Integrand size = 14, antiderivative size = 101 \[ \int x \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\frac {b x^3}{6}+\frac {1}{2} x^2 \text {arctanh}(1+d+d \tanh (a+b x))-\frac {1}{4} x^2 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac {x \operatorname {PolyLog}\left (2,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {\operatorname {PolyLog}\left (3,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{8 b^2} \]
1/6*b*x^3+1/2*x^2*arctanh(1+d+d*tanh(b*x+a))-1/4*x^2*ln(1+(1+d)*exp(2*b*x+ 2*a))-1/4*x*polylog(2,-(1+d)*exp(2*b*x+2*a))/b+1/8*polylog(3,-(1+d)*exp(2* b*x+2*a))/b^2
Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90 \[ \int x \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\frac {2 b^2 x^2 \left (2 \text {arctanh}(1+d+d \tanh (a+b x))-\log \left (1+\frac {e^{-2 (a+b x)}}{1+d}\right )\right )+2 b x \operatorname {PolyLog}\left (2,-\frac {e^{-2 (a+b x)}}{1+d}\right )+\operatorname {PolyLog}\left (3,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{8 b^2} \]
(2*b^2*x^2*(2*ArcTanh[1 + d + d*Tanh[a + b*x]] - Log[1 + 1/((1 + d)*E^(2*( a + b*x)))]) + 2*b*x*PolyLog[2, -(1/((1 + d)*E^(2*(a + b*x))))] + PolyLog[ 3, -(1/((1 + d)*E^(2*(a + b*x))))])/(8*b^2)
Time = 0.65 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.30, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6793, 2615, 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.
| \(\displaystyle \int x \text {arctanh}(d \tanh (a+b x)+d+1) \, dx\) |
| \(\Big \downarrow \) 6793 |
| \(\displaystyle \frac {1}{2} b \int \frac {x^2}{e^{2 a+2 b x} (d+1)+1}dx+\frac {1}{2} x^2 \text {arctanh}(d \tanh (a+b x)+d+1)\) |
| \(\Big \downarrow \) 2615 |
| \(\displaystyle \frac {1}{2} b \left (\frac {x^3}{3}-(d+1) \int \frac {e^{2 a+2 b x} x^2}{e^{2 a+2 b x} (d+1)+1}dx\right )+\frac {1}{2} x^2 \text {arctanh}(d \tanh (a+b x)+d+1)\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {1}{2} b \left (\frac {x^3}{3}-(d+1) \left (\frac {x^2 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {\int x \log \left (e^{2 a+2 b x} (d+1)+1\right )dx}{b (d+1)}\right )\right )+\frac {1}{2} x^2 \text {arctanh}(d \tanh (a+b x)+d+1)\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {1}{2} b \left (\frac {x^3}{3}-(d+1) \left (\frac {x^2 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {\frac {\int \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )dx}{2 b}-\frac {x \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}}{b (d+1)}\right )\right )+\frac {1}{2} x^2 \text {arctanh}(d \tanh (a+b x)+d+1)\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {1}{2} b \left (\frac {x^3}{3}-(d+1) \left (\frac {x^2 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )de^{2 a+2 b x}}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}}{b (d+1)}\right )\right )+\frac {1}{2} x^2 \text {arctanh}(d \tanh (a+b x)+d+1)\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {1}{2} x^2 \text {arctanh}(d \tanh (a+b x)+d+1)+\frac {1}{2} b \left (\frac {x^3}{3}-(d+1) \left (\frac {x^2 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {\frac {\operatorname {PolyLog}\left (3,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}}{b (d+1)}\right )\right )\) |
(x^2*ArcTanh[1 + d + d*Tanh[a + b*x]])/2 + (b*(x^3/3 - (1 + d)*((x^2*Log[1 + (1 + d)*E^(2*a + 2*b*x)])/(2*b*(1 + d)) - (-1/2*(x*PolyLog[2, -((1 + d) *E^(2*a + 2*b*x))])/b + PolyLog[3, -((1 + d)*E^(2*a + 2*b*x))]/(4*b^2))/(b *(1 + d)))))/2
3.3.90.3.1 Defintions of rubi rules used
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] - Simp[ 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]
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_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[c + d*Tanh[a + b*x]]/(f*( m + 1))), x] + Simp[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]
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 = 0.86 (sec) , antiderivative size = 1579, normalized size of antiderivative = 15.63
