Integrand size = 13, antiderivative size = 64 \[ \int \frac {\text {sech}^2(x)}{a+b \cosh (x)} \, dx=-\frac {b \arctan (\sinh (x))}{a^2}+\frac {2 b^2 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b}}+\frac {\tanh (x)}{a} \] Output:
-b*arctan(sinh(x))/a^2+2*b^2*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/ a^2/(a-b)^(1/2)/(a+b)^(1/2)+tanh(x)/a
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98 \[ \int \frac {\text {sech}^2(x)}{a+b \cosh (x)} \, dx=\frac {-2 b \arctan \left (\tanh \left (\frac {x}{2}\right )\right )-\frac {2 b^2 \arctan \left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+a \tanh (x)}{a^2} \] Input:
Integrate[Sech[x]^2/(a + b*Cosh[x]),x]
Output:
(-2*b*ArcTan[Tanh[x/2]] - (2*b^2*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^ 2]])/Sqrt[-a^2 + b^2] + a*Tanh[x])/a^2
Time = 0.45 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {3042, 3281, 25, 27, 3042, 3226, 3042, 3138, 221, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\text {sech}^2(x)}{a+b \cosh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin \left (\frac {\pi }{2}+i x\right )^2 \left (a+b \sin \left (\frac {\pi }{2}+i x\right )\right )}dx\) |
\(\Big \downarrow \) 3281 |
\(\displaystyle \frac {\int -\frac {b \text {sech}(x)}{a+b \cosh (x)}dx}{a}+\frac {\tanh (x)}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tanh (x)}{a}-\frac {\int \frac {b \text {sech}(x)}{a+b \cosh (x)}dx}{a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\tanh (x)}{a}-\frac {b \int \frac {\text {sech}(x)}{a+b \cosh (x)}dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tanh (x)}{a}-\frac {b \int \frac {1}{\sin \left (i x+\frac {\pi }{2}\right ) \left (a+b \sin \left (i x+\frac {\pi }{2}\right )\right )}dx}{a}\) |
\(\Big \downarrow \) 3226 |
\(\displaystyle \frac {\tanh (x)}{a}-\frac {b \left (\frac {\int \text {sech}(x)dx}{a}-\frac {b \int \frac {1}{a+b \cosh (x)}dx}{a}\right )}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tanh (x)}{a}-\frac {b \left (\frac {\int \csc \left (i x+\frac {\pi }{2}\right )dx}{a}-\frac {b \int \frac {1}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\tanh (x)}{a}-\frac {b \left (-\frac {2 b \int \frac {1}{-\left ((a-b) \tanh ^2\left (\frac {x}{2}\right )\right )+a+b}d\tanh \left (\frac {x}{2}\right )}{a}+\frac {\int \csc \left (i x+\frac {\pi }{2}\right )dx}{a}\right )}{a}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\tanh (x)}{a}-\frac {b \left (-\frac {2 b \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}}+\frac {\int \csc \left (i x+\frac {\pi }{2}\right )dx}{a}\right )}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\tanh (x)}{a}-\frac {b \left (\frac {\arctan (\sinh (x))}{a}-\frac {2 b \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}}\right )}{a}\) |
Input:
Int[Sech[x]^2/(a + b*Cosh[x]),x]
Output:
-((b*(ArcTan[Sinh[x]]/a - (2*b*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b] ])/(a*Sqrt[a - b]*Sqrt[a + b])))/a) + Tanh[x]/a
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[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]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[ {a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-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] + Simp[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)*Si n[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])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.76 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.14
method | result | size |
default | \(\frac {2 b^{2} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \left (-\frac {a \tanh \left (\frac {x}{2}\right )}{\tanh \left (\frac {x}{2}\right )^{2}+1}+b \arctan \left (\tanh \left (\frac {x}{2}\right )\right )\right )}{a^{2}}\) | \(73\) |
risch | \(-\frac {2}{a \left ({\mathrm e}^{2 x}+1\right )}+\frac {b^{2} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, a^{2}}-\frac {b^{2} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, a^{2}}+\frac {i b \ln \left ({\mathrm e}^{x}-i\right )}{a^{2}}-\frac {i b \ln \left ({\mathrm e}^{x}+i\right )}{a^{2}}\) | \(160\) |
Input:
int(sech(x)^2/(a+b*cosh(x)),x,method=_RETURNVERBOSE)
Output:
2*b^2/a^2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2 ))-2/a^2*(-a*tanh(1/2*x)/(tanh(1/2*x)^2+1)+b*arctan(tanh(1/2*x)))
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (54) = 108\).
