3.476 \(\int x \text{csch}(a+b x) \text{sech}^2(a+b x) \, dx\)

Optimal. Leaf size=67 \[ -\frac{\text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{\text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}-\frac{\tan ^{-1}(\sinh (a+b x))}{b^2}-\frac{2 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x \text{sech}(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.114413, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {2622, 321, 207, 5462, 6271, 12, 4182, 2279, 2391, 3770} \[ -\frac{\text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{\text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}-\frac{\tan ^{-1}(\sinh (a+b x))}{b^2}-\frac{2 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x \text{sech}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[x*Csch[a + b*x]*Sech[a + b*x]^2,x]

[Out]

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

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 5462

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Wit
h[{u = IntHide[Csch[a + b*x]^n*Sech[a + b*x]^p, x]}, Dist[(c + d*x)^m, u, x] - Dist[d*m, Int[(c + d*x)^(m - 1)
*u, x], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n, p]

Rule 6271

Int[ArcTanh[u_], x_Symbol] :> Simp[x*ArcTanh[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/(1 - u^2), x], x] /; I
nverseFunctionFreeQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x \text{csch}(a+b x) \text{sech}^2(a+b x) \, dx &=-\frac{x \tanh ^{-1}(\cosh (a+b x))}{b}+\frac{x \text{sech}(a+b x)}{b}-\int \left (-\frac{\tanh ^{-1}(\cosh (a+b x))}{b}+\frac{\text{sech}(a+b x)}{b}\right ) \, dx\\ &=-\frac{x \tanh ^{-1}(\cosh (a+b x))}{b}+\frac{x \text{sech}(a+b x)}{b}+\frac{\int \tanh ^{-1}(\cosh (a+b x)) \, dx}{b}-\frac{\int \text{sech}(a+b x) \, dx}{b}\\ &=-\frac{\tan ^{-1}(\sinh (a+b x))}{b^2}+\frac{x \text{sech}(a+b x)}{b}+\frac{\int b x \text{csch}(a+b x) \, dx}{b}\\ &=-\frac{\tan ^{-1}(\sinh (a+b x))}{b^2}+\frac{x \text{sech}(a+b x)}{b}+\int x \text{csch}(a+b x) \, dx\\ &=-\frac{\tan ^{-1}(\sinh (a+b x))}{b^2}-\frac{2 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x \text{sech}(a+b x)}{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{\tan ^{-1}(\sinh (a+b x))}{b^2}-\frac{2 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x \text{sech}(a+b x)}{b}-\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{\tan ^{-1}(\sinh (a+b x))}{b^2}-\frac{2 x \tanh ^{-1}\left (e^{a+b x}\right )}{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{x \text{sech}(a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.282995, size = 106, normalized size = 1.58 \[ \frac{\text{PolyLog}\left (2,-e^{-a-b x}\right )-\text{PolyLog}\left (2,e^{-a-b x}\right )+(a+b x) \left (\log \left (1-e^{-a-b x}\right )-\log \left (e^{-a-b x}+1\right )\right )+b x \text{sech}(a+b x)-a \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )-2 \tan ^{-1}\left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Csch[a + b*x]*Sech[a + b*x]^2,x]

[Out]

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

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Maple [A]  time = 0.085, size = 95, normalized size = 1.4 \begin{align*} 2\,{\frac{x{{\rm e}^{bx+a}}}{b \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }}-2\,{\frac{\arctan \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{{\it dilog} \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{{\it dilog} \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{b}}-{\frac{a\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x*exp(b*x+a)/b/(1+exp(2*b*x+2*a))-2/b^2*arctan(exp(b*x+a))-1/b^2*dilog(exp(b*x+a))-1/b^2*dilog(1+exp(b*x+a))
-1/b*ln(1+exp(b*x+a))*x-1/b^2*a*ln(exp(b*x+a)-1)

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Maxima [A]  time = 1.72526, size = 122, normalized size = 1.82 \begin{align*} \frac{2 \, x e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} - \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}} - \frac{2 \, \arctan \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*csch(b*x+a)*sech(b*x+a)^2,x, algorithm="maxima")

[Out]

2*x*e^(b*x + a)/(b*e^(2*b*x + 2*a) + b) - (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 - 2*arctan(e^(b*x + a))/b^2

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Fricas [B]  time = 2.50582, size = 1166, normalized size = 17.4 \begin{align*} \frac{2 \, b x \cosh \left (b x + a\right ) + 2 \, b x \sinh \left (b x + a\right ) - 2 \,{\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) +{\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )}{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) -{\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )}{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) -{\left (b x \cosh \left (b x + a\right )^{2} + 2 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b x \sinh \left (b x + a\right )^{2} + b x\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) -{\left (a \cosh \left (b x + a\right )^{2} + 2 \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a \sinh \left (b x + a\right )^{2} + a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) +{\left ({\left (b x + a\right )} \cosh \left (b x + a\right )^{2} + 2 \,{\left (b x + a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (b x + a\right )} \sinh \left (b x + a\right )^{2} + b x + a\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right )}{b^{2} \cosh \left (b x + a\right )^{2} + 2 \, b^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )^{2} + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{csch}{\left (a + b x \right )} \operatorname{sech}^{2}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{csch}\left (b x + a\right ) \operatorname{sech}\left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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