Integrand size = 25, antiderivative size = 74 \[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a \arctan (\sinh (c+d x))}{\left (a^2+b^2\right ) d}+\frac {b \log (\cosh (c+d x))}{\left (a^2+b^2\right ) d}+\frac {a^2 \log (a+b \sinh (c+d x))}{b \left (a^2+b^2\right ) d} \]
-a*arctan(sinh(d*x+c))/(a^2+b^2)/d+b*ln(cosh(d*x+c))/(a^2+b^2)/d+a^2*ln(a+ b*sinh(d*x+c))/b/(a^2+b^2)/d
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.05 \[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b (i a+b) \log (i-\sinh (c+d x))+b (-i a+b) \log (i+\sinh (c+d x))+2 a^2 \log (a+b \sinh (c+d x))}{2 b \left (a^2+b^2\right ) d} \]
(b*(I*a + b)*Log[I - Sinh[c + d*x]] + b*((-I)*a + b)*Log[I + Sinh[c + d*x] ] + 2*a^2*Log[a + b*Sinh[c + d*x]])/(2*b*(a^2 + b^2)*d)
Time = 0.34 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 25, 3316, 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.
\(\displaystyle \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin (i c+i d x)^2}{\cos (i c+i d x) (a-i b \sin (i c+i d x))}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\sin (i c+i d x)^2}{\cos (i c+i d x) (a-i b \sin (i c+i d x))}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {b \int \frac {\sinh ^2(c+d x)}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {b^2 \sinh ^2(c+d x)}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{b d}\) |
\(\Big \downarrow \) 615 |
\(\displaystyle \frac {\int \left (\frac {a^2}{\left (a^2+b^2\right ) (a+b \sinh (c+d x))}-\frac {b^2 (a-b \sinh (c+d x))}{\left (a^2+b^2\right ) \left (\sinh ^2(c+d x) b^2+b^2\right )}\right )d(b \sinh (c+d x))}{b d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a b \arctan (\sinh (c+d x))}{a^2+b^2}+\frac {b^2 \log \left (b^2 \sinh ^2(c+d x)+b^2\right )}{2 \left (a^2+b^2\right )}+\frac {a^2 \log (a+b \sinh (c+d x))}{a^2+b^2}}{b d}\) |
(-((a*b*ArcTan[Sinh[c + d*x]])/(a^2 + b^2)) + (a^2*Log[a + b*Sinh[c + d*x] ])/(a^2 + b^2) + (b^2*Log[b^2 + b^2*Sinh[c + d*x]^2])/(2*(a^2 + b^2)))/(b* d)
3.4.80.3.1 Defintions of rubi rules used
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 = 1.10 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.78
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}+\frac {4 b \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-8 a \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2}+4 b^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}+\frac {a^{2} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{b \left (a^{2}+b^{2}\right )}}{d}\) | \(132\) |
default | \(\frac {-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}+\frac {4 b \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-8 a \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2}+4 b^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}+\frac {a^{2} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{b \left (a^{2}+b^{2}\right )}}{d}\) | \(132\) |
risch | \(\frac {x}{b}-\frac {2 b \,d^{2} x}{a^{2} d^{2}+b^{2} d^{2}}-\frac {2 b d c}{a^{2} d^{2}+b^{2} d^{2}}-\frac {2 a^{2} x}{b \left (a^{2}+b^{2}\right )}-\frac {2 a^{2} c}{b d \left (a^{2}+b^{2}\right )}+\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) a}{\left (a^{2}+b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{d x +c}-i\right ) b}{\left (a^{2}+b^{2}\right ) d}-\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) a}{\left (a^{2}+b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{d x +c}+i\right ) b}{\left (a^{2}+b^{2}\right ) d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{b d \left (a^{2}+b^{2}\right )}\) | \(235\) |
1/d*(-1/b*ln(tanh(1/2*d*x+1/2*c)-1)+8/(4*a^2+4*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)))-1/b*ln(tanh(1/2*d*x+1/2*c)+1)+ a^2/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 = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.50 \[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {{\left (a^{2} + b^{2}\right )} d x + 2 \, a b \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) - a^{2} \log \left (\frac {2 \, {\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - b^{2} \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{{\left (a^{2} b + b^{3}\right )} d} \]
-((a^2 + b^2)*d*x + 2*a*b*arctan(cosh(d*x + c) + sinh(d*x + c)) - a^2*log( 2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) - b^2*log(2*cosh( d*x + c)/(cosh(d*x + c) - sinh(d*x + c))))/((a^2*b + b^3)*d)
\[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\sinh {\left (c + d x \right )} \tanh {\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]
Time = 0.38 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.49 \[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {a^{2} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} b + b^{3}\right )} d} + \frac {2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {d x + c}{b d} \]
a^2*log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/((a^2*b + b^3)*d) + 2* a*arctan(e^(-d*x - c))/((a^2 + b^2)*d) + b*log(e^(-2*d*x - 2*c) + 1)/((a^2 + b^2)*d) + (d*x + c)/(b*d)
Time = 0.34 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.64 \[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {2 \, a^{2} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{2} b + b^{3}} - \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} a}{a^{2} + b^{2}} + \frac {b \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{a^{2} + b^{2}}}{2 \, d} \]
1/2*(2*a^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 = 2.09 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.35 \[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\ln \left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}{b\,d+a\,d\,1{}\mathrm {i}}-\frac {x}{b}+\frac {a^2\,\ln \left (a^2\,b^3-b^5-a^4\,b+2\,a^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+a^4\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-2\,a^3\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-a^2\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+2\,a\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{d\,a^2\,b+d\,b^3}+\frac {\ln \left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a\,d+b\,d\,1{}\mathrm {i}} \]
log(exp(c + d*x) + 1i)/(a*d*1i + b*d) - x/b + (log(exp(c + d*x)*1i + 1)*1i )/(a*d + b*d*1i) + (a^2*log(a^2*b^3 - b^5 - a^4*b + 2*a^5*exp(d*x)*exp(c) + b^5*exp(2*c)*exp(2*d*x) + a^4*b*exp(2*c)*exp(2*d*x) - 2*a^3*b^2*exp(d*x) *exp(c) - a^2*b^3*exp(2*c)*exp(2*d*x) + 2*a*b^4*exp(d*x)*exp(c)))/(b^3*d + a^2*b*d)