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(1/2*x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(3/2)-sech(x)*(b-a*sinh(x))/(a^2+b^2)

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Rubi [A]  time = 0.14, antiderivative size = 60, normalized size of antiderivative = 1.00, 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 12

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

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])

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 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 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 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]

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*}

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Mathematica [A]  time = 0.18, 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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fricas [B]  time = 1.33, size = 256, normalized size = 4.27 \[ -\frac {2 \, a^{3} + 2 \, a b^{2} - {\left (a b \cosh \relax (x)^{2} + 2 \, a b \cosh \relax (x) \sinh \relax (x) + a b \sinh \relax (x)^{2} + a b\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {a^{2} \cosh \relax (x)^{2} + a^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cosh \relax (x) + a \sinh \relax (x) + b\right )}}{a \cosh \relax (x)^{2} + a \sinh \relax (x)^{2} + 2 \, b \cosh \relax (x) + 2 \, {\left (a \cosh \relax (x) + b\right )} \sinh \relax (x) - a}\right ) + 2 \, {\left (a^{2} b + b^{3}\right )} \cosh \relax (x) + 2 \, {\left (a^{2} b + b^{3}\right )} \sinh \relax (x)}{a^{4} + 2 \, a^{2} b^{2} + b^{4} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)^{2}} \]

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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giac [A]  time = 0.15, size = 85, normalized size = 1.42 \[ -\frac {a b \log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b e^{x} + a\right )}}{{\left (a^{2} + b^{2}\right )} {\left (e^{\left (2 \, x\right )} + 1\right )}} \]

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))

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maple [A]  time = 0.17, size = 81, normalized size = 1.35 \[ -\frac {4 a b \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {a^{2}+b^{2}}}-\frac {2 \left (b -a \tanh \left (\frac {x}{2}\right )\right )}{\left (a^{2}+b^{2}\right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )} \]

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

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maxima [A]  time = 0.41, size = 91, normalized size = 1.52 \[ -\frac {a b \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b e^{\left (-x\right )} - a\right )}}{a^{2} + b^{2} + {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.66, size = 133, normalized size = 2.22 \[ \frac {a\,b\,\ln \left (\frac {2\,b\,\left (a-b\,{\mathrm {e}}^x\right )}{{\left (a^2+b^2\right )}^{3/2}}+\frac {2\,b\,{\mathrm {e}}^x}{a^2+b^2}\right )}{{\left (a^2+b^2\right )}^{3/2}}-\frac {a\,b\,\ln \left (\frac {2\,b\,{\mathrm {e}}^x}{a^2+b^2}-\frac {2\,b\,\left (a-b\,{\mathrm {e}}^x\right )}{{\left (a^2+b^2\right )}^{3/2}}\right )}{{\left (a^2+b^2\right )}^{3/2}}-\frac {\frac {2\,a}{a^2+b^2}+\frac {2\,b\,{\mathrm {e}}^x}{a^2+b^2}}{{\mathrm {e}}^{2\,x}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^2*(a + b/sinh(x))),x)

[Out]

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

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

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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