Integrand size = 17, antiderivative size = 109 \[ \int x \coth ^{-1}(1-d-d \coth (a+b x)) \, dx=\frac {b x^3}{6}+\frac {1}{2} x^2 \coth ^{-1}(1-d-d \coth (a+b x))-\frac {1}{4} x^2 \log \left (1-(1-d) e^{2 a+2 b x}\right )-\frac {x \operatorname {PolyLog}\left (2,(1-d) e^{2 a+2 b x}\right )}{4 b}+\frac {\operatorname {PolyLog}\left (3,(1-d) e^{2 a+2 b x}\right )}{8 b^2} \]
1/6*b*x^3+1/2*x^2*arccoth(1-d-d*coth(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 = 0.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86 \[ \int x \coth ^{-1}(1-d-d \coth (a+b x)) \, dx=\frac {2 b^2 x^2 \left (2 \coth ^{-1}(1-d-d \coth (a+b x))-\log \left (1+\frac {e^{-2 (a+b x)}}{-1+d}\right )\right )+2 b x \operatorname {PolyLog}\left (2,-\frac {e^{-2 (a+b x)}}{-1+d}\right )+\operatorname {PolyLog}\left (3,-\frac {e^{-2 (a+b x)}}{-1+d}\right )}{8 b^2} \]
(2*b^2*x^2*(2*ArcCoth[1 - d - d*Coth[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))))] + PolyLo g[3, -(1/((-1 + d)*E^(2*(a + b*x))))])/(8*b^2)
Time = 0.69 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.31, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {6796, 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.
| \(\displaystyle \int x \coth ^{-1}(d (-\coth (a+b x))-d+1) \, dx\) |
| \(\Big \downarrow \) 6796 |
| \(\displaystyle \frac {1}{2} b \int \frac {x^2}{1-(1-d) e^{2 a+2 b x}}dx+\frac {1}{2} x^2 \coth ^{-1}(d (-\coth (a+b x))-d+1)\) |
| \(\Big \downarrow \) 2615 |
| \(\displaystyle \frac {1}{2} b \left ((1-d) \int \frac {e^{2 a+2 b x} x^2}{1-(1-d) e^{2 a+2 b x}}dx+\frac {x^3}{3}\right )+\frac {1}{2} x^2 \coth ^{-1}(d (-\coth (a+b x))-d+1)\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {1}{2} b \left ((1-d) \left (\frac {\int x \log \left (1-(1-d) e^{2 a+2 b x}\right )dx}{b (1-d)}-\frac {x^2 \log \left (1-(1-d) e^{2 a+2 b x}\right )}{2 b (1-d)}\right )+\frac {x^3}{3}\right )+\frac {1}{2} x^2 \coth ^{-1}(d (-\coth (a+b x))-d+1)\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {1}{2} b \left ((1-d) \left (\frac {\frac {\int \operatorname {PolyLog}\left (2,(1-d) e^{2 a+2 b x}\right )dx}{2 b}-\frac {x \operatorname {PolyLog}\left (2,(1-d) e^{2 a+2 b x}\right )}{2 b}}{b (1-d)}-\frac {x^2 \log \left (1-(1-d) e^{2 a+2 b x}\right )}{2 b (1-d)}\right )+\frac {x^3}{3}\right )+\frac {1}{2} x^2 \coth ^{-1}(d (-\coth (a+b x))-d+1)\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {1}{2} b \left ((1-d) \left (\frac {\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (2,(1-d) e^{2 a+2 b x}\right )de^{2 a+2 b x}}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,(1-d) e^{2 a+2 b x}\right )}{2 b}}{b (1-d)}-\frac {x^2 \log \left (1-(1-d) e^{2 a+2 b x}\right )}{2 b (1-d)}\right )+\frac {x^3}{3}\right )+\frac {1}{2} x^2 \coth ^{-1}(d (-\coth (a+b x))-d+1)\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {1}{2} b \left ((1-d) \left (\frac {\frac {\operatorname {PolyLog}\left (3,(1-d) e^{2 a+2 b x}\right )}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,(1-d) e^{2 a+2 b x}\right )}{2 b}}{b (1-d)}-\frac {x^2 \log \left (1-(1-d) e^{2 a+2 b x}\right )}{2 b (1-d)}\right )+\frac {x^3}{3}\right )+\frac {1}{2} x^2 \coth ^{-1}(d (-\coth (a+b x))-d+1)\) |
(x^2*ArcCoth[1 - d - d*Coth[a + b*x]])/2 + (b*(x^3/3 + (1 - d)*(-1/2*(x^2* Log[1 - (1 - d)*E^(2*a + 2*b*x)])/(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
3.3.28.3.1 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[ArcCoth[(c_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcCoth[c + d*Coth[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 = 1.57 (sec) , antiderivative size = 1640, normalized size of antiderivative = 15.05
