Integrand size = 19, antiderivative size = 64 \[ \int \frac {\text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\text {arctanh}(\cosh (c+d x))}{a d}+\frac {2 b \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d} \] Output:
-arctanh(cosh(d*x+c))/a/d+2*b*arctanh((b-a*tanh(1/2*d*x+1/2*c))/(a^2+b^2)^ (1/2))/a/(a^2+b^2)^(1/2)/d
Time = 0.87 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.28 \[ \int \frac {\text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-\frac {2 b \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-\log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\sinh \left (\frac {1}{2} (c+d x)\right )\right )}{a d} \] Input:
Integrate[Csch[c + d*x]/(a + b*Sinh[c + d*x]),x]
Output:
((-2*b*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2 ] - Log[Cosh[(c + d*x)/2]] + Log[Sinh[(c + d*x)/2]])/(a*d)
Result contains complex when optimal does not.
Time = 0.40 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {3042, 26, 3226, 26, 3042, 26, 3139, 1083, 217, 4257}
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.
\(\displaystyle \int \frac {\text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i}{\sin (i c+i d x) (a-i b \sin (i c+i d x))}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {1}{\sin (i c+i d x) (a-i b \sin (i c+i d x))}dx\) |
\(\Big \downarrow \) 3226 |
\(\displaystyle i \left (\frac {i b \int \frac {1}{a+b \sinh (c+d x)}dx}{a}+\frac {\int -i \text {csch}(c+d x)dx}{a}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \left (\frac {i b \int \frac {1}{a+b \sinh (c+d x)}dx}{a}-\frac {i \int \text {csch}(c+d x)dx}{a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle i \left (\frac {i b \int \frac {1}{a-i b \sin (i c+i d x)}dx}{a}-\frac {i \int i \csc (i c+i d x)dx}{a}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \left (\frac {i b \int \frac {1}{a-i b \sin (i c+i d x)}dx}{a}+\frac {\int \csc (i c+i d x)dx}{a}\right )\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle i \left (\frac {2 b \int \frac {1}{-a \tanh ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tanh \left (\frac {1}{2} (c+d x)\right )+a}d\left (i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{a d}+\frac {\int \csc (i c+i d x)dx}{a}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle i \left (\frac {\int \csc (i c+i d x)dx}{a}-\frac {4 b \int \frac {1}{\tanh ^2\left (\frac {1}{2} (c+d x)\right )-4 \left (a^2+b^2\right )}d\left (2 i a \tanh \left (\frac {1}{2} (c+d x)\right )-2 i b\right )}{a d}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle i \left (\frac {\int \csc (i c+i d x)dx}{a}+\frac {2 i b \text {arctanh}\left (\frac {\tanh \left (\frac {1}{2} (c+d x)\right )}{2 \sqrt {a^2+b^2}}\right )}{a d \sqrt {a^2+b^2}}\right )\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle i \left (\frac {2 i b \text {arctanh}\left (\frac {\tanh \left (\frac {1}{2} (c+d x)\right )}{2 \sqrt {a^2+b^2}}\right )}{a d \sqrt {a^2+b^2}}+\frac {i \text {arctanh}(\cosh (c+d x))}{a d}\right )\) |
Input:
Int[Csch[c + d*x]/(a + b*Sinh[c + d*x]),x]
Output:
I*((I*ArcTanh[Cosh[c + d*x]])/(a*d) + ((2*I)*b*ArcTanh[Tanh[(c + d*x)/2]/( 2*Sqrt[a^2 + b^2])])/(a*Sqrt[a^2 + b^2]*d))
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[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[ {a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 b \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \sqrt {a^{2}+b^{2}}}}{d}\) | \(63\) |
default | \(\frac {\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 b \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \sqrt {a^{2}+b^{2}}}}{d}\) | \(63\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{d x +c}+1\right )}{d a}+\frac {b \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d a}-\frac {b \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d a}+\frac {\ln \left ({\mathrm e}^{d x +c}-1\right )}{d a}\) | \(152\) |
Input:
int(csch(d*x+c)/(a+b*sinh(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/a*ln(tanh(1/2*d*x+1/2*c))-2*b/a/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*ta nh(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (61) = 122\).
