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

Optimal. Leaf size=66 \[ \frac {\text {csch}(x) (b-a \cosh (x))}{a^2-b^2}+\frac {2 a b \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}} \]

[Out]

2*a*b*arctan((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/(a+b)^(3/2)/(a-b)^(3/2)+(b-a*cosh(x))*csch(x)/(a^2-b^2)

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Rubi [A]  time = 0.13, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3872, 2866, 12, 2659, 205} \[ \frac {\text {csch}(x) (b-a \cosh (x))}{a^2-b^2}+\frac {2 a b \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^2/(a + b*Sech[x]),x]

[Out]

(2*a*b*ArcTan[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)) + ((b - a*Cosh[x])*Csch[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 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 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 {csch}^2(x)}{a+b \text {sech}(x)} \, dx &=-\int \frac {\coth (x) \text {csch}(x)}{-b-a \cosh (x)} \, dx\\ &=\frac {(b-a \cosh (x)) \text {csch}(x)}{a^2-b^2}-\frac {\int \frac {a b}{-b-a \cosh (x)} \, dx}{a^2-b^2}\\ &=\frac {(b-a \cosh (x)) \text {csch}(x)}{a^2-b^2}-\frac {(a b) \int \frac {1}{-b-a \cosh (x)} \, dx}{a^2-b^2}\\ &=\frac {(b-a \cosh (x)) \text {csch}(x)}{a^2-b^2}-\frac {(2 a b) \operatorname {Subst}\left (\int \frac {1}{-a-b-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^2-b^2}\\ &=\frac {2 a b \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}}+\frac {(b-a \cosh (x)) \text {csch}(x)}{a^2-b^2}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 75, normalized size = 1.14 \[ \frac {1}{2} \left (-\frac {4 a b \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {\tanh \left (\frac {x}{2}\right )}{b-a}-\frac {\coth \left (\frac {x}{2}\right )}{a+b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*a*b*ArcTan[((-a + b)*Tanh[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) - Coth[x/2]/(a + b) + Tanh[x/2]/(-a +
b))/2

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fricas [B]  time = 0.42, size = 452, normalized size = 6.85 \[ \left [\frac {2 \, a^{3} - 2 \, a b^{2} - {\left (a b \cosh \relax (x)^{2} + 2 \, a b \cosh \relax (x) \sinh \relax (x) + a b \sinh \relax (x)^{2} - a b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {a^{2} \cosh \relax (x)^{2} + a^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) - a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cosh \relax (x) + a \sinh \relax (x) + b\right )}}{a \cosh \relax (x)^{2} + a \sinh \relax (x)^{2} + 2 \, b \cosh \relax (x) + 2 \, {\left (a \cosh \relax (x) + b\right )} \sinh \relax (x) + a}\right ) - 2 \, {\left (a^{2} b - b^{3}\right )} \cosh \relax (x) - 2 \, {\left (a^{2} b - b^{3}\right )} \sinh \relax (x)}{a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) \sinh \relax (x) - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)^{2}}, \frac {2 \, {\left (a^{3} - a b^{2} + {\left (a b \cosh \relax (x)^{2} + 2 \, a b \cosh \relax (x) \sinh \relax (x) + a b \sinh \relax (x)^{2} - a b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cosh \relax (x) + a \sinh \relax (x) + b}{\sqrt {a^{2} - b^{2}}}\right ) - {\left (a^{2} b - b^{3}\right )} \cosh \relax (x) - {\left (a^{2} b - b^{3}\right )} \sinh \relax (x)\right )}}{a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) \sinh \relax (x) - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^2/(a+b*sech(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*co
sh(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)) - 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), 2*(a^3 - 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)*arctan(-(a*cosh(x) + a*sinh(x) + b)/sqrt(a^2 - b^2
)) - (a^2*b - b^3)*cosh(x) - (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 = 0.12, size = 64, normalized size = 0.97 \[ \frac {2 \, a b \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (b e^{x} - a\right )}}{{\left (a^{2} - b^{2}\right )} {\left (e^{\left (2 \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*b*arctan((a*e^x + b)/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 = 0.16, size = 77, normalized size = 1.17 \[ -\frac {\tanh \left (\frac {x}{2}\right )}{2 \left (a -b \right )}-\frac {1}{2 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )}+\frac {2 a b \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(a-b)*tanh(1/2*x)-1/2/(a+b)/tanh(1/2*x)+2/(a+b)/(a-b)*a*b/((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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(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 = 1.56, size = 151, normalized size = 2.29 \[ \frac {a\,b\,\ln \left (-\frac {2\,b\,{\mathrm {e}}^x}{a^2-b^2}-\frac {2\,b\,\left (a+b\,{\mathrm {e}}^x\right )}{{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}\right )}{{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}-\frac {\frac {2\,a}{a^2-b^2}-\frac {2\,b\,{\mathrm {e}}^x}{a^2-b^2}}{{\mathrm {e}}^{2\,x}-1}-\frac {a\,b\,\ln \left (\frac {2\,b\,\left (a+b\,{\mathrm {e}}^x\right )}{{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}-\frac {2\,b\,{\mathrm {e}}^x}{a^2-b^2}\right )}{{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^2*(a + b/cosh(x))),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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