3.3.90 \(\int x \tanh ^{-1}(1+d+d \tanh (a+b x)) \, dx\) [290]

Optimal. Leaf size=101 \[ \frac {b x^3}{6}+\frac {1}{2} x^2 \tanh ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{4} x^2 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac {x \text {PolyLog}\left (2,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {\text {PolyLog}\left (3,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{8 b^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.16, antiderivative size = 101, normalized size of antiderivative = 1.00, 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 = {6374, 2215, 2221, 2611, 2320, 6724} \begin {gather*} \frac {\text {Li}_3\left (-\left ((d+1) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {x \text {Li}_2\left (-\left ((d+1) e^{2 a+2 b x}\right )\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 \tanh ^{-1}(d \tanh (a+b x)+d+1)+\frac {b x^3}{6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[1 + d + d*Tanh[a + b*x]],x]

[Out]

(b*x^3)/6 + (x^2*ArcTanh[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 2215

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 2221

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]
- 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 2320

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 2611

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 6374

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] + 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 6724

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 \tanh ^{-1}(1+d+d \tanh (a+b x)) \, dx &=\frac {1}{2} x^2 \tanh ^{-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 \tanh ^{-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 \tanh ^{-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 \tanh ^{-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 \tanh ^{-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 {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 \tanh ^{-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 = 3.61, size = 91, normalized size = 0.90 \begin {gather*} \frac {2 b^2 x^2 \left (2 \tanh ^{-1}(1+d+d \tanh (a+b x))-\log \left (1+\frac {e^{-2 (a+b x)}}{1+d}\right )\right )+2 b x \text {PolyLog}\left (2,-\frac {e^{-2 (a+b x)}}{1+d}\right )+\text {PolyLog}\left (3,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{8 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTanh[1 + d + d*Tanh[a + b*x]],x]

[Out]

(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)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.88, size = 1627, normalized size = 16.11

method	result	size
risch	\(\text {Expression too large to display}\)	\(1627\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctanh(1+d+d*tanh(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

-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+1/4*I*x^2*Pi*csgn(I*d*exp(2*b*x
+2*a)/(exp(2*b*x+2*a)+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)*(-1-d)^(1/2))+1/2
/b^2*a^2/(1+d)*ln(1-exp(b*x+a)*(-1-d)^(1/2))+1/8*I*x^2*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))*csgn(I*d)*
csgn(I*d*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-1/2*x^2*ln(exp(b*x+a))+1/6*b*x^3-1/4*I*Pi*x^2+1/8/b^2*d/(1+d)*poly
log(3,-(1+d)*exp(2*b*x+2*a))+1/2/b^2*a/(1+d)*dilog(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*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x^2+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)/(exp(2*b*x+2*a)+1))^2-1/4/b^2*d*a^2/(1+d)*ln(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1)-1/2/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)*(-1-d)^(1/2))*x+1/2/b*d*a/(1+d)*ln(1-exp(b*x+a)*(-1-d)
^(1/2))*x-1/8*I*x^2*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+1/8*I*x^2*Pi*csg
n(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*e
xp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-1/4*I*x^2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2-1/8*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+1/8*I*x^2*Pi*csgn(I*exp(2*b*x+2*a))^3-1/4/(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))-1/2/b/(1+d)*ln(1+(1+d)*exp(2*b*
x+2*a))*x*a+1/2/b^2*d*a/(1+d)*dilog(1+exp(b*x+a)*(-1-d)^(1/2))+1/2/b^2*d*a/(1+d)*dilog(1-exp(b*x+a)*(-1-d)^(1/
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-1/4/b^
2*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)/(exp(2*b*x+2*a)+1))^3+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)/(exp(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))-1/4/b^2*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*a^2+1/2/b*a/(1+d)*ln(1+exp(b*x+a)*(-1-
d)^(1/2))*x+1/2/b*a/(1+d)*ln(1-exp(b*x+a)*(-1-d)^(1/2))*x+1/2/b^2*d*a^2/(1+d)*ln(1+exp(b*x+a)*(-1-d)^(1/2))+1/
2/b^2*d*a^2/(1+d)*ln(1-exp(b*x+a)*(-1-d)^(1/2))-1/8*I*x^2*Pi*csgn(I*d*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)/(exp(2*b*x+2*a)+1))*csgn(I*d*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-1/4*x^2*ln(d
)+1/4*ln(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1)*x^2

________________________________________________________________________________________

Maxima [A]
time = 0.72, size = 101, normalized size = 1.00 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(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*arctanh(d*tanh(b*x + a) + d + 1)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (87) = 174\).
time = 0.45, size = 323, normalized size = 3.20 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \operatorname {atanh}{\left (d \tanh {\left (a + b x \right )} + d + 1 \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(1+d+d*tanh(b*x+a)),x)

[Out]

Integral(x*atanh(d*tanh(a + b*x) + d + 1), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(1+d+d*tanh(b*x+a)),x, algorithm="giac")

[Out]

integrate(x*arctanh(d*tanh(b*x + a) + d + 1), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,\mathrm {atanh}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )+1\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(d + d*tanh(a + b*x) + 1),x)

[Out]

int(x*atanh(d + d*tanh(a + b*x) + 1), x)

________________________________________________________________________________________