) - 2*c - 1)/((c^ 2 + d^2 + 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*c + 1) + 1) + 6*I*b^2*x^ 2*dilog(2*((-I*(c + 1)*d - d^2)*tan(b*x + a)^2 - c^2 + I*(c + 1)*d + (-I*c ^2 - 2*(c + 1)*d + I*d^2 - 2*I*c - I)*tan(b*x + a) - 2*c - 1)/((c^2 + d^2 + 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*c + 1) + 1) + 6*I*b^2*x^2*dilog( 2*((I*(c - 1)*d - d^2)*tan(b*x + a)^2 - c^2 - I*(c - 1)*d + (I*c^2 - 2*(c - 1)*d - I*d^2 - 2*I*c + I)*tan(b*x + a) + 2*c - 1)/((c^2 + d^2 - 2*c + 1) *tan(b*x + a)^2 + c^2 + d^2 - 2*c + 1) + 1) - 6*I*b^2*x^2*dilog(2*((-I*(c - 1)*d - d^2)*tan(b*x + a)^2 - c^2 + I*(c - 1)*d + (-I*c^2 - 2*(c - 1)*d + I*d^2 + 2*I*c - I)*tan(b*x + a) + 2*c - 1)/((c^2 + d^2 - 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*c + 1) + 1) + 4*a^3*log(((I*(c + 1)*d + d^2)*tan(b *x + a)^2 - c^2 + I*(c + 1)*d + (I*c^2 + I*d^2 + 2*I*c + I)*tan(b*x + a) - 2*c - 1)/(tan(b*x + a)^2 + 1)) + 4*a^3*log(((I*(c + 1)*d - d^2)*tan(b*x + a)^2 + c^2 + I*(c + 1)*d + (I*c^2 + I*d^2 + 2*I*c + I)*tan(b*x + a) + 2*c + 1)/(tan(b*x + a)^2 + 1)) - 4*a^3*log(((I*(c - 1)*d + d^2)*tan(b*x + a)^ 2 - c^2 + I*(c - 1)*d + (I*c^2 + I*d^2 - 2*I*c + I)*tan(b*x + a) + 2*c - 1 )/(tan(b*x + a)^2 + 1)) - 4*a^3*log(((I*(c - 1)*d - d^2)*tan(b*x + a)^2 + c^2 + I*(c - 1)*d + (I*c^2 + I*d^2 - 2*I*c + I)*tan(b*x + a) - 2*c + 1)...
\[ \int x^2 \text {arctanh}(c+d \tan (a+b x)) \, dx=\int x^{2} \operatorname {atanh}{\left (c + d \tan {\left (a + b x \right )} \right )}\, dx \]
\[ \int x^2 \text {arctanh}(c+d \tan (a+b x)) \, dx=\int { x^{2} \operatorname {artanh}\left (d \tan \left (b x + a\right ) + c\right ) \,d x } \]
1/12*x^3*log((c^2 + d^2 + 2*c + 1)*cos(2*b*x + 2*a)^2 + 4*(c + 1)*d*sin(2* b*x + 2*a) + (c^2 + d^2 + 2*c + 1)*sin(2*b*x + 2*a)^2 + c^2 + d^2 + 2*(c^2 - d^2 + 2*c + 1)*cos(2*b*x + 2*a) + 2*c + 1) - 1/12*x^3*log((c^2 + d^2 - 2*c + 1)*cos(2*b*x + 2*a)^2 + 4*(c - 1)*d*sin(2*b*x + 2*a) + (c^2 + d^2 - 2*c + 1)*sin(2*b*x + 2*a)^2 + c^2 + d^2 + 2*(c^2 - d^2 - 2*c + 1)*cos(2*b* x + 2*a) - 2*c + 1) - 4*b*d*integrate(-1/3*(2*(c^2 + d^2 - 1)*x^3*cos(2*b* x + 2*a)^2 + 2*c*d*x^3*sin(2*b*x + 2*a) + 2*(c^2 + d^2 - 1)*x^3*sin(2*b*x + 2*a)^2 + (c^2 - d^2 - 1)*x^3*cos(2*b*x + 2*a) - (2*c*d*x^3*sin(2*b*x + 2 *a) - (c^2 - d^2 - 1)*x^3*cos(2*b*x + 2*a))*cos(4*b*x + 4*a) + (2*c*d*x^3* cos(2*b*x + 2*a) + (c^2 - d^2 - 1)*x^3*sin(2*b*x + 2*a))*sin(4*b*x + 4*a)) /(c^4 + d^4 + 2*(c^2 + 1)*d^2 + (c^4 + d^4 + 2*(c^2 + 1)*d^2 - 2*c^2 + 1)* cos(4*b*x + 4*a)^2 + 4*(c^4 + d^4 + 2*(c^2 - 1)*d^2 - 2*c^2 + 1)*cos(2*b*x + 2*a)^2 + (c^4 + d^4 + 2*(c^2 + 1)*d^2 - 2*c^2 + 1)*sin(4*b*x + 4*a)^2 + 4*(c^4 + d^4 + 2*(c^2 - 1)*d^2 - 2*c^2 + 1)*sin(2*b*x + 2*a)^2 - 2*c^2 + 2*(c^4 + d^4 - 2*(3*c^2 - 1)*d^2 - 2*c^2 + 2*(c^4 - d^4 - 2*c^2 + 1)*cos(2 *b*x + 2*a) - 4*(c*d^3 + (c^3 - c)*d)*sin(2*b*x + 2*a) + 1)*cos(4*b*x + 4* a) + 4*(c^4 - d^4 - 2*c^2 + 1)*cos(2*b*x + 2*a) - 4*(2*c*d^3 - 2*(c^3 - c) *d - 2*(c*d^3 + (c^3 - c)*d)*cos(2*b*x + 2*a) - (c^4 - d^4 - 2*c^2 + 1)*si n(2*b*x + 2*a))*sin(4*b*x + 4*a) + 8*(c*d^3 + (c^3 - c)*d)*sin(2*b*x + 2*a ) + 1), x)
\[ \int x^2 \text {arctanh}(c+d \tan (a+b x)) \, dx=\int { x^{2} \operatorname {artanh}\left (d \tan \left (b x + a\right ) + c\right ) \,d x } \]
Timed out. \[ \int x^2 \text {arctanh}(c+d \tan (a+b x)) \, dx=\int x^2\,\mathrm {atanh}\left (c+d\,\mathrm {tan}\left (a+b\,x\right )\right ) \,d x \]