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

Optimal. Leaf size=47 \[ -\frac{\cosh (a+b x)}{b^2}-\frac{\tanh ^{-1}(\cosh (a+b x))}{b^2}+\frac{x \sinh (a+b x)}{b}-\frac{x \text{csch}(a+b x)}{b} \]

[Out]

-(ArcTanh[Cosh[a + b*x]]/b^2) - Cosh[a + b*x]/b^2 - (x*Csch[a + b*x])/b + (x*Sinh[a + b*x])/b

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Rubi [A]  time = 0.0541248, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {5450, 3296, 2638, 5419, 3770} \[ -\frac{\cosh (a+b x)}{b^2}-\frac{\tanh ^{-1}(\cosh (a+b x))}{b^2}+\frac{x \sinh (a+b x)}{b}-\frac{x \text{csch}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Cosh[a + b*x]]/b^2) - Cosh[a + b*x]/b^2 - (x*Csch[a + b*x])/b + (x*Sinh[a + b*x])/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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

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

Rule 5419

Int[Coth[(a_.) + (b_.)*(x_)^(n_.)]^(q_.)*Csch[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.), x_Symbol] :> -Simp[(
x^(m - n + 1)*Csch[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Csch[a + b*x^n]^p, x],
x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 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 \cosh (a+b x) \coth ^2(a+b x) \, dx &=\int x \cosh (a+b x) \, dx+\int x \coth (a+b x) \text{csch}(a+b x) \, dx\\ &=-\frac{x \text{csch}(a+b x)}{b}+\frac{x \sinh (a+b x)}{b}+\frac{\int \text{csch}(a+b x) \, dx}{b}-\frac{\int \sinh (a+b x) \, dx}{b}\\ &=-\frac{\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac{\cosh (a+b x)}{b^2}-\frac{x \text{csch}(a+b x)}{b}+\frac{x \sinh (a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.218691, size = 66, normalized size = 1.4 \[ \frac{2 b x \sinh (a+b x)-2 \cosh (a+b x)+b x \tanh \left (\frac{1}{2} (a+b x)\right )-b x \coth \left (\frac{1}{2} (a+b x)\right )+2 \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cosh[a + b*x] - b*x*Coth[(a + b*x)/2] + 2*Log[Tanh[(a + b*x)/2]] + 2*b*x*Sinh[a + b*x] + b*x*Tanh[(a + b*x
)/2])/(2*b^2)

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Maple [A]  time = 0.077, size = 89, normalized size = 1.9 \begin{align*}{\frac{ \left ( bx-1 \right ){{\rm e}^{bx+a}}}{2\,{b}^{2}}}-{\frac{ \left ( bx+1 \right ){{\rm e}^{-bx-a}}}{2\,{b}^{2}}}-2\,{\frac{{{\rm e}^{bx+a}}x}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}-{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{\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*cosh(b*x+a)^3*csch(b*x+a)^2,x)

[Out]

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

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Maxima [B]  time = 1.36774, size = 147, normalized size = 3.13 \begin{align*} -\frac{6 \, b x e^{\left (b x + 2 \, a\right )} -{\left (b x e^{\left (4 \, a\right )} - e^{\left (4 \, a\right )}\right )} e^{\left (3 \, b x\right )} -{\left (b x + 1\right )} e^{\left (-b x\right )}}{2 \,{\left (b^{2} e^{\left (2 \, b x + 3 \, a\right )} - b^{2} e^{a}\right )}} - \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*b*x*e^(b*x + 2*a) - (b*x*e^(4*a) - e^(4*a))*e^(3*b*x) - (b*x + 1)*e^(-b*x))/(b^2*e^(2*b*x + 3*a) - b^2
*e^a) - log((e^(b*x + a) + 1)*e^(-a))/b^2 + log((e^(b*x + a) - 1)*e^(-a))/b^2

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Fricas [B]  time = 2.29503, size = 1008, normalized size = 21.45 \begin{align*} \frac{{\left (b x - 1\right )} \cosh \left (b x + a\right )^{4} + 4 \,{\left (b x - 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} +{\left (b x - 1\right )} \sinh \left (b x + a\right )^{4} - 6 \, b x \cosh \left (b x + a\right )^{2} + 6 \,{\left ({\left (b x - 1\right )} \cosh \left (b x + a\right )^{2} - b x\right )} \sinh \left (b x + a\right )^{2} + b x - 2 \,{\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} +{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - \cosh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 2 \,{\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} +{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - \cosh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 4 \,{\left ({\left (b x - 1\right )} \cosh \left (b x + a\right )^{3} - 3 \, b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1}{2 \,{\left (b^{2} \cosh \left (b x + a\right )^{3} + 3 \, b^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + b^{2} \sinh \left (b x + a\right )^{3} - b^{2} \cosh \left (b x + a\right ) +{\left (3 \, b^{2} \cosh \left (b x + a\right )^{2} - b^{2}\right )} \sinh \left (b x + a\right )\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)^2,x, algorithm="fricas")

[Out]

1/2*((b*x - 1)*cosh(b*x + a)^4 + 4*(b*x - 1)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*x - 1)*sinh(b*x + a)^4 - 6*b*x
*cosh(b*x + a)^2 + 6*((b*x - 1)*cosh(b*x + a)^2 - b*x)*sinh(b*x + a)^2 + b*x - 2*(cosh(b*x + a)^3 + 3*cosh(b*x
+ a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 + (3*cosh(b*x + a)^2 - 1)*sinh(b*x + a) - cosh(b*x + a))*log(cosh(b*x
+ a) + sinh(b*x + a) + 1) + 2*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 + (3*cosh(b
*x + a)^2 - 1)*sinh(b*x + a) - cosh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 4*((b*x - 1)*cosh(b*x +
a)^3 - 3*b*x*cosh(b*x + a))*sinh(b*x + a) + 1)/(b^2*cosh(b*x + a)^3 + 3*b^2*cosh(b*x + a)*sinh(b*x + a)^2 + b
^2*sinh(b*x + a)^3 - b^2*cosh(b*x + a) + (3*b^2*cosh(b*x + a)^2 - b^2)*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)**3*csch(b*x+a)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.23585, size = 194, normalized size = 4.13 \begin{align*} \frac{b x e^{\left (4 \, b x + 4 \, a\right )} - 6 \, b x e^{\left (2 \, b x + 2 \, a\right )} + b x - 2 \, e^{\left (3 \, b x + 3 \, a\right )} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, e^{\left (b x + a\right )} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, e^{\left (3 \, b x + 3 \, a\right )} \log \left (e^{\left (b x + a\right )} - 1\right ) - 2 \, e^{\left (b x + a\right )} \log \left (e^{\left (b x + a\right )} - 1\right ) - e^{\left (4 \, b x + 4 \, a\right )} + 1}{2 \,{\left (b^{2} e^{\left (3 \, b x + 3 \, a\right )} - b^{2} e^{\left (b x + a\right )}\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)^2,x, algorithm="giac")

[Out]

1/2*(b*x*e^(4*b*x + 4*a) - 6*b*x*e^(2*b*x + 2*a) + b*x - 2*e^(3*b*x + 3*a)*log(e^(b*x + a) + 1) + 2*e^(b*x + a
)*log(e^(b*x + a) + 1) + 2*e^(3*b*x + 3*a)*log(e^(b*x + a) - 1) - 2*e^(b*x + a)*log(e^(b*x + a) - 1) - e^(4*b*
x + 4*a) + 1)/(b^2*e^(3*b*x + 3*a) - b^2*e^(b*x + a))