Integrand size = 14, antiderivative size = 100 \[ \int x \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\frac {b x^3}{6}+\frac {1}{2} x^2 \text {arctanh}(1+d+d \coth (a+b x))-\frac {1}{4} x^2 \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac {x \operatorname {PolyLog}\left (2,(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac {\operatorname {PolyLog}\left (3,(1+d) e^{2 a+2 b x}\right )}{8 b^2} \]
1/6*b*x^3+1/2*x^2*arctanh(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 = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.90 \[ \int x \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\frac {2 b^2 x^2 \left (2 \text {arctanh}(1+d+d \coth (a+b x))-\log \left (1-\frac {e^{-2 (a+b x)}}{1+d}\right )\right )+2 b x \operatorname {PolyLog}\left (2,\frac {e^{-2 (a+b x)}}{1+d}\right )+\operatorname {PolyLog}\left (3,\frac {e^{-2 (a+b x)}}{1+d}\right )}{8 b^2} \]
(2*b^2*x^2*(2*ArcTanh[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)))] + PolyLog[3, 1/((1 + d)*E^(2*(a + b*x)))])/(8*b^2)
Time = 0.66 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.28, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6795, 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.
\(\displaystyle \int x \text {arctanh}(d \coth (a+b x)+d+1) \, dx\) |
\(\Big \downarrow \) 6795 |
\(\displaystyle \frac {1}{2} b \int \frac {x^2}{1-(d+1) e^{2 a+2 b x}}dx+\frac {1}{2} x^2 \text {arctanh}(d \coth (a+b x)+d+1)\) |
\(\Big \downarrow \) 2615 |
\(\displaystyle \frac {1}{2} b \left ((d+1) \int \frac {e^{2 a+2 b x} x^2}{1-(d+1) e^{2 a+2 b x}}dx+\frac {x^3}{3}\right )+\frac {1}{2} x^2 \text {arctanh}(d \coth (a+b x)+d+1)\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {1}{2} b \left ((d+1) \left (\frac {\int x \log \left (1-(d+1) e^{2 a+2 b x}\right )dx}{b (d+1)}-\frac {x^2 \log \left (1-(d+1) e^{2 a+2 b x}\right )}{2 b (d+1)}\right )+\frac {x^3}{3}\right )+\frac {1}{2} x^2 \text {arctanh}(d \coth (a+b x)+d+1)\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {1}{2} b \left ((d+1) \left (\frac {\frac {\int \operatorname {PolyLog}\left (2,(d+1) e^{2 a+2 b x}\right )dx}{2 b}-\frac {x \operatorname {PolyLog}\left (2,(d+1) e^{2 a+2 b x}\right )}{2 b}}{b (d+1)}-\frac {x^2 \log \left (1-(d+1) e^{2 a+2 b x}\right )}{2 b (d+1)}\right )+\frac {x^3}{3}\right )+\frac {1}{2} x^2 \text {arctanh}(d \coth (a+b x)+d+1)\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {1}{2} b \left ((d+1) \left (\frac {\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (2,(d+1) e^{2 a+2 b x}\right )de^{2 a+2 b x}}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,(d+1) e^{2 a+2 b x}\right )}{2 b}}{b (d+1)}-\frac {x^2 \log \left (1-(d+1) e^{2 a+2 b x}\right )}{2 b (d+1)}\right )+\frac {x^3}{3}\right )+\frac {1}{2} x^2 \text {arctanh}(d \coth (a+b x)+d+1)\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {1}{2} x^2 \text {arctanh}(d \coth (a+b x)+d+1)+\frac {1}{2} b \left ((d+1) \left (\frac {\frac {\operatorname {PolyLog}\left (3,(d+1) e^{2 a+2 b x}\right )}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,(d+1) e^{2 a+2 b x}\right )}{2 b}}{b (d+1)}-\frac {x^2 \log \left (1-(d+1) e^{2 a+2 b x}\right )}{2 b (d+1)}\right )+\frac {x^3}{3}\right )\) |
(x^2*ArcTanh[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
3.4.4.3.1 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_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[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 = 0.92 (sec) , antiderivative size = 1555, normalized size of antiderivative = 15.55
