Integrand size = 15, antiderivative size = 62 \[ \int \frac {A+B \text {sech}(x)}{a+b \cosh (x)} \, dx=\frac {B \arctan (\sinh (x))}{a}+\frac {2 (a A-b B) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}} \]
B*arctan(sinh(x))/a+2*(A*a-B*b)*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2 ))/a/(a-b)^(1/2)/(a+b)^(1/2)
Time = 0.20 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.02 \[ \int \frac {A+B \text {sech}(x)}{a+b \cosh (x)} \, dx=\frac {2 \left (B \arctan \left (\tanh \left (\frac {x}{2}\right )\right )+\frac {(-a A+b B) \arctan \left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}\right )}{a} \]
(2*(B*ArcTan[Tanh[x/2]] + ((-(a*A) + b*B)*ArcTan[((a - b)*Tanh[x/2])/Sqrt[ -a^2 + b^2]])/Sqrt[-a^2 + b^2]))/a
Time = 0.44 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 3307, 3042, 3480, 3042, 3138, 221, 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.
\(\displaystyle \int \frac {A+B \text {sech}(x)}{a+b \cosh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc \left (\frac {\pi }{2}+i x\right )}{a+b \sin \left (\frac {\pi }{2}+i x\right )}dx\) |
\(\Big \downarrow \) 3307 |
\(\displaystyle \int \frac {\text {sech}(x) (A \cosh (x)+B)}{a+b \cosh (x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B+A \sin \left (\frac {\pi }{2}+i x\right )}{\sin \left (\frac {\pi }{2}+i x\right ) \left (a+b \sin \left (\frac {\pi }{2}+i x\right )\right )}dx\) |
\(\Big \downarrow \) 3480 |
\(\displaystyle \frac {(a A-b B) \int \frac {1}{a+b \cosh (x)}dx}{a}+\frac {B \int \text {sech}(x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(a A-b B) \int \frac {1}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{a}+\frac {B \int \csc \left (i x+\frac {\pi }{2}\right )dx}{a}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {2 (a A-b B) \int \frac {1}{-\left ((a-b) \tanh ^2\left (\frac {x}{2}\right )\right )+a+b}d\tanh \left (\frac {x}{2}\right )}{a}+\frac {B \int \csc \left (i x+\frac {\pi }{2}\right )dx}{a}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 (a A-b B) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}}+\frac {B \int \csc \left (i x+\frac {\pi }{2}\right )dx}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {2 (a A-b B) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}}+\frac {B \arctan (\sinh (x))}{a}\) |
(B*ArcTan[Sinh[x]])/a + (2*(a*A - b*B)*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqr t[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b])
3.3.4.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*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[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])^n/Sin[e + f*x]^n), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IntegerQ [n]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.95
method | result | size |
default | \(\frac {2 B \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a}-\frac {2 \left (-A a +B b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}}\) | \(59\) |
risch | \(\frac {i B \ln \left ({\mathrm e}^{x}+i\right )}{a}-\frac {i B \ln \left ({\mathrm e}^{x}-i\right )}{a}+\frac {\ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) B b}{\sqrt {a^{2}-b^{2}}\, a}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) B b}{\sqrt {a^{2}-b^{2}}\, a}\) | \(254\) |
2*B/a*arctan(tanh(1/2*x))-2*(-A*a+B*b)/a/((a+b)*(a-b))^(1/2)*arctanh((a-b) *tanh(1/2*x)/((a+b)*(a-b))^(1/2))
Time = 0.48 (sec) , antiderivative size = 249, normalized size of antiderivative = 4.02 \[ \int \frac {A+B \text {sech}(x)}{a+b \cosh (x)} \, dx=\left [-\frac {{\left (A a - B b\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) + b}\right ) - 2 \, {\left (B a^{2} - B b^{2}\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}{a^{3} - a b^{2}}, -\frac {2 \, {\left ({\left (A a - B b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{a^{2} - b^{2}}\right ) - {\left (B a^{2} - B b^{2}\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )\right )}}{a^{3} - a b^{2}}\right ] \]
