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

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Rubi [A]  time = 0.17, antiderivative size = 62, normalized size of antiderivative = 1.00, 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 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]

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 3770

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

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 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 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 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]

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*}

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Mathematica [A]  time = 0.09, size = 62, normalized size = 1.00 \[ \frac {-2 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )-2 a \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )+b x}{a b} \]

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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fricas [A]  time = 0.45, size = 193, normalized size = 3.11 \[ \left [\frac {b x - 2 \, a \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) + \sqrt {-a^{2} + b^{2}} \log \left (\frac {a^{2} \cosh \relax (x)^{2} + a^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) - a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cosh \relax (x) + a \sinh \relax (x) + b\right )}}{a \cosh \relax (x)^{2} + a \sinh \relax (x)^{2} + 2 \, b \cosh \relax (x) + 2 \, {\left (a \cosh \relax (x) + b\right )} \sinh \relax (x) + a}\right )}{a b}, \frac {b x - 2 \, a \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) - 2 \, \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cosh \relax (x) + a \sinh \relax (x) + b}{\sqrt {a^{2} - b^{2}}}\right )}{a b}\right ] \]

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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giac [A]  time = 0.15, size = 52, normalized size = 0.84 \[ \frac {x}{a} - \frac {2 \, \arctan \left (e^{x}\right )}{b} + \frac {2 \, \sqrt {a^{2} - b^{2}} \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{a b} \]

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)

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maple [B]  time = 0.12, size = 113, normalized size = 1.82 \[ -\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}+\frac {2 a \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 b \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}-\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{b} \]

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

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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(tanh(x)^2/(a+b*sech(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 = 3.92, size = 273, normalized size = 4.40 \[ \frac {\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}+\frac {\ln \left (2\,a\,b^3-2\,a^3\,b+a^3\,\sqrt {b^2-a^2}+a^4\,{\mathrm {e}}^x+4\,b^4\,{\mathrm {e}}^x-2\,a\,b^2\,\sqrt {b^2-a^2}-4\,b^3\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}-5\,a^2\,b^2\,{\mathrm {e}}^x+3\,a^2\,b\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}\right )\,\sqrt {b^2-a^2}-\ln \left (2\,a\,b^3-2\,a^3\,b-a^3\,\sqrt {b^2-a^2}+a^4\,{\mathrm {e}}^x+4\,b^4\,{\mathrm {e}}^x+2\,a\,b^2\,\sqrt {b^2-a^2}+4\,b^3\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}-5\,a^2\,b^2\,{\mathrm {e}}^x-3\,a^2\,b\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}\right )\,\sqrt {b^2-a^2}+b\,x}{a\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^2/(a + b/cosh(x)),x)

[Out]

(log(exp(x) - 1i)*1i - log(exp(x) + 1i)*1i)/b + (log(2*a*b^3 - 2*a^3*b + a^3*(b^2 - a^2)^(1/2) + a^4*exp(x) +
4*b^4*exp(x) - 2*a*b^2*(b^2 - a^2)^(1/2) - 4*b^3*exp(x)*(b^2 - a^2)^(1/2) - 5*a^2*b^2*exp(x) + 3*a^2*b*exp(x)*
(b^2 - a^2)^(1/2))*(b^2 - a^2)^(1/2) - log(2*a*b^3 - 2*a^3*b - a^3*(b^2 - a^2)^(1/2) + a^4*exp(x) + 4*b^4*exp(
x) + 2*a*b^2*(b^2 - a^2)^(1/2) + 4*b^3*exp(x)*(b^2 - a^2)^(1/2) - 5*a^2*b^2*exp(x) - 3*a^2*b*exp(x)*(b^2 - a^2
)^(1/2))*(b^2 - a^2)^(1/2) + b*x)/(a*b)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{2}{\relax (x )}}{a + b \operatorname {sech}{\relax (x )}}\, dx \]

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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