3.185 \(\int \frac{\coth ^3(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=94 \[ -\frac{a^3 \log (a+b \cosh (x))}{\left (a^2-b^2\right )^2}-\frac{\text{csch}^2(x) (a-b \cosh (x))}{2 \left (a^2-b^2\right )}+\frac{(2 a+b) \log (1-\cosh (x))}{4 (a+b)^2}+\frac{(2 a-b) \log (\cosh (x)+1)}{4 (a-b)^2} \]

[Out]

-((a - b*Cosh[x])*Csch[x]^2)/(2*(a^2 - b^2)) + ((2*a + b)*Log[1 - Cosh[x]])/(4*(a + b)^2) + ((2*a - b)*Log[1 +
Cosh[x]])/(4*(a - b)^2) - (a^3*Log[a + b*Cosh[x]])/(a^2 - b^2)^2

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Rubi [A]  time = 0.199426, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2721, 1647, 801} \[ -\frac{a^3 \log (a+b \cosh (x))}{\left (a^2-b^2\right )^2}-\frac{\text{csch}^2(x) (a-b \cosh (x))}{2 \left (a^2-b^2\right )}+\frac{(2 a+b) \log (1-\cosh (x))}{4 (a+b)^2}+\frac{(2 a-b) \log (\cosh (x)+1)}{4 (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^3/(a + b*Cosh[x]),x]

[Out]

-((a - b*Cosh[x])*Csch[x]^2)/(2*(a^2 - b^2)) + ((2*a + b)*Log[1 - Cosh[x]])/(4*(a + b)^2) + ((2*a - b)*Log[1 +
Cosh[x]])/(4*(a - b)^2) - (a^3*Log[a + b*Cosh[x]])/(a^2 - b^2)^2

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

\begin{align*} \int \frac{\coth ^3(x)}{a+b \cosh (x)} \, dx &=\operatorname{Subst}\left (\int \frac{x^3}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \cosh (x)\right )\\ &=-\frac{(a-b \cosh (x)) \text{csch}^2(x)}{2 \left (a^2-b^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{\frac{a b^4}{a^2-b^2}-\frac{b^2 \left (2 a^2-b^2\right ) x}{a^2-b^2}}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \cosh (x)\right )}{2 b^2}\\ &=-\frac{(a-b \cosh (x)) \text{csch}^2(x)}{2 \left (a^2-b^2\right )}+\frac{\operatorname{Subst}\left (\int \left (-\frac{b^2 (2 a+b)}{2 (a+b)^2 (b-x)}-\frac{2 a^3 b^2}{(a-b)^2 (a+b)^2 (a+x)}+\frac{(2 a-b) b^2}{2 (a-b)^2 (b+x)}\right ) \, dx,x,b \cosh (x)\right )}{2 b^2}\\ &=-\frac{(a-b \cosh (x)) \text{csch}^2(x)}{2 \left (a^2-b^2\right )}+\frac{(2 a+b) \log (1-\cosh (x))}{4 (a+b)^2}+\frac{(2 a-b) \log (1+\cosh (x))}{4 (a-b)^2}-\frac{a^3 \log (a+b \cosh (x))}{\left (a^2-b^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.210392, size = 101, normalized size = 1.07 \[ \frac{-12 a^2 b \log \left (\tanh \left (\frac{x}{2}\right )\right )-8 a^3 \log (a+b \cosh (x))+8 a^3 \log (\sinh (x))-(a-b)^2 (a+b) \text{csch}^2\left (\frac{x}{2}\right )+(a-b) (a+b)^2 \text{sech}^2\left (\frac{x}{2}\right )+4 b^3 \log \left (\tanh \left (\frac{x}{2}\right )\right )}{8 (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^3/(a + b*Cosh[x]),x]

[Out]

(-((a - b)^2*(a + b)*Csch[x/2]^2) - 8*a^3*Log[a + b*Cosh[x]] + 8*a^3*Log[Sinh[x]] - 12*a^2*b*Log[Tanh[x/2]] +
4*b^3*Log[Tanh[x/2]] + (a - b)*(a + b)^2*Sech[x/2]^2)/(8*(a - b)^2*(a + b)^2)

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Maple [A]  time = 0.029, size = 97, normalized size = 1. \begin{align*} -{\frac{1}{8\,a-8\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\frac{{a}^{3}}{ \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-a-b \right ) }-{\frac{1}{8\,a+8\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{\frac{a}{ \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }+{\frac{b}{2\, \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*tanh(1/2*x)^2/(a-b)-a^3/(a+b)^2/(a-b)^2*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-a-b)-1/8/(a+b)/tanh(1/2*x)^2+1
/(a+b)^2*ln(tanh(1/2*x))*a+1/2/(a+b)^2*ln(tanh(1/2*x))*b

