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

Optimal. Leaf size=88 \[ \frac {\text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}-\frac {\sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac {x \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {x \sinh ^2(a+b x)}{2 b}+\frac {x}{4 b}-\frac {x^2}{2} \]

[Out]

1/4*x/b-1/2*x^2+x*ln(1-exp(2*b*x+2*a))/b+1/2*polylog(2,exp(2*b*x+2*a))/b^2-1/4*cosh(b*x+a)*sinh(b*x+a)/b^2+1/2
*x*sinh(b*x+a)^2/b

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Rubi [A]  time = 0.12, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 8

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

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]

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 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 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 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]

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*}

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Mathematica [A]  time = 0.25, size = 82, normalized size = 0.93 \[ -\frac {4 \left (\text {Li}_2\left (e^{-2 (a+b x)}\right )-(a+b x)^2\right )-8 (a+b x) \log \left (1-e^{-2 (a+b x)}\right )+\sinh (2 (a+b x))-2 b x \cosh (2 (a+b x))+8 a \log (\sinh (a+b x))}{8 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(-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)])/b^2

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fricas [B]  time = 1.11, size = 488, normalized size = 5.55 \[ \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 )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cosh \left (b x + a\right )^{3} \operatorname {csch}\left (b x + a\right )\,{d x} \]

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)

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maple [B]  time = 0.61, size = 162, normalized size = 1.84 \[ -\frac {x^{2}}{2}+\frac {\left (2 b x -1\right ) {\mathrm e}^{2 b x +2 a}}{16 b^{2}}+\frac {\left (2 b x +1\right ) {\mathrm e}^{-2 b x -2 a}}{16 b^{2}}-\frac {2 a x}{b}-\frac {a^{2}}{b^{2}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x}{b}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a}{b^{2}}+\frac {\polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) x}{b}+\frac {\polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {2 a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {a \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.41, size = 113, normalized size = 1.28 \[ -x^{2} + \frac {{\left (8 \, b^{2} x^{2} e^{\left (2 \, a\right )} + {\left (2 \, b x e^{\left (4 \, a\right )} - e^{\left (4 \, a\right )}\right )} e^{\left (2 \, b x\right )} + {\left (2 \, b x + 1\right )} e^{\left (-2 \, b x\right )}\right )} e^{\left (-2 \, a\right )}}{16 \, b^{2}} + \frac {b x \log \left (e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (b x + a\right )}\right )}{b^{2}} + \frac {b x \log \left (-e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (b x + a\right )}\right )}{b^{2}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,{\mathrm {cosh}\left (a+b\,x\right )}^3}{\mathrm {sinh}\left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*cosh(a + b*x)^3)/sinh(a + b*x),x)

[Out]

int((x*cosh(a + b*x)^3)/sinh(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cosh ^{3}{\left (a + b x \right )} \operatorname {csch}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*cosh(a + b*x)**3*csch(a + b*x), x)

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