3.197 \(\int \frac{\text{sech}^4(x)}{a+b \sinh (x)} \, dx\)

Optimal. Leaf size=100 \[ -\frac{2 b^4 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac{\text{sech}^3(x) (a \sinh (x)+b)}{3 \left (a^2+b^2\right )}+\frac{\text{sech}(x) \left (a \left (2 a^2+5 b^2\right ) \sinh (x)+3 b^3\right )}{3 \left (a^2+b^2\right )^2} \]

[Out]

(-2*b^4*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) + (Sech[x]^3*(b + a*Sinh[x]))/(3*(a^2 +
b^2)) + (Sech[x]*(3*b^3 + a*(2*a^2 + 5*b^2)*Sinh[x]))/(3*(a^2 + b^2)^2)

________________________________________________________________________________________

Rubi [A]  time = 0.213126, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2696, 2866, 12, 2660, 618, 206} \[ -\frac{2 b^4 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac{\text{sech}^3(x) (a \sinh (x)+b)}{3 \left (a^2+b^2\right )}+\frac{\text{sech}(x) \left (a \left (2 a^2+5 b^2\right ) \sinh (x)+3 b^3\right )}{3 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^4/(a + b*Sinh[x]),x]

[Out]

(-2*b^4*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) + (Sech[x]^3*(b + a*Sinh[x]))/(3*(a^2 +
b^2)) + (Sech[x]*(3*b^3 + a*(2*a^2 + 5*b^2)*Sinh[x]))/(3*(a^2 + b^2)^2)

Rule 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rule 12

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\text{sech}^4(x)}{a+b \sinh (x)} \, dx &=\frac{\text{sech}^3(x) (b+a \sinh (x))}{3 \left (a^2+b^2\right )}-\frac{\int \frac{\text{sech}^2(x) \left (-2 a^2-3 b^2-2 a b \sinh (x)\right )}{a+b \sinh (x)} \, dx}{3 \left (a^2+b^2\right )}\\ &=\frac{\text{sech}^3(x) (b+a \sinh (x))}{3 \left (a^2+b^2\right )}+\frac{\text{sech}(x) \left (3 b^3+a \left (2 a^2+5 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}+\frac{\int \frac{3 b^4}{a+b \sinh (x)} \, dx}{3 \left (a^2+b^2\right )^2}\\ &=\frac{\text{sech}^3(x) (b+a \sinh (x))}{3 \left (a^2+b^2\right )}+\frac{\text{sech}(x) \left (3 b^3+a \left (2 a^2+5 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}+\frac{b^4 \int \frac{1}{a+b \sinh (x)} \, dx}{\left (a^2+b^2\right )^2}\\ &=\frac{\text{sech}^3(x) (b+a \sinh (x))}{3 \left (a^2+b^2\right )}+\frac{\text{sech}(x) \left (3 b^3+a \left (2 a^2+5 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}+\frac{\left (2 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{\left (a^2+b^2\right )^2}\\ &=\frac{\text{sech}^3(x) (b+a \sinh (x))}{3 \left (a^2+b^2\right )}+\frac{\text{sech}(x) \left (3 b^3+a \left (2 a^2+5 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}-\frac{\left (4 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac{x}{2}\right )\right )}{\left (a^2+b^2\right )^2}\\ &=-\frac{2 b^4 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac{\text{sech}^3(x) (b+a \sinh (x))}{3 \left (a^2+b^2\right )}+\frac{\text{sech}(x) \left (3 b^3+a \left (2 a^2+5 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.332811, size = 102, normalized size = 1.02 \[ \frac{a \left (2 a^2+5 b^2\right ) \tanh (x)+\frac{6 b^4 \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}+\left (a^2+b^2\right ) \text{sech}^3(x) (a \sinh (x)+b)+3 b^3 \text{sech}(x)}{3 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^4/(a + b*Sinh[x]),x]

[Out]

((6*b^4*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] + 3*b^3*Sech[x] + (a^2 + b^2)*Sech[x]^3*(
b + a*Sinh[x]) + a*(2*a^2 + 5*b^2)*Tanh[x])/(3*(a^2 + b^2)^2)

________________________________________________________________________________________

Maple [B]  time = 0.034, size = 182, normalized size = 1.8 \begin{align*} -2\,{\frac{ \left ( -{a}^{3}-2\,a{b}^{2} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{5}+ \left ( -{a}^{2}b-2\,{b}^{3} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{4}+ \left ( -2/3\,{a}^{3}-8/3\,a{b}^{2} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{3}-2\,{b}^{3} \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+ \left ( -{a}^{3}-2\,a{b}^{2} \right ) \tanh \left ( x/2 \right ) -1/3\,{a}^{2}b-4/3\,{b}^{3}}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+2\,{\frac{{b}^{4}}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^4/(a+b*sinh(x)),x)

[Out]

