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

Optimal. Leaf size=85 \[ -\frac{a^2 b \log (a \cosh (x)+b)}{\left (a^2-b^2\right )^2}+\frac{\text{csch}^2(x) (b-a \cosh (x))}{2 \left (a^2-b^2\right )}-\frac{a \log (1-\cosh (x))}{4 (a+b)^2}+\frac{a \log (\cosh (x)+1)}{4 (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.238127, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {3872, 2837, 12, 823, 801} \[ -\frac{a^2 b \log (a \cosh (x)+b)}{\left (a^2-b^2\right )^2}+\frac{\text{csch}^2(x) (b-a \cosh (x))}{2 \left (a^2-b^2\right )}-\frac{a \log (1-\cosh (x))}{4 (a+b)^2}+\frac{a \log (\cosh (x)+1)}{4 (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(a + b*Sech[x]),x]

[Out]

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2837

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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

Mathematica [A]  time = 0.329667, size = 86, normalized size = 1.01 \[ \frac{1}{8} \left (-\frac{4 a \left (\left (a^2+b^2\right ) \log \left (\tanh \left (\frac{x}{2}\right )\right )-2 a b \log (\sinh (x))+2 a b \log (a \cosh (x)+b)\right )}{(a-b)^2 (a+b)^2}-\frac{\text{csch}^2\left (\frac{x}{2}\right )}{a+b}-\frac{\text{sech}^2\left (\frac{x}{2}\right )}{a-b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.029, size = 82, normalized size = 1. \begin{align*}{\frac{1}{8\,a-8\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\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{{a}^{2}b}{ \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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)^3/(a+b*sech(x)),x)

[Out]

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

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Maxima [A]  time = 1.11189, size = 200, normalized size = 2.35 \begin{align*} -\frac{a^{2} b \log \left (2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac{a \log \left (e^{\left (-x\right )} + 1\right )}{2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac{a \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*sech(x)),x, algorithm="maxima")

[Out]

-a^2*b*log(2*b*e^(-x) + a*e^(-2*x) + a)/(a^4 - 2*a^2*b^2 + b^4) + 1/2*a*log(e^(-x) + 1)/(a^2 - 2*a*b + b^2) -
1/2*a*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 = 2.74301, size = 2097, normalized size = 24.67 \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*sech(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*(a^2*b*cosh(x)^4 + 4*a^2*b*cosh(x)*sinh(x)^3
+ a^2*b*sinh(x)^4 - 2*a^2*b*cosh(x)^2 + a^2*b + 2*(3*a^2*b*cosh(x)^2 - a^2*b)*sinh(x)^2 + 4*(a^2*b*cosh(x)^3
- a^2*b*cosh(x))*sinh(x))*log(2*(a*cosh(x) + b)/(cosh(x) - sinh(x))) - ((a^3 + 2*a^2*b + a*b^2)*cosh(x)^4 + 4*
(a^3 + 2*a^2*b + a*b^2)*cosh(x)*sinh(x)^3 + (a^3 + 2*a^2*b + a*b^2)*sinh(x)^4 + a^3 + 2*a^2*b + a*b^2 - 2*(a^3
+ 2*a^2*b + a*b^2)*cosh(x)^2 - 2*(a^3 + 2*a^2*b + a*b^2 - 3*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^2)*sinh(x)^2 + 4*
((a^3 + 2*a^2*b + a*b^2)*cosh(x)^3 - (a^3 + 2*a^2*b + a*b^2)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) + 1) + ((
a^3 - 2*a^2*b + a*b^2)*cosh(x)^4 + 4*(a^3 - 2*a^2*b + a*b^2)*cosh(x)*sinh(x)^3 + (a^3 - 2*a^2*b + a*b^2)*sinh(
x)^4 + a^3 - 2*a^2*b + a*b^2 - 2*(a^3 - 2*a^2*b + a*b^2)*cosh(x)^2 - 2*(a^3 - 2*a^2*b + a*b^2 - 3*(a^3 - 2*a^2
*b + a*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((a^3 - 2*a^2*b + a*b^2)*cosh(x)^3 - (a^3 - 2*a^2*b + a*b^2)*cosh(x))*sin
h(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 \operatorname{sech}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)**3/(a+b*sech(x)),x)

[Out]

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

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Giac [B]  time = 1.14697, size = 235, normalized size = 2.76 \begin{align*} -\frac{a^{3} b \log \left ({\left | a{\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b \right |}\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac{a \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{4 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac{a \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac{a^{2} b{\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 )} - 8 \, a^{2} b + 4 \, 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*sech(x)),x, algorithm="giac")

[Out]

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