### 3.490 $$\int x \text{csch}^2(a+b x) \text{sech}(a+b x) \, dx$$

Optimal. Leaf size=79 $\frac{i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{i \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x \text{csch}(a+b x)}{b}$

[Out]

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

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Rubi [A]  time = 0.11245, antiderivative size = 79, 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 = {2621, 321, 207, 5462, 5203, 12, 4180, 2279, 2391, 3770} $\frac{i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{i \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x \text{csch}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(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 5203

Int[ArcTan[u_], x_Symbol] :> Simp[x*ArcTan[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/(1 + u^2), x], x] /; Inv
erseFunctionFreeQ[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 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && 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}^2(a+b x) \text{sech}(a+b x) \, dx &=-\frac{x \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x \text{csch}(a+b x)}{b}-\int \left (-\frac{\tan ^{-1}(\sinh (a+b x))}{b}-\frac{\text{csch}(a+b x)}{b}\right ) \, dx\\ &=-\frac{x \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x \text{csch}(a+b x)}{b}+\frac{\int \tan ^{-1}(\sinh (a+b x)) \, dx}{b}+\frac{\int \text{csch}(a+b x) \, dx}{b}\\ &=-\frac{\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac{x \text{csch}(a+b x)}{b}-\frac{\int b x \text{sech}(a+b x) \, dx}{b}\\ &=-\frac{\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac{x \text{csch}(a+b x)}{b}-\int x \text{sech}(a+b x) \, dx\\ &=-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac{x \text{csch}(a+b x)}{b}+\frac{i \int \log \left (1-i e^{a+b x}\right ) \, dx}{b}-\frac{i \int \log \left (1+i e^{a+b x}\right ) \, dx}{b}\\ &=-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac{x \text{csch}(a+b x)}{b}+\frac{i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^2}-\frac{i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{a+b x}\right )}{b^2}\\ &=-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac{x \text{csch}(a+b x)}{b}+\frac{i \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{i \text{Li}_2\left (i e^{a+b x}\right )}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.805017, size = 112, normalized size = 1.42 $\frac{2 i \text{PolyLog}(2,-i (\sinh (a+b x)+\cosh (a+b x)))-2 i \text{PolyLog}(2,i (\sinh (a+b x)+\cosh (a+b x)))+b x \tanh \left (\frac{1}{2} (a+b x)\right )-b x \coth \left (\frac{1}{2} (a+b x)\right )+2 \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )-4 b x \tan ^{-1}(\sinh (a+b x)+\cosh (a+b x))}{2 b^2}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*b*x*ArcTan[Cosh[a + b*x] + Sinh[a + b*x]] - b*x*Coth[(a + b*x)/2] + 2*Log[Tanh[(a + b*x)/2]] + (2*I)*PolyL
og[2, (-I)*(Cosh[a + b*x] + Sinh[a + b*x])] - (2*I)*PolyLog[2, I*(Cosh[a + b*x] + Sinh[a + b*x])] + b*x*Tanh[(
a + b*x)/2])/(2*b^2)

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Maple [B]  time = 0.041, size = 179, normalized size = 2.3 \begin{align*} -2\,{\frac{x{{\rm e}^{bx+a}}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}+2\,{\frac{a\arctan \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{i{\it dilog} \left ( 1-i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{i{\it dilog} \left ( 1+i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{i\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{i\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}-{\frac{i\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) x}{b}}-{\frac{i\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}+{\frac{\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{2}}}-{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2 \, x e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} - \frac{\log \left ({\left (e^{\left (b x + a\right )} + 1\right )} e^{\left (-a\right )}\right )}{b^{2}} + \frac{\log \left ({\left (e^{\left (b x + a\right )} - 1\right )} e^{\left (-a\right )}\right )}{b^{2}} - 8 \, \int \frac{x e^{\left (b x + a\right )}}{4 \,{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x*e^(b*x + a)/(b*e^(2*b*x + 2*a) - b) - log((e^(b*x + a) + 1)*e^(-a))/b^2 + log((e^(b*x + a) - 1)*e^(-a))/b
^2 - 8*integrate(1/4*x*e^(b*x + a)/(e^(2*b*x + 2*a) + 1), x)

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Fricas [B]  time = 2.67801, size = 1624, normalized size = 20.56 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*b*x*cosh(b*x + a) + 2*b*x*sinh(b*x + a) - (-I*cosh(b*x + a)^2 - 2*I*cosh(b*x + a)*sinh(b*x + a) - I*sinh(b
*x + a)^2 + I)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - (I*cosh(b*x + a)^2 + 2*I*cosh(b*x + a)*sinh(b*x + a)
+ I*sinh(b*x + a)^2 - I)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + (cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(
b*x + a) + sinh(b*x + a)^2 - 1)*log(cosh(b*x + a) + sinh(b*x + a) + 1) - (I*a*cosh(b*x + a)^2 + 2*I*a*cosh(b*x
+ a)*sinh(b*x + a) + I*a*sinh(b*x + a)^2 - I*a)*log(cosh(b*x + a) + sinh(b*x + a) + I) - (-I*a*cosh(b*x + a)^
2 - 2*I*a*cosh(b*x + a)*sinh(b*x + a) - I*a*sinh(b*x + a)^2 + I*a)*log(cosh(b*x + a) + sinh(b*x + a) - I) - (c
osh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)*log(cosh(b*x + a) + sinh(b*x + a) - 1) -
((I*b*x + I*a)*cosh(b*x + a)^2 + (2*I*b*x + 2*I*a)*cosh(b*x + a)*sinh(b*x + a) + (I*b*x + I*a)*sinh(b*x + a)^
2 - I*b*x - I*a)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) - ((-I*b*x - I*a)*cosh(b*x + a)^2 + (-2*I*b*x - 2*
I*a)*cosh(b*x + a)*sinh(b*x + a) + (-I*b*x - I*a)*sinh(b*x + a)^2 + I*b*x + I*a)*log(-I*cosh(b*x + a) - I*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}^{2}{\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*csch(b*x+a)**2*sech(b*x+a),x)

[Out]

Integral(x*csch(a + b*x)**2*sech(a + b*x), 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 )^{2} \operatorname{sech}\left (b x + a\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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