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

Optimal. Leaf size=87 \[ -\frac {2 b^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b}}-\frac {b \tanh (x)}{a^2}+\frac {\left (a^2+2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^3}+\frac {\tanh (x) \text {sech}(x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 87, 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 = {2802, 3055, 3001, 3770, 2659, 208} \[ -\frac {2 b^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b}}+\frac {\left (a^2+2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^3}-\frac {b \tanh (x)}{a^2}+\frac {\tanh (x) \text {sech}(x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(a + b*Cosh[x]),x]

[Out]

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[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 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

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

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Mathematica [A]  time = 0.21, size = 82, normalized size = 0.94 \[ \frac {2 \left (a^2+2 b^2\right ) \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )+\frac {4 b^3 \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+a \tanh (x) (a \text {sech}(x)-2 b)}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.85, size = 1370, normalized size = 15.75 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(2*a^3*b - 2*a*b^3 + (a^4 - a^2*b^2)*cosh(x)^3 + (a^4 - a^2*b^2)*sinh(x)^3 + 2*(a^3*b - a*b^3)*cosh(x)^2 + (2
*a^3*b - 2*a*b^3 + 3*(a^4 - a^2*b^2)*cosh(x))*sinh(x)^2 + (b^3*cosh(x)^4 + 4*b^3*cosh(x)*sinh(x)^3 + b^3*sinh(
x)^4 + 2*b^3*cosh(x)^2 + b^3 + 2*(3*b^3*cosh(x)^2 + b^3)*sinh(x)^2 + 4*(b^3*cosh(x)^3 + b^3*cosh(x))*sinh(x))*
sqrt(a^2 - b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 - b^2 + 2*(b^2*cosh(x) + a*b)*sinh(
x) + 2*sqrt(a^2 - b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) +
a)*sinh(x) + b)) + ((a^4 + a^2*b^2 - 2*b^4)*cosh(x)^4 + 4*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)*sinh(x)^3 + (a^4 + a
^2*b^2 - 2*b^4)*sinh(x)^4 + a^4 + a^2*b^2 - 2*b^4 + 2*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)^2 + 2*(a^4 + a^2*b^2 - 2
*b^4 + 3*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)^2)*sinh(x)^2 + 4*((a^4 + a^2*b^2 - 2*b^4)*cosh(x)^3 + (a^4 + a^2*b^2
- 2*b^4)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) - (a^4 - a^2*b^2)*cosh(x) - (a^4 - a^2*b^2 - 3*(a^4 - a^2
*b^2)*cosh(x)^2 - 4*(a^3*b - a*b^3)*cosh(x))*sinh(x))/(a^5 - a^3*b^2 + (a^5 - a^3*b^2)*cosh(x)^4 + 4*(a^5 - a^
3*b^2)*cosh(x)*sinh(x)^3 + (a^5 - a^3*b^2)*sinh(x)^4 + 2*(a^5 - a^3*b^2)*cosh(x)^2 + 2*(a^5 - a^3*b^2 + 3*(a^5
- a^3*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((a^5 - a^3*b^2)*cosh(x)^3 + (a^5 - a^3*b^2)*cosh(x))*sinh(x)), (2*a^3*b
- 2*a*b^3 + (a^4 - a^2*b^2)*cosh(x)^3 + (a^4 - a^2*b^2)*sinh(x)^3 + 2*(a^3*b - a*b^3)*cosh(x)^2 + (2*a^3*b - 2
*a*b^3 + 3*(a^4 - a^2*b^2)*cosh(x))*sinh(x)^2 + 2*(b^3*cosh(x)^4 + 4*b^3*cosh(x)*sinh(x)^3 + b^3*sinh(x)^4 + 2
*b^3*cosh(x)^2 + b^3 + 2*(3*b^3*cosh(x)^2 + b^3)*sinh(x)^2 + 4*(b^3*cosh(x)^3 + b^3*cosh(x))*sinh(x))*sqrt(-a^
2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + ((a^4 + a^2*b^2 - 2*b^4)*cosh(x)^
4 + 4*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)*sinh(x)^3 + (a^4 + a^2*b^2 - 2*b^4)*sinh(x)^4 + a^4 + a^2*b^2 - 2*b^4 +
2*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)^2 + 2*(a^4 + a^2*b^2 - 2*b^4 + 3*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)^2)*sinh(x)^
2 + 4*((a^4 + a^2*b^2 - 2*b^4)*cosh(x)^3 + (a^4 + a^2*b^2 - 2*b^4)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x))
- (a^4 - a^2*b^2)*cosh(x) - (a^4 - a^2*b^2 - 3*(a^4 - a^2*b^2)*cosh(x)^2 - 4*(a^3*b - a*b^3)*cosh(x))*sinh(x)
)/(a^5 - a^3*b^2 + (a^5 - a^3*b^2)*cosh(x)^4 + 4*(a^5 - a^3*b^2)*cosh(x)*sinh(x)^3 + (a^5 - a^3*b^2)*sinh(x)^4
+ 2*(a^5 - a^3*b^2)*cosh(x)^2 + 2*(a^5 - a^3*b^2 + 3*(a^5 - a^3*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((a^5 - a^3*b^2
)*cosh(x)^3 + (a^5 - a^3*b^2)*cosh(x))*sinh(x))]

