Integrand size = 11, antiderivative size = 48 \[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=\frac {b \arctan (\sinh (x))}{a^2+b^2}+\frac {a \log (\cosh (x))}{a^2+b^2}-\frac {a \log (a+b \sinh (x))}{a^2+b^2} \] Output:
b*arctan(sinh(x))/(a^2+b^2)+a*ln(cosh(x))/(a^2+b^2)-a*ln(a+b*sinh(x))/(a^2 +b^2)
Result contains complex when optimal does not.
Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.17 \[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=\frac {(a-i b) \log (i-\sinh (x))+(a+i b) \log (i+\sinh (x))-2 a \log (a+b \sinh (x))}{2 \left (a^2+b^2\right )} \] Input:
Integrate[Tanh[x]/(a + b*Sinh[x]),x]
Output:
((a - I*b)*Log[I - Sinh[x]] + (a + I*b)*Log[I + Sinh[x]] - 2*a*Log[a + b*S inh[x]])/(2*(a^2 + b^2))
Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.818, Rules used = {3042, 26, 3200, 25, 587, 16, 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.
\(\displaystyle \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \tan (i x)}{a-i b \sin (i x)}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\tan (i x)}{a-i b \sin (i x)}dx\) |
\(\Big \downarrow \) 3200 |
\(\displaystyle -\int -\frac {b \sinh (x)}{(a+b \sinh (x)) \left (\sinh ^2(x) b^2+b^2\right )}d(b \sinh (x))\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {b \sinh (x)}{\left (b^2 \sinh ^2(x)+b^2\right ) (a+b \sinh (x))}d(b \sinh (x))\) |
\(\Big \downarrow \) 587 |
\(\displaystyle \frac {\int \frac {b^2+a \sinh (x) b}{\sinh ^2(x) b^2+b^2}d(b \sinh (x))}{a^2+b^2}-\frac {a \int \frac {1}{a+b \sinh (x)}d(b \sinh (x))}{a^2+b^2}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\int \frac {b^2+a \sinh (x) b}{\sinh ^2(x) b^2+b^2}d(b \sinh (x))}{a^2+b^2}-\frac {a \log (a+b \sinh (x))}{a^2+b^2}\) |
\(\Big \downarrow \) 452 |
\(\displaystyle \frac {a \int \frac {b \sinh (x)}{\sinh ^2(x) b^2+b^2}d(b \sinh (x))+b^2 \int \frac {1}{\sinh ^2(x) b^2+b^2}d(b \sinh (x))}{a^2+b^2}-\frac {a \log (a+b \sinh (x))}{a^2+b^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {a \int \frac {b \sinh (x)}{\sinh ^2(x) b^2+b^2}d(b \sinh (x))+b \arctan (\sinh (x))}{a^2+b^2}-\frac {a \log (a+b \sinh (x))}{a^2+b^2}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {\frac {1}{2} a \log \left (b^2 \sinh ^2(x)+b^2\right )+b \arctan (\sinh (x))}{a^2+b^2}-\frac {a \log (a+b \sinh (x))}{a^2+b^2}\) |
Input:
Int[Tanh[x]/(a + b*Sinh[x]),x]
Output:
-((a*Log[a + b*Sinh[x]])/(a^2 + b^2)) + (b*ArcTan[Sinh[x]] + (a*Log[b^2 + b^2*Sinh[x]^2])/2)/(a^2 + b^2)
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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_) + (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[(x_.)/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)), x_Symbol] :> Simp[(- c)*(d/(b*c^2 + a*d^2)) Int[1/(c + d*x), x], x] + Simp[1/(b*c^2 + a*d^2) Int[(a*d + b*c*x)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c ^2 + a*d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[(x^p*(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] && NeQ[a^2 - b ^2, 0] && IntegerQ[(p + 1)/2]
Time = 0.33 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.52
method | result | size |
default | \(-\frac {2 a \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a \right )}{2 a^{2}+2 b^{2}}+\frac {2 a \ln \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )+4 b \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{2 a^{2}+2 b^{2}}\) | \(73\) |
