Integrand size = 11, antiderivative size = 53 \[ \int \frac {\text {csch}(x)}{a+b \text {sech}(x)} \, dx=\frac {\log (1-\cosh (x))}{2 (a+b)}-\frac {\log (1+\cosh (x))}{2 (a-b)}+\frac {b \log (b+a \cosh (x))}{a^2-b^2} \] Output:
ln(1-cosh(x))/(2*a+2*b)-ln(1+cosh(x))/(2*a-2*b)+b*ln(b+a*cosh(x))/(a^2-b^2 )
Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94 \[ \int \frac {\text {csch}(x)}{a+b \text {sech}(x)} \, dx=\frac {\log \left (\cosh \left (\frac {x}{2}\right )\right )}{-a+b}-\frac {b \log (b+a \cosh (x))}{-a^2+b^2}+\frac {\log \left (\sinh \left (\frac {x}{2}\right )\right )}{a+b} \] Input:
Integrate[Csch[x]/(a + b*Sech[x]),x]
Output:
Log[Cosh[x/2]]/(-a + b) - (b*Log[b + a*Cosh[x]])/(-a^2 + b^2) + Log[Sinh[x /2]]/(a + b)
Time = 0.38 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.11, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.182, Rules used = {3042, 26, 4359, 26, 25, 3042, 26, 3200, 587, 16, 452, 219, 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.
\(\displaystyle \int \frac {\text {csch}(x)}{a+b \text {sech}(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i}{\cos \left (-\frac {\pi }{2}+i x\right ) \left (a-b \csc \left (-\frac {\pi }{2}+i x\right )\right )}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {1}{\cos \left (i x-\frac {\pi }{2}\right ) \left (a-b \csc \left (i x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 4359 |
\(\displaystyle i \int \frac {i \coth (x)}{-b-a \cosh (x)}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -\int -\frac {\coth (x)}{b+a \cosh (x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\coth (x)}{a \cosh (x)+b}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \tan \left (-\frac {\pi }{2}+i x\right )}{b-a \sin \left (-\frac {\pi }{2}+i x\right )}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\tan \left (i x-\frac {\pi }{2}\right )}{b-a \sin \left (i x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3200 |
\(\displaystyle -\int \frac {a \cosh (x)}{(b+a \cosh (x)) \left (a^2-a^2 \cosh ^2(x)\right )}d(a \cosh (x))\) |
\(\Big \downarrow \) 587 |
\(\displaystyle \frac {b \int \frac {1}{b+a \cosh (x)}d(a \cosh (x))}{a^2-b^2}-\frac {\int \frac {a^2-a b \cosh (x)}{a^2-a^2 \cosh ^2(x)}d(a \cosh (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {b \log (a \cosh (x)+b)}{a^2-b^2}-\frac {\int \frac {a^2-a b \cosh (x)}{a^2-a^2 \cosh ^2(x)}d(a \cosh (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 452 |
\(\displaystyle \frac {b \log (a \cosh (x)+b)}{a^2-b^2}-\frac {a^2 \int \frac {1}{a^2-a^2 \cosh ^2(x)}d(a \cosh (x))-b \int \frac {a \cosh (x)}{a^2-a^2 \cosh ^2(x)}d(a \cosh (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {b \log (a \cosh (x)+b)}{a^2-b^2}-\frac {a \text {arctanh}(\cosh (x))-b \int \frac {a \cosh (x)}{a^2-a^2 \cosh ^2(x)}d(a \cosh (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {b \log (a \cosh (x)+b)}{a^2-b^2}-\frac {\frac {1}{2} b \log \left (a^2-a^2 \cosh ^2(x)\right )+a \text {arctanh}(\cosh (x))}{a^2-b^2}\) |
Input:
Int[Csch[x]/(a + b*Sech[x]),x]
Output:
(b*Log[b + a*Cosh[x]])/(a^2 - b^2) - (a*ArcTanh[Cosh[x]] + (b*Log[a^2 - a^ 2*Cosh[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]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[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]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m _.), x_Symbol] :> Int[Cot[e + f*x]^p*(b + a*Sin[e + f*x])^m, x] /; FreeQ[{a , b, e, f, p}, x] && IntegerQ[m] && EqQ[m, p]
Time = 0.24 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91
method | result | size |
default | \(\frac {b \ln \left (a \tanh \left (\frac {x}{2}\right )^{2}-b \tanh \left (\frac {x}{2}\right )^{2}+a +b \right )}{\left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a +b}\) | \(48\) |
