### 3.400 $$\int x \coth (a+b x) \, dx$$

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

[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 = 0.0806948, 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 = {3716, 2190, 2279, 2391} $\frac{\text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{2 b^2}+\frac{x \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac{x^2}{2}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Coth[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 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && 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 \coth (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 [A]  time = 0.0061351, size = 47, normalized size = 1.04 $\frac{\text{PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}+\frac{x \log \left (1-e^{2 a+2 b x}\right )}{b}-\frac{x^2}{2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x*Coth[a + b*x],x]

[Out]

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

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Maple [B]  time = 0.023, size = 122, normalized size = 2.7 \begin{align*} -{\frac{{x}^{2}}{2}}-2\,{\frac{ax}{b}}-{\frac{{a}^{2}}{{b}^{2}}}+{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}+{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{a\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{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*cosh(b*x+a)*csch(b*x+a),x)

[Out]

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

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Maxima [A]  time = 1.29799, size = 78, normalized size = 1.73 \begin{align*} -\frac{1}{2} \, x^{2} + \frac{b x \log \left (e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (b x + a\right )}\right )}{b^{2}} + \frac{b x \log \left (-e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (b x + a\right )}\right )}{b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)*csch(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.07054, size = 336, normalized size = 7.47 \begin{align*} -\frac{b^{2} x^{2} - 2 \, b x \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 2 \, a \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - 2 \,{\left (b x + a\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) - 2 \,{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 2 \,{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \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*cosh(b*x+a)*csch(b*x+a),x, algorithm="fricas")

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)*csch(b*x+a),x)

[Out]

Integral(x*cosh(a + b*x)*csch(a + b*x), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)*csch(b*x+a),x, algorithm="giac")

[Out]

integrate(x*cosh(b*x + a)*csch(b*x + a), x)