Time = 0.12 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.48 \[ \int \frac {\text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\sqrt {a^{2} + b^{2}} b \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) - {\left (a^{2} + b^{2}\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1\right ) + {\left (a^{2} + b^{2}\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1\right )}{{\left (a^{3} + a b^{2}\right )} d} \] Input:
integrate(csch(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")
Output:
(sqrt(a^2 + b^2)*b*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b* cosh(d*x + c) + 2*a^2 + b^2 + 2*(b^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a))/(b*cosh(d*x + c )^2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh (d*x + c) - b)) - (a^2 + b^2)*log(cosh(d*x + c) + sinh(d*x + c) + 1) + (a^ 2 + b^2)*log(cosh(d*x + c) + sinh(d*x + c) - 1))/((a^3 + a*b^2)*d)
\[ \int \frac {\text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\operatorname {csch}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \] Input:
integrate(csch(d*x+c)/(a+b*sinh(d*x+c)),x)
Output:
Integral(csch(c + d*x)/(a + b*sinh(c + d*x)), x)
Time = 0.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.75 \[ \int \frac {\text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {b \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a d} - \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} \] Input:
integrate(csch(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima")
Output:
-b*log((b*e^(-d*x - c) - a - sqrt(a^2 + b^2))/(b*e^(-d*x - c) - a + sqrt(a ^2 + b^2)))/(sqrt(a^2 + b^2)*a*d) - log(e^(-d*x - c) + 1)/(a*d) + log(e^(- d*x - c) - 1)/(a*d)
Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.59 \[ \int \frac {\text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {b \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a} + \frac {\log \left (e^{\left (d x + c\right )} + 1\right )}{a} - \frac {\log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a}}{d} \] Input:
integrate(csch(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")
Output:
-(b*log(abs(2*b*e^(d*x + c) + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^(d*x + c) + 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a) + log(e^(d*x + c) + 1)/a - log(abs(e^(d*x + c) - 1))/a)/d
Time = 1.39 (sec) , antiderivative size = 347, normalized size of antiderivative = 5.42 \[ \int \frac {\text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\ln \left (32\,a-32\,a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a\,d}-\frac {\ln \left (32\,a+32\,a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a\,d}+\frac {b\,\ln \left (128\,a^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-64\,a^2\,b^3-64\,a^4\,b+32\,a\,b^3\,\sqrt {a^2+b^2}+64\,a^3\,b\,\sqrt {a^2+b^2}+160\,a^3\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-128\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}+32\,a\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-96\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{d\,a^3+d\,a\,b^2}-\frac {b\,\ln \left (64\,a^4\,b+64\,a^2\,b^3-128\,a^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+32\,a\,b^3\,\sqrt {a^2+b^2}+64\,a^3\,b\,\sqrt {a^2+b^2}-160\,a^3\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-128\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}-32\,a\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-96\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{d\,a^3+d\,a\,b^2} \] Input:
int(1/(sinh(c + d*x)*(a + b*sinh(c + d*x))),x)
Output:
log(32*a - 32*a*exp(d*x)*exp(c))/(a*d) - log(32*a + 32*a*exp(d*x)*exp(c))/ (a*d) + (b*log(128*a^5*exp(d*x)*exp(c) - 64*a^2*b^3 - 64*a^4*b + 32*a*b^3* (a^2 + b^2)^(1/2) + 64*a^3*b*(a^2 + b^2)^(1/2) + 160*a^3*b^2*exp(d*x)*exp( c) - 128*a^4*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2) + 32*a*b^4*exp(d*x)*exp(c) - 96*a^2*b^2*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a^3*d + a*b^2*d) - (b*log(64*a^4*b + 64*a^2*b^3 - 128*a^5*exp(d*x)*exp(c) + 32*a *b^3*(a^2 + b^2)^(1/2) + 64*a^3*b*(a^2 + b^2)^(1/2) - 160*a^3*b^2*exp(d*x) *exp(c) - 128*a^4*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2) - 32*a*b^4*exp(d*x)*ex p(c) - 96*a^2*b^2*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a ^3*d + a*b^2*d)
Time = 0.26 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.77 \[ \int \frac {\text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b i +\mathrm {log}\left (e^{d x +c}-1\right ) a^{2}+\mathrm {log}\left (e^{d x +c}-1\right ) b^{2}-\mathrm {log}\left (e^{d x +c}+1\right ) a^{2}-\mathrm {log}\left (e^{d x +c}+1\right ) b^{2}}{a d \left (a^{2}+b^{2}\right )} \] Input:
int(csch(d*x+c)/(a+b*sinh(d*x+c)),x)
Output:
( - 2*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sqrt(a**2 + b**2))*b *i + log(e**(c + d*x) - 1)*a**2 + log(e**(c + d*x) - 1)*b**2 - log(e**(c + d*x) + 1)*a**2 - log(e**(c + d*x) + 1)*b**2)/(a*d*(a**2 + b**2))