3.407 \(\int x \cosh (a+b x) \coth (a+b x) \, dx\)

Optimal. Leaf size=66 \[ -\frac {\text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {\text {Li}_2\left (e^{a+b x}\right )}{b^2}-\frac {\sinh (a+b x)}{b^2}+\frac {x \cosh (a+b x)}{b}-\frac {2 x \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

-2*x*arctanh(exp(b*x+a))/b+x*cosh(b*x+a)/b-polylog(2,-exp(b*x+a))/b^2+polylog(2,exp(b*x+a))/b^2-sinh(b*x+a)/b^
2

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Rubi [A]  time = 0.06, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5450, 3296, 2637, 4182, 2279, 2391} \[ -\frac {\text {PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac {\text {PolyLog}\left (2,e^{a+b x}\right )}{b^2}-\frac {\sinh (a+b x)}{b^2}+\frac {x \cosh (a+b x)}{b}-\frac {2 x \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[x*Cosh[a + b*x]*Coth[a + b*x],x]

[Out]

(-2*x*ArcTanh[E^(a + b*x)])/b + (x*Cosh[a + b*x])/b - PolyLog[2, -E^(a + b*x)]/b^2 + PolyLog[2, E^(a + b*x)]/b
^2 - Sinh[a + b*x]/b^2

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]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4182

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

Rule 5450

Int[Cosh[(a_.) + (b_.)*(x_)]^(n_.)*Coth[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int
[(c + d*x)^m*Cosh[a + b*x]^n*Coth[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Cosh[a + b*x]^(n - 2)*Coth[a + b*x]^p
, x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 131, normalized size = 1.98 \[ -\frac {-\text {Li}_2\left (-e^{-a-b x}\right )+\text {Li}_2\left (e^{-a-b x}\right )-a \log \left (1-e^{-a-b x}\right )-b x \log \left (1-e^{-a-b x}\right )+a \log \left (e^{-a-b x}+1\right )+b x \log \left (e^{-a-b x}+1\right )+\sinh (a+b x)-b x \cosh (a+b x)+a \log \left (\tanh \left (\frac {1}{2} (a+b x)\right )\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cosh[a + b*x]*Coth[a + b*x],x]

[Out]

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

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fricas [B]  time = 0.83, size = 255, normalized size = 3.86 \[ \frac {{\left (b x - 1\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b x - 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b x - 1\right )} \sinh \left (b x + a\right )^{2} + b x + 2 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 2 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - 2 \, {\left (b x \cosh \left (b x + a\right ) + b x \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 2 \, {\left (a \cosh \left (b x + a\right ) + a \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \, {\left ({\left (b x + a\right )} \cosh \left (b x + a\right ) + {\left (b x + a\right )} \sinh \left (b x + a\right )\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 1}{2 \, {\left (b^{2} \cosh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)^2*csch(b*x+a),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cosh \left (b x + a\right )^{2} \operatorname {csch}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)^2*csch(b*x+a),x, algorithm="giac")

[Out]

integrate(x*cosh(b*x + a)^2*csch(b*x + a), x)

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maple [B]  time = 0.52, size = 139, normalized size = 2.11 \[ \frac {\left (b x -1\right ) {\mathrm e}^{b x +a}}{2 b^{2}}+\frac {\left (b x +1\right ) {\mathrm e}^{-b x -a}}{2 b^{2}}-\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) x}{b}-\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) a}{b^{2}}-\frac {\polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x}{b}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a}{b^{2}}+\frac {\polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {2 a \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(b*x+a)^2*csch(b*x+a),x)

[Out]

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

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maxima [A]  time = 0.43, size = 94, normalized size = 1.42 \[ \frac {{\left ({\left (b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (b x\right )} + {\left (b x + 1\right )} e^{\left (-b x\right )}\right )} e^{\left (-a\right )}}{2 \, b^{2}} - \frac {b x \log \left (e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (b x + a\right )}\right )}{b^{2}} + \frac {b x \log \left (-e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (b x + a\right )}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)^2*csch(b*x+a),x, algorithm="maxima")

[Out]

1/2*((b*x*e^(2*a) - e^(2*a))*e^(b*x) + (b*x + 1)*e^(-b*x))*e^(-a)/b^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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x\,{\mathrm {cosh}\left (a+b\,x\right )}^2}{\mathrm {sinh}\left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*cosh(a + b*x)^2)/sinh(a + b*x),x)

[Out]

int((x*cosh(a + b*x)^2)/sinh(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cosh ^{2}{\left (a + b x \right )} \operatorname {csch}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)**2*csch(b*x+a),x)

[Out]

Integral(x*cosh(a + b*x)**2*csch(a + b*x), x)

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