Integrand size = 19, antiderivative size = 69 \[ \int \frac {\text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {a \arctan (\sinh (c+d x))}{\left (a^2+b^2\right ) d}-\frac {b \log (\cosh (c+d x))}{\left (a^2+b^2\right ) d}+\frac {b \log (a+b \sinh (c+d x))}{\left (a^2+b^2\right ) d} \]
a*arctan(sinh(d*x+c))/(a^2+b^2)/d-b*ln(cosh(d*x+c))/(a^2+b^2)/d+b*ln(a+b*s inh(d*x+c))/(a^2+b^2)/d
Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.65 \[ \int \frac {\text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {b \left (\left (-a+\sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}-b \sinh (c+d x)\right )-2 \sqrt {-b^2} \log (a+b \sinh (c+d x))+\left (a+\sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}+b \sinh (c+d x)\right )\right )}{2 \sqrt {-b^2} \left (a^2+b^2\right ) d} \]
-1/2*(b*((-a + Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Sinh[c + d*x]] - 2*Sqrt[-b^2 ]*Log[a + b*Sinh[c + d*x]] + (a + Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Sinh[c + d*x]]))/(Sqrt[-b^2]*(a^2 + b^2)*d)
Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, 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.
\(\displaystyle \int \frac {\text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (i c+i d x) (a-i b \sin (i c+i d x))}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle -\frac {b \int -\frac {1}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b \int \frac {1}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{d}\) |
\(\Big \downarrow \) 479 |
\(\displaystyle -\frac {b \left (-\frac {\int \frac {a-b \sinh (c+d x)}{\sinh ^2(c+d x) b^2+b^2}d(b \sinh (c+d x))}{a^2+b^2}-\frac {\log (a+b \sinh (c+d x))}{a^2+b^2}\right )}{d}\) |
\(\Big \downarrow \) 452 |
\(\displaystyle -\frac {b \left (-\frac {a \int \frac {1}{\sinh ^2(c+d x) b^2+b^2}d(b \sinh (c+d x))-\int \frac {b \sinh (c+d x)}{\sinh ^2(c+d x) b^2+b^2}d(b \sinh (c+d x))}{a^2+b^2}-\frac {\log (a+b \sinh (c+d x))}{a^2+b^2}\right )}{d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {b \left (-\frac {\frac {a \arctan (\sinh (c+d x))}{b}-\int \frac {b \sinh (c+d x)}{\sinh ^2(c+d x) b^2+b^2}d(b \sinh (c+d x))}{a^2+b^2}-\frac {\log (a+b \sinh (c+d x))}{a^2+b^2}\right )}{d}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle -\frac {b \left (-\frac {\frac {a \arctan (\sinh (c+d x))}{b}-\frac {1}{2} \log \left (b^2 \sinh ^2(c+d x)+b^2\right )}{a^2+b^2}-\frac {\log (a+b \sinh (c+d x))}{a^2+b^2}\right )}{d}\) |
-((b*(-(Log[a + b*Sinh[c + d*x]]/(a^2 + b^2)) - ((a*ArcTan[Sinh[c + d*x]]) /b - Log[b^2 + b^2*Sinh[c + d*x]^2]/2)/(a^2 + b^2)))/d)
3.4.7.3.1 Defintions of rubi rules used
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 = 2.57 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.28
method | result | size |
derivativedivides | \(\frac {\frac {-b \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+2 a \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}+b^{2}}+\frac {b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{a^{2}+b^{2}}}{d}\) | \(88\) |
default | \(\frac {\frac {-b \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+2 a \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}+b^{2}}+\frac {b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{a^{2}+b^{2}}}{d}\) | \(88\) |
risch | \(\frac {2 b \,d^{2} x}{a^{2} d^{2}+b^{2} d^{2}}+\frac {2 b d c}{a^{2} d^{2}+b^{2} d^{2}}-\frac {2 b x}{a^{2}+b^{2}}-\frac {2 b c}{d \left (a^{2}+b^{2}\right )}+\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) a}{\left (a^{2}+b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{d x +c}+i\right ) b}{\left (a^{2}+b^{2}\right ) d}-\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) a}{\left (a^{2}+b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{d x +c}-i\right ) b}{\left (a^{2}+b^{2}\right ) d}+\frac {b \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{d \left (a^{2}+b^{2}\right )}\) | \(217\) |
1/d*(2/(a^2+b^2)*(-1/2*b*ln(1+tanh(1/2*d*x+1/2*c)^2)+a*arctan(tanh(1/2*d*x +1/2*c)))+b/(a^2+b^2)*ln(tanh(1/2*d*x+1/2*c)^2*a-2*b*tanh(1/2*d*x+1/2*c)-a ))
Time = 0.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.33 \[ \int \frac {\text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {2 \, a \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + b \log \left (\frac {2 \, {\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - b \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} \]
(2*a*arctan(cosh(d*x + c) + sinh(d*x + c)) + b*log(2*(b*sinh(d*x + c) + a) /(cosh(d*x + c) - sinh(d*x + c))) - b*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))))/((a^2 + b^2)*d)
\[ \int \frac {\text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\operatorname {sech}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.38 \[ \int \frac {\text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {b \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac {b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} \]
-2*a*arctan(e^(-d*x - c))/((a^2 + b^2)*d) + b*log(-2*a*e^(-d*x - c) + b*e^ (-2*d*x - 2*c) - b)/((a^2 + b^2)*d) - b*log(e^(-2*d*x - 2*c) + 1)/((a^2 + b^2)*d)
Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.75 \[ \int \frac {\text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {2 \, b^{2} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{2} b + b^{3}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} a}{a^{2} + b^{2}} - \frac {b \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{a^{2} + b^{2}}}{2 \, d} \]
1/2*(2*b^2*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/(a^2*b + b^3) + (pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*a/(a^2 + b^2) - b* log((e^(d*x + c) - e^(-d*x - c))^2 + 4)/(a^2 + b^2))/d
Time = 2.00 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.87 \[ \int \frac {\text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b\,\ln \left (2\,a^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-4\,b^3-a^2\,b+4\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+a^2\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+8\,a\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{d\,a^2+d\,b^2}-\frac {\ln \left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}{b\,d+a\,d\,1{}\mathrm {i}}-\frac {\ln \left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a\,d+b\,d\,1{}\mathrm {i}} \]