Integrand size = 11, antiderivative size = 48 \[ \int \frac {\text {sech}(x)}{a+b \sinh (x)} \, dx=\frac {a \arctan (\sinh (x))}{a^2+b^2}-\frac {b \log (\cosh (x))}{a^2+b^2}+\frac {b \log (a+b \sinh (x))}{a^2+b^2} \] Output:
a*arctan(sinh(x))/(a^2+b^2)-b*ln(cosh(x))/(a^2+b^2)+b*ln(a+b*sinh(x))/(a^2 +b^2)
Leaf count is larger than twice the leaf count of optimal. \(99\) vs. \(2(48)=96\).
Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.06 \[ \int \frac {\text {sech}(x)}{a+b \sinh (x)} \, dx=-\frac {b \left (\left (-a+\sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}-b \sinh (x)\right )-2 \sqrt {-b^2} \log (a+b \sinh (x))+\left (a+\sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}+b \sinh (x)\right )\right )}{2 \sqrt {-b^2} \left (a^2+b^2\right )} \] Input:
Integrate[Sech[x]/(a + b*Sinh[x]),x]
Output:
-1/2*(b*((-a + Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Sinh[x]] - 2*Sqrt[-b^2]*Log[ a + b*Sinh[x]] + (a + Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Sinh[x]]))/(Sqrt[-b^2 ]*(a^2 + b^2))
Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.23, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {3042, 3147, 25, 479, 452, 216, 240}
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.
\(\displaystyle \int \frac {\text {sech}(x)}{a+b \sinh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (i x) (a-i b \sin (i x))}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle -b \int -\frac {1}{(a+b \sinh (x)) \left (\sinh ^2(x) b^2+b^2\right )}d(b \sinh (x))\) |
\(\Big \downarrow \) 25 |
\(\displaystyle b \int \frac {1}{(a+b \sinh (x)) \left (\sinh ^2(x) b^2+b^2\right )}d(b \sinh (x))\) |
\(\Big \downarrow \) 479 |
\(\displaystyle -b \left (-\frac {\int \frac {a-b \sinh (x)}{\sinh ^2(x) b^2+b^2}d(b \sinh (x))}{a^2+b^2}-\frac {\log (a+b \sinh (x))}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 452 |
\(\displaystyle -b \left (-\frac {a \int \frac {1}{\sinh ^2(x) b^2+b^2}d(b \sinh (x))-\int \frac {b \sinh (x)}{\sinh ^2(x) b^2+b^2}d(b \sinh (x))}{a^2+b^2}-\frac {\log (a+b \sinh (x))}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -b \left (-\frac {\frac {a \arctan (\sinh (x))}{b}-\int \frac {b \sinh (x)}{\sinh ^2(x) b^2+b^2}d(b \sinh (x))}{a^2+b^2}-\frac {\log (a+b \sinh (x))}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 240 |
\(\displaystyle -b \left (-\frac {\frac {a \arctan (\sinh (x))}{b}-\frac {1}{2} \log \left (b^2 \sinh ^2(x)+b^2\right )}{a^2+b^2}-\frac {\log (a+b \sinh (x))}{a^2+b^2}\right )\) |
Input:
Int[Sech[x]/(a + b*Sinh[x]),x]
Output:
-(b*(-(Log[a + b*Sinh[x]]/(a^2 + b^2)) - ((a*ArcTan[Sinh[x]])/b - Log[b^2 + b^2*Sinh[x]^2]/2)/(a^2 + b^2)))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c Int[1/ (a + b*x^2), x], x] + Simp[d Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c^2 + a*d^2, 0]
Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)), x_Symbol] :> Simp[d*(Log [RemoveContent[c + d*x, x]]/(b*c^2 + a*d^2)), x] + Simp[b/(b*c^2 + a*d^2) Int[(c - d*x)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 1.96 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.33
method | result | size |
default | \(\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a \right )}{a^{2}+b^{2}}+\frac {-b \ln \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )+2 a \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a^{2}+b^{2}}\) | \(64\) |
