3.204 \(\int \frac{\text{sech}(x)}{(a+b \sinh (x))^2} \, dx\)

Optimal. Leaf size=79 \[ -\frac{b}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac{2 a b \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac{\left (a^2-b^2\right ) \tan ^{-1}(\sinh (x))}{\left (a^2+b^2\right )^2}-\frac{2 a b \log (\cosh (x))}{\left (a^2+b^2\right )^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.102935, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.546, Rules used = {2668, 710, 801, 635, 203, 260} \[ -\frac{b}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac{2 a b \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac{\left (a^2-b^2\right ) \tan ^{-1}(\sinh (x))}{\left (a^2+b^2\right )^2}-\frac{2 a b \log (\cosh (x))}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]/(a + b*Sinh[x])^2,x]

[Out]

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

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 710

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

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]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.442941, size = 121, normalized size = 1.53 \[ -\frac{b \left (\frac{2 \left (a^2+b^2\right )}{a+b \sinh (x)}+\left (\frac{b^2-a^2}{\sqrt{-b^2}}+2 a\right ) \log \left (\sqrt{-b^2}-b \sinh (x)\right )+\left (\frac{a^2-b^2}{\sqrt{-b^2}}+2 a\right ) \log \left (\sqrt{-b^2}+b \sinh (x)\right )-4 a \log (a+b \sinh (x))\right )}{2 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]/(a + b*Sinh[x])^2,x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.046, size = 201, normalized size = 2.5 \begin{align*} -2\,{\frac{ab\ln \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}}+2\,{\frac{\arctan \left ( \tanh \left ( x/2 \right ) \right ){a}^{2}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}}-2\,{\frac{\arctan \left ( \tanh \left ( x/2 \right ) \right ){b}^{2}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}}-2\,{\frac{a{b}^{2}\tanh \left ( x/2 \right ) }{ \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }}-2\,{\frac{{b}^{4}\tanh \left ( x/2 \right ) }{a \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }}+2\,{\frac{ab\ln \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(a^4+2*a^2*b^2+b^4)*a*b*ln(tanh(1/2*x)^2+1)+2/(a^4+2*a^2*b^2+b^4)*arctan(tanh(1/2*x))*a^2-2/(a^4+2*a^2*b^2+
b^4)*arctan(tanh(1/2*x))*b^2-2*b^2/(a^2+b^2)^2*a*tanh(1/2*x)/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)-2*b^4/(a^2+b^
2)^2/a*tanh(1/2*x)/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)+2*b/(a^2+b^2)^2*a*ln(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)

________________________________________________________________________________________

Maxima [A]  time = 1.74772, size = 201, normalized size = 2.54 \begin{align*} \frac{2 \, a b \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{2 \, a b \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (a^{2} - b^{2}\right )} \arctan \left (e^{\left (-x\right )}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \, b e^{\left (-x\right )}}{a^{2} b + b^{3} + 2 \,{\left (a^{3} + a b^{2}\right )} e^{\left (-x\right )} -{\left (a^{2} b + b^{3}\right )} e^{\left (-2 \, x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)/(a+b*sinh(x))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.22788, size = 963, normalized size = 12.19 \begin{align*} \frac{2 \,{\left ({\left (a^{2} b - b^{3} -{\left (a^{2} b - b^{3}\right )} \cosh \left (x\right )^{2} -{\left (a^{2} b - b^{3}\right )} \sinh \left (x\right )^{2} - 2 \,{\left (a^{3} - a b^{2}\right )} \cosh \left (x\right ) - 2 \,{\left (a^{3} - a b^{2} +{\left (a^{2} b - b^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) +{\left (a^{2} b + b^{3}\right )} \cosh \left (x\right ) -{\left (a b^{2} \cosh \left (x\right )^{2} + a b^{2} \sinh \left (x\right )^{2} + 2 \, a^{2} b \cosh \left (x\right ) - a b^{2} + 2 \,{\left (a b^{2} \cosh \left (x\right ) + a^{2} b\right )} \sinh \left (x\right )\right )} \log \left (\frac{2 \,{\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) +{\left (a b^{2} \cosh \left (x\right )^{2} + a b^{2} \sinh \left (x\right )^{2} + 2 \, a^{2} b \cosh \left (x\right ) - a b^{2} + 2 \,{\left (a b^{2} \cosh \left (x\right ) + a^{2} b\right )} \sinh \left (x\right )\right )} \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) +{\left (a^{2} b + b^{3}\right )} \sinh \left (x\right )\right )}}{a^{4} b + 2 \, a^{2} b^{3} + b^{5} -{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right )^{2} -{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \sinh \left (x\right )^{2} - 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cosh \left (x\right ) - 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)/(a+b*sinh(x))^2,x, algorithm="fricas")

[Out]

2*((a^2*b - b^3 - (a^2*b - b^3)*cosh(x)^2 - (a^2*b - b^3)*sinh(x)^2 - 2*(a^3 - a*b^2)*cosh(x) - 2*(a^3 - a*b^2
+ (a^2*b - b^3)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) + (a^2*b + b^3)*cosh(x) - (a*b^2*cosh(x)^2 + a*b^
2*sinh(x)^2 + 2*a^2*b*cosh(x) - a*b^2 + 2*(a*b^2*cosh(x) + a^2*b)*sinh(x))*log(2*(b*sinh(x) + a)/(cosh(x) - si
nh(x))) + (a*b^2*cosh(x)^2 + a*b^2*sinh(x)^2 + 2*a^2*b*cosh(x) - a*b^2 + 2*(a*b^2*cosh(x) + a^2*b)*sinh(x))*lo
g(2*cosh(x)/(cosh(x) - sinh(x))) + (a^2*b + b^3)*sinh(x))/(a^4*b + 2*a^2*b^3 + b^5 - (a^4*b + 2*a^2*b^3 + b^5)
*cosh(x)^2 - (a^4*b + 2*a^2*b^3 + b^5)*sinh(x)^2 - 2*(a^5 + 2*a^3*b^2 + a*b^4)*cosh(x) - 2*(a^5 + 2*a^3*b^2 +
a*b^4 + (a^4*b + 2*a^2*b^3 + b^5)*cosh(x))*sinh(x))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}{\left (x \right )}}{\left (a + b \sinh{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)/(a+b*sinh(x))**2,x)

[Out]

Integral(sech(x)/(a + b*sinh(x))**2, x)

________________________________________________________________________________________

Giac [B]  time = 1.13246, size = 251, normalized size = 3.18 \begin{align*} \frac{2 \, a b^{2} \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{a b \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \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^{2} - b^{2}\right )}}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac{2 \,{\left (a b^{2}{\left (e^{\left (-x\right )} - e^{x}\right )} - 3 \, a^{2} b - b^{3}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b{\left (e^{\left (-x\right )} - e^{x}\right )} - 2 \, a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)/(a+b*sinh(x))^2,x, algorithm="giac")

[Out]

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