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

Optimal. Leaf size=42 \[ \frac {\log (\sinh (a+b x))}{b^3}-\frac {x \coth (a+b x)}{b^2}-\frac {x^2 \text {csch}^2(a+b x)}{2 b} \]

[Out]

-x*coth(b*x+a)/b^2-1/2*x^2*csch(b*x+a)^2/b+ln(sinh(b*x+a))/b^3

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Rubi [A]  time = 0.06, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5419, 4184, 3475} \[ -\frac {x \coth (a+b x)}{b^2}+\frac {\log (\sinh (a+b x))}{b^3}-\frac {x^2 \text {csch}^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((x*Coth[a + b*x])/b^2) - (x^2*Csch[a + b*x]^2)/(2*b) + Log[Sinh[a + b*x]]/b^3

Rule 3475

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

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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]

Rubi steps

\begin {align*} \int x^2 \coth (a+b x) \text {csch}^2(a+b x) \, dx &=-\frac {x^2 \text {csch}^2(a+b x)}{2 b}+\frac {\int x \text {csch}^2(a+b x) \, dx}{b}\\ &=-\frac {x \coth (a+b x)}{b^2}-\frac {x^2 \text {csch}^2(a+b x)}{2 b}+\frac {\int \coth (a+b x) \, dx}{b^2}\\ &=-\frac {x \coth (a+b x)}{b^2}-\frac {x^2 \text {csch}^2(a+b x)}{2 b}+\frac {\log (\sinh (a+b x))}{b^3}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 55, normalized size = 1.31 \[ \frac {\log (\sinh (a+b x))}{b^3}-\frac {x \coth (a)}{b^2}+\frac {x \text {csch}(a) \sinh (b x) \text {csch}(a+b x)}{b^2}-\frac {x^2 \text {csch}^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((x*Coth[a])/b^2) - (x^2*Csch[a + b*x]^2)/(2*b) + Log[Sinh[a + b*x]]/b^3 + (x*Csch[a]*Csch[a + b*x]*Sinh[b*x]
)/b^2

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fricas [B]  time = 0.41, size = 383, normalized size = 9.12 \[ -\frac {2 \, b x \cosh \left (b x + a\right )^{4} + 8 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + 2 \, b x \sinh \left (b x + a\right )^{4} + 2 \, {\left (b^{2} x^{2} - b x\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b^{2} x^{2} + 6 \, b x \cosh \left (b x + a\right )^{2} - b x\right )} \sinh \left (b x + a\right )^{2} - {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\frac {2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + 4 \, {\left (2 \, b x \cosh \left (b x + a\right )^{3} + {\left (b^{2} x^{2} - b x\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{b^{3} \cosh \left (b x + a\right )^{4} + 4 \, b^{3} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b^{3} \sinh \left (b x + a\right )^{4} - 2 \, b^{3} \cosh \left (b x + a\right )^{2} + b^{3} + 2 \, {\left (3 \, b^{3} \cosh \left (b x + a\right )^{2} - b^{3}\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b^{3} \cosh \left (b x + a\right )^{3} - b^{3} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*b*x*cosh(b*x + a)^4 + 8*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + 2*b*x*sinh(b*x + a)^4 + 2*(b^2*x^2 - b*x)*cosh
(b*x + a)^2 + 2*(b^2*x^2 + 6*b*x*cosh(b*x + a)^2 - b*x)*sinh(b*x + a)^2 - (cosh(b*x + a)^4 + 4*cosh(b*x + a)*s
inh(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)*log(2*sinh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 4*(2*b*x*co
sh(b*x + a)^3 + (b^2*x^2 - b*x)*cosh(b*x + a))*sinh(b*x + a))/(b^3*cosh(b*x + a)^4 + 4*b^3*cosh(b*x + a)*sinh(
b*x + a)^3 + b^3*sinh(b*x + a)^4 - 2*b^3*cosh(b*x + a)^2 + b^3 + 2*(3*b^3*cosh(b*x + a)^2 - b^3)*sinh(b*x + a)
^2 + 4*(b^3*cosh(b*x + a)^3 - b^3*cosh(b*x + a))*sinh(b*x + a))

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giac [B]  time = 0.13, size = 139, normalized size = 3.31 \[ -\frac {2 \, b^{2} x^{2} e^{\left (2 \, b x + 2 \, a\right )} + 2 \, b x e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b x e^{\left (2 \, b x + 2 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} \log \left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right ) + 2 \, e^{\left (2 \, b x + 2 \, a\right )} \log \left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right ) - \log \left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}{b^{3} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{3} e^{\left (2 \, b x + 2 \, a\right )} + b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.19, size = 72, normalized size = 1.71 \[ -\frac {2 x}{b^{2}}-\frac {2 a}{b^{3}}-\frac {2 x \left (b x \,{\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 b x +2 a}-1\right )}{b^{2} \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 b x +2 a}-1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x/b^2-2/b^3*a-2*x*(b*x*exp(2*b*x+2*a)+exp(2*b*x+2*a)-1)/b^2/(exp(2*b*x+2*a)-1)^2+1/b^3*ln(exp(2*b*x+2*a)-1)

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maxima [B]  time = 0.60, size = 107, normalized size = 2.55 \[ -\frac {2 \, {\left ({\left (b x^{2} e^{\left (2 \, a\right )} - x e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + x e^{\left (4 \, b x + 4 \, a\right )}\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + \frac {\log \left ({\left (e^{\left (b x + a\right )} + 1\right )} e^{\left (-a\right )}\right )}{b^{3}} + \frac {\log \left ({\left (e^{\left (b x + a\right )} - 1\right )} e^{\left (-a\right )}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.47, size = 101, normalized size = 2.40 \[ \frac {\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1\right )}{b^3}-\frac {\frac {x^2}{b}+\frac {x^2\,{\mathrm {e}}^{2\,a+2\,b\,x}}{b}}{{\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1}-\frac {2\,x}{b^2}-\frac {b\,x^2+2\,x}{b^2\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*cosh(a + b*x))/sinh(a + b*x)^3,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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