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

Optimal. Leaf size=56 \[ -\frac{\sqrt{a^2-b^2} \tan ^{-1}\left (\frac{\cosh (x) (a \tanh (x)+b)}{\sqrt{a^2-b^2}}\right )}{b^2}+\frac{a \tan ^{-1}(\sinh (x))}{b^2}+\frac{\text{sech}(x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.0881527, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {3510, 3486, 3770, 3509, 206} \[ -\frac{\sqrt{a^2-b^2} \tan ^{-1}\left (\frac{\cosh (x) (a \tanh (x)+b)}{\sqrt{a^2-b^2}}\right )}{b^2}+\frac{a \tan ^{-1}(\sinh (x))}{b^2}+\frac{\text{sech}(x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(a + b*Tanh[x]),x]

[Out]

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

Rule 3510

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Dist[d^2/b^2, I
nt[(d*Sec[e + f*x])^(m - 2)*(a - b*Tan[e + f*x]), x], x] + Dist[(d^2*(a^2 + b^2))/b^2, Int[(d*Sec[e + f*x])^(m
- 2)/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 + b^2, 0] && IGtQ[m, 1]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3509

Int[sec[(e_.) + (f_.)*(x_)]/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Dist[f^(-1), Subst[Int[1/(a^
2 + b^2 - x^2), x], x, (b - a*Tan[e + f*x])/Sec[e + f*x]], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 + b^2, 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}^3(x)}{a+b \tanh (x)} \, dx &=\frac{\int \text{sech}(x) (a-b \tanh (x)) \, dx}{b^2}-\frac{\left (a^2-b^2\right ) \int \frac{\text{sech}(x)}{a+b \tanh (x)} \, dx}{b^2}\\ &=\frac{\text{sech}(x)}{b}+\frac{a \int \text{sech}(x) \, dx}{b^2}-\frac{\left (i \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a^2-b^2-x^2} \, dx,x,\cosh (x) (-i b-i a \tanh (x))\right )}{b^2}\\ &=\frac{a \tan ^{-1}(\sinh (x))}{b^2}-\frac{\sqrt{a^2-b^2} \tan ^{-1}\left (\frac{\cosh (x) (b+a \tanh (x))}{\sqrt{a^2-b^2}}\right )}{b^2}+\frac{\text{sech}(x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.084032, size = 65, normalized size = 1.16 \[ \frac{-2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{a \tanh \left (\frac{x}{2}\right )+b}{\sqrt{a-b} \sqrt{a+b}}\right )+2 a \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+b \text{sech}(x)}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*ArcTan[Tanh[x/2]] - 2*Sqrt[a - b]*Sqrt[a + b]*ArcTan[(b + a*Tanh[x/2])/(Sqrt[a - b]*Sqrt[a + b])] + b*Sec
h[x])/b^2

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Maple [B]  time = 0.03, size = 110, normalized size = 2. \begin{align*} -2\,{\frac{{a}^{2}}{{b}^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+2\,{\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+2\,{\frac{1}{b \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}+2\,{\frac{\arctan \left ( \tanh \left ( x/2 \right ) \right ) a}{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^3/(a+b*tanh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.56028, size = 1000, normalized size = 17.86 \begin{align*} \left [\frac{\sqrt{-a^{2} + b^{2}}{\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1\right )} \log \left (\frac{{\left (a + b\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a + b\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a + b\right )} \sinh \left (x\right )^{2} - 2 \, \sqrt{-a^{2} + b^{2}}{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} - a + b}{{\left (a + b\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a + b\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a + b\right )} \sinh \left (x\right )^{2} + a - b}\right ) + 2 \,{\left (a \cosh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) \sinh \left (x\right ) + a \sinh \left (x\right )^{2} + a\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + 2 \, b \cosh \left (x\right ) + 2 \, b \sinh \left (x\right )}{b^{2} \cosh \left (x\right )^{2} + 2 \, b^{2} \cosh \left (x\right ) \sinh \left (x\right ) + b^{2} \sinh \left (x\right )^{2} + b^{2}}, \frac{2 \,{\left (\sqrt{a^{2} - b^{2}}{\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1\right )} \arctan \left (\frac{\sqrt{a^{2} - b^{2}}}{{\left (a + b\right )} \cosh \left (x\right ) +{\left (a + b\right )} \sinh \left (x\right )}\right ) +{\left (a \cosh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) \sinh \left (x\right ) + a \sinh \left (x\right )^{2} + a\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + b \cosh \left (x\right ) + b \sinh \left (x\right )\right )}}{b^{2} \cosh \left (x\right )^{2} + 2 \, b^{2} \cosh \left (x\right ) \sinh \left (x\right ) + b^{2} \sinh \left (x\right )^{2} + b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(sqrt(-a^2 + b^2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*log(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*
sinh(x) + (a + b)*sinh(x)^2 - 2*sqrt(-a^2 + b^2)*(cosh(x) + sinh(x)) - a + b)/((a + b)*cosh(x)^2 + 2*(a + b)*c
osh(x)*sinh(x) + (a + b)*sinh(x)^2 + a - b)) + 2*(a*cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2 + a)*arctan(
cosh(x) + sinh(x)) + 2*b*cosh(x) + 2*b*sinh(x))/(b^2*cosh(x)^2 + 2*b^2*cosh(x)*sinh(x) + b^2*sinh(x)^2 + b^2),
2*(sqrt(a^2 - b^2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*arctan(sqrt(a^2 - b^2)/((a + b)*cosh(x) +
(a + b)*sinh(x))) + (a*cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2 + a)*arctan(cosh(x) + sinh(x)) + b*cosh(x
) + b*sinh(x))/(b^2*cosh(x)^2 + 2*b^2*cosh(x)*sinh(x) + b^2*sinh(x)^2 + b^2)]

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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 \tanh{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.24752, size = 85, normalized size = 1.52 \begin{align*} \frac{2 \, a \arctan \left (e^{x}\right )}{b^{2}} - \frac{2 \, \sqrt{a^{2} - b^{2}} \arctan \left (\frac{a e^{x} + b e^{x}}{\sqrt{a^{2} - b^{2}}}\right )}{b^{2}} + \frac{2 \, e^{x}}{b{\left (e^{\left (2 \, x\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*arctan(e^x)/b^2 - 2*sqrt(a^2 - b^2)*arctan((a*e^x + b*e^x)/sqrt(a^2 - b^2))/b^2 + 2*e^x/(b*(e^(2*x) + 1))