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

Optimal. Leaf size=74 \[ \frac {2 \left (a c^2+b d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tanh \left (\frac {x}{2}\right )}{\sqrt {c+d}}\right )}{c^2 \sqrt {c-d} \sqrt {c+d}}-\frac {b d \tan ^{-1}(\sinh (x))}{c^2}+\frac {b \tanh (x)}{c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {4234, 3056, 3001, 3770, 2659, 208} \[ \frac {2 \left (a c^2+b d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tanh \left (\frac {x}{2}\right )}{\sqrt {c+d}}\right )}{c^2 \sqrt {c-d} \sqrt {c+d}}-\frac {b d \tan ^{-1}(\sinh (x))}{c^2}+\frac {b \tanh (x)}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[x]^2)/(c + d*Cosh[x]),x]

[Out]

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

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 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 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + 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] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[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]

Rule 4234

Int[(u_)*((A_) + (C_.)*sec[(a_.) + (b_.)*(x_)]^2), x_Symbol] :> Int[(ActivateTrig[u]*(C + A*Cos[a + b*x]^2))/C
os[a + b*x]^2, x] /; FreeQ[{a, b, A, C}, x] && KnownSineIntegrandQ[u, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.22, size = 127, normalized size = 1.72 \[ -\frac {2 \text {sech}(x) \left (a \cosh ^2(x)+b\right ) \left (2 \cosh (x) \left (\left (a c^2+b d^2\right ) \tan ^{-1}\left (\frac {(c-d) \tanh \left (\frac {x}{2}\right )}{\sqrt {d^2-c^2}}\right )+b d \sqrt {d^2-c^2} \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )\right )-b c \sqrt {d^2-c^2} \sinh (x)\right )}{c^2 \sqrt {d^2-c^2} (a \cosh (2 x)+a+2 b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b + a*Cosh[x]^2)*Sech[x]*(2*(b*d*Sqrt[-c^2 + d^2]*ArcTan[Tanh[x/2]] + (a*c^2 + b*d^2)*ArcTan[((c - d)*Tan
h[x/2])/Sqrt[-c^2 + d^2]])*Cosh[x] - b*c*Sqrt[-c^2 + d^2]*Sinh[x]))/(c^2*Sqrt[-c^2 + d^2]*(a + 2*b + a*Cosh[2*
x]))

