3.209 \(\int x \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx\)
Optimal. Leaf size=101 \[ \frac{\text{PolyLog}\left (3,-(d+1) e^{2 a+2 b x}\right )}{8 b^2}-\frac{x \text{PolyLog}\left (2,-(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{4} x^2 \log \left ((d+1) e^{2 a+2 b x}+1\right )+\frac{1}{2} x^2 \coth ^{-1}(d \tanh (a+b x)+d+1)+\frac{b x^3}{6} \]
[Out]
(b*x^3)/6 + (x^2*ArcCoth[1 + d + d*Tanh[a + b*x]])/2 - (x^2*Log[1 + (1 + d)*E^(2*a + 2*b*x)])/4 - (x*PolyLog[2
, -((1 + d)*E^(2*a + 2*b*x))])/(4*b) + PolyLog[3, -((1 + d)*E^(2*a + 2*b*x))]/(8*b^2)
________________________________________________________________________________________
Rubi [A] time = 0.230903, antiderivative size = 101, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used =
{6240, 2184, 2190, 2531, 2282, 6589} \[ \frac{\text{PolyLog}\left (3,-(d+1) e^{2 a+2 b x}\right )}{8 b^2}-\frac{x \text{PolyLog}\left (2,-(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{4} x^2 \log \left ((d+1) e^{2 a+2 b x}+1\right )+\frac{1}{2} x^2 \coth ^{-1}(d \tanh (a+b x)+d+1)+\frac{b x^3}{6} \]
Antiderivative was successfully verified.
[In]
Int[x*ArcCoth[1 + d + d*Tanh[a + b*x]],x]
[Out]
(b*x^3)/6 + (x^2*ArcCoth[1 + d + d*Tanh[a + b*x]])/2 - (x^2*Log[1 + (1 + d)*E^(2*a + 2*b*x)])/4 - (x*PolyLog[2
, -((1 + d)*E^(2*a + 2*b*x))])/(4*b) + PolyLog[3, -((1 + d)*E^(2*a + 2*b*x))]/(8*b^2)
Rule 6240
Int[ArcCoth[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(
m + 1)*ArcCoth[c + d*Tanh[a + b*x]])/(f*(m + 1)), x] + Dist[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]
Rule 2184
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] - Dist[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]
Rule 2190
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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(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]
Rule 2531
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] + Dist[(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]
Rule 2282
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]
Rule 6589
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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]
Rubi steps
\begin{align*} \int x \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx &=\frac{1}{2} x^2 \coth ^{-1}(1+d+d \tanh (a+b x))+\frac{1}{2} b \int \frac{x^2}{1+(1+d) e^{2 a+2 b x}} \, dx\\ &=\frac{b x^3}{6}+\frac{1}{2} x^2 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac{1}{2} (b (1+d)) \int \frac{e^{2 a+2 b x} x^2}{1+(1+d) e^{2 a+2 b x}} \, dx\\ &=\frac{b x^3}{6}+\frac{1}{2} x^2 \coth ^{-1}(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{1}{2} \int x \log \left (1+(1+d) e^{2 a+2 b x}\right ) \, dx\\ &=\frac{b x^3}{6}+\frac{1}{2} x^2 \coth ^{-1}(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 \text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{\int \text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right ) \, dx}{4 b}\\ &=\frac{b x^3}{6}+\frac{1}{2} x^2 \coth ^{-1}(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 \text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2((-1-d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^2}\\ &=\frac{b x^3}{6}+\frac{1}{2} x^2 \coth ^{-1}(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 \text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{\text{Li}_3\left (-(1+d) e^{2 a+2 b x}\right )}{8 b^2}\\ \end{align*}
Mathematica [A] time = 0.0958579, size = 91, normalized size = 0.9 \[ \frac{2 b x \text{PolyLog}\left (2,-\frac{e^{-2 (a+b x)}}{d+1}\right )+\text{PolyLog}\left (3,-\frac{e^{-2 (a+b x)}}{d+1}\right )+2 b^2 x^2 \left (2 \coth ^{-1}(d \tanh (a+b x)+d+1)-\log \left (\frac{e^{-2 (a+b x)}}{d+1}+1\right )\right )}{8 b^2} \]
Antiderivative was successfully verified.