-1/2*x^2*ln(exp(b*x+a))-1/4/b^2*a^2*d/(1+d)*ln(d*exp(2*b*x+2*a)+exp(2*b*x+ 2*a)+1)+1/2/b^2*a^2*d/(1+d)*ln(1+exp(b*x+a)*(-d-1)^(1/2))+1/2/b^2*a^2*d/(1 +d)*ln(1-exp(b*x+a)*(-d-1)^(1/2))+1/2/b^2*a*d/(1+d)*dilog(1+exp(b*x+a)*(-d -1)^(1/2))+1/2/b^2*a*d/(1+d)*dilog(1-exp(b*x+a)*(-d-1)^(1/2))-1/2/b/(1+d)* ln(1+(1+d)*exp(2*b*x+2*a))*a*x-1/4/b^2*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))* a^2+1/6*b*x^3-1/2/b*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*a*x+1/2/b*a*d/(1+d) *x*ln(1+exp(b*x+a)*(-d-1)^(1/2))+1/2/b*a*d/(1+d)*x*ln(1-exp(b*x+a)*(-d-1)^ (1/2))-1/4*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x^2-1/4/b^2*a^2/(1+d)*ln(d*e xp(2*b*x+2*a)+exp(2*b*x+2*a)+1)-1/4/b^2/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*a ^2-1/4/b/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2*a))*x-1/4/b^2/(1+d)*polylog(2, -(1+d)*exp(2*b*x+2*a))*a+1/8/b^2*d/(1+d)*polylog(3,-(1+d)*exp(2*b*x+2*a))+ 1/2/b^2*a^2/(1+d)*ln(1+exp(b*x+a)*(-d-1)^(1/2))+1/2/b^2*a^2/(1+d)*ln(1-exp (b*x+a)*(-d-1)^(1/2))+1/2/b^2*a/(1+d)*dilog(1+exp(b*x+a)*(-d-1)^(1/2))+1/2 /b^2*a/(1+d)*dilog(1-exp(b*x+a)*(-d-1)^(1/2))-1/4/(1+d)*ln(1+(1+d)*exp(2*b *x+2*a))*x^2+1/8/b^2/(1+d)*polylog(3,-(1+d)*exp(2*b*x+2*a))-1/4/b*d/(1+d)* polylog(2,-(1+d)*exp(2*b*x+2*a))*x-1/4/b^2*d/(1+d)*polylog(2,-(1+d)*exp(2* b*x+2*a))*a+1/2/b*a/(1+d)*x*ln(1+exp(b*x+a)*(-d-1)^(1/2))+1/2/b*a/(1+d)*x* ln(1-exp(b*x+a)*(-d-1)^(1/2))+1/4*x^2*ln(d*exp(2*b*x+2*a)+exp(2*b*x+2*a)+1 )-1/8*(-I*Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))+I*Pi*csgn(I/(exp( 2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-2*I*Pi*csgn(...
Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (87) = 174\).
Time = 0.27 (sec) , antiderivative size = 323, normalized size of antiderivative = 3.20 \[ \int x \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\frac {2 \, b^{3} x^{3} + 3 \, b^{2} x^{2} \log \left (-\frac {{\left (d + 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b x {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 3 \, a^{2} \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) + \sqrt {-4 \, d - 4}\right ) - 3 \, a^{2} \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) - \sqrt {-4 \, d - 4}\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + 6 \, {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 6 \, {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{12 \, b^{2}} \]
1/12*(2*b^3*x^3 + 3*b^2*x^2*log(-((d + 2)*cosh(b*x + a) + d*sinh(b*x + a)) /(d*cosh(b*x + a) + d*sinh(b*x + a))) - 6*b*x*dilog(1/2*sqrt(-4*d - 4)*(co sh(b*x + a) + sinh(b*x + a))) - 6*b*x*dilog(-1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a))) - 3*a^2*log(2*(d + 1)*cosh(b*x + a) + 2*(d + 1)*sin h(b*x + a) + sqrt(-4*d - 4)) - 3*a^2*log(2*(d + 1)*cosh(b*x + a) + 2*(d + 1)*sinh(b*x + a) - sqrt(-4*d - 4)) - 3*(b^2*x^2 - a^2)*log(1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - 3*(b^2*x^2 - a^2)*log(-1/2*sqrt (-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + 6*polylog(3, 1/2*sqrt(-4 *d - 4)*(cosh(b*x + a) + sinh(b*x + a))) + 6*polylog(3, -1/2*sqrt(-4*d - 4 )*(cosh(b*x + a) + sinh(b*x + a))))/b^2
\[ \int x \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\int x \operatorname {atanh}{\left (d \tanh {\left (a + b x \right )} + d + 1 \right )}\, dx \]
Time = 0.74 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00 \[ \int x \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\frac {1}{24} \, {\left (\frac {4 \, x^{3}}{d} - \frac {3 \, {\left (2 \, b^{2} x^{2} \log \left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{3} d}\right )} b d + \frac {1}{2} \, x^{2} \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + d + 1\right ) \]
1/24*(4*x^3/d - 3*(2*b^2*x^2*log((d + 1)*e^(2*b*x + 2*a) + 1) + 2*b*x*dilo g(-(d + 1)*e^(2*b*x + 2*a)) - polylog(3, -(d + 1)*e^(2*b*x + 2*a)))/(b^3*d ))*b*d + 1/2*x^2*arctanh(d*tanh(b*x + a) + d + 1)
\[ \int x \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\int { x \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + d + 1\right ) \,d x } \]
Timed out. \[ \int x \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\int x\,\mathrm {atanh}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )+1\right ) \,d x \]