________________________________________________________________________________________

fricas [B]  time = 0.79, size = 598, normalized size = 8.08 \[ \left [-\frac {2 \, b c^{3} - 2 \, b c d^{2} - {\left (a c^{2} + b d^{2} + {\left (a c^{2} + b d^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a c^{2} + b d^{2}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a c^{2} + b d^{2}\right )} \sinh \relax (x)^{2}\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {d^{2} \cosh \relax (x)^{2} + d^{2} \sinh \relax (x)^{2} + 2 \, c d \cosh \relax (x) + 2 \, c^{2} - d^{2} + 2 \, {\left (d^{2} \cosh \relax (x) + c d\right )} \sinh \relax (x) - 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cosh \relax (x) + d \sinh \relax (x) + c\right )}}{d \cosh \relax (x)^{2} + d \sinh \relax (x)^{2} + 2 \, c \cosh \relax (x) + 2 \, {\left (d \cosh \relax (x) + c\right )} \sinh \relax (x) + d}\right ) + 2 \, {\left (b c^{2} d - b d^{3} + {\left (b c^{2} d - b d^{3}\right )} \cosh \relax (x)^{2} + 2 \, {\left (b c^{2} d - b d^{3}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (b c^{2} d - b d^{3}\right )} \sinh \relax (x)^{2}\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right )}{c^{4} - c^{2} d^{2} + {\left (c^{4} - c^{2} d^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (c^{4} - c^{2} d^{2}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (c^{4} - c^{2} d^{2}\right )} \sinh \relax (x)^{2}}, -\frac {2 \, {\left (b c^{3} - b c d^{2} + {\left (a c^{2} + b d^{2} + {\left (a c^{2} + b d^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a c^{2} + b d^{2}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a c^{2} + b d^{2}\right )} \sinh \relax (x)^{2}\right )} \sqrt {-c^{2} + d^{2}} \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cosh \relax (x) + d \sinh \relax (x) + c\right )}}{c^{2} - d^{2}}\right ) + {\left (b c^{2} d - b d^{3} + {\left (b c^{2} d - b d^{3}\right )} \cosh \relax (x)^{2} + 2 \, {\left (b c^{2} d - b d^{3}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (b c^{2} d - b d^{3}\right )} \sinh \relax (x)^{2}\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right )\right )}}{c^{4} - c^{2} d^{2} + {\left (c^{4} - c^{2} d^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (c^{4} - c^{2} d^{2}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (c^{4} - c^{2} d^{2}\right )} \sinh \relax (x)^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.14, size = 71, normalized size = 0.96 \[ -\frac {2 \, b d \arctan \left (e^{x}\right )}{c^{2}} + \frac {2 \, {\left (a c^{2} + b d^{2}\right )} \arctan \left (\frac {d e^{x} + c}{\sqrt {-c^{2} + d^{2}}}\right )}{\sqrt {-c^{2} + d^{2}} c^{2}} - \frac {2 \, b}{c {\left (e^{\left (2 \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.18, size = 112, normalized size = 1.51 \[ \frac {2 \arctanh \left (\frac {\left (c -d \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right ) a}{\sqrt {\left (c +d \right ) \left (c -d \right )}}+\frac {2 \arctanh \left (\frac {\left (c -d \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right ) b \,d^{2}}{c^{2} \sqrt {\left (c +d \right ) \left (c -d \right )}}+\frac {2 b \tanh \left (\frac {x}{2}\right )}{c \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )}-\frac {2 b d \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/((c+d)*(c-d))^(1/2)*arctanh((c-d)*tanh(1/2*x)/((c+d)*(c-d))^(1/2))*a+2/c^2/((c+d)*(c-d))^(1/2)*arctanh((c-d)
*tanh(1/2*x)/((c+d)*(c-d))^(1/2))*b*d^2+2*b/c*tanh(1/2*x)/(tanh(1/2*x)^2+1)-2*b/c^2*d*arctan(tanh(1/2*x))