-2/(a^4+2*a^2*b^2+b^4)*((-a^3-2*a*b^2)*tanh(1/2*x)^5+(-a^2*b-2*b^3)*tanh(1/2*x)^4+(-2/3*a^3-8/3*a*b^2)*tanh(1/
2*x)^3-2*b^3*tanh(1/2*x)^2+(-a^3-2*a*b^2)*tanh(1/2*x)-1/3*a^2*b-4/3*b^3)/(tanh(1/2*x)^2+1)^3+2*b^4/(a^4+2*a^2*
b^2+b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(a+b*sinh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.22651, size = 2807, normalized size = 28.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(a+b*sinh(x)),x, algorithm="fricas")

[Out]

1/3*(6*(a^2*b^3 + b^5)*cosh(x)^5 + 6*(a^2*b^3 + b^5)*sinh(x)^5 - 4*a^5 - 14*a^3*b^2 - 10*a*b^4 - 6*(a^3*b^2 +
a*b^4)*cosh(x)^4 - 6*(a^3*b^2 + a*b^4 - 5*(a^2*b^3 + b^5)*cosh(x))*sinh(x)^4 + 4*(2*a^4*b + 7*a^2*b^3 + 5*b^5)
*cosh(x)^3 + 4*(2*a^4*b + 7*a^2*b^3 + 5*b^5 + 15*(a^2*b^3 + b^5)*cosh(x)^2 - 6*(a^3*b^2 + a*b^4)*cosh(x))*sinh
(x)^3 - 12*(a^5 + 3*a^3*b^2 + 2*a*b^4)*cosh(x)^2 - 12*(a^5 + 3*a^3*b^2 + 2*a*b^4 - 5*(a^2*b^3 + b^5)*cosh(x)^3
+ 3*(a^3*b^2 + a*b^4)*cosh(x)^2 - (2*a^4*b + 7*a^2*b^3 + 5*b^5)*cosh(x))*sinh(x)^2 + 3*(b^4*cosh(x)^6 + 6*b^4
*cosh(x)*sinh(x)^5 + b^4*sinh(x)^6 + 3*b^4*cosh(x)^4 + 3*b^4*cosh(x)^2 + 3*(5*b^4*cosh(x)^2 + b^4)*sinh(x)^4 +
b^4 + 4*(5*b^4*cosh(x)^3 + 3*b^4*cosh(x))*sinh(x)^3 + 3*(5*b^4*cosh(x)^4 + 6*b^4*cosh(x)^2 + b^4)*sinh(x)^2 +
6*(b^4*cosh(x)^5 + 2*b^4*cosh(x)^3 + b^4*cosh(x))*sinh(x))*sqrt(a^2 + b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2
+ 2*a*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)
)/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) - b)) + 6*(a^2*b^3 + b^5)*cosh(x) + 6*(
a^2*b^3 + b^5 + 5*(a^2*b^3 + b^5)*cosh(x)^4 - 4*(a^3*b^2 + a*b^4)*cosh(x)^3 + 2*(2*a^4*b + 7*a^2*b^3 + 5*b^5)*
cosh(x)^2 - 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*cosh(x))*sinh(x))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^6 + 6
*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)*sinh(x)^5 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sinh(x)^6 + a^6 +
3*a^4*b^2 + 3*a^2*b^4 + b^6 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^4 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^
4 + b^6 + 5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6
)*cosh(x)^3 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x))*sinh(x)^3 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)
*cosh(x)^2 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + 5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^4 + 6*(a^6 +
3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^2)*sinh(x)^2 + 6*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^5 + 2*(a^6
+ 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^3 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x))*sinh(x))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{4}{\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)**4/(a+b*sinh(x)),x)

[Out]

Integral(sech(x)**4/(a + b*sinh(x)), x)

________________________________________________________________________________________

Giac [A]  time = 1.36113, size = 243, normalized size = 2.43 \begin{align*} \frac{b^{4} \log \left (\frac{{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} + \frac{2 \,{\left (3 \, b^{3} e^{\left (5 \, x\right )} - 3 \, a b^{2} e^{\left (4 \, x\right )} + 4 \, a^{2} b e^{\left (3 \, x\right )} + 10 \, b^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} - 12 \, a b^{2} e^{\left (2 \, x\right )} + 3 \, b^{3} e^{x} - 2 \, a^{3} - 5 \, a b^{2}\right )}}{3 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(a+b*sinh(x)),x, algorithm="giac")

[Out]

b^4*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4
)*sqrt(a^2 + b^2)) + 2/3*(3*b^3*e^(5*x) - 3*a*b^2*e^(4*x) + 4*a^2*b*e^(3*x) + 10*b^3*e^(3*x) - 6*a^3*e^(2*x) -
12*a*b^2*e^(2*x) + 3*b^3*e^x - 2*a^3 - 5*a*b^2)/((a^4 + 2*a^2*b^2 + b^4)*(e^(2*x) + 1)^3)