1/8/b^2/(1+d)*polylog(3,(1+d)*exp(2*b*x+2*a))-1/4/(1+d)*ln(1-(1+d)*exp(2*b *x+2*a))*x^2-1/2*x^2*ln(exp(b*x+a))+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)*(1+d)^(1/2))+1/2/b*a*d/ (1+d)*x*ln(1+exp(b*x+a)*(1+d)^(1/2))+1/2/b^2*a^2/(1+d)*ln(1-exp(b*x+a)*(1+ d)^(1/2))+1/2/b^2*a^2/(1+d)*ln(1+exp(b*x+a)*(1+d)^(1/2))+1/2/b^2*a/(1+d)*d ilog(1-exp(b*x+a)*(1+d)^(1/2))+1/2/b^2*a/(1+d)*dilog(1+exp(b*x+a)*(1+d)^(1 /2))-1/4/b^2*a^2/(1+d)*ln(d*exp(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-1/4*d/(1+d)*ln(1-(1+d) *exp(2*b*x+2*a))*x^2+1/8/b^2*d/(1+d)*polylog(3,(1+d)*exp(2*b*x+2*a))-1/4/b ^2*d/(1+d)*ln(1-(1+d)*exp(2*b*x+2*a))*a^2-1/4/b*d/(1+d)*polylog(2,(1+d)*ex p(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/(1 +d)*ln(1-(1+d)*exp(2*b*x+2*a))*a*x+1/2/b^2*a*d/(1+d)*dilog(1-exp(b*x+a)*(1 +d)^(1/2))+1/2/b^2*a*d/(1+d)*dilog(1+exp(b*x+a)*(1+d)^(1/2))+1/2/b*a/(1+d) *x*ln(1-exp(b*x+a)*(1+d)^(1/2))+1/2/b*a/(1+d)*x*ln(1+exp(b*x+a)*(1+d)^(1/2 ))+1/2/b^2*a^2*d/(1+d)*ln(1-exp(b*x+a)*(1+d)^(1/2))+1/2/b^2*a^2*d/(1+d)*ln (1+exp(b*x+a)*(1+d)^(1/2))-1/4/b^2*a^2*d/(1+d)*ln(d*exp(2*b*x+2*a)+exp(2*b *x+2*a)-1)+1/4*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)/(exp(2*b*x+2*a)-1))*csgn(I*d/(exp(2*b*x+2*a)-1)*exp(2*b*x+2*...
Leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (86) = 172\).
Time = 0.27 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.06 \[ \int x \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\frac {2 \, b^{3} x^{3} + 3 \, b^{2} x^{2} \log \left (-\frac {d \cosh \left (b x + a\right ) + {\left (d + 2\right )} \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_2\left (\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b x {\rm Li}_2\left (-\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 3 \, a^{2} \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) + 2 \, \sqrt {d + 1}\right ) - 3 \, a^{2} \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) - 2 \, \sqrt {d + 1}\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + 6 \, {\rm polylog}\left (3, \sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 6 \, {\rm polylog}\left (3, -\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{12 \, b^{2}} \]
1/12*(2*b^3*x^3 + 3*b^2*x^2*log(-(d*cosh(b*x + a) + (d + 2)*sinh(b*x + a)) /(d*cosh(b*x + a) + d*sinh(b*x + a))) - 6*b*x*dilog(sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a))) - 6*b*x*dilog(-sqrt(d + 1)*(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) + 2 *sqrt(d + 1)) - 3*a^2*log(2*(d + 1)*cosh(b*x + a) + 2*(d + 1)*sinh(b*x + a ) - 2*sqrt(d + 1)) - 3*(b^2*x^2 - a^2)*log(sqrt(d + 1)*(cosh(b*x + a) + si nh(b*x + a)) + 1) - 3*(b^2*x^2 - a^2)*log(-sqrt(d + 1)*(cosh(b*x + a) + si nh(b*x + a)) + 1) + 6*polylog(3, sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a ))) + 6*polylog(3, -sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a))))/b^2
\[ \int x \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\int x \operatorname {atanh}{\left (d \coth {\left (a + b x \right )} + d + 1 \right )}\, dx \]
Time = 0.76 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int x \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\frac {1}{24} \, {\left (\frac {4 \, x^{3}}{d} - \frac {3 \, {\left (2 \, b^{2} x^{2} \log \left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{3} d}\right )} b d + \frac {1}{2} \, x^{2} \operatorname {artanh}\left (d \coth \left (b x + a\right ) + d + 1\right ) \]
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*dil og((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*coth(b*x + a) + d + 1)
\[ \int x \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\int { x \operatorname {artanh}\left (d \coth \left (b x + a\right ) + d + 1\right ) \,d x } \]
Timed out. \[ \int x \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\int x\,\mathrm {atanh}\left (d+d\,\mathrm {coth}\left (a+b\,x\right )+1\right ) \,d x \]