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

Optimal. Leaf size=114 \[ -\frac{b^2 x}{a \left (a^2-b^2\right )}+\frac{a x}{a^2-b^2}-\frac{a \coth (x)}{a^2-b^2}+\frac{b \text{csch}(x)}{a^2-b^2}+\frac{2 b^3 \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a (a-b)^{3/2} (a+b)^{3/2}} \]

[Out]

(a*x)/(a^2 - b^2) - (b^2*x)/(a*(a^2 - b^2)) + (2*b^3*ArcTan[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(a*(a - b)^(
3/2)*(a + b)^(3/2)) - (a*Coth[x])/(a^2 - b^2) + (b*Csch[x])/(a^2 - b^2)

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Rubi [A]  time = 0.20499, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {3898, 2902, 2606, 8, 3473, 2735, 2659, 205} \[ -\frac{b^2 x}{a \left (a^2-b^2\right )}+\frac{a x}{a^2-b^2}-\frac{a \coth (x)}{a^2-b^2}+\frac{b \text{csch}(x)}{a^2-b^2}+\frac{2 b^3 \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a (a-b)^{3/2} (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^2/(a + b*Sech[x]),x]

[Out]

(a*x)/(a^2 - b^2) - (b^2*x)/(a*(a^2 - b^2)) + (2*b^3*ArcTan[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(a*(a - b)^(
3/2)*(a + b)^(3/2)) - (a*Coth[x])/(a^2 - b^2) + (b*Csch[x])/(a^2 - b^2)

Rule 3898

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

Rule 2902

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[(a*d^2)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[(b*d)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a^2*d^2)/(g^2*(a^2 - b^
2)), Int[((g*Cos[e + f*x])^(p + 2)*(d*Sin[e + f*x])^(n - 2))/(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 2735

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

Mathematica [A]  time = 0.334968, size = 81, normalized size = 0.71 \[ \frac{\frac{2 b^3 \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+a^2 x-a^2 \coth (x)+a b \text{csch}(x)-b^2 x}{a^3-a b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x - b^2*x + (2*b^3*ArcTan[((a - b)*Tanh[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - a^2*Coth[x] + a*b*Csch[
x])/(a^3 - a*b^2)

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Maple [A]  time = 0.03, size = 104, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,a-2\,b}\tanh \left ({\frac{x}{2}} \right ) }+{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{2\,b+2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+2\,{\frac{{b}^{3}}{ \left ( a-b \right ) a \left ( a+b \right ) \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(coth(x)^2/(a+b*sech(x)),x)

[Out]

-1/2/(a-b)*tanh(1/2*x)+1/a*ln(tanh(1/2*x)+1)-1/2/(a+b)/tanh(1/2*x)-1/a*ln(tanh(1/2*x)-1)+2/(a-b)/a/(a+b)*b^3/(
(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(coth(x)^2/(a+b*sech(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.35958, size = 1582, normalized size = 13.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(2*a^4 - 2*a^2*b^2 - (a^4 - 2*a^2*b^2 + b^4)*x*cosh(x)^2 - (a^4 - 2*a^2*b^2 + b^4)*x*sinh(x)^2 - (b^3*cosh(x)
^2 + 2*b^3*cosh(x)*sinh(x) + b^3*sinh(x)^2 - b^3)*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*cos
h(x)^2 + a*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)*sinh(x) + a)) + (a^4 - 2*a^2*b^2 + b^4)*x - 2*(a^3*b -
a*b^3)*cosh(x) - 2*(a^3*b - a*b^3 + (a^4 - 2*a^2*b^2 + b^4)*x*cosh(x))*sinh(x))/(a^5 - 2*a^3*b^2 + a*b^4 - (a^
5 - 2*a^3*b^2 + a*b^4)*cosh(x)^2 - 2*(a^5 - 2*a^3*b^2 + a*b^4)*cosh(x)*sinh(x) - (a^5 - 2*a^3*b^2 + a*b^4)*sin
h(x)^2), (2*a^4 - 2*a^2*b^2 - (a^4 - 2*a^2*b^2 + b^4)*x*cosh(x)^2 - (a^4 - 2*a^2*b^2 + b^4)*x*sinh(x)^2 + 2*(b
^3*cosh(x)^2 + 2*b^3*cosh(x)*sinh(x) + b^3*sinh(x)^2 - b^3)*sqrt(a^2 - b^2)*arctan(-(a*cosh(x) + a*sinh(x) + b
)/sqrt(a^2 - b^2)) + (a^4 - 2*a^2*b^2 + b^4)*x - 2*(a^3*b - a*b^3)*cosh(x) - 2*(a^3*b - a*b^3 + (a^4 - 2*a^2*b
^2 + b^4)*x*cosh(x))*sinh(x))/(a^5 - 2*a^3*b^2 + a*b^4 - (a^5 - 2*a^3*b^2 + a*b^4)*cosh(x)^2 - 2*(a^5 - 2*a^3*
b^2 + a*b^4)*cosh(x)*sinh(x) - (a^5 - 2*a^3*b^2 + a*b^4)*sinh(x)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{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(coth(x)**2/(a+b*sech(x)),x)

[Out]

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

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Giac [A]  time = 1.15148, size = 111, normalized size = 0.97 \begin{align*} \frac{2 \, b^{3} \arctan \left (\frac{a e^{x} + b}{\sqrt{a^{2} - b^{2}}}\right )}{{\left (a^{3} - a b^{2}\right )} \sqrt{a^{2} - b^{2}}} + \frac{x}{a} + \frac{2 \,{\left (b e^{x} - a\right )}}{{\left (a^{2} - b^{2}\right )}{\left (e^{\left (2 \, x\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^3*arctan((a*e^x + b)/sqrt(a^2 - b^2))/((a^3 - a*b^2)*sqrt(a^2 - b^2)) + x/a + 2*(b*e^x - a)/((a^2 - b^2)*(
e^(2*x) - 1))