3.350 \(\int x \tanh ^{-1}(e^{a+b x}) \, dx\)

Optimal. Leaf size=71 \[ \frac {\text {Li}_3\left (-e^{a+b x}\right )}{2 b^2}-\frac {\text {Li}_3\left (e^{a+b x}\right )}{2 b^2}-\frac {x \text {Li}_2\left (-e^{a+b x}\right )}{2 b}+\frac {x \text {Li}_2\left (e^{a+b x}\right )}{2 b} \]

[Out]

-1/2*x*polylog(2,-exp(b*x+a))/b+1/2*x*polylog(2,exp(b*x+a))/b+1/2*polylog(3,-exp(b*x+a))/b^2-1/2*polylog(3,exp
(b*x+a))/b^2

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Rubi [A]  time = 0.06, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6213, 2531, 2282, 6589} \[ \frac {\text {PolyLog}\left (3,-e^{a+b x}\right )}{2 b^2}-\frac {\text {PolyLog}\left (3,e^{a+b x}\right )}{2 b^2}-\frac {x \text {PolyLog}\left (2,-e^{a+b x}\right )}{2 b}+\frac {x \text {PolyLog}\left (2,e^{a+b x}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[E^(a + b*x)],x]

[Out]

-(x*PolyLog[2, -E^(a + b*x)])/(2*b) + (x*PolyLog[2, E^(a + b*x)])/(2*b) + PolyLog[3, -E^(a + b*x)]/(2*b^2) - P
olyLog[3, E^(a + b*x)]/(2*b^2)

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 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 6213

Int[ArcTanh[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Dist[1/2, Int[x^m*Log[1 + a + b*
f^(c + d*x)], x], x] - Dist[1/2, Int[x^m*Log[1 - a - b*f^(c + d*x)], x], x] /; FreeQ[{a, b, c, d, f}, x] && IG
tQ[m, 0]

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 \tanh ^{-1}\left (e^{a+b x}\right ) \, dx &=-\left (\frac {1}{2} \int x \log \left (1-e^{a+b x}\right ) \, dx\right )+\frac {1}{2} \int x \log \left (1+e^{a+b x}\right ) \, dx\\ &=-\frac {x \text {Li}_2\left (-e^{a+b x}\right )}{2 b}+\frac {x \text {Li}_2\left (e^{a+b x}\right )}{2 b}+\frac {\int \text {Li}_2\left (-e^{a+b x}\right ) \, dx}{2 b}-\frac {\int \text {Li}_2\left (e^{a+b x}\right ) \, dx}{2 b}\\ &=-\frac {x \text {Li}_2\left (-e^{a+b x}\right )}{2 b}+\frac {x \text {Li}_2\left (e^{a+b x}\right )}{2 b}+\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}-\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}\\ &=-\frac {x \text {Li}_2\left (-e^{a+b x}\right )}{2 b}+\frac {x \text {Li}_2\left (e^{a+b x}\right )}{2 b}+\frac {\text {Li}_3\left (-e^{a+b x}\right )}{2 b^2}-\frac {\text {Li}_3\left (e^{a+b x}\right )}{2 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 113, normalized size = 1.59 \[ \frac {b^2 x^2 \log \left (1-e^{a+b x}\right )-b^2 x^2 \log \left (e^{a+b x}+1\right )+2 b^2 x^2 \tanh ^{-1}\left (e^{a+b x}\right )-2 b x \text {Li}_2\left (-e^{a+b x}\right )+2 b x \text {Li}_2\left (e^{a+b x}\right )+2 \text {Li}_3\left (-e^{a+b x}\right )-2 \text {Li}_3\left (e^{a+b x}\right )}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTanh[E^(a + b*x)],x]

[Out]

(2*b^2*x^2*ArcTanh[E^(a + b*x)] + b^2*x^2*Log[1 - E^(a + b*x)] - b^2*x^2*Log[1 + E^(a + b*x)] - 2*b*x*PolyLog[
2, -E^(a + b*x)] + 2*b*x*PolyLog[2, E^(a + b*x)] + 2*PolyLog[3, -E^(a + b*x)] - 2*PolyLog[3, E^(a + b*x)])/(4*
b^2)