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giac [A]  time = 0.15, size = 89, normalized size = 1.02 \[ -\frac {2 \, b^{3} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} a^{3}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} \arctan \left (e^{x}\right )}{a^{3}} + \frac {a e^{\left (3 \, x\right )} + 2 \, b e^{\left (2 \, x\right )} - a e^{x} + 2 \, b}{a^{2} {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 146, normalized size = 1.68 \[ -\frac {2 b^{3} \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {\tanh ^{3}\left (\frac {x}{2}\right )}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) b}{a^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\tanh \left (\frac {x}{2}\right )}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 \tanh \left (\frac {x}{2}\right ) b}{a^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a}+\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right ) b^{2}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b^3/a^3/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(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*tanh(1/2*x)^3*b+1/a/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)-2/a^2/(tanh(1/2*x)^2+1)^2*
tanh(1/2*x)*b+1/a*arctan(tanh(1/2*x))+2/a^3*arctan(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*cosh(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 = 4.13, size = 476, normalized size = 5.47 \[ \frac {{\mathrm {e}}^x}{a+a\,{\mathrm {e}}^{2\,x}}-\frac {2\,{\mathrm {e}}^x}{a+2\,a\,{\mathrm {e}}^{2\,x}+a\,{\mathrm {e}}^{4\,x}}+\frac {2\,b}{a^2\,{\mathrm {e}}^{2\,x}+a^2}-\frac {\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a}-\frac {b^2\,\left (\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{a^3}-\frac {b^3\,\ln \left (16\,a^5\,b-48\,a\,b^5-24\,b^5\,\sqrt {a^2-b^2}+32\,a^3\,b^3+32\,a^6\,{\mathrm {e}}^x+24\,b^6\,{\mathrm {e}}^x+16\,a^4\,b\,\sqrt {a^2-b^2}+40\,a^2\,b^3\,\sqrt {a^2-b^2}+32\,a^5\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-112\,a^2\,b^4\,{\mathrm {e}}^x+56\,a^4\,b^2\,{\mathrm {e}}^x+72\,a^3\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-72\,a\,b^4\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a^3\,\sqrt {a^2-b^2}}+\frac {b^3\,\ln \left (16\,a^5\,b-48\,a\,b^5+24\,b^5\,\sqrt {a^2-b^2}+32\,a^3\,b^3+32\,a^6\,{\mathrm {e}}^x+24\,b^6\,{\mathrm {e}}^x-16\,a^4\,b\,\sqrt {a^2-b^2}-40\,a^2\,b^3\,\sqrt {a^2-b^2}-32\,a^5\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-112\,a^2\,b^4\,{\mathrm {e}}^x+56\,a^4\,b^2\,{\mathrm {e}}^x-72\,a^3\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}+72\,a\,b^4\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a^3\,\sqrt {a^2-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^3*(a + b*cosh(x))),x)

[Out]

exp(x)/(a + a*exp(2*x)) - (2*exp(x))/(a + 2*a*exp(2*x) + a*exp(4*x)) + (2*b)/(a^2*exp(2*x) + a^2) - (log(exp(x
)*1i + 1)*1i - log(exp(x) + 1i)*1i)/(2*a) - (b^2*(log(exp(x)*1i + 1)*1i - log(exp(x) + 1i)*1i))/a^3 - (b^3*log
(16*a^5*b - 48*a*b^5 - 24*b^5*(a^2 - b^2)^(1/2) + 32*a^3*b^3 + 32*a^6*exp(x) + 24*b^6*exp(x) + 16*a^4*b*(a^2 -
b^2)^(1/2) + 40*a^2*b^3*(a^2 - b^2)^(1/2) + 32*a^5*exp(x)*(a^2 - b^2)^(1/2) - 112*a^2*b^4*exp(x) + 56*a^4*b^2
*exp(x) + 72*a^3*b^2*exp(x)*(a^2 - b^2)^(1/2) - 72*a*b^4*exp(x)*(a^2 - b^2)^(1/2)))/(a^3*(a^2 - b^2)^(1/2)) +
(b^3*log(16*a^5*b - 48*a*b^5 + 24*b^5*(a^2 - b^2)^(1/2) + 32*a^3*b^3 + 32*a^6*exp(x) + 24*b^6*exp(x) - 16*a^4*
b*(a^2 - b^2)^(1/2) - 40*a^2*b^3*(a^2 - b^2)^(1/2) - 32*a^5*exp(x)*(a^2 - b^2)^(1/2) - 112*a^2*b^4*exp(x) + 56
*a^4*b^2*exp(x) - 72*a^3*b^2*exp(x)*(a^2 - b^2)^(1/2) + 72*a*b^4*exp(x)*(a^2 - b^2)^(1/2)))/(a^3*(a^2 - b^2)^(
1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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