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((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/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 = 0.20, antiderivative size = 114, normalized size of antiderivative = 1.00, 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 8

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

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 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 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 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 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 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 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])

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*}

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Mathematica [A]  time = 0.35, 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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fricas [B]  time = 0.42, size = 646, normalized size = 5.67 \[ \left [\frac {2 \, a^{4} - 2 \, a^{2} b^{2} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x \cosh \relax (x)^{2} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x \sinh \relax (x)^{2} - {\left (b^{3} \cosh \relax (x)^{2} + 2 \, b^{3} \cosh \relax (x) \sinh \relax (x) + b^{3} \sinh \relax (x)^{2} - b^{3}\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 ) + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x - 2 \, {\left (a^{3} b - a b^{3}\right )} \cosh \relax (x) - 2 \, {\left (a^{3} b - a b^{3} + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x \cosh \relax (x)\right )} \sinh \relax (x)}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cosh \relax (x) \sinh \relax (x) - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sinh \relax (x)^{2}}, \frac {2 \, a^{4} - 2 \, a^{2} b^{2} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x \cosh \relax (x)^{2} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x \sinh \relax (x)^{2} + 2 \, {\left (b^{3} \cosh \relax (x)^{2} + 2 \, b^{3} \cosh \relax (x) \sinh \relax (x) + b^{3} \sinh \relax (x)^{2} - b^{3}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cosh \relax (x) + a \sinh \relax (x) + b}{\sqrt {a^{2} - b^{2}}}\right ) + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x - 2 \, {\left (a^{3} b - a b^{3}\right )} \cosh \relax (x) - 2 \, {\left (a^{3} b - a b^{3} + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x \cosh \relax (x)\right )} \sinh \relax (x)}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cosh \relax (x) \sinh \relax (x) - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sinh \relax (x)^{2}}\right ] \]

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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giac [A]  time = 0.13, size = 82, normalized size = 0.72 \[ \frac {2 \, b^{3} \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{3} - a b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {x}{a} + \frac {2 \, {\left (b e^{x} - a\right )}}{{\left (a^{2} - b^{2}\right )} {\left (e^{\left (2 \, x\right )} - 1\right )}} \]

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))

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maple [A]  time = 0.18, size = 104, normalized size = 0.91 \[ -\frac {\tanh \left (\frac {x}{2}\right )}{2 \left (a -b \right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}+\frac {2 b^{3} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right ) a \left (a +b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}-\frac {1}{2 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )} \]

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)+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))+1/a*ln(tanh(1/2*x)+1)-1/2/(a+b)/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(coth(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 = 1.67, size = 383, normalized size = 3.36 \[ \frac {x}{a}-\frac {\frac {2\,a}{a^2-b^2}-\frac {2\,b\,{\mathrm {e}}^x}{a^2-b^2}}{{\mathrm {e}}^{2\,x}-1}-\frac {2\,\mathrm {atan}\left (\left ({\mathrm {e}}^x\,\left (\frac {2\,b^3}{a^3\,\left (a\,b^2-a^3\right )\,\left (a^2-b^2\right )\,\sqrt {b^6}}-\frac {2\,\left (a\,b^3\,\sqrt {b^6}-a^3\,b\,\sqrt {b^6}\right )}{a^2\,b^2\,\left (a\,b^2-a^3\right )\,\sqrt {a^2\,{\left (a^2-b^2\right )}^3}\,\sqrt {a^8-3\,a^6\,b^2+3\,a^4\,b^4-a^2\,b^6}}\right )+\frac {2\,\left (a^4\,\sqrt {b^6}-a^2\,b^2\,\sqrt {b^6}\right )}{a^2\,b^2\,\left (a\,b^2-a^3\right )\,\sqrt {a^2\,{\left (a^2-b^2\right )}^3}\,\sqrt {a^8-3\,a^6\,b^2+3\,a^4\,b^4-a^2\,b^6}}\right )\,\left (\frac {a^4\,\sqrt {a^8-3\,a^6\,b^2+3\,a^4\,b^4-a^2\,b^6}}{2}-\frac {a^2\,b^2\,\sqrt {a^8-3\,a^6\,b^2+3\,a^4\,b^4-a^2\,b^6}}{2}\right )\right )\,\sqrt {b^6}}{\sqrt {a^8-3\,a^6\,b^2+3\,a^4\,b^4-a^2\,b^6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^2/(a + b/cosh(x)),x)

[Out]

x/a - ((2*a)/(a^2 - b^2) - (2*b*exp(x))/(a^2 - b^2))/(exp(2*x) - 1) - (2*atan((exp(x)*((2*b^3)/(a^3*(a*b^2 - a
^3)*(a^2 - b^2)*(b^6)^(1/2)) - (2*(a*b^3*(b^6)^(1/2) - a^3*b*(b^6)^(1/2)))/(a^2*b^2*(a*b^2 - a^3)*(a^2*(a^2 -
b^2)^3)^(1/2)*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)^(1/2))) + (2*(a^4*(b^6)^(1/2) - a^2*b^2*(b^6)^(1/2)))/(a
^2*b^2*(a*b^2 - a^3)*(a^2*(a^2 - b^2)^3)^(1/2)*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)^(1/2)))*((a^4*(a^8 - a^
2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)^(1/2))/2 - (a^2*b^2*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)^(1/2))/2))*(b^6)^(1
/2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)^(1/2)

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

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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