### 3.337 $$\int x \tanh (a+b x) \, dx$$

Optimal. Leaf size=45 $\frac{\text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}+\frac{x \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac{x^2}{2}$

[Out]

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

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Rubi [A]  time = 0.0824533, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {3718, 2190, 2279, 2391} $\frac{\text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}+\frac{x \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac{x^2}{2}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Tanh[a + b*x],x]

[Out]

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

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, 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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

Mathematica [C]  time = 3.52623, size = 149, normalized size = 3.31 $\frac{1}{2} \left (x^2 \tanh (a)+\frac{-\text{PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}(\coth (a))+b x\right )}\right )-b^2 x^2 \tanh (a) \sqrt{-\text{csch}^2(a)} e^{-\tanh ^{-1}(\coth (a))}+2 b x \log \left (1-e^{-2 \left (\tanh ^{-1}(\coth (a))+b x\right )}\right )+2 \tanh ^{-1}(\coth (a)) \left (\log \left (1-e^{-2 \left (\tanh ^{-1}(\coth (a))+b x\right )}\right )-\log \left (i \sinh \left (\tanh ^{-1}(\coth (a))+b x\right )\right )+b x\right )+i \pi b x-i \pi \log \left (e^{2 b x}+1\right )+i \pi \log (\cosh (b x))}{b^2}\right )$

Warning: Unable to verify antiderivative.

[In]

Integrate[x*Tanh[a + b*x],x]

[Out]

(x^2*Tanh[a] + (I*b*Pi*x - I*Pi*Log[1 + E^(2*b*x)] + 2*b*x*Log[1 - E^(-2*(b*x + ArcTanh[Coth[a]]))] + I*Pi*Log
[Cosh[b*x]] + 2*ArcTanh[Coth[a]]*(b*x + Log[1 - E^(-2*(b*x + ArcTanh[Coth[a]]))] - Log[I*Sinh[b*x + ArcTanh[Co
th[a]]]]) - PolyLog[2, E^(-2*(b*x + ArcTanh[Coth[a]]))] - (b^2*x^2*Sqrt[-Csch[a]^2]*Tanh[a])/E^ArcTanh[Coth[a]
])/b^2)/2

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Maple [A]  time = 0.026, size = 70, normalized size = 1.6 \begin{align*} -{\frac{{x}^{2}}{2}}-2\,{\frac{ax}{b}}-{\frac{{a}^{2}}{{b}^{2}}}+{\frac{x\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{b}}+{\frac{{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{2}}}+2\,{\frac{a\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x*sech(b*x+a)*sinh(b*x+a),x)

[Out]

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

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Maxima [A]  time = 1.1975, size = 54, normalized size = 1.2 \begin{align*} -\frac{1}{2} \, x^{2} + \frac{2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )}{2 \, b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(b*x+a)*sinh(b*x+a),x, algorithm="maxima")

[Out]

-1/2*x^2 + 1/2*(2*b*x*log(e^(2*b*x + 2*a) + 1) + dilog(-e^(2*b*x + 2*a)))/b^2

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Fricas [C]  time = 1.91707, size = 427, normalized size = 9.49 \begin{align*} -\frac{b^{2} x^{2} + 2 \, a \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + 2 \, a \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) - 2 \,{\left (b x + a\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) - 2 \,{\left (b x + a\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) - 2 \,{\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 2 \,{\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right )}{2 \, b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(b*x+a)*sinh(b*x+a),x, algorithm="fricas")

[Out]

-1/2*(b^2*x^2 + 2*a*log(cosh(b*x + a) + sinh(b*x + a) + I) + 2*a*log(cosh(b*x + a) + sinh(b*x + a) - I) - 2*(b
*x + a)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) - 2*(b*x + a)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) -
2*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - 2*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)))/b^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sinh{\left (a + b x \right )} \operatorname{sech}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(b*x+a)*sinh(b*x+a),x)

[Out]

Integral(x*sinh(a + b*x)*sech(a + b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{sech}\left (b x + a\right ) \sinh \left (b x + a\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(b*x+a)*sinh(b*x+a),x, algorithm="giac")

[Out]

integrate(x*sech(b*x + a)*sinh(b*x + a), x)