3.174 \(\int \frac{\text{csch}^3(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=91 \[ \frac{b^3 \log (a+b \cosh (x))}{\left (a^2-b^2\right )^2}+\frac{\text{csch}^2(x) (b-a \cosh (x))}{2 \left (a^2-b^2\right )}-\frac{(a+2 b) \log (1-\cosh (x))}{4 (a+b)^2}+\frac{(a-2 b) \log (\cosh (x)+1)}{4 (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.15923, antiderivative size = 91, 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 = {2668, 741, 801} \[ \frac{b^3 \log (a+b \cosh (x))}{\left (a^2-b^2\right )^2}+\frac{\text{csch}^2(x) (b-a \cosh (x))}{2 \left (a^2-b^2\right )}-\frac{(a+2 b) \log (1-\cosh (x))}{4 (a+b)^2}+\frac{(a-2 b) \log (\cosh (x)+1)}{4 (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(a + b*Cosh[x]),x]

[Out]

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(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] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

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

Mathematica [A]  time = 0.269375, size = 100, normalized size = 1.1 \[ -\frac{4 a^3 \log \left (\tanh \left (\frac{x}{2}\right )\right )-12 a b^2 \log \left (\tanh \left (\frac{x}{2}\right )\right )-8 b^3 \log (a+b \cosh (x))+(a-b)^2 (a+b) \text{csch}^2\left (\frac{x}{2}\right )+(a-b) (a+b)^2 \text{sech}^2\left (\frac{x}{2}\right )+8 b^3 \log (\sinh (x))}{8 (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a - b)^2*(a + b)*Csch[x/2]^2 - 8*b^3*Log[a + b*Cosh[x]] + 8*b^3*Log[Sinh[x]] + 4*a^3*Log[Tanh[x/2]] - 12*a*
b^2*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 = 0.023, size = 97, normalized size = 1.1 \begin{align*}{\frac{1}{8\,a-8\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}+{\frac{{b}^{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}{2\, \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-{\frac{b}{ \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(csch(x)^3/(a+b*cosh(x)),x)

[Out]

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

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Maxima [A]  time = 1.10484, size = 208, normalized size = 2.29 \begin{align*} \frac{b^{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 (a - 2 \, b\right )} \log \left (e^{\left (-x\right )} + 1\right )}{2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac{{\left (a + 2 \, b\right )} \log \left (e^{\left (-x\right )} - 1\right )}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac{a e^{\left (-x\right )} - 2 \, b e^{\left (-2 \, x\right )} + a 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(csch(x)^3/(a+b*cosh(x)),x, algorithm="maxima")

[Out]

b^3*log(2*a*e^(-x) + b*e^(-2*x) + b)/(a^4 - 2*a^2*b^2 + b^4) + 1/2*(a - 2*b)*log(e^(-x) + 1)/(a^2 - 2*a*b + b^
2) - 1/2*(a + 2*b)*log(e^(-x) - 1)/(a^2 + 2*a*b + b^2) - (a*e^(-x) - 2*b*e^(-2*x) + a*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 = 2.25945, size = 2072, normalized size = 22.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(a^3 - a*b^2)*cosh(x)^3 + 2*(a^3 - a*b^2)*sinh(x)^3 - 4*(a^2*b - b^3)*cosh(x)^2 - 2*(2*a^2*b - 2*b^3 -
3*(a^3 - a*b^2)*cosh(x))*sinh(x)^2 + 2*(a^3 - a*b^2)*cosh(x) - 2*(b^3*cosh(x)^4 + 4*b^3*cosh(x)*sinh(x)^3 + b
^3*sinh(x)^4 - 2*b^3*cosh(x)^2 + b^3 + 2*(3*b^3*cosh(x)^2 - b^3)*sinh(x)^2 + 4*(b^3*cosh(x)^3 - b^3*cosh(x))*s
inh(x))*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) - ((a^3 - 3*a*b^2 - 2*b^3)*cosh(x)^4 + 4*(a^3 - 3*a*b^2 - 2
*b^3)*cosh(x)*sinh(x)^3 + (a^3 - 3*a*b^2 - 2*b^3)*sinh(x)^4 + a^3 - 3*a*b^2 - 2*b^3 - 2*(a^3 - 3*a*b^2 - 2*b^3
)*cosh(x)^2 - 2*(a^3 - 3*a*b^2 - 2*b^3 - 3*(a^3 - 3*a*b^2 - 2*b^3)*cosh(x)^2)*sinh(x)^2 + 4*((a^3 - 3*a*b^2 -
2*b^3)*cosh(x)^3 - (a^3 - 3*a*b^2 - 2*b^3)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) + 1) + ((a^3 - 3*a*b^2 + 2*
b^3)*cosh(x)^4 + 4*(a^3 - 3*a*b^2 + 2*b^3)*cosh(x)*sinh(x)^3 + (a^3 - 3*a*b^2 + 2*b^3)*sinh(x)^4 + a^3 - 3*a*b
^2 + 2*b^3 - 2*(a^3 - 3*a*b^2 + 2*b^3)*cosh(x)^2 - 2*(a^3 - 3*a*b^2 + 2*b^3 - 3*(a^3 - 3*a*b^2 + 2*b^3)*cosh(x
)^2)*sinh(x)^2 + 4*((a^3 - 3*a*b^2 + 2*b^3)*cosh(x)^3 - (a^3 - 3*a*b^2 + 2*b^3)*cosh(x))*sinh(x))*log(cosh(x)
+ sinh(x) - 1) + 2*(a^3 - a*b^2 + 3*(a^3 - a*b^2)*cosh(x)^2 - 4*(a^2*b - b^3)*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)*
cosh(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{\operatorname{csch}^{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(csch(x)**3/(a+b*cosh(x)),x)

[Out]

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

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Giac [B]  time = 1.15935, size = 242, normalized size = 2.66 \begin{align*} \frac{b^{4} \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 (a - 2 \, b\right )} \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{4 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac{{\left (a + 2 \, b\right )} \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac{b^{3}{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 2 \, a^{3}{\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a b^{2}{\left (e^{\left (-x\right )} + e^{x}\right )} + 4 \, a^{2} b - 8 \, b^{3}}{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(csch(x)^3/(a+b*cosh(x)),x, algorithm="giac")

[Out]

b^4*log(abs(b*(e^(-x) + e^x) + 2*a))/(a^4*b - 2*a^2*b^3 + b^5) + 1/4*(a - 2*b)*log(e^(-x) + e^x + 2)/(a^2 - 2*
a*b + b^2) - 1/4*(a + 2*b)*log(e^(-x) + e^x - 2)/(a^2 + 2*a*b + b^2) + 1/2*(b^3*(e^(-x) + e^x)^2 - 2*a^3*(e^(-
x) + e^x) + 2*a*b^2*(e^(-x) + e^x) + 4*a^2*b - 8*b^3)/((a^4 - 2*a^2*b^2 + b^4)*((e^(-x) + e^x)^2 - 4))