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

Optimal. Leaf size=87 \[ \frac{b^3 \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac{a \left (a^2+3 b^2\right ) \tan ^{-1}(\sinh (x))}{2 \left (a^2+b^2\right )^2}-\frac{b^3 \log (\cosh (x))}{\left (a^2+b^2\right )^2}+\frac{\text{sech}^2(x) (a \sinh (x)+b)}{2 \left (a^2+b^2\right )} \]

[Out]

(a*(a^2 + 3*b^2)*ArcTan[Sinh[x]])/(2*(a^2 + b^2)^2) - (b^3*Log[Cosh[x]])/(a^2 + b^2)^2 + (b^3*Log[a + b*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 = 0.122492, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2668, 741, 801, 635, 203, 260} \[ \frac{b^3 \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac{a \left (a^2+3 b^2\right ) \tan ^{-1}(\sinh (x))}{2 \left (a^2+b^2\right )^2}-\frac{b^3 \log (\cosh (x))}{\left (a^2+b^2\right )^2}+\frac{\text{sech}^2(x) (a \sinh (x)+b)}{2 \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(a + b*Sinh[x]),x]

[Out]

(a*(a^2 + 3*b^2)*ArcTan[Sinh[x]])/(2*(a^2 + b^2)^2) - (b^3*Log[Cosh[x]])/(a^2 + b^2)^2 + (b^3*Log[a + b*Sinh[x
]])/(a^2 + b^2)^2 + (Sech[x]^2*(b + a*Sinh[x]))/(2*(a^2 + b^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]

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}^3(x)}{a+b \sinh (x)} \, dx &=b^3 \operatorname{Subst}\left (\int \frac{1}{(a+x) \left (-b^2-x^2\right )^2} \, dx,x,b \sinh (x)\right )\\ &=\frac{\text{sech}^2(x) (b+a \sinh (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 \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{b \operatorname{Subst}\left (\int \left (-\frac{2 b^2}{\left (a^2+b^2\right ) (a+x)}+\frac{-a^3-3 a b^2+2 b^2 x}{\left (a^2+b^2\right ) \left (b^2+x^2\right )}\right ) \, dx,x,b \sinh (x)\right )}{2 \left (a^2+b^2\right )}\\ &=\frac{b^3 \log (a+b \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 )}-\frac{b \operatorname{Subst}\left (\int \frac{-a^3-3 a b^2+2 b^2 x}{b^2+x^2} \, dx,x,b \sinh (x)\right )}{2 \left (a^2+b^2\right )^2}\\ &=\frac{b^3 \log (a+b \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 )}-\frac{b^3 \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 (a b \left (a^2+3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2+x^2} \, dx,x,b \sinh (x)\right )}{2 \left (a^2+b^2\right )^2}\\ &=\frac{a \left (a^2+3 b^2\right ) \tan ^{-1}(\sinh (x))}{2 \left (a^2+b^2\right )^2}-\frac{b^3 \log (\cosh (x))}{\left (a^2+b^2\right )^2}+\frac{b^3 \log (a+b \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 = 0.124474, size = 77, normalized size = 0.89 \[ \frac{b \left (a^2+b^2\right ) \text{sech}^2(x)+2 a \left (a^2+3 b^2\right ) \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+a \left (a^2+b^2\right ) \tanh (x) \text{sech}(x)+2 b^3 (\log (a+b \sinh (x))-\log (\cosh (x)))}{2 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(a^2 + 3*b^2)*ArcTan[Tanh[x/2]] + 2*b^3*(-Log[Cosh[x]] + Log[a + b*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 = 0.033, size = 353, normalized size = 4.1 \begin{align*} -{\frac{{a}^{3}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}} \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}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}} \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}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \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}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{{a}^{3}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+{\frac{a{b}^{2}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-{\frac{{b}^{3}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }+{\frac{{a}^{3}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }+3\,{\frac{\arctan \left ( \tanh \left ( x/2 \right ) \right ) a{b}^{2}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}}+{\frac{{b}^{3}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a^4+2*a^2*b^2+b^4)/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3*a^3-1/(a^4+2*a^2*b^2+b^4)/(tanh(1/2*x)^2+1)^2*tanh(1/
2*x)^3*a*b^2-2/(a^4+2*a^2*b^2+b^4)/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^2*a^2*b-2/(a^4+2*a^2*b^2+b^4)/(tanh(1/2*x)^
2+1)^2*tanh(1/2*x)^2*b^3+1/(a^4+2*a^2*b^2+b^4)/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)*a^3+1/(a^4+2*a^2*b^2+b^4)/(tanh
(1/2*x)^2+1)^2*tanh(1/2*x)*a*b^2-1/(a^4+2*a^2*b^2+b^4)*b^3*ln(tanh(1/2*x)^2+1)+1/(a^4+2*a^2*b^2+b^4)*arctan(ta
nh(1/2*x))*a^3+3/(a^4+2*a^2*b^2+b^4)*arctan(tanh(1/2*x))*a*b^2+b^3/(a^4+2*a^2*b^2+b^4)*ln(a*tanh(1/2*x)^2-2*ta
nh(1/2*x)*b-a)

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Maxima [A]  time = 1.84185, size = 215, normalized size = 2.47 \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{b^{3} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (a^{3} + 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*sinh(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) - b^3*log(e^(-2*x) + 1)/(a^4 + 2*a^2*b^2 + b^4)
- (a^3 + 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 = 2.31088, size = 1724, normalized size = 19.82 \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*sinh(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 + 3*a*b^2)*cosh(x)^4 + 4*(a^3 + 3*a*b^2)*cosh(x)*sinh(x)^3 + (a^3 + 3*a*b^2)*
sinh(x)^4 + a^3 + 3*a*b^2 + 2*(a^3 + 3*a*b^2)*cosh(x)^2 + 2*(a^3 + 3*a*b^2 + 3*(a^3 + 3*a*b^2)*cosh(x)^2)*sinh
(x)^2 + 4*((a^3 + 3*a*b^2)*cosh(x)^3 + (a^3 + 3*a*b^2)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) - (a^3 + a*
b^2)*cosh(x) + (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*cos
h(x)^2 + b^3)*sinh(x)^2 + 4*(b^3*cosh(x)^3 + b^3*cosh(x))*sinh(x))*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x)))
- (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))*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 \sinh{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.24736, size = 289, normalized size = 3.32 \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{b^{3} \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} + 3 \, a b^{2}\right )}}{4 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\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(sech(x)^3/(a+b*sinh(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) - 1/2*b^3*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 + 3*a*b^2)/(a^4 + 2*a^2*b^2 + b^4) + 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))