Integrand size = 13, antiderivative size = 267 \[ \int x \arctan (c+d \tanh (a+b x)) \, dx=\frac {1}{2} x^2 \arctan (c+d \tanh (a+b x))+\frac {1}{4} i x^2 \log \left (1+\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )+\frac {i x \operatorname {PolyLog}\left (2,-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}-\frac {i x \operatorname {PolyLog}\left (2,-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}-\frac {i \operatorname {PolyLog}\left (3,-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{8 b^2}+\frac {i \operatorname {PolyLog}\left (3,-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{8 b^2} \]
1/2*x^2*arctan(c+d*tanh(b*x+a))+1/4*I*x^2*ln(1+(I-c-d)*exp(2*b*x+2*a)/(I-c +d))-1/4*I*x^2*ln(1+(I+c+d)*exp(2*b*x+2*a)/(I+c-d))+1/4*I*x*polylog(2,-(I- c-d)*exp(2*b*x+2*a)/(I-c+d))/b-1/4*I*x*polylog(2,-(I+c+d)*exp(2*b*x+2*a)/( I+c-d))/b-1/8*I*polylog(3,-(I-c-d)*exp(2*b*x+2*a)/(I-c+d))/b^2+1/8*I*polyl og(3,-(I+c+d)*exp(2*b*x+2*a)/(I+c-d))/b^2
Time = 0.78 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.24 \[ \int x \arctan (c+d \tanh (a+b x)) \, dx=\frac {1}{2} x^2 \arctan (c+d \tanh (a+b x))-\frac {d \left (2 b^2 x^2 \log \left (1+\frac {2 \left (1+(c+d)^2\right ) e^{2 (a+b x)}}{2+2 c^2-2 d^2-4 \sqrt {-d^2}}\right )-2 b^2 x^2 \log \left (1+\frac {\left (1+(c+d)^2\right ) e^{2 (a+b x)}}{1+c^2-d^2+2 \sqrt {-d^2}}\right )+2 b x \operatorname {PolyLog}\left (2,\frac {\left (1+c^2+2 c d+d^2\right ) e^{2 (a+b x)}}{-1-c^2+d^2+2 \sqrt {-d^2}}\right )-2 b x \operatorname {PolyLog}\left (2,-\frac {\left (1+(c+d)^2\right ) e^{2 (a+b x)}}{1+c^2-d^2+2 \sqrt {-d^2}}\right )-\operatorname {PolyLog}\left (3,-\frac {2 \left (1+(c+d)^2\right ) e^{2 (a+b x)}}{2+2 c^2-2 d^2-4 \sqrt {-d^2}}\right )+\operatorname {PolyLog}\left (3,-\frac {\left (1+(c+d)^2\right ) e^{2 (a+b x)}}{1+c^2-d^2+2 \sqrt {-d^2}}\right )\right )}{8 b^2 \sqrt {-d^2}} \]
(x^2*ArcTan[c + d*Tanh[a + b*x]])/2 - (d*(2*b^2*x^2*Log[1 + (2*(1 + (c + d )^2)*E^(2*(a + b*x)))/(2 + 2*c^2 - 2*d^2 - 4*Sqrt[-d^2])] - 2*b^2*x^2*Log[ 1 + ((1 + (c + d)^2)*E^(2*(a + b*x)))/(1 + c^2 - d^2 + 2*Sqrt[-d^2])] + 2* b*x*PolyLog[2, ((1 + c^2 + 2*c*d + d^2)*E^(2*(a + b*x)))/(-1 - c^2 + d^2 + 2*Sqrt[-d^2])] - 2*b*x*PolyLog[2, -(((1 + (c + d)^2)*E^(2*(a + b*x)))/(1 + c^2 - d^2 + 2*Sqrt[-d^2]))] - PolyLog[3, (-2*(1 + (c + d)^2)*E^(2*(a + b *x)))/(2 + 2*c^2 - 2*d^2 - 4*Sqrt[-d^2])] + PolyLog[3, -(((1 + (c + d)^2)* E^(2*(a + b*x)))/(1 + c^2 - d^2 + 2*Sqrt[-d^2]))]))/(8*b^2*Sqrt[-d^2])
Time = 1.05 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.28, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {5722, 2620, 3011, 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.