risch | \(-\frac {x}{a +b}+\frac {x}{a -b}-\frac {2 b x}{a^{2}-b^{2}}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{a +b}-\frac {\ln \left (1+{\mathrm e}^{x}\right )}{a -b}+\frac {b \ln \left ({\mathrm e}^{2 x}+\frac {2 b \,{\mathrm e}^{x}}{a}+1\right )}{a^{2}-b^{2}}\) | \(87\) |
Input:
int(csch(x)/(a+b*sech(x)),x,method=_RETURNVERBOSE)
Output:
b/(a+b)/(a-b)*ln(a*tanh(1/2*x)^2-b*tanh(1/2*x)^2+a+b)+1/(a+b)*ln(tanh(1/2* x))
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.09 \[ \int \frac {\text {csch}(x)}{a+b \text {sech}(x)} \, dx=\frac {b \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - {\left (a + b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + {\left (a - b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{a^{2} - b^{2}} \] Input:
integrate(csch(x)/(a+b*sech(x)),x, algorithm="fricas")
Output:
(b*log(2*(a*cosh(x) + b)/(cosh(x) - sinh(x))) - (a + b)*log(cosh(x) + sinh (x) + 1) + (a - b)*log(cosh(x) + sinh(x) - 1))/(a^2 - b^2)
\[ \int \frac {\text {csch}(x)}{a+b \text {sech}(x)} \, dx=\int \frac {\operatorname {csch}{\left (x \right )}}{a + b \operatorname {sech}{\left (x \right )}}\, dx \] Input:
integrate(csch(x)/(a+b*sech(x)),x)
Output:
Integral(csch(x)/(a + b*sech(x)), x)
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.11 \[ \int \frac {\text {csch}(x)}{a+b \text {sech}(x)} \, dx=\frac {b \log \left (2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a\right )}{a^{2} - b^{2}} - \frac {\log \left (e^{\left (-x\right )} + 1\right )}{a - b} + \frac {\log \left (e^{\left (-x\right )} - 1\right )}{a + b} \] Input:
integrate(csch(x)/(a+b*sech(x)),x, algorithm="maxima")
Output:
b*log(2*b*e^(-x) + a*e^(-2*x) + a)/(a^2 - b^2) - log(e^(-x) + 1)/(a - b) + log(e^(-x) - 1)/(a + b)
Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.23 \[ \int \frac {\text {csch}(x)}{a+b \text {sech}(x)} \, dx=\frac {a b \log \left ({\left | a {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b \right |}\right )}{a^{3} - a b^{2}} - \frac {\log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{2 \, {\left (a - b\right )}} + \frac {\log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{2 \, {\left (a + b\right )}} \] Input:
integrate(csch(x)/(a+b*sech(x)),x, algorithm="giac")
Output:
a*b*log(abs(a*(e^(-x) + e^x) + 2*b))/(a^3 - a*b^2) - 1/2*log(e^(-x) + e^x + 2)/(a - b) + 1/2*log(e^(-x) + e^x - 2)/(a + b)
Time = 2.94 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.79 \[ \int \frac {\text {csch}(x)}{a+b \text {sech}(x)} \, dx=\frac {\ln \left (128\,a\,b-32\,a^2-128\,b^2+32\,a^2\,{\mathrm {e}}^x+128\,b^2\,{\mathrm {e}}^x-128\,a\,b\,{\mathrm {e}}^x\right )}{a+b}-\frac {\ln \left (-128\,a\,b-32\,a^2-128\,b^2-32\,a^2\,{\mathrm {e}}^x-128\,b^2\,{\mathrm {e}}^x-128\,a\,b\,{\mathrm {e}}^x\right )}{a-b}+\frac {b\,\ln \left (16\,a\,b^2-4\,a^3\,{\mathrm {e}}^{2\,x}-4\,a^3+32\,b^3\,{\mathrm {e}}^x-8\,a^2\,b\,{\mathrm {e}}^x+16\,a\,b^2\,{\mathrm {e}}^{2\,x}\right )}{a^2-b^2} \] Input:
int(1/(sinh(x)*(a + b/cosh(x))),x)
Output:
log(128*a*b - 32*a^2 - 128*b^2 + 32*a^2*exp(x) + 128*b^2*exp(x) - 128*a*b* exp(x))/(a + b) - log(- 128*a*b - 32*a^2 - 128*b^2 - 32*a^2*exp(x) - 128*b ^2*exp(x) - 128*a*b*exp(x))/(a - b) + (b*log(16*a*b^2 - 4*a^3*exp(2*x) - 4 *a^3 + 32*b^3*exp(x) - 8*a^2*b*exp(x) + 16*a*b^2*exp(2*x)))/(a^2 - b^2)
Time = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.25 \[ \int \frac {\text {csch}(x)}{a+b \text {sech}(x)} \, dx=\frac {\mathrm {log}\left (e^{x}-1\right ) a -\mathrm {log}\left (e^{x}-1\right ) b -\mathrm {log}\left (e^{x}+1\right ) a -\mathrm {log}\left (e^{x}+1\right ) b +\mathrm {log}\left (e^{2 x} a +2 e^{x} b +a \right ) b}{a^{2}-b^{2}} \] Input:
int(csch(x)/(a+b*sech(x)),x)
Output:
(log(e**x - 1)*a - log(e**x - 1)*b - log(e**x + 1)*a - log(e**x + 1)*b + l og(e**(2*x)*a + 2*e**x*b + a)*b)/(a**2 - b**2)