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

Optimal. Leaf size=95 \[ -\frac{a^2 b \log (a \sinh (x)+b)}{\left (a^2+b^2\right )^2}-\frac{\text{sech}^2(x) (b-a \sinh (x))}{2 \left (a^2+b^2\right )}-\frac{i a \log (-\sinh (x)+i)}{4 (a-i b)^2}+\frac{i a \log (\sinh (x)+i)}{4 (a+i b)^2} \]

[Out]

((-I/4)*a*Log[I - Sinh[x]])/(a - I*b)^2 + ((I/4)*a*Log[I + Sinh[x]])/(a + I*b)^2 - (a^2*b*Log[b + a*Sinh[x]])/
(a^2 + b^2)^2 - (Sech[x]^2*(b - a*Sinh[x]))/(2*(a^2 + b^2))

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Rubi [A]  time = 0.220397, antiderivative size = 95, 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} \[ -\frac{a^2 b \log (a \sinh (x)+b)}{\left (a^2+b^2\right )^2}-\frac{\text{sech}^2(x) (b-a \sinh (x))}{2 \left (a^2+b^2\right )}-\frac{i a \log (-\sinh (x)+i)}{4 (a-i b)^2}+\frac{i a \log (\sinh (x)+i)}{4 (a+i b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(a + b*Csch[x]),x]

[Out]

((-I/4)*a*Log[I - Sinh[x]])/(a - I*b)^2 + ((I/4)*a*Log[I + Sinh[x]])/(a + I*b)^2 - (a^2*b*Log[b + a*Sinh[x]])/
(a^2 + b^2)^2 - (Sech[x]^2*(b - a*Sinh[x]))/(2*(a^2 + b^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{sech}^3(x)}{a+b \text{csch}(x)} \, dx &=i \int \frac{\text{sech}^2(x) \tanh (x)}{i b+i a \sinh (x)} \, dx\\ &=-\left (\left (i a^3\right ) \operatorname{Subst}\left (\int \frac{x}{a (i b+x) \left (a^2-x^2\right )^2} \, dx,x,i a \sinh (x)\right )\right )\\ &=-\left (\left (i a^2\right ) \operatorname{Subst}\left (\int \frac{x}{(i b+x) \left (a^2-x^2\right )^2} \, dx,x,i a \sinh (x)\right )\right )\\ &=-\frac{\text{sech}^2(x) (b-a \sinh (x))}{2 \left (a^2+b^2\right )}-\frac{i \operatorname{Subst}\left (\int \frac{-i a^2 b+a^2 x}{(i b+x) \left (a^2-x^2\right )} \, dx,x,i a \sinh (x)\right )}{2 \left (a^2+b^2\right )}\\ &=-\frac{\text{sech}^2(x) (b-a \sinh (x))}{2 \left (a^2+b^2\right )}-\frac{i \operatorname{Subst}\left (\int \left (\frac{a (a-i b)}{2 (a+i b) (a-x)}-\frac{2 a^2 b}{\left (a^2+b^2\right ) (b-i x)}+\frac{a (a+i b)}{2 (a-i b) (a+x)}\right ) \, dx,x,i a \sinh (x)\right )}{2 \left (a^2+b^2\right )}\\ &=-\frac{i a \log (i-\sinh (x))}{4 (a-i b)^2}+\frac{i a \log (i+\sinh (x))}{4 (a+i b)^2}-\frac{a^2 b \log (b+a \sinh (x))}{\left (a^2+b^2\right )^2}-\frac{\text{sech}^2(x) (b-a \sinh (x))}{2 \left (a^2+b^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.148813, size = 78, normalized size = 0.82 \[ \frac{-b \left (a^2+b^2\right ) \text{sech}^2(x)+a \left (a^2+b^2\right ) \tanh (x) \text{sech}(x)+2 a \left (\left (a^2-b^2\right ) \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+a b (\log (\cosh (x))-\log (a \sinh (x)+b))\right )}{2 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.036, size = 275, normalized size = 2.9 \begin{align*} -{\frac{{a}^{2}b}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-2\,a\tanh \left ( x/2 \right ) -b \right ) }-{\frac{{a}^{3}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-{\frac{a{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}{a}^{2}b}{ \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}{b}^{3}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{{a}^{3}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+{\frac{a{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+{\frac{{a}^{2}b}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }+{\frac{{a}^{3}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-{\frac{a{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.51223, size = 217, normalized size = 2.28 \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^{2} b \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (a^{3} - a b^{2}\right )} \arctan \left (e^{\left (-x\right )}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \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(sech(x)^3/(a+b*csch(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) + a^2*b*log(e^(-2*x) + 1)/(a^4 + 2*a^2*b^2 +
b^4) - (a^3 - a*b^2)*arctan(e^(-x))/(a^4 + 2*a^2*b^2 + b^4) + (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 = 1.73911, size = 1748, normalized size = 18.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^3 + a*b^2)*cosh(x)^3 + (a^3 + a*b^2)*sinh(x)^3 - 2*(a^2*b + b^3)*cosh(x)^2 - (2*a^2*b + 2*b^3 - 3*(a^3 + a
*b^2)*cosh(x))*sinh(x)^2 + ((a^3 - a*b^2)*cosh(x)^4 + 4*(a^3 - a*b^2)*cosh(x)*sinh(x)^3 + (a^3 - a*b^2)*sinh(x
)^4 + a^3 - a*b^2 + 2*(a^3 - a*b^2)*cosh(x)^2 + 2*(a^3 - a*b^2 + 3*(a^3 - a*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((a^
3 - a*b^2)*cosh(x)^3 + (a^3 - a*b^2)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) - (a^3 + a*b^2)*cosh(x) - (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*sinh(x) + b)/(cosh(x) - sinh(x)))
+ (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*cosh(x)/(cosh(x) - sinh(x))) - (
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{sech}^{3}{\left (x \right )}}{a + b \operatorname{csch}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.1754, size = 294, normalized size = 3.09 \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^{2} b \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )}{\left (a^{3} - a b^{2}\right )}}{4 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\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(sech(x)^3/(a+b*csch(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) + 1/2*a^2*b*log((e^(-x) - e^x)^2 + 4)/(a^4
+ 2*a^2*b^2 + b^4) + 1/4*(pi + 2*arctan(1/2*(e^(2*x) - 1)*e^(-x)))*(a^3 - a*b^2)/(a^4 + 2*a^2*b^2 + b^4) - 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))