risch | \(\frac {i \ln \left ({\mathrm e}^{x}+i\right ) b}{a^{2}+b^{2}}+\frac {\ln \left ({\mathrm e}^{x}+i\right ) a}{a^{2}+b^{2}}-\frac {i \ln \left ({\mathrm e}^{x}-i\right ) b}{a^{2}+b^{2}}+\frac {\ln \left ({\mathrm e}^{x}-i\right ) a}{a^{2}+b^{2}}-\frac {a \ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}-1\right )}{a^{2}+b^{2}}\) | \(101\) |
Input:
int(tanh(x)/(a+b*sinh(x)),x,method=_RETURNVERBOSE)
Output:
-2*a/(2*a^2+2*b^2)*ln(tanh(1/2*x)^2*a-2*b*tanh(1/2*x)-a)+4/(2*a^2+2*b^2)*( 1/2*a*ln(1+tanh(1/2*x)^2)+b*arctan(tanh(1/2*x)))
Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=\frac {2 \, b \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - a \log \left (\frac {2 \, {\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + a \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} + b^{2}} \] Input:
integrate(tanh(x)/(a+b*sinh(x)),x, algorithm="fricas")
Output:
(2*b*arctan(cosh(x) + sinh(x)) - a*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x ))) + a*log(2*cosh(x)/(cosh(x) - sinh(x))))/(a^2 + b^2)
\[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=\int \frac {\tanh {\left (x \right )}}{a + b \sinh {\left (x \right )}}\, dx \] Input:
integrate(tanh(x)/(a+b*sinh(x)),x)
Output:
Integral(tanh(x)/(a + b*sinh(x)), x)
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.38 \[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=-\frac {2 \, b \arctan \left (e^{\left (-x\right )}\right )}{a^{2} + b^{2}} - \frac {a \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{2} + b^{2}} + \frac {a \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{2} + b^{2}} \] Input:
integrate(tanh(x)/(a+b*sinh(x)),x, algorithm="maxima")
Output:
-2*b*arctan(e^(-x))/(a^2 + b^2) - a*log(-2*a*e^(-x) + b*e^(-2*x) - b)/(a^2 + b^2) + a*log(e^(-2*x) + 1)/(a^2 + b^2)
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.85 \[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=-\frac {a b \log \left ({\left | -b {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{2} b + b^{3}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} b}{2 \, {\left (a^{2} + b^{2}\right )}} + \frac {a \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \, {\left (a^{2} + b^{2}\right )}} \] Input:
integrate(tanh(x)/(a+b*sinh(x)),x, algorithm="giac")
Output:
-a*b*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)))*b/(a^2 + b^2) + 1/2*a*log((e^(-x) - e^x)^2 + 4) /(a^2 + b^2)
Time = 2.34 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.98 \[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=\frac {\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}{a-b\,1{}\mathrm {i}}-\frac {a\,\ln \left (b^3\,{\mathrm {e}}^{2\,x}-4\,a^2\,b-b^3+8\,a^3\,{\mathrm {e}}^x+2\,a\,b^2\,{\mathrm {e}}^x+4\,a^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a^2+b^2}+\frac {\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{-b+a\,1{}\mathrm {i}} \] Input:
int(tanh(x)/(a + b*sinh(x)),x)
Output:
(log(exp(x)*1i + 1)*1i)/(a*1i - b) + log(exp(x) + 1i)/(a - b*1i) - (a*log( b^3*exp(2*x) - 4*a^2*b - b^3 + 8*a^3*exp(x) + 2*a*b^2*exp(x) + 4*a^2*b*exp (2*x)))/(a^2 + b^2)
Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.02 \[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=\frac {2 \mathit {atan} \left (e^{x}\right ) b +\mathrm {log}\left (e^{2 x}+1\right ) a -\mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) a}{a^{2}+b^{2}} \] Input:
int(tanh(x)/(a+b*sinh(x)),x)
Output:
(2*atan(e**x)*b + log(e**(2*x) + 1)*a - log(e**(2*x)*b + 2*e**x*a - b)*a)/ (a**2 + b**2)