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

Optimal. Leaf size=85 \[ -\frac {2 b^3 \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b}}-\frac {b \sinh (x)}{a^2}+\frac {x \left (a^2+2 b^2\right )}{2 a^3}+\frac {\sinh (x) \cosh (x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 85, 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 = {3853, 4104, 3919, 3831, 2659, 205} \[ \frac {x \left (a^2+2 b^2\right )}{2 a^3}-\frac {2 b^3 \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b}}-\frac {b \sinh (x)}{a^2}+\frac {\sinh (x) \cosh (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^2/(a + b*Sech[x]),x]

[Out]

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

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 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 3853

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(a*f*n), x] - Dist[1/(a*d*n), Int[((d*Csc[e + f*x])^(n + 1)*Simp[b*n - a*(n + 1)*Csc[e
+ f*x] - b*(n + 1)*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] && LeQ[n, -1] && IntegerQ[2*n]

Rule 3919

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

Rule 4104

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 78, normalized size = 0.92 \[ \frac {\frac {8 b^3 \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+2 a^2 x+a^2 \sinh (2 x)-4 a b \sinh (x)+4 b^2 x}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*x + 4*b^2*x + (8*b^3*ArcTan[((-a + b)*Tanh[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - 4*a*b*Sinh[x] + a^
2*Sinh[2*x])/(4*a^3)

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fricas [B]  time = 0.44, size = 860, normalized size = 10.12 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*((a^4 - a^2*b^2)*cosh(x)^4 + (a^4 - a^2*b^2)*sinh(x)^4 - a^4 + a^2*b^2 + 4*(a^4 + a^2*b^2 - 2*b^4)*x*cosh
(x)^2 - 4*(a^3*b - a*b^3)*cosh(x)^3 - 4*(a^3*b - a*b^3 - (a^4 - a^2*b^2)*cosh(x))*sinh(x)^3 + 2*(3*(a^4 - a^2*
b^2)*cosh(x)^2 + 2*(a^4 + a^2*b^2 - 2*b^4)*x - 6*(a^3*b - a*b^3)*cosh(x))*sinh(x)^2 - 8*(b^3*cosh(x)^2 + 2*b^3
*cosh(x)*sinh(x) + b^3*sinh(x)^2)*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)) + 4*(a^3*b - a*b^3)*cosh(x) + 4*(a^3*b - a*b^3 + (a^4 - a
^2*b^2)*cosh(x)^3 + 2*(a^4 + a^2*b^2 - 2*b^4)*x*cosh(x) - 3*(a^3*b - a*b^3)*cosh(x)^2)*sinh(x))/((a^5 - a^3*b^
2)*cosh(x)^2 + 2*(a^5 - a^3*b^2)*cosh(x)*sinh(x) + (a^5 - a^3*b^2)*sinh(x)^2), 1/8*((a^4 - a^2*b^2)*cosh(x)^4
+ (a^4 - a^2*b^2)*sinh(x)^4 - a^4 + a^2*b^2 + 4*(a^4 + a^2*b^2 - 2*b^4)*x*cosh(x)^2 - 4*(a^3*b - a*b^3)*cosh(x
)^3 - 4*(a^3*b - a*b^3 - (a^4 - a^2*b^2)*cosh(x))*sinh(x)^3 + 2*(3*(a^4 - a^2*b^2)*cosh(x)^2 + 2*(a^4 + a^2*b^
2 - 2*b^4)*x - 6*(a^3*b - a*b^3)*cosh(x))*sinh(x)^2 + 16*(b^3*cosh(x)^2 + 2*b^3*cosh(x)*sinh(x) + b^3*sinh(x)^
2)*sqrt(a^2 - b^2)*arctan(-(a*cosh(x) + a*sinh(x) + b)/sqrt(a^2 - b^2)) + 4*(a^3*b - a*b^3)*cosh(x) + 4*(a^3*b
- a*b^3 + (a^4 - a^2*b^2)*cosh(x)^3 + 2*(a^4 + a^2*b^2 - 2*b^4)*x*cosh(x) - 3*(a^3*b - a*b^3)*cosh(x)^2)*sinh
(x))/((a^5 - a^3*b^2)*cosh(x)^2 + 2*(a^5 - a^3*b^2)*cosh(x)*sinh(x) + (a^5 - a^3*b^2)*sinh(x)^2)]