________________________________________________________________________________________

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((a+b*sech(x)^2)/(c+d*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*c^2-4*d^2>0)', see `assume?`
for more details)Is 4*c^2-4*d^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 6.90, size = 704, normalized size = 9.51 \[ \frac {\ln \left (\frac {\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (\frac {32\,\left (a^2\,c^4+2\,a\,b\,c^2\,d^2-4\,{\mathrm {e}}^x\,b^2\,c^3\,d-2\,b^2\,c^2\,d^2+3\,{\mathrm {e}}^x\,b^2\,c\,d^3+2\,b^2\,d^4\right )}{c^2\,d^4}-\frac {\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (a\,c^2+b\,d^2\right )\,\left (\frac {32\,c\,\left (2\,b\,d^3+4\,a\,c^3\,{\mathrm {e}}^x+2\,a\,c^2\,d-a\,c\,d^2\,{\mathrm {e}}^x+3\,b\,c\,d^2\,{\mathrm {e}}^x\right )}{d^5}+\frac {32\,\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (a\,c^2+b\,d^2\right )\,\left (4\,{\mathrm {e}}^x\,c^3+3\,c^2\,d-3\,{\mathrm {e}}^x\,c\,d^2-2\,d^3\right )}{d^5\,\left (c^2-d^2\right )}\right )}{c^2\,\left (c^2-d^2\right )}\right )\,\left (a\,c^2+b\,d^2\right )}{c^2\,\left (c^2-d^2\right )}-\frac {32\,b\,\left (a\,c^2+b\,d^2\right )\,\left (2\,b\,d+a\,c\,{\mathrm {e}}^x+4\,b\,c\,{\mathrm {e}}^x\right )}{c^3\,d^3}\right )\,\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (a\,c^2+b\,d^2\right )}{c^4-c^2\,d^2}-\frac {2\,b}{c\,\left ({\mathrm {e}}^{2\,x}+1\right )}-\frac {\ln \left (-\frac {32\,b\,\left (a\,c^2+b\,d^2\right )\,\left (2\,b\,d+a\,c\,{\mathrm {e}}^x+4\,b\,c\,{\mathrm {e}}^x\right )}{c^3\,d^3}-\frac {\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (\frac {32\,\left (a^2\,c^4+2\,a\,b\,c^2\,d^2-4\,{\mathrm {e}}^x\,b^2\,c^3\,d-2\,b^2\,c^2\,d^2+3\,{\mathrm {e}}^x\,b^2\,c\,d^3+2\,b^2\,d^4\right )}{c^2\,d^4}+\frac {\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (a\,c^2+b\,d^2\right )\,\left (\frac {32\,c\,\left (2\,b\,d^3+4\,a\,c^3\,{\mathrm {e}}^x+2\,a\,c^2\,d-a\,c\,d^2\,{\mathrm {e}}^x+3\,b\,c\,d^2\,{\mathrm {e}}^x\right )}{d^5}-\frac {32\,\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (a\,c^2+b\,d^2\right )\,\left (4\,{\mathrm {e}}^x\,c^3+3\,c^2\,d-3\,{\mathrm {e}}^x\,c\,d^2-2\,d^3\right )}{d^5\,\left (c^2-d^2\right )}\right )}{c^2\,\left (c^2-d^2\right )}\right )\,\left (a\,c^2+b\,d^2\right )}{c^2\,\left (c^2-d^2\right )}\right )\,\sqrt {\left (c+d\right )\,\left (c-d\right )}\,\left (a\,c^2+b\,d^2\right )}{c^4-c^2\,d^2}+\frac {b\,d\,\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,1{}\mathrm {i}}{c^2}-\frac {b\,d\,\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((((c + d)*(c - d))^(1/2)*((32*(a^2*c^4 + 2*b^2*d^4 - 2*b^2*c^2*d^2 + 3*b^2*c*d^3*exp(x) - 4*b^2*c^3*d*exp
(x) + 2*a*b*c^2*d^2))/(c^2*d^4) - (((c + d)*(c - d))^(1/2)*(a*c^2 + b*d^2)*((32*c*(2*b*d^3 + 4*a*c^3*exp(x) +
2*a*c^2*d - a*c*d^2*exp(x) + 3*b*c*d^2*exp(x)))/d^5 + (32*((c + d)*(c - d))^(1/2)*(a*c^2 + b*d^2)*(3*c^2*d - 2
*d^3 + 4*c^3*exp(x) - 3*c*d^2*exp(x)))/(d^5*(c^2 - d^2))))/(c^2*(c^2 - d^2)))*(a*c^2 + b*d^2))/(c^2*(c^2 - d^2
)) - (32*b*(a*c^2 + b*d^2)*(2*b*d + a*c*exp(x) + 4*b*c*exp(x)))/(c^3*d^3))*((c + d)*(c - d))^(1/2)*(a*c^2 + b*
d^2))/(c^4 - c^2*d^2) - (2*b)/(c*(exp(2*x) + 1)) - (log(- (32*b*(a*c^2 + b*d^2)*(2*b*d + a*c*exp(x) + 4*b*c*ex
p(x)))/(c^3*d^3) - (((c + d)*(c - d))^(1/2)*((32*(a^2*c^4 + 2*b^2*d^4 - 2*b^2*c^2*d^2 + 3*b^2*c*d^3*exp(x) - 4
*b^2*c^3*d*exp(x) + 2*a*b*c^2*d^2))/(c^2*d^4) + (((c + d)*(c - d))^(1/2)*(a*c^2 + b*d^2)*((32*c*(2*b*d^3 + 4*a
*c^3*exp(x) + 2*a*c^2*d - a*c*d^2*exp(x) + 3*b*c*d^2*exp(x)))/d^5 - (32*((c + d)*(c - d))^(1/2)*(a*c^2 + b*d^2
)*(3*c^2*d - 2*d^3 + 4*c^3*exp(x) - 3*c*d^2*exp(x)))/(d^5*(c^2 - d^2))))/(c^2*(c^2 - d^2)))*(a*c^2 + b*d^2))/(
c^2*(c^2 - d^2)))*((c + d)*(c - d))^(1/2)*(a*c^2 + b*d^2))/(c^4 - c^2*d^2) + (b*d*log(exp(x) - 1i)*1i)/c^2 - (
b*d*log(exp(x) + 1i)*1i)/c^2

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________