\(\displaystyle \int x \arctan (d \tanh (a+b x)+c) \, dx\) |
\(\Big \downarrow \) 5722 |
\(\displaystyle -\frac {1}{2} b (1+i (c+d)) \int \frac {e^{2 a+2 b x} x^2}{-c+(-c-d+i) e^{2 a+2 b x}+d+i}dx+\frac {1}{2} b (1-i (c+d)) \int \frac {e^{2 a+2 b x} x^2}{c+(c+d+i) e^{2 a+2 b x}-d+i}dx+\frac {1}{2} x^2 \arctan (d \tanh (a+b x)+c)\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -\frac {1}{2} b (1+i (c+d)) \left (\frac {x^2 \log \left (1+\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{2 b (-c-d+i)}-\frac {\int x \log \left (\frac {e^{2 a+2 b x} (-c-d+i)}{-c+d+i}+1\right )dx}{b (-c-d+i)}\right )+\frac {1}{2} b (1-i (c+d)) \left (\frac {x^2 \log \left (1+\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{2 b (c+d+i)}-\frac {\int x \log \left (\frac {e^{2 a+2 b x} (c+d+i)}{c-d+i}+1\right )dx}{b (c+d+i)}\right )+\frac {1}{2} x^2 \arctan (d \tanh (a+b x)+c)\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle -\frac {1}{2} b (1+i (c+d)) \left (\frac {x^2 \log \left (1+\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{2 b (-c-d+i)}-\frac {\frac {\int \operatorname {PolyLog}\left (2,-\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )dx}{2 b}-\frac {x \operatorname {PolyLog}\left (2,-\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{2 b}}{b (-c-d+i)}\right )+\frac {1}{2} b (1-i (c+d)) \left (\frac {x^2 \log \left (1+\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{2 b (c+d+i)}-\frac {\frac {\int \operatorname {PolyLog}\left (2,-\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )dx}{2 b}-\frac {x \operatorname {PolyLog}\left (2,-\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{2 b}}{b (c+d+i)}\right )+\frac {1}{2} x^2 \arctan (d \tanh (a+b x)+c)\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle -\frac {1}{2} b (1+i (c+d)) \left (\frac {x^2 \log \left (1+\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{2 b (-c-d+i)}-\frac {\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (2,-\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )de^{2 a+2 b x}}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{2 b}}{b (-c-d+i)}\right )+\frac {1}{2} b (1-i (c+d)) \left (\frac {x^2 \log \left (1+\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{2 b (c+d+i)}-\frac {\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (2,-\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )de^{2 a+2 b x}}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{2 b}}{b (c+d+i)}\right )+\frac {1}{2} x^2 \arctan (d \tanh (a+b x)+c)\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {1}{2} x^2 \arctan (d \tanh (a+b x)+c)-\frac {1}{2} b (1+i (c+d)) \left (\frac {x^2 \log \left (1+\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{2 b (-c-d+i)}-\frac {\frac {\operatorname {PolyLog}\left (3,-\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{2 b}}{b (-c-d+i)}\right )+\frac {1}{2} b (1-i (c+d)) \left (\frac {x^2 \log \left (1+\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{2 b (c+d+i)}-\frac {\frac {\operatorname {PolyLog}\left (3,-\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{2 b}}{b (c+d+i)}\right )\) |
(x^2*ArcTan[c + d*Tanh[a + b*x]])/2 - (b*(1 + I*(c + d))*((x^2*Log[1 + ((I - c - d)*E^(2*a + 2*b*x))/(I - c + d)])/(2*b*(I - c - d)) - (-1/2*(x*Poly Log[2, -(((I - c - d)*E^(2*a + 2*b*x))/(I - c + d))])/b + PolyLog[3, -(((I - c - d)*E^(2*a + 2*b*x))/(I - c + d))]/(4*b^2))/(b*(I - c - d))))/2 + (b *(1 - I*(c + d))*((x^2*Log[1 + ((I + c + d)*E^(2*a + 2*b*x))/(I + c - d)]) /(2*b*(I + c + d)) - (-1/2*(x*PolyLog[2, -(((I + c + d)*E^(2*a + 2*b*x))/( I + c - d))])/b + PolyLog[3, -(((I + c + d)*E^(2*a + 2*b*x))/(I + c - d))] /(4*b^2))/(b*(I + c + d))))/2
3.1.82.3.1 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[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] + (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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 3.00 (sec) , antiderivative size = 6567, normalized size of antiderivative = 24.60
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1067 vs. \(2 (197) = 394\).
Time = 0.36 (sec) , antiderivative size = 1067, normalized size of antiderivative = 4.00 \[ \int x \arctan (c+d \tanh (a+b x)) \, dx=\text {Too large to display} \]
1/4*(2*b^2*x^2*arctan((c*cosh(b*x + a) + d*sinh(b*x + a))/cosh(b*x + a)) + 2*I*b*x*dilog(sqrt(-(c^2 - d^2 + 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 1))*(cos h(b*x + a) + sinh(b*x + a))) + 2*I*b*x*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))) - 2*I*b*x*dilo g(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))) - 2*I*b*x*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^2*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^2*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^2*log(2*(c^2 + 2*c*d + d^2 + 1)*cos h(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^2*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*b^2*x^2 - I*a^2)*log(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)) + 1) + (I*b^2*x^ 2 - I*a^2)*log(-sqrt(-(c^2 - d^2 + 2*I*d + 1)/(c^2 - 2*c*d + d^2 + 1))*(co sh(b*x + a) + sinh(b*x + a)) + 1) + (-I*b^2*x^2 + I*a^2)*log(sqrt(-(c^2...
\[ \int x \arctan (c+d \tanh (a+b x)) \, dx=\int x \operatorname {atan}{\left (c + d \tanh {\left (a + b x \right )} \right )}\, dx \]
\[ \int x \arctan (c+d \tanh (a+b x)) \, dx=\int { x \arctan \left (d \tanh \left (b x + a\right ) + c\right ) \,d x } \]
1/2*x^2*arctan(((c*e^(2*a) + d*e^(2*a))*e^(2*b*x) + c - d)/(e^(2*b*x + 2*a ) + 1)) - 2*b*d*integrate(x^2*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)
\[ \int x \arctan (c+d \tanh (a+b x)) \, dx=\int { x \arctan \left (d \tanh \left (b x + a\right ) + c\right ) \,d x } \]
Timed out. \[ \int x \arctan (c+d \tanh (a+b x)) \, dx=\int x\,\mathrm {atan}\left (c+d\,\mathrm {tanh}\left (a+b\,x\right )\right ) \,d x \]