Integrand size = 25, antiderivative size = 90 \[ \int \frac {\text {csch}(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {b \arctan (\sinh (c+d x))}{\left (a^2+b^2\right ) d}-\frac {a \log (\cosh (c+d x))}{\left (a^2+b^2\right ) d}+\frac {\log (\sinh (c+d x))}{a d}-\frac {b^2 \log (a+b \sinh (c+d x))}{a \left (a^2+b^2\right ) d} \]
-b*arctan(sinh(d*x+c))/(a^2+b^2)/d-a*ln(cosh(d*x+c))/(a^2+b^2)/d+ln(sinh(d *x+c))/a/d-b^2*ln(a+b*sinh(d*x+c))/a/(a^2+b^2)/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {\text {csch}(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {\log (i-\sinh (c+d x))}{a+i b}-\frac {2 \log (\sinh (c+d x))}{a}+\frac {\log (i+\sinh (c+d x))}{a-i b}+\frac {2 b^2 \log (a+b \sinh (c+d x))}{a \left (a^2+b^2\right )}}{2 d} \]
-1/2*(Log[I - Sinh[c + d*x]]/(a + I*b) - (2*Log[Sinh[c + d*x]])/a + Log[I + Sinh[c + d*x]]/(a - I*b) + (2*b^2*Log[a + b*Sinh[c + d*x]])/(a*(a^2 + b^ 2)))/d
Time = 0.36 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 26, 3316, 26, 27, 615, 2009}
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.
\(\displaystyle \int \frac {\text {csch}(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i}{\sin (i c+i d x) \cos (i c+i d x) (a-i b \sin (i c+i d x))}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {1}{\cos (i c+i d x) \sin (i c+i d x) (a-i b \sin (i c+i d x))}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle -\frac {i b \int \frac {i \text {csch}(c+d x)}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{d}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {b \int \frac {\text {csch}(c+d x)}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b^2 \int \frac {\text {csch}(c+d x)}{b (a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{d}\) |
\(\Big \downarrow \) 615 |
\(\displaystyle \frac {b^2 \int \left (\frac {\text {csch}(c+d x)}{a b^3}-\frac {1}{a \left (a^2+b^2\right ) (a+b \sinh (c+d x))}+\frac {-b^2-a \sinh (c+d x) b}{b^2 \left (a^2+b^2\right ) \left (\sinh ^2(c+d x) b^2+b^2\right )}\right )d(b \sinh (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^2 \left (-\frac {\arctan (\sinh (c+d x))}{b \left (a^2+b^2\right )}-\frac {a \log \left (b^2 \sinh ^2(c+d x)+b^2\right )}{2 b^2 \left (a^2+b^2\right )}-\frac {\log (a+b \sinh (c+d x))}{a \left (a^2+b^2\right )}+\frac {\log (b \sinh (c+d x))}{a b^2}\right )}{d}\) |
(b^2*(-(ArcTan[Sinh[c + d*x]]/(b*(a^2 + b^2))) + Log[b*Sinh[c + d*x]]/(a*b ^2) - Log[a + b*Sinh[c + d*x]]/(a*(a^2 + b^2)) - (a*Log[b^2 + b^2*Sinh[c + d*x]^2])/(2*b^2*(a^2 + b^2))))/d
3.5.38.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 1.71 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.20
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {-a \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-2 b \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}+b^{2}}-\frac {b^{2} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{\left (a^{2}+b^{2}\right ) a}}{d}\) | \(108\) |
default | \(\frac {\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {-a \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-2 b \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}+b^{2}}-\frac {b^{2} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{\left (a^{2}+b^{2}\right ) a}}{d}\) | \(108\) |
risch | \(\frac {2 a \,d^{2} x}{a^{2} d^{2}+b^{2} d^{2}}+\frac {2 a d c}{a^{2} d^{2}+b^{2} d^{2}}-\frac {2 x}{a}-\frac {2 c}{d a}+\frac {2 b^{2} x}{a \left (a^{2}+b^{2}\right )}+\frac {2 b^{2} c}{a d \left (a^{2}+b^{2}\right )}+\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) b}{\left (a^{2}+b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{d x +c}-i\right ) a}{\left (a^{2}+b^{2}\right ) d}-\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) b}{\left (a^{2}+b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{d x +c}+i\right ) a}{\left (a^{2}+b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d a}-\frac {b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{a d \left (a^{2}+b^{2}\right )}\) | \(267\) |
1/d*(1/a*ln(tanh(1/2*d*x+1/2*c))+1/(a^2+b^2)*(-a*ln(1+tanh(1/2*d*x+1/2*c)^ 2)-2*b*arctan(tanh(1/2*d*x+1/2*c)))-b^2/(a^2+b^2)/a*ln(tanh(1/2*d*x+1/2*c) ^2*a-2*b*tanh(1/2*d*x+1/2*c)-a))
Time = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.49 \[ \int \frac {\text {csch}(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {2 \, a b \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + b^{2} \log \left (\frac {2 \, {\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + a^{2} \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - {\left (a^{2} + b^{2}\right )} \log \left (\frac {2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{{\left (a^{3} + a b^{2}\right )} d} \]
-(2*a*b*arctan(cosh(d*x + c) + sinh(d*x + c)) + b^2*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) + a^2*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) - (a^2 + b^2)*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))))/((a^3 + a*b^2)*d)
\[ \int \frac {\text {csch}(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\operatorname {csch}{\left (c + d x \right )} \operatorname {sech}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.53 \[ \int \frac {\text {csch}(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {b^{2} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{3} + a b^{2}\right )} d} + \frac {2 \, b \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac {a \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} \]
-b^2*log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/((a^3 + a*b^2)*d) + 2 *b*arctan(e^(-d*x - c))/((a^2 + b^2)*d) - a*log(e^(-2*d*x - 2*c) + 1)/((a^ 2 + b^2)*d) + log(e^(-d*x - c) + 1)/(a*d) + log(e^(-d*x - c) - 1)/(a*d)
Time = 0.29 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.63 \[ \int \frac {\text {csch}(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {2 \, b^{3} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{3} b + a b^{3}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} b}{a^{2} + b^{2}} + \frac {a \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{a^{2} + b^{2}} - \frac {2 \, \log \left ({\left | e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} \right |}\right )}{a}}{2 \, d} \]
-1/2*(2*b^3*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/(a^3*b + a*b^3) + (pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*b/(a^2 + b^2) + a*log((e^(d*x + c) - e^(-d*x - c))^2 + 4)/(a^2 + b^2) - 2*log(abs(e^(d*x + c) - e^(-d*x - c)))/a)/d
Timed out. \[ \int \frac {\text {csch}(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {1}{\mathrm {cosh}\left (c+d\,x\right )\,\mathrm {sinh}\left (c+d\,x\right )\,\left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )} \,d x \]