Time = 0.12 (sec) , antiderivative size = 515, normalized size of antiderivative = 8.05 \[ \int \frac {\text {sech}^2(x)}{a+b \cosh (x)} \, dx =\text {Too large to display} \] Input:
integrate(sech(x)^2/(a+b*cosh(x)),x, algorithm="fricas")
Output:
[-(2*a^3 - 2*a*b^2 - (b^2*cosh(x)^2 + 2*b^2*cosh(x)*sinh(x) + b^2*sinh(x)^ 2 + b^2)*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*cos h(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cos h(x) + a)*sinh(x) + b)) + 2*(a^2*b - b^3 + (a^2*b - b^3)*cosh(x)^2 + 2*(a^ 2*b - b^3)*cosh(x)*sinh(x) + (a^2*b - b^3)*sinh(x)^2)*arctan(cosh(x) + sin h(x)))/(a^4 - a^2*b^2 + (a^4 - a^2*b^2)*cosh(x)^2 + 2*(a^4 - a^2*b^2)*cosh (x)*sinh(x) + (a^4 - a^2*b^2)*sinh(x)^2), -2*(a^3 - a*b^2 + (b^2*cosh(x)^2 + 2*b^2*cosh(x)*sinh(x) + b^2*sinh(x)^2 + b^2)*sqrt(-a^2 + b^2)*arctan(-s qrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + (a^2*b - b^3 + (a^2*b - b^3)*cosh(x)^2 + 2*(a^2*b - b^3)*cosh(x)*sinh(x) + (a^2*b - b^3)* sinh(x)^2)*arctan(cosh(x) + sinh(x)))/(a^4 - a^2*b^2 + (a^4 - a^2*b^2)*cos h(x)^2 + 2*(a^4 - a^2*b^2)*cosh(x)*sinh(x) + (a^4 - a^2*b^2)*sinh(x)^2)]
\[ \int \frac {\text {sech}^2(x)}{a+b \cosh (x)} \, dx=\int \frac {\operatorname {sech}^{2}{\left (x \right )}}{a + b \cosh {\left (x \right )}}\, dx \] Input:
integrate(sech(x)**2/(a+b*cosh(x)),x)
Output:
Integral(sech(x)**2/(a + b*cosh(x)), x)
Exception generated. \[ \int \frac {\text {sech}^2(x)}{a+b \cosh (x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sech(x)^2/(a+b*cosh(x)),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.95 \[ \int \frac {\text {sech}^2(x)}{a+b \cosh (x)} \, dx=\frac {2 \, b^{2} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} a^{2}} - \frac {2 \, b \arctan \left (e^{x}\right )}{a^{2}} - \frac {2}{a {\left (e^{\left (2 \, x\right )} + 1\right )}} \] Input:
integrate(sech(x)^2/(a+b*cosh(x)),x, algorithm="giac")
Output:
2*b^2*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/(sqrt(-a^2 + b^2)*a^2) - 2*b*ar ctan(e^x)/a^2 - 2/(a*(e^(2*x) + 1))
