3.469 \(\int x \text {csch}(a+b x) \text {sech}(a+b x) \, dx\)

Optimal. Leaf size=58 \[ -\frac {\text {Li}_2\left (-e^{2 a+2 b x}\right )}{2 b^2}+\frac {\text {Li}_2\left (e^{2 a+2 b x}\right )}{2 b^2}-\frac {2 x \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5461, 4182, 2279, 2391} \[ -\frac {\text {PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}+\frac {\text {PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}-\frac {2 x \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[x*Csch[a + b*x]*Sech[a + b*x],x]

[Out]

(-2*x*ArcTanh[E^(2*a + 2*b*x)])/b - PolyLog[2, -E^(2*a + 2*b*x)]/(2*b^2) + PolyLog[2, E^(2*a + 2*b*x)]/(2*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 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 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 110, normalized size = 1.90 \[ \frac {\text {Li}_2\left (-e^{-2 (a+b x)}\right )-\text {Li}_2\left (e^{-2 (a+b x)}\right )+2 a \log \left (1-e^{-2 (a+b x)}\right )+2 b x \log \left (1-e^{-2 (a+b x)}\right )-2 a \log \left (e^{-2 (a+b x)}+1\right )-2 b x \log \left (e^{-2 (a+b x)}+1\right )-2 a \log (\tanh (a+b x))}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Csch[a + b*x]*Sech[a + b*x],x]

[Out]

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

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fricas [C]  time = 0.43, size = 224, normalized size = 3.86 \[ \frac {b x \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + a \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + a \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) - a \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - {\left (b x + a\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) - {\left (b x + a\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) + {\left (b x + a\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x*log(cosh(b*x + a) + sinh(b*x + a) + 1) + a*log(cosh(b*x + a) + sinh(b*x + a) + I) + a*log(cosh(b*x + a) +
sinh(b*x + a) - I) - a*log(cosh(b*x + a) + sinh(b*x + a) - 1) - (b*x + a)*log(I*cosh(b*x + a) + I*sinh(b*x +
a) + 1) - (b*x + a)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) + (b*x + a)*log(-cosh(b*x + a) - sinh(b*x + a)
+ 1) + dilog(cosh(b*x + a) + sinh(b*x + a)) - dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - dilog(-I*cosh(b*x +
a) - I*sinh(b*x + a)) + dilog(-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 \[ \int x \operatorname {csch}\left (b x + a\right ) \operatorname {sech}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.44, size = 125, normalized size = 2.16 \[ -\frac {x \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b}-\frac {\polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{2}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x}{b}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a}{b^{2}}+\frac {\polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) x}{b}+\frac {\polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {a \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.62, size = 87, normalized size = 1.50 \[ -\frac {2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )}{2 \, b^{2}} + \frac {b x \log \left (e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (b x + a\right )}\right )}{b^{2}} + \frac {b x \log \left (-e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (b x + a\right )}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*x*log(e^(2*b*x + 2*a) + 1) + dilog(-e^(2*b*x + 2*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 \[ \int \frac {x}{\mathrm {cosh}\left (a+b\,x\right )\,\mathrm {sinh}\left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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