Integrand size = 14, antiderivative size = 223 \[ \int \frac {1}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\frac {4 c \text {arctanh}\left (\frac {\sqrt {b-2 c-\sqrt {b^2-4 a c}} \tanh \left (\frac {x}{2}\right )}{\sqrt {b+2 c-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b-2 c-\sqrt {b^2-4 a c}} \sqrt {b+2 c-\sqrt {b^2-4 a c}}}-\frac {4 c \text {arctanh}\left (\frac {\sqrt {b-2 c+\sqrt {b^2-4 a c}} \tanh \left (\frac {x}{2}\right )}{\sqrt {b+2 c+\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b-2 c+\sqrt {b^2-4 a c}} \sqrt {b+2 c+\sqrt {b^2-4 a c}}} \] Output:
4*c*arctanh((b-2*c-(-4*a*c+b^2)^(1/2))^(1/2)*tanh(1/2*x)/(b+2*c-(-4*a*c+b^ 2)^(1/2))^(1/2))/(-4*a*c+b^2)^(1/2)/(b-2*c-(-4*a*c+b^2)^(1/2))^(1/2)/(b+2* c-(-4*a*c+b^2)^(1/2))^(1/2)-4*c*arctanh((b-2*c+(-4*a*c+b^2)^(1/2))^(1/2)*t anh(1/2*x)/(b+2*c+(-4*a*c+b^2)^(1/2))^(1/2))/(-4*a*c+b^2)^(1/2)/(b-2*c+(-4 *a*c+b^2)^(1/2))^(1/2)/(b+2*c+(-4*a*c+b^2)^(1/2))^(1/2)
Time = 0.72 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.89 \[ \int \frac {1}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\frac {2 \sqrt {2} c \left (\frac {\arctan \left (\frac {\left (b-2 c+\sqrt {b^2-4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)-2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {-b^2+2 c (a+c)-b \sqrt {b^2-4 a c}}}+\frac {\arctan \left (\frac {\left (-b+2 c+\sqrt {b^2-4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)+2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {-b^2+2 c (a+c)+b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c}} \] Input:
Integrate[(a + b*Cosh[x] + c*Cosh[x]^2)^(-1),x]
Output:
(2*Sqrt[2]*c*(ArcTan[((b - 2*c + Sqrt[b^2 - 4*a*c])*Tanh[x/2])/Sqrt[-2*b^2 + 4*c*(a + c) - 2*b*Sqrt[b^2 - 4*a*c]]]/Sqrt[-b^2 + 2*c*(a + c) - b*Sqrt[ b^2 - 4*a*c]] + ArcTan[((-b + 2*c + Sqrt[b^2 - 4*a*c])*Tanh[x/2])/Sqrt[-2* b^2 + 4*c*(a + c) + 2*b*Sqrt[b^2 - 4*a*c]]]/Sqrt[-b^2 + 2*c*(a + c) + b*Sq rt[b^2 - 4*a*c]]))/Sqrt[b^2 - 4*a*c]
Time = 0.62 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3730, 3042, 3138, 221}
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.
\(\displaystyle \int \frac {1}{a+b \cosh (x)+c \cosh ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{a+b \cos (i x)+c \cos (i x)^2}dx\) |
\(\Big \downarrow \) 3730 |
\(\displaystyle \frac {2 c \int \frac {1}{b+2 c \cosh (x)-\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {1}{b+2 c \cosh (x)+\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 c \int \frac {1}{b+2 c \sin \left (i x+\frac {\pi }{2}\right )-\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {1}{b+2 c \sin \left (i x+\frac {\pi }{2}\right )+\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {4 c \int \frac {1}{-\left (\left (b-2 c-\sqrt {b^2-4 a c}\right ) \tanh ^2\left (\frac {x}{2}\right )\right )+b+2 c-\sqrt {b^2-4 a c}}d\tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-4 a c}}-\frac {4 c \int \frac {1}{-\left (\left (b-2 c+\sqrt {b^2-4 a c}\right ) \tanh ^2\left (\frac {x}{2}\right )\right )+b+2 c+\sqrt {b^2-4 a c}}d\tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-4 a c}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {4 c \text {arctanh}\left (\frac {\tanh \left (\frac {x}{2}\right ) \sqrt {-\sqrt {b^2-4 a c}+b-2 c}}{\sqrt {-\sqrt {b^2-4 a c}+b+2 c}}\right )}{\sqrt {b^2-4 a c} \sqrt {-\sqrt {b^2-4 a c}+b-2 c} \sqrt {-\sqrt {b^2-4 a c}+b+2 c}}-\frac {4 c \text {arctanh}\left (\frac {\tanh \left (\frac {x}{2}\right ) \sqrt {\sqrt {b^2-4 a c}+b-2 c}}{\sqrt {\sqrt {b^2-4 a c}+b+2 c}}\right )}{\sqrt {b^2-4 a c} \sqrt {\sqrt {b^2-4 a c}+b-2 c} \sqrt {\sqrt {b^2-4 a c}+b+2 c}}\) |
Input:
Int[(a + b*Cosh[x] + c*Cosh[x]^2)^(-1),x]
Output:
(4*c*ArcTanh[(Sqrt[b - 2*c - Sqrt[b^2 - 4*a*c]]*Tanh[x/2])/Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[b - 2*c - Sqrt[b^2 - 4*a*c]]* Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]]) - (4*c*ArcTanh[(Sqrt[b - 2*c + Sqrt[b^2 - 4*a*c]]*Tanh[x/2])/Sqrt[b + 2*c + Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a* c]*Sqrt[b - 2*c + Sqrt[b^2 - 4*a*c]]*Sqrt[b + 2*c + Sqrt[b^2 - 4*a*c]])