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fricas [C]  time = 0.90, size = 199, normalized size = 2.80 \[ \frac {b^{2} x^{2} \log \left (-\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1}{\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1}\right ) - b^{2} x^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 2 \, b x {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 2 \, b x {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + a^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) - 2 \, {\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 2 \, {\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(exp(b*x+a)),x, algorithm="fricas")

[Out]

1/4*(b^2*x^2*log(-(cosh(b*x + a) + sinh(b*x + a) + 1)/(cosh(b*x + a) + sinh(b*x + a) - 1)) - b^2*x^2*log(cosh(
b*x + a) + sinh(b*x + a) + 1) + 2*b*x*dilog(cosh(b*x + a) + sinh(b*x + a)) - 2*b*x*dilog(-cosh(b*x + a) - sinh
(b*x + a)) + a^2*log(cosh(b*x + a) + sinh(b*x + a) - 1) + (b^2*x^2 - a^2)*log(-cosh(b*x + a) - sinh(b*x + a) +
1) - 2*polylog(3, cosh(b*x + a) + sinh(b*x + a)) + 2*polylog(3, -cosh(b*x + a) - sinh(b*x + a)))/b^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {artanh}\left (e^{\left (b x + a\right )}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(exp(b*x+a)),x, algorithm="giac")

[Out]

integrate(x*arctanh(e^(b*x + a)), x)

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maple [B]  time = 0.05, size = 153, normalized size = 2.15 \[ \frac {x^{2} \arctanh \left ({\mathrm e}^{b x +a}\right )}{2}-\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) x^{2}}{4}+\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) a^{2}}{4 b^{2}}-\frac {x \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{2 b}+\frac {\polylog \left (3, -{\mathrm e}^{b x +a}\right )}{2 b^{2}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{4}-\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{4 b^{2}}+\frac {x \polylog \left (2, {\mathrm e}^{b x +a}\right )}{2 b}-\frac {\polylog \left (3, {\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {a^{2} \arctanh \left ({\mathrm e}^{b x +a}\right )}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctanh(exp(b*x+a)),x)

[Out]

1/2*x^2*arctanh(exp(b*x+a))-1/4*ln(exp(b*x+a)+1)*x^2+1/4/b^2*ln(exp(b*x+a)+1)*a^2-1/2*x*polylog(2,-exp(b*x+a))
/b+1/2*polylog(3,-exp(b*x+a))/b^2+1/4*ln(1-exp(b*x+a))*x^2-1/4/b^2*ln(1-exp(b*x+a))*a^2+1/2*x*polylog(2,exp(b*
x+a))/b-1/2*polylog(3,exp(b*x+a))/b^2-1/2/b^2*a^2*arctanh(exp(b*x+a))

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maxima [A]  time = 0.33, size = 108, normalized size = 1.52 \[ \frac {1}{2} \, x^{2} \operatorname {artanh}\left (e^{\left (b x + a\right )}\right ) - \frac {1}{4} \, b {\left (\frac {b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})}{b^{3}} - \frac {b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})}{b^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(exp(b*x+a)),x, algorithm="maxima")

[Out]

1/2*x^2*arctanh(e^(b*x + a)) - 1/4*b*((b^2*x^2*log(e^(b*x + a) + 1) + 2*b*x*dilog(-e^(b*x + a)) - 2*polylog(3,
-e^(b*x + a)))/b^3 - (b^2*x^2*log(-e^(b*x + a) + 1) + 2*b*x*dilog(e^(b*x + a)) - 2*polylog(3, e^(b*x + a)))/b
^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {atanh}\left ({\mathrm {e}}^{a+b\,x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(exp(a + b*x)),x)

[Out]

int(x*atanh(exp(a + b*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {atanh}{\left (e^{a} e^{b x} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(exp(b*x+a)),x)

[Out]

Integral(x*atanh(exp(a)*exp(b*x)), x)

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