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

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Rubi [A]  time = 0.09, antiderivative size = 56, normalized size of antiderivative = 1.00, 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 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])

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 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 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 3770

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

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*}

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Mathematica [A]  time = 0.09, 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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fricas [B]  time = 0.85, size = 309, normalized size = 5.52 \[ \left [\frac {\sqrt {-a^{2} + b^{2}} {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} + 1\right )} \log \left (\frac {{\left (a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (a + b\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a + b\right )} \sinh \relax (x)^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} - a + b}{{\left (a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (a + b\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a + b\right )} \sinh \relax (x)^{2} + a - b}\right ) + 2 \, {\left (a \cosh \relax (x)^{2} + 2 \, a \cosh \relax (x) \sinh \relax (x) + a \sinh \relax (x)^{2} + a\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) + 2 \, b \cosh \relax (x) + 2 \, b \sinh \relax (x)}{b^{2} \cosh \relax (x)^{2} + 2 \, b^{2} \cosh \relax (x) \sinh \relax (x) + b^{2} \sinh \relax (x)^{2} + b^{2}}, \frac {2 \, {\left (\sqrt {a^{2} - b^{2}} {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} + 1\right )} \arctan \left (\frac {\sqrt {a^{2} - b^{2}}}{{\left (a + b\right )} \cosh \relax (x) + {\left (a + b\right )} \sinh \relax (x)}\right ) + {\left (a \cosh \relax (x)^{2} + 2 \, a \cosh \relax (x) \sinh \relax (x) + a \sinh \relax (x)^{2} + a\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) + b \cosh \relax (x) + b \sinh \relax (x)\right )}}{b^{2} \cosh \relax (x)^{2} + 2 \, b^{2} \cosh \relax (x) \sinh \relax (x) + b^{2} \sinh \relax (x)^{2} + b^{2}}\right ] \]

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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giac [A]  time = 0.12, size = 63, normalized size = 1.12 \[ \frac {2 \, a \arctan \left (e^{x}\right )}{b^{2}} - \frac {2 \, \sqrt {a^{2} - b^{2}} \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{b^{2}} + \frac {2 \, e^{x}}{b {\left (e^{\left (2 \, x\right )} + 1\right )}} \]

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))

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maple [B]  time = 0.11, size = 110, normalized size = 1.96 \[ -\frac {2 \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{b^{2} \sqrt {a^{2}-b^{2}}}+\frac {2 \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}+\frac {2}{b \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )}+\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right ) a}{b^{2}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 3.81, size = 119, normalized size = 2.12 \[ \frac {\ln \left (a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^x-\sqrt {b^2-a^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}}{b^2}-\frac {\ln \left (a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^x+\sqrt {b^2-a^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}}{b^2}+\frac {2\,{\mathrm {e}}^x}{b\,\left ({\mathrm {e}}^{2\,x}+1\right )}-\frac {a\,\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,1{}\mathrm {i}}{b^2}+\frac {a\,\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^3*(a + b*tanh(x))),x)

[Out]

(a*log(exp(x) + 1i)*1i)/b^2 - (a*log(exp(x) - 1i)*1i)/b^2 - (log(a*exp(x) + b*exp(x) + (b^2 - a^2)^(1/2))*(-(a
+ b)*(a - b))^(1/2))/b^2 + (log(a*exp(x) + b*exp(x) - (b^2 - a^2)^(1/2))*(-(a + b)*(a - b))^(1/2))/b^2 + (2*e
xp(x))/(b*(exp(2*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{3}{\relax (x )}}{a + b \tanh {\relax (x )}}\, dx \]

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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