3.120 \(\int \frac {\text {sech}^3(x)}{a+b \coth (x)} \, dx\)

Optimal. Leaf size=83 \[ -\frac {b^2 \tan ^{-1}(\sinh (x))}{a^3}-\frac {b \text {sech}(x)}{a^2}+\frac {b \sqrt {a^2-b^2} \tanh ^{-1}\left (\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}\right )}{a^3}+\frac {\tan ^{-1}(\sinh (x))}{2 a}+\frac {\tanh (x) \text {sech}(x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3518, 3110, 3768, 3770, 3104, 3074, 206} \[ -\frac {b^2 \tan ^{-1}(\sinh (x))}{a^3}+\frac {b \sqrt {a^2-b^2} \tanh ^{-1}\left (\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}\right )}{a^3}-\frac {b \text {sech}(x)}{a^2}+\frac {\tan ^{-1}(\sinh (x))}{2 a}+\frac {\tanh (x) \text {sech}(x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(a + b*Coth[x]),x]

[Out]

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

Rule 206

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

Rule 3074

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

Rule 3104

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

Rule 3110

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(cos[c + d*x]^m*sin[c + d*x]^n)/(a*cos[c + d*x] + b*sin[c + d
*x]), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 85, normalized size = 1.02 \[ \frac {2 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )+4 b \sqrt {b-a} \sqrt {a+b} \tan ^{-1}\left (\frac {a+b \tanh \left (\frac {x}{2}\right )}{\sqrt {b-a} \sqrt {a+b}}\right )+a \text {sech}(x) (a \tanh (x)-2 b)}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^3/(a + b*Coth[x]),x]

[Out]

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

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fricas [B]  time = 0.47, size = 856, normalized size = 10.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+b*coth(x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.12, size = 102, normalized size = 1.23 \[ \frac {{\left (a^{2} - 2 \, b^{2}\right )} \arctan \left (e^{x}\right )}{a^{3}} - \frac {2 \, {\left (a^{2} b - b^{3}\right )} \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} a^{3}} + \frac {a e^{\left (3 \, x\right )} - 2 \, b e^{\left (3 \, x\right )} - a e^{x} - 2 \, b e^{x}}{a^{2} {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+b*coth(x)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.18, size = 187, normalized size = 2.25 \[ -\frac {2 b \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a \sqrt {-a^{2}+b^{2}}}+\frac {2 b^{3} \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a^{3} \sqrt {-a^{2}+b^{2}}}-\frac {\tanh ^{3}\left (\frac {x}{2}\right )}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b}{a^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\tanh \left (\frac {x}{2}\right )}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 b}{a^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a}-\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right ) b^{2}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^3/(a+b*coth(x)),x)

[Out]

-2/a*b/(-a^2+b^2)^(1/2)*arctan(1/2*(2*tanh(1/2*x)*b+2*a)/(-a^2+b^2)^(1/2))+2*b^3/a^3/(-a^2+b^2)^(1/2)*arctan(1
/2*(2*tanh(1/2*x)*b+2*a)/(-a^2+b^2)^(1/2))-1/a/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3-2/a^2/(tanh(1/2*x)^2+1)^2*tan
h(1/2*x)^2*b+1/a/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)-2/a^2/(tanh(1/2*x)^2+1)^2*b+1/a*arctan(tanh(1/2*x))-2/a^3*arc
tan(tanh(1/2*x))*b^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(sech(x)^3/(a+b*coth(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*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 3.94, size = 166, normalized size = 2.00 \[ \frac {{\mathrm {e}}^x\,\left (a-2\,b\right )}{a^2\,\left ({\mathrm {e}}^{2\,x}+1\right )}+\frac {\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,\left (a^2\,1{}\mathrm {i}-b^2\,2{}\mathrm {i}\right )}{2\,a^3}-\frac {2\,{\mathrm {e}}^x}{a\,\left (2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1\right )}-\frac {\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,\left (a^2\,1{}\mathrm {i}-b^2\,2{}\mathrm {i}\right )}{2\,a^3}+\frac {b\,\ln \left (a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^x+\sqrt {a^2-b^2}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}}{a^3}-\frac {b\,\ln \left (a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^x-\sqrt {a^2-b^2}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^3*(a + b*coth(x))),x)

[Out]

(log(exp(x) + 1i)*(a^2*1i - b^2*2i))/(2*a^3) - (log(exp(x) - 1i)*(a^2*1i - b^2*2i))/(2*a^3) - (2*exp(x))/(a*(2
*exp(2*x) + exp(4*x) + 1)) + (exp(x)*(a - 2*b))/(a^2*(exp(2*x) + 1)) + (b*log(a*exp(x) + b*exp(x) + (a^2 - b^2
)^(1/2))*((a + b)*(a - b))^(1/2))/a^3 - (b*log(a*exp(x) + b*exp(x) - (a^2 - b^2)^(1/2))*((a + b)*(a - b))^(1/2
))/a^3

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{3}{\relax (x )}}{a + b \coth {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**3/(a+b*coth(x)),x)

[Out]

Integral(sech(x)**3/(a + b*coth(x)), x)

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