risch | \(\frac {i \ln \left ({\mathrm e}^{x}+i\right ) a}{a^{2}+b^{2}}-\frac {\ln \left ({\mathrm e}^{x}+i\right ) b}{a^{2}+b^{2}}-\frac {i \ln \left ({\mathrm e}^{x}-i\right ) a}{a^{2}+b^{2}}-\frac {\ln \left ({\mathrm e}^{x}-i\right ) b}{a^{2}+b^{2}}+\frac {b \ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}-1\right )}{a^{2}+b^{2}}\) | \(102\) |
Input:
int(sech(x)/(a+b*sinh(x)),x,method=_RETURNVERBOSE)
Output:
b/(a^2+b^2)*ln(tanh(1/2*x)^2*a-2*b*tanh(1/2*x)-a)+2/(a^2+b^2)*(-1/2*b*ln(1 +tanh(1/2*x)^2)+a*arctan(tanh(1/2*x)))
Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int \frac {\text {sech}(x)}{a+b \sinh (x)} \, dx=\frac {2 \, a \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + b \log \left (\frac {2 \, {\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - b \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} + b^{2}} \] Input:
integrate(sech(x)/(a+b*sinh(x)),x, algorithm="fricas")
Output:
(2*a*arctan(cosh(x) + sinh(x)) + b*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x ))) - b*log(2*cosh(x)/(cosh(x) - sinh(x))))/(a^2 + b^2)
\[ \int \frac {\text {sech}(x)}{a+b \sinh (x)} \, dx=\int \frac {\operatorname {sech}{\left (x \right )}}{a + b \sinh {\left (x \right )}}\, dx \] Input:
integrate(sech(x)/(a+b*sinh(x)),x)
Output:
Integral(sech(x)/(a + b*sinh(x)), x)
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.38 \[ \int \frac {\text {sech}(x)}{a+b \sinh (x)} \, dx=-\frac {2 \, a \arctan \left (e^{\left (-x\right )}\right )}{a^{2} + b^{2}} + \frac {b \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{2} + b^{2}} - \frac {b \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{2} + b^{2}} \] Input:
integrate(sech(x)/(a+b*sinh(x)),x, algorithm="maxima")
Output:
-2*a*arctan(e^(-x))/(a^2 + b^2) + b*log(-2*a*e^(-x) + b*e^(-2*x) - b)/(a^2 + b^2) - b*log(e^(-2*x) + 1)/(a^2 + b^2)
Time = 0.14 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.85 \[ \int \frac {\text {sech}(x)}{a+b \sinh (x)} \, dx=\frac {b^{2} \log \left ({\left | -b {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{2} b + b^{3}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} a}{2 \, {\left (a^{2} + b^{2}\right )}} - \frac {b \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \, {\left (a^{2} + b^{2}\right )}} \] Input:
integrate(sech(x)/(a+b*sinh(x)),x, algorithm="giac")
Output:
b^2*log(abs(-b*(e^(-x) - e^x) + 2*a))/(a^2*b + b^3) + 1/2*(pi + 2*arctan(1 /2*(e^(2*x) - 1)*e^(-x)))*a/(a^2 + b^2) - 1/2*b*log((e^(-x) - e^x)^2 + 4)/ (a^2 + b^2)
Time = 2.24 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.94 \[ \int \frac {\text {sech}(x)}{a+b \sinh (x)} \, dx=\frac {b\,\ln \left (4\,b^3\,{\mathrm {e}}^{2\,x}-a^2\,b-4\,b^3+2\,a^3\,{\mathrm {e}}^x+8\,a\,b^2\,{\mathrm {e}}^x+a^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a^2+b^2}-\frac {\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}{b+a\,1{}\mathrm {i}}-\frac {\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a+b\,1{}\mathrm {i}} \] Input:
int(1/(cosh(x)*(a + b*sinh(x))),x)
Output:
(b*log(4*b^3*exp(2*x) - a^2*b - 4*b^3 + 2*a^3*exp(x) + 8*a*b^2*exp(x) + a^ 2*b*exp(2*x)))/(a^2 + b^2) - log(exp(x) + 1i)/(a*1i + b) - (log(exp(x)*1i + 1)*1i)/(a + b*1i)
Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.02 \[ \int \frac {\text {sech}(x)}{a+b \sinh (x)} \, dx=\frac {2 \mathit {atan} \left (e^{x}\right ) a -\mathrm {log}\left (e^{2 x}+1\right ) b +\mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) b}{a^{2}+b^{2}} \] Input:
int(sech(x)/(a+b*sinh(x)),x)
Output:
(2*atan(e**x)*a - log(e**(2*x) + 1)*b + log(e**(2*x)*b + 2*e**x*a - b)*b)/ (a**2 + b**2)