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giac [A]  time = 0.14, size = 92, normalized size = 1.08 \[ -\frac {2 \, b^{3} \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}} a^{3}} + \frac {a e^{\left (2 \, x\right )} - 4 \, b e^{x}}{8 \, a^{2}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} x}{2 \, a^{3}} + \frac {{\left (4 \, a b e^{x} - a^{2}\right )} e^{\left (-2 \, x\right )}}{8 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b^3*arctan((a*e^x + b)/sqrt(a^2 - b^2))/(sqrt(a^2 - b^2)*a^3) + 1/8*(a*e^(2*x) - 4*b*e^x)/a^2 + 1/2*(a^2 +
2*b^2)*x/a^3 + 1/8*(4*a*b*e^x - a^2)*e^(-2*x)/a^3

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maple [B]  time = 0.15, size = 174, normalized size = 2.05 \[ \frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b}{a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 a}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) b^{2}}{a^{3}}-\frac {2 b^{3} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b}{a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) b^{2}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a/(tanh(1/2*x)-1)^2+1/2/a/(tanh(1/2*x)-1)+1/a^2/(tanh(1/2*x)-1)*b-1/2/a*ln(tanh(1/2*x)-1)-1/a^3*ln(tanh(1/
2*x)-1)*b^2-2*b^3/a^3/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))-1/2/a/(tanh(1/2*x)+1)^
2+1/2/a/(tanh(1/2*x)+1)+1/a^2/(tanh(1/2*x)+1)*b+1/2/a*ln(tanh(1/2*x)+1)+1/a^3*ln(tanh(1/2*x)+1)*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(cosh(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 = 1.58, size = 167, normalized size = 1.96 \[ \frac {{\mathrm {e}}^{2\,x}}{8\,a}-\frac {{\mathrm {e}}^{-2\,x}}{8\,a}-\frac {b\,{\mathrm {e}}^x}{2\,a^2}+\frac {b\,{\mathrm {e}}^{-x}}{2\,a^2}+\frac {x\,\left (a^2+2\,b^2\right )}{2\,a^3}+\frac {b^3\,\ln \left (\frac {2\,b^3\,{\mathrm {e}}^x}{a^4}-\frac {2\,b^3\,\left (a+b\,{\mathrm {e}}^x\right )}{a^4\,\sqrt {a+b}\,\sqrt {b-a}}\right )}{a^3\,\sqrt {a+b}\,\sqrt {b-a}}-\frac {b^3\,\ln \left (\frac {2\,b^3\,{\mathrm {e}}^x}{a^4}+\frac {2\,b^3\,\left (a+b\,{\mathrm {e}}^x\right )}{a^4\,\sqrt {a+b}\,\sqrt {b-a}}\right )}{a^3\,\sqrt {a+b}\,\sqrt {b-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^2/(a + b/cosh(x)),x)

[Out]

exp(2*x)/(8*a) - exp(-2*x)/(8*a) - (b*exp(x))/(2*a^2) + (b*exp(-x))/(2*a^2) + (x*(a^2 + 2*b^2))/(2*a^3) + (b^3
*log((2*b^3*exp(x))/a^4 - (2*b^3*(a + b*exp(x)))/(a^4*(a + b)^(1/2)*(b - a)^(1/2))))/(a^3*(a + b)^(1/2)*(b - a
)^(1/2)) - (b^3*log((2*b^3*exp(x))/a^4 + (2*b^3*(a + b*exp(x)))/(a^4*(a + b)^(1/2)*(b - a)^(1/2))))/(a^3*(a +
b)^(1/2)*(b - a)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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