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Maxima [A]  time = 1.09367, size = 211, normalized size = 2.24 \begin{align*} -\frac{a^{3} \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (2 \, a - b\right )} \log \left (e^{\left (-x\right )} + 1\right )}{2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{{\left (2 \, a + b\right )} \log \left (e^{\left (-x\right )} - 1\right )}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac{b e^{\left (-x\right )} - 2 \, a e^{\left (-2 \, x\right )} + b e^{\left (-3 \, x\right )}}{a^{2} - b^{2} - 2 \,{\left (a^{2} - b^{2}\right )} e^{\left (-2 \, x\right )} +{\left (a^{2} - b^{2}\right )} e^{\left (-4 \, x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.24211, size = 2071, normalized size = 22.03 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a^2*b - b^3)*cosh(x)^3 + 2*(a^2*b - b^3)*sinh(x)^3 - 4*(a^3 - a*b^2)*cosh(x)^2 - 2*(2*a^3 - 2*a*b^2 -
3*(a^2*b - b^3)*cosh(x))*sinh(x)^2 + 2*(a^2*b - b^3)*cosh(x) - 2*(a^3*cosh(x)^4 + 4*a^3*cosh(x)*sinh(x)^3 + a^
3*sinh(x)^4 - 2*a^3*cosh(x)^2 + a^3 + 2*(3*a^3*cosh(x)^2 - a^3)*sinh(x)^2 + 4*(a^3*cosh(x)^3 - a^3*cosh(x))*si
nh(x))*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) + ((2*a^3 + 3*a^2*b - b^3)*cosh(x)^4 + 4*(2*a^3 + 3*a^2*b -
b^3)*cosh(x)*sinh(x)^3 + (2*a^3 + 3*a^2*b - b^3)*sinh(x)^4 + 2*a^3 + 3*a^2*b - b^3 - 2*(2*a^3 + 3*a^2*b - b^3)
*cosh(x)^2 - 2*(2*a^3 + 3*a^2*b - b^3 - 3*(2*a^3 + 3*a^2*b - b^3)*cosh(x)^2)*sinh(x)^2 + 4*((2*a^3 + 3*a^2*b -
b^3)*cosh(x)^3 - (2*a^3 + 3*a^2*b - b^3)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) + 1) + ((2*a^3 - 3*a^2*b + b
^3)*cosh(x)^4 + 4*(2*a^3 - 3*a^2*b + b^3)*cosh(x)*sinh(x)^3 + (2*a^3 - 3*a^2*b + b^3)*sinh(x)^4 + 2*a^3 - 3*a^
2*b + b^3 - 2*(2*a^3 - 3*a^2*b + b^3)*cosh(x)^2 - 2*(2*a^3 - 3*a^2*b + b^3 - 3*(2*a^3 - 3*a^2*b + b^3)*cosh(x)
^2)*sinh(x)^2 + 4*((2*a^3 - 3*a^2*b + b^3)*cosh(x)^3 - (2*a^3 - 3*a^2*b + b^3)*cosh(x))*sinh(x))*log(cosh(x) +
sinh(x) - 1) + 2*(a^2*b - b^3 + 3*(a^2*b - b^3)*cosh(x)^2 - 4*(a^3 - a*b^2)*cosh(x))*sinh(x))/((a^4 - 2*a^2*b
^2 + b^4)*cosh(x)^4 + 4*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)*sinh(x)^3 + (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^4 + a^4 -
2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2 - 2*(a^4 - 2*a^2*b^2 + b^4 - 3*(a^4 - 2*a^2*b^2 + b^4)*c
osh(x)^2)*sinh(x)^2 + 4*((a^4 - 2*a^2*b^2 + b^4)*cosh(x)^3 - (a^4 - 2*a^2*b^2 + b^4)*cosh(x))*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{3}{\left (x \right )}}{a + b \cosh{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**3/(a+b*cosh(x)),x)

[Out]

Integral(coth(x)**3/(a + b*cosh(x)), x)

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Giac [A]  time = 1.19589, size = 240, normalized size = 2.55 \begin{align*} -\frac{a^{3} b \log \left ({\left | b{\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} + \frac{{\left (2 \, a - b\right )} \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{4 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{{\left (2 \, a + b\right )} \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac{a^{3}{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 2 \, a^{2} b{\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b^{3}{\left (e^{\left (-x\right )} + e^{x}\right )} - 4 \, a b^{2}}{2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3*b*log(abs(b*(e^(-x) + e^x) + 2*a))/(a^4*b - 2*a^2*b^3 + b^5) + 1/4*(2*a - b)*log(e^(-x) + e^x + 2)/(a^2 -
2*a*b + b^2) + 1/4*(2*a + b)*log(e^(-x) + e^x - 2)/(a^2 + 2*a*b + b^2) - 1/2*(a^3*(e^(-x) + e^x)^2 - 2*a^2*b*
(e^(-x) + e^x) + 2*b^3*(e^(-x) + e^x) - 4*a*b^2)/((a^4 - 2*a^2*b^2 + b^4)*((e^(-x) + e^x)^2 - 4))