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

Optimal. Leaf size=60 \[ \frac{2 a b \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac{\text{sech}(x) (b-a \sinh (x))}{a^2+b^2} \]

[Out]

(2*a*b*ArcTanh[(a - b*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(3/2) - (Sech[x]*(b - a*Sinh[x]))/(a^2 + b^2)

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Rubi [A]  time = 0.14187, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {3872, 2866, 12, 2660, 618, 204} \[ \frac{2 a b \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac{\text{sech}(x) (b-a \sinh (x))}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(a + b*Csch[x]),x]

[Out]

(2*a*b*ArcTanh[(a - b*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(3/2) - (Sech[x]*(b - a*Sinh[x]))/(a^2 + b^2)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\text{sech}^2(x)}{a+b \text{csch}(x)} \, dx &=i \int \frac{\text{sech}(x) \tanh (x)}{i b+i a \sinh (x)} \, dx\\ &=-\frac{\text{sech}(x) (b-a \sinh (x))}{a^2+b^2}-\frac{i \int \frac{a b}{i b+i a \sinh (x)} \, dx}{a^2+b^2}\\ &=-\frac{\text{sech}(x) (b-a \sinh (x))}{a^2+b^2}-\frac{(i a b) \int \frac{1}{i b+i a \sinh (x)} \, dx}{a^2+b^2}\\ &=-\frac{\text{sech}(x) (b-a \sinh (x))}{a^2+b^2}-\frac{(2 i a b) \operatorname{Subst}\left (\int \frac{1}{i b+2 i a x-i b x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a^2+b^2}\\ &=-\frac{\text{sech}(x) (b-a \sinh (x))}{a^2+b^2}+\frac{(4 i a b) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2+b^2\right )-x^2} \, dx,x,2 i a-2 i b \tanh \left (\frac{x}{2}\right )\right )}{a^2+b^2}\\ &=\frac{2 a b \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac{\text{sech}(x) (b-a \sinh (x))}{a^2+b^2}\\ \end{align*}

Mathematica [A]  time = 0.176511, size = 67, normalized size = 1.12 \[ \frac{a \left (\tanh (x)-\frac{2 b \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}\right )-b \text{sech}(x)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.029, size = 81, normalized size = 1.4 \begin{align*} -4\,{\frac{ab}{ \left ( 2\,{a}^{2}+2\,{b}^{2} \right ) \sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,\tanh \left ( x/2 \right ) b-2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-2\,{\frac{-a\tanh \left ( x/2 \right ) +b}{ \left ({a}^{2}+{b}^{2} \right ) \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*a*b/(2*a^2+2*b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*tanh(1/2*x)*b-2*a)/(a^2+b^2)^(1/2))-2/(a^2+b^2)*(-a*tanh(1
/2*x)+b)/(tanh(1/2*x)^2+1)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.62838, size = 693, normalized size = 11.55 \begin{align*} -\frac{2 \, a^{3} + 2 \, a b^{2} -{\left (a b \cosh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) \sinh \left (x\right ) + a b \sinh \left (x\right )^{2} + a b\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 ) + 2 \,{\left (a^{2} b + b^{3}\right )} \cosh \left (x\right ) + 2 \,{\left (a^{2} b + b^{3}\right )} \sinh \left (x\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4} +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^3 + 2*a*b^2 - (a*b*cosh(x)^2 + 2*a*b*cosh(x)*sinh(x) + a*b*sinh(x)^2 + a*b)*sqrt(a^2 + b^2)*log((a^2*cos
h(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*co
sh(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)) + 2*(a^2*b
+ b^3)*cosh(x) + 2*(a^2*b + b^3)*sinh(x))/(a^4 + 2*a^2*b^2 + b^4 + (a^4 + 2*a^2*b^2 + b^4)*cosh(x)^2 + 2*(a^4
+ 2*a^2*b^2 + b^4)*cosh(x)*sinh(x) + (a^4 + 2*a^2*b^2 + b^4)*sinh(x)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15072, size = 115, normalized size = 1.92 \begin{align*} -\frac{a b \log \left (\frac{{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac{3}{2}}} - \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(sech(x)^2/(a+b*csch(x)),x, algorithm="giac")

[Out]

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