### 3.455 $$\int x \coth ^2(a+b x) \text{csch}(a+b x) \, dx$$

Optimal. Leaf size=82 $-\frac{\text{PolyLog}\left (2,-e^{a+b x}\right )}{2 b^2}+\frac{\text{PolyLog}\left (2,e^{a+b x}\right )}{2 b^2}-\frac{\text{csch}(a+b x)}{2 b^2}-\frac{x \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x \coth (a+b x) \text{csch}(a+b x)}{2 b}$

[Out]

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

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Rubi [A]  time = 0.124972, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.312, Rules used = {5457, 4182, 2279, 2391, 4185} $-\frac{\text{PolyLog}\left (2,-e^{a+b x}\right )}{2 b^2}+\frac{\text{PolyLog}\left (2,e^{a+b x}\right )}{2 b^2}-\frac{\text{csch}(a+b x)}{2 b^2}-\frac{x \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x \coth (a+b x) \text{csch}(a+b x)}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 5457

Int[Coth[(a_.) + (b_.)*(x_)]^(p_)*Csch[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(c + d
*x)^m*Csch[a + b*x]*Coth[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Csch[a + b*x]^3*Coth[a + b*x]^(p - 2), x] /; F
reeQ[{a, b, c, d, m}, x] && IGtQ[p/2, 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 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 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

Rubi steps

\begin{align*} \int x \coth ^2(a+b x) \text{csch}(a+b x) \, dx &=\int x \text{csch}(a+b x) \, dx+\int x \text{csch}^3(a+b x) \, dx\\ &=-\frac{2 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\text{csch}(a+b x)}{2 b^2}-\frac{x \coth (a+b x) \text{csch}(a+b x)}{2 b}-\frac{1}{2} \int x \text{csch}(a+b x) \, dx-\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{x \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\text{csch}(a+b x)}{2 b^2}-\frac{x \coth (a+b x) \text{csch}(a+b x)}{2 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{\int \log \left (1-e^{a+b x}\right ) \, dx}{2 b}-\frac{\int \log \left (1+e^{a+b x}\right ) \, dx}{2 b}\\ &=-\frac{x \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\text{csch}(a+b x)}{2 b^2}-\frac{x \coth (a+b x) \text{csch}(a+b x)}{2 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{\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}-\frac{\operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}\\ &=-\frac{x \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\text{csch}(a+b x)}{2 b^2}-\frac{x \coth (a+b x) \text{csch}(a+b x)}{2 b}-\frac{\text{Li}_2\left (-e^{a+b x}\right )}{2 b^2}+\frac{\text{Li}_2\left (e^{a+b x}\right )}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 2.07128, size = 144, normalized size = 1.76 $-\frac{-4 \text{PolyLog}\left (2,-e^{-a-b x}\right )+4 \text{PolyLog}\left (2,e^{-a-b x}\right )-4 (a+b x) \left (\log \left (1-e^{-a-b x}\right )-\log \left (e^{-a-b x}+1\right )\right )-2 \tanh \left (\frac{1}{2} (a+b x)\right )+2 \coth \left (\frac{1}{2} (a+b x)\right )+b x \text{csch}^2\left (\frac{1}{2} (a+b x)\right )+b x \text{sech}^2\left (\frac{1}{2} (a+b x)\right )+4 a \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{8 b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(2*Coth[(a + b*x)/2] + b*x*Csch[(a + b*x)/2]^2 - 4*(a + b*x)*(Log[1 - E^(-a - b*x)] - Log[1 + E^(-a - b*x)])
+ 4*a*Log[Tanh[(a + b*x)/2]] - 4*PolyLog[2, -E^(-a - b*x)] + 4*PolyLog[2, E^(-a - b*x)] + b*x*Sech[(a + b*x)/2
]^2 - 2*Tanh[(a + b*x)/2])/(8*b^2)

