3.414 $$\int x \cosh ^2(a+b x) \coth (a+b x) \, dx$$

Optimal. Leaf size=88 $\frac{\text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{2 b^2}-\frac{\sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac{x \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{x}{4 b}-\frac{x^2}{2}$

[Out]

x/(4*b) - x^2/2 + (x*Log[1 - E^(2*(a + b*x))])/b + PolyLog[2, E^(2*(a + b*x))]/(2*b^2) - (Cosh[a + b*x]*Sinh[a
+ b*x])/(4*b^2) + (x*Sinh[a + b*x]^2)/(2*b)

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Rubi [A]  time = 0.120577, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {5450, 5372, 2635, 8, 3716, 2190, 2279, 2391} $\frac{\text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{2 b^2}-\frac{\sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac{x \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{x}{4 b}-\frac{x^2}{2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

x/(4*b) - x^2/2 + (x*Log[1 - E^(2*(a + b*x))])/b + PolyLog[2, E^(2*(a + b*x))]/(2*b^2) - (Cosh[a + b*x]*Sinh[a
+ b*x])/(4*b^2) + (x*Sinh[a + b*x]^2)/(2*b)

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]

Rule 5372

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, 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]

Rubi steps

\begin{align*} \int x \cosh ^2(a+b x) \coth (a+b x) \, dx &=\int x \coth (a+b x) \, dx+\int x \cosh (a+b x) \sinh (a+b x) \, dx\\ &=-\frac{x^2}{2}+\frac{x \sinh ^2(a+b x)}{2 b}-2 \int \frac{e^{2 (a+b x)} x}{1-e^{2 (a+b x)}} \, dx-\frac{\int \sinh ^2(a+b x) \, dx}{2 b}\\ &=-\frac{x^2}{2}+\frac{x \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac{\cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{\int 1 \, dx}{4 b}-\frac{\int \log \left (1-e^{2 (a+b x)}\right ) \, dx}{b}\\ &=\frac{x}{4 b}-\frac{x^2}{2}+\frac{x \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac{\cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{x \sinh ^2(a+b x)}{2 b}-\frac{\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^2}\\ &=\frac{x}{4 b}-\frac{x^2}{2}+\frac{x \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac{\text{Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}-\frac{\cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{x \sinh ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.233824, size = 82, normalized size = 0.93 $-\frac{4 \left (\text{PolyLog}\left (2,e^{-2 (a+b x)}\right )-(a+b x)^2\right )-8 (a+b x) \log \left (1-e^{-2 (a+b x)}\right )+\sinh (2 (a+b x))-2 b x \cosh (2 (a+b x))+8 a \log (\sinh (a+b x))}{8 b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(-2*b*x*Cosh[2*(a + b*x)] - 8*(a + b*x)*Log[1 - E^(-2*(a + b*x))] + 8*a*Log[Sinh[a + b*x]] + 4*(-(a + b*x)^2
+ PolyLog[2, E^(-2*(a + b*x))]) + Sinh[2*(a + b*x)])/(8*b^2)

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Maple [B]  time = 0.07, size = 162, normalized size = 1.8 \begin{align*} -{\frac{{x}^{2}}{2}}+{\frac{ \left ( 2\,bx-1 \right ){{\rm e}^{2\,bx+2\,a}}}{16\,{b}^{2}}}+{\frac{ \left ( 2\,bx+1 \right ){{\rm e}^{-2\,bx-2\,a}}}{16\,{b}^{2}}}-2\,{\frac{ax}{b}}-{\frac{{a}^{2}}{{b}^{2}}}+{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}+{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{a\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{2}}}+2\,{\frac{a\ln \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)^3*csch(b*x+a),x)

[Out]

-1/2*x^2+1/16*(2*b*x-1)/b^2*exp(2*b*x+2*a)+1/16*(2*b*x+1)/b^2*exp(-2*b*x-2*a)-2/b*a*x-a^2/b^2+1/b*ln(1+exp(b*x
+a))*x+1/b^2*polylog(2,-exp(b*x+a))+1/b*ln(1-exp(b*x+a))*x+1/b^2*ln(1-exp(b*x+a))*a+1/b^2*polylog(2,exp(b*x+a)
)-1/b^2*a*ln(exp(b*x+a)-1)+2/b^2*a*ln(exp(b*x+a))

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Maxima [A]  time = 1.36419, size = 153, normalized size = 1.74 \begin{align*} -x^{2} + \frac{{\left (8 \, b^{2} x^{2} e^{\left (2 \, a\right )} +{\left (2 \, b x e^{\left (4 \, a\right )} - e^{\left (4 \, a\right )}\right )} e^{\left (2 \, b x\right )} +{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x\right )}\right )} e^{\left (-2 \, a\right )}}{16 \, 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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2 + 1/16*(8*b^2*x^2*e^(2*a) + (2*b*x*e^(4*a) - e^(4*a))*e^(2*b*x) + (2*b*x + 1)*e^(-2*b*x))*e^(-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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Fricas [B]  time = 2.16096, size = 1355, normalized size = 15.4 \begin{align*} \frac{{\left (2 \, b x - 1\right )} \cosh \left (b x + a\right )^{4} + 4 \,{\left (2 \, b x - 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} +{\left (2 \, b x - 1\right )} \sinh \left (b x + a\right )^{4} - 8 \,{\left (b^{2} x^{2} - 2 \, a^{2}\right )} \cosh \left (b x + a\right )^{2} - 2 \,{\left (4 \, b^{2} x^{2} - 3 \,{\left (2 \, b x - 1\right )} \cosh \left (b x + a\right )^{2} - 8 \, a^{2}\right )} \sinh \left (b x + a\right )^{2} + 2 \, b x + 16 \,{\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}\right )}{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 16 \,{\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}\right )}{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + 16 \,{\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}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 16 \,{\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}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 16 \,{\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}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 4 \,{\left ({\left (2 \, b x - 1\right )} \cosh \left (b x + a\right )^{3} - 4 \,{\left (b^{2} x^{2} - 2 \, a^{2}\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1}{16 \,{\left (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}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((2*b*x - 1)*cosh(b*x + a)^4 + 4*(2*b*x - 1)*cosh(b*x + a)*sinh(b*x + a)^3 + (2*b*x - 1)*sinh(b*x + a)^4
- 8*(b^2*x^2 - 2*a^2)*cosh(b*x + a)^2 - 2*(4*b^2*x^2 - 3*(2*b*x - 1)*cosh(b*x + a)^2 - 8*a^2)*sinh(b*x + a)^2
+ 2*b*x + 16*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2)*dilog(cosh(b*x + a) + sinh(b*
x + a)) + 16*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2)*dilog(-cosh(b*x + a) - sinh(b
*x + a)) + 16*(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)*log(cosh(b*x + a
) + sinh(b*x + a) + 1) - 16*(a*cosh(b*x + a)^2 + 2*a*cosh(b*x + a)*sinh(b*x + a) + a*sinh(b*x + a)^2)*log(cosh
(b*x + a) + sinh(b*x + a) - 1) + 16*((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)*log(-cosh(b*x + a) - sinh(b*x + a) + 1) + 4*((2*b*x - 1)*cosh(b*x + a)^3 - 4*(b^2*x^2
- 2*a^2)*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)

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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)**3*csch(b*x+a),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 )^{3} \operatorname{csch}\left (b x + a\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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