[-((A*a - B*b)*sqrt(a^2 - b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b* cosh(x) + 2*a^2 - b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) + 2*sqrt(a^2 - b^2)* (b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2* (b*cosh(x) + a)*sinh(x) + b)) - 2*(B*a^2 - B*b^2)*arctan(cosh(x) + sinh(x) ))/(a^3 - a*b^2), -2*((A*a - B*b)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2 )*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) - (B*a^2 - B*b^2)*arctan(cosh(x ) + sinh(x)))/(a^3 - a*b^2)]
\[ \int \frac {A+B \text {sech}(x)}{a+b \cosh (x)} \, dx=\int \frac {A + B \operatorname {sech}{\left (x \right )}}{a + b \cosh {\left (x \right )}}\, dx \]
Exception generated. \[ \int \frac {A+B \text {sech}(x)}{a+b \cosh (x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.85 \[ \int \frac {A+B \text {sech}(x)}{a+b \cosh (x)} \, dx=\frac {2 \, B \arctan \left (e^{x}\right )}{a} + \frac {2 \, {\left (A a - B b\right )} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} a} \]
Time = 7.26 (sec) , antiderivative size = 636, normalized size of antiderivative = 10.26 \[ \int \frac {A+B \text {sech}(x)}{a+b \cosh (x)} \, dx=\frac {\ln \left (\frac {\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (A\,a-B\,b\right )\,\left (\frac {32\,\left (A^2\,a^2\,b-2\,A\,B\,a\,b^2-4\,{\mathrm {e}}^x\,B^2\,a^3-2\,B^2\,a^2\,b+3\,{\mathrm {e}}^x\,B^2\,a\,b^2+2\,B^2\,b^3\right )}{b^5}+\frac {\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (A\,a-B\,b\right )\,\left (\frac {32\,a^2\,\left (2\,B\,b^2-4\,A\,a^2\,{\mathrm {e}}^x+A\,b^2\,{\mathrm {e}}^x-2\,A\,a\,b+3\,B\,a\,b\,{\mathrm {e}}^x\right )}{b^5}-\frac {32\,a^2\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (A\,a-B\,b\right )\,\left (4\,{\mathrm {e}}^x\,a^3+3\,a^2\,b-3\,{\mathrm {e}}^x\,a\,b^2-2\,b^3\right )}{b^5\,\left (a\,b^2-a^3\right )}\right )}{a\,b^2-a^3}\right )}{a\,b^2-a^3}-\frac {32\,B\,\left (A\,a-B\,b\right )\,\left (2\,B\,b-A\,b\,{\mathrm {e}}^x+4\,B\,a\,{\mathrm {e}}^x\right )}{b^5}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (A\,a-B\,b\right )}{a\,b^2-a^3}-\frac {\ln \left (-\frac {\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (A\,a-B\,b\right )\,\left (\frac {32\,\left (A^2\,a^2\,b-2\,A\,B\,a\,b^2-4\,{\mathrm {e}}^x\,B^2\,a^3-2\,B^2\,a^2\,b+3\,{\mathrm {e}}^x\,B^2\,a\,b^2+2\,B^2\,b^3\right )}{b^5}-\frac {\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (A\,a-B\,b\right )\,\left (\frac {32\,a^2\,\left (2\,B\,b^2-4\,A\,a^2\,{\mathrm {e}}^x+A\,b^2\,{\mathrm {e}}^x-2\,A\,a\,b+3\,B\,a\,b\,{\mathrm {e}}^x\right )}{b^5}+\frac {32\,a^2\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (A\,a-B\,b\right )\,\left (4\,{\mathrm {e}}^x\,a^3+3\,a^2\,b-3\,{\mathrm {e}}^x\,a\,b^2-2\,b^3\right )}{b^5\,\left (a\,b^2-a^3\right )}\right )}{a\,b^2-a^3}\right )}{a\,b^2-a^3}-\frac {32\,B\,\left (A\,a-B\,b\right )\,\left (2\,B\,b-A\,b\,{\mathrm {e}}^x+4\,B\,a\,{\mathrm {e}}^x\right )}{b^5}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}\,\left (A\,a-B\,b\right )}{a\,b^2-a^3}-\frac {B\,\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,1{}\mathrm {i}}{a}+\frac {B\,\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a} \]
(B*log(exp(x) + 1i)*1i)/a - (B*log(exp(x) - 1i)*1i)/a + (log((((a + b)*(a - b))^(1/2)*(A*a - B*b)*((32*(2*B^2*b^3 + A^2*a^2*b - 2*B^2*a^2*b - 4*B^2* a^3*exp(x) + 3*B^2*a*b^2*exp(x) - 2*A*B*a*b^2))/b^5 + (((a + b)*(a - b))^( 1/2)*(A*a - B*b)*((32*a^2*(2*B*b^2 - 4*A*a^2*exp(x) + A*b^2*exp(x) - 2*A*a *b + 3*B*a*b*exp(x)))/b^5 - (32*a^2*((a + b)*(a - b))^(1/2)*(A*a - B*b)*(3 *a^2*b - 2*b^3 + 4*a^3*exp(x) - 3*a*b^2*exp(x)))/(b^5*(a*b^2 - a^3))))/(a* b^2 - a^3)))/(a*b^2 - a^3) - (32*B*(A*a - B*b)*(2*B*b - A*b*exp(x) + 4*B*a *exp(x)))/b^5)*((a + b)*(a - b))^(1/2)*(A*a - B*b))/(a*b^2 - a^3) - (log(- (((a + b)*(a - b))^(1/2)*(A*a - B*b)*((32*(2*B^2*b^3 + A^2*a^2*b - 2*B^2* a^2*b - 4*B^2*a^3*exp(x) + 3*B^2*a*b^2*exp(x) - 2*A*B*a*b^2))/b^5 - (((a + b)*(a - b))^(1/2)*(A*a - B*b)*((32*a^2*(2*B*b^2 - 4*A*a^2*exp(x) + A*b^2* exp(x) - 2*A*a*b + 3*B*a*b*exp(x)))/b^5 + (32*a^2*((a + b)*(a - b))^(1/2)* (A*a - B*b)*(3*a^2*b - 2*b^3 + 4*a^3*exp(x) - 3*a*b^2*exp(x)))/(b^5*(a*b^2 - a^3))))/(a*b^2 - a^3)))/(a*b^2 - a^3) - (32*B*(A*a - B*b)*(2*B*b - A*b* exp(x) + 4*B*a*exp(x)))/b^5)*((a + b)*(a - b))^(1/2)*(A*a - B*b))/(a*b^2 - a^3)