1/2/b*a*d/(d-1)*x*ln(1+exp(b*x+a)*(1-d)^(1/2))+1/2/b*a*d/(d-1)*x*ln(1-exp( b*x+a)*(1-d)^(1/2))-1/2/b*d/(d-1)*ln(1+(d-1)*exp(2*b*x+2*a))*a*x-1/2*x^2*l n(exp(b*x+a))+1/6*b*x^3+1/2/b^2*a^2*d/(d-1)*ln(1+exp(b*x+a)*(1-d)^(1/2))+1 /2/b^2*a^2*d/(d-1)*ln(1-exp(b*x+a)*(1-d)^(1/2))+1/2/b^2*a*d/(d-1)*dilog(1+ exp(b*x+a)*(1-d)^(1/2))+1/2/b^2*a*d/(d-1)*dilog(1-exp(b*x+a)*(1-d)^(1/2))- 1/8/b^2/(d-1)*polylog(3,-(d-1)*exp(2*b*x+2*a))+1/4/(d-1)*ln(1+(d-1)*exp(2* b*x+2*a))*x^2+1/8/b^2*d/(d-1)*polylog(3,-(d-1)*exp(2*b*x+2*a))+1/4/b^2/(d- 1)*ln(1+(d-1)*exp(2*b*x+2*a))*a^2+1/4/b/(d-1)*polylog(2,-(d-1)*exp(2*b*x+2 *a))*x+1/4/b^2/(d-1)*polylog(2,-(d-1)*exp(2*b*x+2*a))*a-1/2/b^2*a^2/(d-1)* ln(1+exp(b*x+a)*(1-d)^(1/2))-1/2/b^2*a^2/(d-1)*ln(1-exp(b*x+a)*(1-d)^(1/2) )-1/2/b^2*a/(d-1)*dilog(1+exp(b*x+a)*(1-d)^(1/2))-1/2/b^2*a/(d-1)*dilog(1- exp(b*x+a)*(1-d)^(1/2))+1/4/b^2*a^2/(d-1)*ln(d*exp(2*b*x+2*a)-exp(2*b*x+2* a)+1)-1/4*d/(d-1)*ln(1+(d-1)*exp(2*b*x+2*a))*x^2-1/4/b^2*d/(d-1)*ln(1+(d-1 )*exp(2*b*x+2*a))*a^2-1/4/b^2*a^2*d/(d-1)*ln(d*exp(2*b*x+2*a)-exp(2*b*x+2* a)+1)-1/4/b*d/(d-1)*polylog(2,-(d-1)*exp(2*b*x+2*a))*x-1/4/b^2*d/(d-1)*pol ylog(2,-(d-1)*exp(2*b*x+2*a))*a+1/2/b/(d-1)*ln(1+(d-1)*exp(2*b*x+2*a))*a*x -1/2/b*a/(d-1)*x*ln(1+exp(b*x+a)*(1-d)^(1/2))-1/2/b*a/(d-1)*x*ln(1-exp(b*x +a)*(1-d)^(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(2*b*x+2*a)-1)*(d*exp(2*b*x+2*a)-exp(2*b*x+2*a)+1))^3-I*Pi*csgn (I/(exp(2*b*x+2*a)-1))*csgn(I/(exp(2*b*x+2*a)-1)*(d*exp(2*b*x+2*a)-exp(...
Leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (90) = 180\).
Time = 0.26 (sec) , antiderivative size = 322, normalized size of antiderivative = 2.95 \[ \int x \coth ^{-1}(1-d-d \coth (a+b x)) \, dx=\frac {2 \, b^{3} x^{3} - 3 \, b^{2} x^{2} \log \left (\frac {d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + {\left (d - 2\right )} \sinh \left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b x {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 3 \, a^{2} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) + \sqrt {-4 \, d + 4}\right ) - 3 \, a^{2} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) - \sqrt {-4 \, d + 4}\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + 6 \, {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 6 \, {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{12 \, b^{2}} \]
1/12*(2*b^3*x^3 - 3*b^2*x^2*log((d*cosh(b*x + a) + d*sinh(b*x + a))/(d*cos h(b*x + a) + (d - 2)*sinh(b*x + a))) - 6*b*x*dilog(1/2*sqrt(-4*d + 4)*(cos h(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)*sinh (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
\[ \int x \coth ^{-1}(1-d-d \coth (a+b x)) \, dx=- \int x \operatorname {acoth}{\left (d \coth {\left (a + b x \right )} + d - 1 \right )}\, dx \]
Time = 0.83 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.93 \[ \int x \coth ^{-1}(1-d-d \coth (a+b x)) \, dx=\frac {1}{24} \, {\left (\frac {4 \, x^{3}}{d} - \frac {3 \, {\left (2 \, b^{2} x^{2} \log \left ({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}(-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{3} d}\right )} b d - \frac {1}{2} \, x^{2} \operatorname {arcoth}\left (d \coth \left (b x + a\right ) + d - 1\right ) \]
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*arccoth(d*coth(b*x + a) + d - 1)
\[ \int x \coth ^{-1}(1-d-d \coth (a+b x)) \, dx=\int { x \operatorname {arcoth}\left (-d \coth \left (b x + a\right ) - d + 1\right ) \,d x } \]
Timed out. \[ \int x \coth ^{-1}(1-d-d \coth (a+b x)) \, dx=\int -x\,\mathrm {acoth}\left (d+d\,\mathrm {coth}\left (a+b\,x\right )-1\right ) \,d x \]