3.102 \(\int \frac {\text {sech}^4(x)}{a+b \text {sech}(x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3851, 4082, 3998, 3770, 3831, 2659, 205} \[ -\frac {2 a^3 \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^3 \sqrt {a-b} \sqrt {a+b}}+\frac {\left (2 a^2+b^2\right ) \tan ^{-1}(\sinh (x))}{2 b^3}-\frac {a \tanh (x)}{b^2}+\frac {\tanh (x) \text {sech}(x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^4/(a + b*Sech[x]),x]

[Out]

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

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 3770

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

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3851

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(d^3*Cot[e
+ f*x]*(d*Csc[e + f*x])^(n - 3))/(b*f*(n - 2)), x] + Dist[d^3/(b*(n - 2)), Int[((d*Csc[e + f*x])^(n - 3)*Simp
[a*(n - 3) + b*(n - 3)*Csc[e + f*x] - a*(n - 2)*Csc[e + f*x]^2, x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a,
b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 3]

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 4082

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m
+ 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 82, normalized size = 0.94 \[ \frac {2 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )+\frac {4 a^3 \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+b \tanh (x) (b \text {sech}(x)-2 a)}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^4/(a + b*Sech[x]),x]

[Out]

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

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fricas [B]  time = 0.53, size = 1444, normalized size = 16.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(a+b*sech(x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.12, size = 89, normalized size = 1.02 \[ -\frac {2 \, a^{3} \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}} b^{3}} + \frac {{\left (2 \, a^{2} + b^{2}\right )} \arctan \left (e^{x}\right )}{b^{3}} + \frac {b e^{\left (3 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - b e^{x} + 2 \, a}{b^{2} {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(a+b*sech(x)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 146, normalized size = 1.68 \[ -\frac {2 a^{3} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) a}{b^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {\tanh ^{3}\left (\frac {x}{2}\right )}{b \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 \tanh \left (\frac {x}{2}\right ) a}{b^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\tanh \left (\frac {x}{2}\right )}{b \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right ) a^{2}}{b^{3}}+\frac {\arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^4/(a+b*sech(x)),x)

[Out]

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

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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)^4/(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 = 5.08, size = 476, normalized size = 5.47 \[ \frac {{\mathrm {e}}^x}{b+b\,{\mathrm {e}}^{2\,x}}-\frac {2\,{\mathrm {e}}^x}{b+2\,b\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{4\,x}}+\frac {2\,a}{b^2\,{\mathrm {e}}^{2\,x}+b^2}-\frac {\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,b}-\frac {a^2\,\left (\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{b^3}-\frac {a^3\,\ln \left (16\,a\,b^5-48\,a^5\,b-24\,a^5\,\sqrt {b^2-a^2}+32\,a^3\,b^3+24\,a^6\,{\mathrm {e}}^x+32\,b^6\,{\mathrm {e}}^x+16\,a\,b^4\,\sqrt {b^2-a^2}+40\,a^3\,b^2\,\sqrt {b^2-a^2}+32\,b^5\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}+56\,a^2\,b^4\,{\mathrm {e}}^x-112\,a^4\,b^2\,{\mathrm {e}}^x+72\,a^2\,b^3\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}-72\,a^4\,b\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}\right )}{b^3\,\sqrt {b^2-a^2}}+\frac {a^3\,\ln \left (16\,a\,b^5-48\,a^5\,b+24\,a^5\,\sqrt {b^2-a^2}+32\,a^3\,b^3+24\,a^6\,{\mathrm {e}}^x+32\,b^6\,{\mathrm {e}}^x-16\,a\,b^4\,\sqrt {b^2-a^2}-40\,a^3\,b^2\,\sqrt {b^2-a^2}-32\,b^5\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}+56\,a^2\,b^4\,{\mathrm {e}}^x-112\,a^4\,b^2\,{\mathrm {e}}^x-72\,a^2\,b^3\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}+72\,a^4\,b\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}\right )}{b^3\,\sqrt {b^2-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^4*(a + b/cosh(x))),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{4}{\relax (x )}}{a + b \operatorname {sech}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**4/(a+b*sech(x)),x)

[Out]

Integral(sech(x)**4/(a + b*sech(x)), x)

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