Time = 4.45 (sec) , antiderivative size = 294, normalized size of antiderivative = 4.59 \[ \int \frac {\text {sech}^2(x)}{a+b \cosh (x)} \, dx=\frac {b^2\,\ln \left (64\,a\,b^3-64\,a^3\,b+32\,b^3\,\sqrt {a^2-b^2}-128\,a^4\,{\mathrm {e}}^x-32\,b^4\,{\mathrm {e}}^x-64\,a^2\,b\,\sqrt {a^2-b^2}-128\,a^3\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}+160\,a^2\,b^2\,{\mathrm {e}}^x+96\,a\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a^2\,\sqrt {a^2-b^2}}+\frac {b\,\left (\ln \left (32\,{\mathrm {e}}^x-32{}\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left (32\,{\mathrm {e}}^x+32{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{a^2}-\frac {b^2\,\ln \left (64\,a^3\,b-64\,a\,b^3+32\,b^3\,\sqrt {a^2-b^2}+128\,a^4\,{\mathrm {e}}^x+32\,b^4\,{\mathrm {e}}^x-64\,a^2\,b\,\sqrt {a^2-b^2}-128\,a^3\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-160\,a^2\,b^2\,{\mathrm {e}}^x+96\,a\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a^2\,\sqrt {a^2-b^2}}-\frac {2}{a+a\,{\mathrm {e}}^{2\,x}} \] Input:
int(1/(cosh(x)^2*(a + b*cosh(x))),x)
Output:
(b*(log(32*exp(x) - 32i)*1i - log(32*exp(x) + 32i)*1i))/a^2 - 2/(a + a*exp (2*x)) - (b^2*log(64*a^3*b - 64*a*b^3 + 32*b^3*(a^2 - b^2)^(1/2) + 128*a^4 *exp(x) + 32*b^4*exp(x) - 64*a^2*b*(a^2 - b^2)^(1/2) - 128*a^3*exp(x)*(a^2 - b^2)^(1/2) - 160*a^2*b^2*exp(x) + 96*a*b^2*exp(x)*(a^2 - b^2)^(1/2)))/( a^2*(a^2 - b^2)^(1/2)) + (b^2*log(64*a*b^3 - 64*a^3*b + 32*b^3*(a^2 - b^2) ^(1/2) - 128*a^4*exp(x) - 32*b^4*exp(x) - 64*a^2*b*(a^2 - b^2)^(1/2) - 128 *a^3*exp(x)*(a^2 - b^2)^(1/2) + 160*a^2*b^2*exp(x) + 96*a*b^2*exp(x)*(a^2 - b^2)^(1/2)))/(a^2*(a^2 - b^2)^(1/2))
Time = 0.22 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.80 \[ \int \frac {\text {sech}^2(x)}{a+b \cosh (x)} \, dx=\frac {-2 e^{2 x} \mathit {atan} \left (e^{x}\right ) a^{2} b +2 e^{2 x} \mathit {atan} \left (e^{x}\right ) b^{3}-2 \mathit {atan} \left (e^{x}\right ) a^{2} b +2 \mathit {atan} \left (e^{x}\right ) b^{3}-2 e^{2 x} \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b +a}{\sqrt {-a^{2}+b^{2}}}\right ) b^{2}-2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b +a}{\sqrt {-a^{2}+b^{2}}}\right ) b^{2}+2 e^{2 x} a^{3}-2 e^{2 x} a \,b^{2}}{a^{2} \left (e^{2 x} a^{2}-e^{2 x} b^{2}+a^{2}-b^{2}\right )} \] Input:
int(sech(x)^2/(a+b*cosh(x)),x)
Output:
(2*( - e**(2*x)*atan(e**x)*a**2*b + e**(2*x)*atan(e**x)*b**3 - atan(e**x)* a**2*b + atan(e**x)*b**3 - e**(2*x)*sqrt( - a**2 + b**2)*atan((e**x*b + a) /sqrt( - a**2 + b**2))*b**2 - sqrt( - a**2 + b**2)*atan((e**x*b + a)/sqrt( - a**2 + b**2))*b**2 + e**(2*x)*a**3 - e**(2*x)*a*b**2))/(a**2*(e**(2*x)* a**2 - e**(2*x)*b**2 + a**2 - b**2))