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

Optimal. Leaf size=62 \[ \frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a b}+\frac{x}{a}-\frac{\tan ^{-1}(\sinh (x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.171266, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {3894, 4051, 3770, 3919, 3831, 2659, 205} \[ \frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a b}+\frac{x}{a}-\frac{\tan ^{-1}(\sinh (x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2/(a + b*Sech[x]),x]

[Out]

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

Rule 3894

Int[cot[(c_.) + (d_.)*(x_)]^2*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[(-1 + Csc[c + d*x]
^2)*(a + b*Csc[c + d*x])^n, x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0]

Rule 4051

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[C/b, I
nt[Csc[e + f*x], x], x] + Dist[1/b, Int[(A*b - a*C*Csc[e + f*x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, A, C}, x]

Rule 3770

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

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*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 205

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

Rubi steps

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

Mathematica [A]  time = 0.0809918, size = 62, normalized size = 1. \[ \frac{-2 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{(b-a) \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )-2 a \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+b x}{a b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.026, size = 113, normalized size = 1.8 \begin{align*}{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-2\,{\frac{\arctan \left ( \tanh \left ( x/2 \right ) \right ) }{b}}+2\,{\frac{a}{b\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{b}{a\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.81215, size = 551, normalized size = 8.89 \begin{align*} \left [\frac{b x - 2 \, a \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + \sqrt{-a^{2} + b^{2}} \log \left (\frac{a^{2} \cosh \left (x\right )^{2} + a^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) - a^{2} + 2 \, b^{2} + 2 \,{\left (a^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cosh \left (x\right ) + a \sinh \left (x\right ) + b\right )}}{a \cosh \left (x\right )^{2} + a \sinh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) + 2 \,{\left (a \cosh \left (x\right ) + b\right )} \sinh \left (x\right ) + a}\right )}{a b}, \frac{b x - 2 \, a \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - 2 \, \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cosh \left (x\right ) + a \sinh \left (x\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )}{a b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(b*x - 2*a*arctan(cosh(x) + sinh(x)) + sqrt(-a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) -
a^2 + 2*b^2 + 2*(a^2*cosh(x) + a*b)*sinh(x) + 2*sqrt(-a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a
*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)*sinh(x) + a)))/(a*b), (b*x - 2*a*arctan(cosh(x) + sinh(x)) - 2*sq
rt(a^2 - b^2)*arctan(-(a*cosh(x) + a*sinh(x) + b)/sqrt(a^2 - b^2)))/(a*b)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17209, size = 70, normalized size = 1.13 \begin{align*} \frac{x}{a} - \frac{2 \, \arctan \left (e^{x}\right )}{b} + \frac{2 \, \sqrt{a^{2} - b^{2}} \arctan \left (\frac{a e^{x} + b}{\sqrt{a^{2} - b^{2}}}\right )}{a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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