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[((a_.) + cos[(d_.) + (e_.)*(x_)]^(n_.)*(b_.) + cos[(d_.) + (e_.)*(x_)]^ (n2_.)*(c_.))^(-1), x_Symbol] :> Module[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c /q) Int[1/(b - q + 2*c*Cos[d + e*x]^n), x], x] - Simp[2*(c/q) Int[1/(b + q + 2*c*Cos[d + e*x]^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n 2, 2*n] && NeQ[b^2 - 4*a*c, 0]
Time = 0.56 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.93
method | result | size |
default | \(2 \left (a -b +c \right ) \left (\frac {\left (b -2 c -\sqrt {-4 a c +b^{2}}\right ) \arctan \left (\frac {\left (a -b +c \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \left (a -b +c \right ) \sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}+\frac {\left (-b +2 c -\sqrt {-4 a c +b^{2}}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b -c \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \left (a -b +c \right ) \sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}\right )\) | \(208\) |
risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (16 a^{4} c^{2}-8 a^{3} b^{2} c +32 a^{3} c^{3}+a^{2} b^{4}-32 a^{2} b^{2} c^{2}+16 a^{2} c^{4}+10 a \,b^{4} c -8 a \,b^{2} c^{3}-b^{6}+c^{2} b^{4}\right ) \textit {\_Z}^{4}+\left (-8 a^{2} c^{2}+6 a \,b^{2} c -8 a \,c^{3}-b^{4}+2 b^{2} c^{2}\right ) \textit {\_Z}^{2}+c^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}+\left (-\frac {8 a^{4}}{b}+22 a^{2} b +\frac {b^{5}}{c^{2}}+\frac {6 b \,a^{3}}{c}-\frac {24 c \,a^{3}}{b}-\frac {b^{3} a^{2}}{c^{2}}-\frac {24 c^{2} a^{2}}{b}-\frac {8 b^{3} a}{c}+18 a b c -\frac {8 c^{3} a}{b}-3 b^{3}+2 b \,c^{2}\right ) \textit {\_R}^{3}+\left (\frac {4 a^{3}}{b}-6 a b -\frac {b \,a^{2}}{c}+\frac {8 c \,a^{2}}{b}+\frac {4 c^{2} a}{b}+\frac {b^{3}}{c}-b c \right ) \textit {\_R}^{2}+\left (\frac {2 a^{2}}{b}-\frac {4 b a}{c}+\frac {4 c a}{b}-2 b +\frac {b^{3}}{c^{2}}+\frac {2 c^{2}}{b}\right ) \textit {\_R} +\frac {b}{c}-\frac {a}{b}-\frac {c}{b}\right )\) | \(353\) |
Input:
int(1/(a+b*cosh(x)+c*cosh(x)^2),x,method=_RETURNVERBOSE)
Output:
2*(a-b+c)*(1/2*(b-2*c-(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2)/(a-b+c)/(((-4 *a*c+b^2)^(1/2)-a+c)*(a-b+c))^(1/2)*arctan((a-b+c)*tanh(1/2*x)/(((-4*a*c+b ^2)^(1/2)-a+c)*(a-b+c))^(1/2))+1/2*(-b+2*c-(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2 )^(1/2)/(a-b+c)/(((-4*a*c+b^2)^(1/2)+a-c)*(a-b+c))^(1/2)*arctanh((-a+b-c)* tanh(1/2*x)/(((-4*a*c+b^2)^(1/2)+a-c)*(a-b+c))^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 3485 vs. \(2 (183) = 366\).
Time = 0.21 (sec) , antiderivative size = 3485, normalized size of antiderivative = 15.63 \[ \int \frac {1}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+b*cosh(x)+c*cosh(x)^2),x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {1}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*cosh(x)+c*cosh(x)**2),x)
Output:
Timed out
\[ \int \frac {1}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\int { \frac {1}{c \cosh \left (x\right )^{2} + b \cosh \left (x\right ) + a} \,d x } \] Input:
integrate(1/(a+b*cosh(x)+c*cosh(x)^2),x, algorithm="maxima")
Output:
integrate(1/(c*cosh(x)^2 + b*cosh(x) + a), x)
Time = 60.75 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.00 \[ \int \frac {1}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=0 \] Input:
integrate(1/(a+b*cosh(x)+c*cosh(x)^2),x, algorithm="giac")
Output:
0
Timed out. \[ \int \frac {1}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\text {Hanged} \] Input:
int(1/(a + b*cosh(x) + c*cosh(x)^2),x)
Output:
\text{Hanged}
\[ \int \frac {1}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\int \frac {1}{\cosh \left (x \right )^{2} c +\cosh \left (x \right ) b +a}d x \] Input:
int(1/(a+b*cosh(x)+c*cosh(x)^2),x)
Output:
int(1/(cosh(x)**2*c + cosh(x)*b + a),x)