[In]
Integrate[x*ArcCoth[1 + d + d*Tanh[a + b*x]],x]
[Out]
(2*b^2*x^2*(2*ArcCoth[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)
________________________________________________________________________________________
Maple [C] time = 10.194, size = 1584, normalized size = 15.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*arccoth(1+d+d*tanh(b*x+a)),x)
[Out]
-1/8*I*x^2*Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+1/6*b*x^3-1/4/b^2*a^2/(1+d)*l
n(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1)+1/2/b^2*a/(1+d)*dilog(1-exp(b*x+a)*(-d-1)^(1/2))-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+1/2/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))+1/8/b^2
*d/(1+d)*polylog(3,-(1+d)*exp(2*b*x+2*a))+1/2/b^2*a/(1+d)*dilog(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*x^2*ln(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1)-1/4/b^2*d*a^2/(1+d)*ln(exp(2*b*x+2*a)*d+
exp(2*b*x+2*a)+1)+1/8*I*x^2*Pi*csgn(I*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))*csgn(I/(exp(2*b*x+2*a)+1)*(exp(2*b*
x+2*a)*d+exp(2*b*x+2*a)+1))^2-1/8*I*x^2*Pi*csgn(I*d)*csgn(I*d/(exp(2*b*x+2*a)+1)*exp(2*b*x+2*a))^2-1/2*x^2*ln(
exp(b*x+a))-1/4*x^2*ln(d)+1/8*I*x^2*Pi*csgn(I*d/(exp(2*b*x+2*a)+1)*exp(2*b*x+2*a))^3-1/8*I*x^2*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*a))^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))+1/8*I*x^2*Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I/(exp(2*b*
x+2*a)+1)*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))^2+1/8*I*x^2*Pi*csgn(I*exp(2*b*x+2*a))^3+1/8*I*x^2*Pi*csgn(I*exp
(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-1/8*I*x^2*Pi*csgn(I/(exp(2*b*x+2*a)+1)*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))^
3+1/2/b^2*d*a/(1+d)*dilog(1+exp(b*x+a)*(-d-1)^(1/2))+1/2/b^2*d*a/(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))*x*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)*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)*ln(1+exp(b*x+a)*(-d-1)
^(1/2))*x+1/2/b*a/(1+d)*ln(1-exp(b*x+a)*(-d-1)^(1/2))*x+1/2/b^2*d*a^2/(1+d)*ln(1+exp(b*x+a)*(-d-1)^(1/2))+1/2/
b^2*d*a^2/(1+d)*ln(1-exp(b*x+a)*(-d-1)^(1/2))-1/8*I*x^2*Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(e
xp(2*b*x+2*a)+1))^2+1/8*I*x^2*Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-1/4*I*x^2*Pi*csgn(I*exp(b*x+a))*c
sgn(I*exp(2*b*x+2*a))^2+1/8*I*x^2*Pi*csgn(I*d)*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*a))+1/8*I*x^2*Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(e
xp(2*b*x+2*a)+1))-1/8*I*x^2*Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))*csgn(I/(
exp(2*b*x+2*a)+1)*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))-1/2/b*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x*a+1/2/b*d*a/
(1+d)*ln(1+exp(b*x+a)*(-d-1)^(1/2))*x+1/2/b*d*a/(1+d)*ln(1-exp(b*x+a)*(-d-1)^(1/2))*x
________________________________________________________________________________________
Maxima [A] time = 3.48612, size = 136, normalized size = 1.35 \begin{align*} \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 \tanh \left (b x + a\right ) + d + 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*arccoth(1+d+d*tanh(b*x+a)),x, algorithm="maxima")
[Out]
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*dilog(-(d + 1)*e^(2*b*x + 2*a)) - polylo
g(3, -(d + 1)*e^(2*b*x + 2*a)))/(b^3*d))*b*d + 1/2*x^2*arccoth(d*tanh(b*x + a) + d + 1)
________________________________________________________________________________________
Fricas [C] time = 1.74289, size = 953, normalized size = 9.44 \begin{align*} \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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*arccoth(1+d+d*tanh(b*x+a)),x, algorithm="fricas")
[Out]
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)*(cosh(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
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*acoth(1+d+d*tanh(b*x+a)),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{arcoth}\left (d \tanh \left (b x + a\right ) + d + 1\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*arccoth(1+d+d*tanh(b*x+a)),x, algorithm="giac")
[Out]
integrate(x*arccoth(d*tanh(b*x + a) + d + 1), x)