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Maple [B]  time = 0.064, size = 156, normalized size = 1.9 \begin{align*} -{\frac{{{\rm e}^{bx+a}} \left ( bx{{\rm e}^{2\,bx+2\,a}}+bx+{{\rm e}^{2\,bx+2\,a}}-1 \right ) }{{b}^{2} \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) ^{2}}}-{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{2\,b}}-{\frac{a\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{2\,{b}^{2}}}-{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{2\,{b}^{2}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{2\,b}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{2\,{b}^{2}}}+{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{2\,{b}^{2}}}+{\frac{a{\it Artanh} \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)^2*csch(b*x+a)^3,x)

[Out]

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

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Maxima [A]  time = 1.59869, size = 167, normalized size = 2.04 \begin{align*} -\frac{{\left (b x e^{\left (3 \, a\right )} + e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )} +{\left (b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\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 )}{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 )}{2 \, b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)^2*csch(b*x+a)^3,x, algorithm="maxima")

[Out]

-((b*x*e^(3*a) + e^(3*a))*e^(3*b*x) + (b*x*e^a - e^a)*e^(b*x))/(b^2*e^(4*b*x + 4*a) - 2*b^2*e^(2*b*x + 2*a) +
b^2) - 1/2*(b*x*log(e^(b*x + a) + 1) + dilog(-e^(b*x + a)))/b^2 + 1/2*(b*x*log(-e^(b*x + a) + 1) + dilog(e^(b*
x + a)))/b^2

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Fricas [B]  time = 2.37047, size = 2294, normalized size = 27.98 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)^2*csch(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/2*(2*(b*x + 1)*cosh(b*x + a)^3 + 6*(b*x + 1)*cosh(b*x + a)*sinh(b*x + a)^2 + 2*(b*x + 1)*sinh(b*x + a)^3 +
2*(b*x - 1)*cosh(b*x + a) - (cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b
*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*di
log(cosh(b*x + a) + sinh(b*x + a)) + (cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*
(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a
) + 1)*dilog(-cosh(b*x + a) - sinh(b*x + a)) + (b*x*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*
x*sinh(b*x + a)^4 - 2*b*x*cosh(b*x + a)^2 + 2*(3*b*x*cosh(b*x + a)^2 - b*x)*sinh(b*x + a)^2 + b*x + 4*(b*x*cos
h(b*x + a)^3 - b*x*cosh(b*x + a))*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) + 1) + (a*cosh(b*x + a)^4 +
4*a*cosh(b*x + a)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 - 2*a*cosh(b*x + a)^2 + 2*(3*a*cosh(b*x + a)^2 - a)*sin
h(b*x + a)^2 + 4*(a*cosh(b*x + a)^3 - a*cosh(b*x + a))*sinh(b*x + a) + a)*log(cosh(b*x + a) + sinh(b*x + a) -
1) - ((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*x + a)*sinh(b*x + a)^4 - 2*(b
*x + a)*cosh(b*x + a)^2 + 2*(3*(b*x + a)*cosh(b*x + a)^2 - b*x - a)*sinh(b*x + a)^2 + b*x + 4*((b*x + a)*cosh(
b*x + a)^3 - (b*x + a)*cosh(b*x + a))*sinh(b*x + a) + a)*log(-cosh(b*x + a) - sinh(b*x + a) + 1) + 2*(3*(b*x +
1)*cosh(b*x + a)^2 + b*x - 1)*sinh(b*x + a))/(b^2*cosh(b*x + a)^4 + 4*b^2*cosh(b*x + a)*sinh(b*x + a)^3 + b^2
*sinh(b*x + a)^4 - 2*b^2*cosh(b*x + a)^2 + 2*(3*b^2*cosh(b*x + a)^2 - b^2)*sinh(b*x + a)^2 + b^2 + 4*(b^2*cosh
(b*x + a)^3 - b^2*cosh(b*x + a))*sinh(b*x + a))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)**2*csch(b*x+a)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \cosh \left (b x + a\right )^{2} \operatorname{csch}\left (b x + a\right )^{3}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)^2*csch(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x*cosh(b*x + a)^2*csch(b*x + a)^3, x)