Integrand size = 21, antiderivative size = 246 \[ \int \frac {d+e \cosh (x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\frac {2 \left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt {b-2 c-\sqrt {b^2-4 a c}} \tanh \left (\frac {x}{2}\right )}{\sqrt {b+2 c-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b-2 c-\sqrt {b^2-4 a c}} \sqrt {b+2 c-\sqrt {b^2-4 a c}}}+\frac {2 \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt {b-2 c+\sqrt {b^2-4 a c}} \tanh \left (\frac {x}{2}\right )}{\sqrt {b+2 c+\sqrt {b^2-4 a c}}}\right )}{\sqrt {b-2 c+\sqrt {b^2-4 a c}} \sqrt {b+2 c+\sqrt {b^2-4 a c}}} \]
2*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))*(e+(-b*e+2*c*d)/(-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)+2*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))*(e+(b*e-2*c*d)/( -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 = 0.77 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.98 \[ \int \frac {d+e \cosh (x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\frac {\sqrt {2} \left (-\frac {\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \arctan \left (\frac {\left (b-2 c+\sqrt {b^2-4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)-2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {-b^2+2 c (a+c)-b \sqrt {b^2-4 a c}}}+\frac {\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) \arctan \left (\frac {\left (-b+2 c+\sqrt {b^2-4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)+2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {-b^2+2 c (a+c)+b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c}} \]
(Sqrt[2]*(-(((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*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]]) + ((2*c*d + (-b + Sqrt [b^2 - 4*a*c])*e)*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]]))/Sqrt[b^2 - 4*a*c]
Time = 0.56 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3774, 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.
\(\displaystyle \int \frac {d+e \cosh (x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {d+e \cos (i x)}{a+b \cos (i x)+c \cos (i x)^2}dx\) |
\(\Big \downarrow \) 3774 |
\(\displaystyle \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \int \frac {1}{b+2 c \cosh (x)-\sqrt {b^2-4 a c}}dx+\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{b+2 c \cosh (x)+\sqrt {b^2-4 a c}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \int \frac {1}{b+2 c \sin \left (i x+\frac {\pi }{2}\right )-\sqrt {b^2-4 a c}}dx+\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{b+2 c \sin \left (i x+\frac {\pi }{2}\right )+\sqrt {b^2-4 a c}}dx\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle 2 \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \int \frac {1}{-\left (\left (b-2 c-\sqrt {b^2-4 a c}\right ) \tanh ^2\left (\frac {x}{2}\right )\right )+b+2 c-\sqrt {b^2-4 a c}}d\tanh \left (\frac {x}{2}\right )+2 \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{-\left (\left (b-2 c+\sqrt {b^2-4 a c}\right ) \tanh ^2\left (\frac {x}{2}\right )\right )+b+2 c+\sqrt {b^2-4 a c}}d\tanh \left (\frac {x}{2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \text {arctanh}\left (\frac {\tanh \left (\frac {x}{2}\right ) \sqrt {-\sqrt {b^2-4 a c}+b-2 c}}{\sqrt {-\sqrt {b^2-4 a c}+b+2 c}}\right )}{\sqrt {-\sqrt {b^2-4 a c}+b-2 c} \sqrt {-\sqrt {b^2-4 a c}+b+2 c}}+\frac {2 \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\tanh \left (\frac {x}{2}\right ) \sqrt {\sqrt {b^2-4 a c}+b-2 c}}{\sqrt {\sqrt {b^2-4 a c}+b+2 c}}\right )}{\sqrt {\sqrt {b^2-4 a c}+b-2 c} \sqrt {\sqrt {b^2-4 a c}+b+2 c}}\) |
(2*(e + (2*c*d - b*e)/Sqrt[b^2 - 4*a*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*c - S qrt[b^2 - 4*a*c]]*Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]]) + (2*(e - (2*c*d - b* e)/Sqrt[b^2 - 4*a*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*c + Sqrt[b^2 - 4*a*c]]*S qrt[b + 2*c + Sqrt[b^2 - 4*a*c]])
3.9.37.3.1 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[(cos[(d_.) + (e_.)*(x_)]*(B_.) + (A_))/((a_.) + cos[(d_.) + (e_.)*(x_)] *(b_.) + cos[(d_.) + (e_.)*(x_)]^2*(c_.)), x_Symbol] :> Module[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(B + (b*B - 2*A*c)/q) Int[1/(b + q + 2*c*Cos[d + e*x]) , x], x] + Simp[(B - (b*B - 2*A*c)/q) Int[1/(b - q + 2*c*Cos[d + e*x]), x ], x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0]
Time = 3.68 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.03
method | result | size |
default | \(2 \left (a -b +c \right ) \left (\frac {\left (-d \sqrt {-4 a c +b^{2}}+e \sqrt {-4 a c +b^{2}}-2 a e +b d +b e -2 c d \right ) \arctan \left (\frac {\left (a -b +c \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \left (a -b +c \right ) \sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}+\frac {\left (-d \sqrt {-4 a c +b^{2}}+e \sqrt {-4 a c +b^{2}}+2 a e -b d -b e +2 c d \right ) \operatorname {arctanh}\left (\frac {\left (-a +b -c \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \left (a -b +c \right ) \sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}\right )\) | \(254\) |
risch | \(\text {Expression too large to display}\) | \(8285\) |
2*(a-b+c)*(1/2*(-d*(-4*a*c+b^2)^(1/2)+e*(-4*a*c+b^2)^(1/2)-2*a*e+b*d+b*e-2 *c*d)/(-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*( -d*(-4*a*c+b^2)^(1/2)+e*(-4*a*c+b^2)^(1/2)+2*a*e-b*d-b*e+2*c*d)/(-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. 6997 vs. \(2 (206) = 412\).
Time = 3.06 (sec) , antiderivative size = 6997, normalized size of antiderivative = 28.44 \[ \int \frac {d+e \cosh (x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {d+e \cosh (x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\text {Timed out} \]
\[ \int \frac {d+e \cosh (x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\int { \frac {e \cosh \left (x\right ) + d}{c \cosh \left (x\right )^{2} + b \cosh \left (x\right ) + a} \,d x } \]
Time = 1.56 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.00 \[ \int \frac {d+e \cosh (x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=0 \]
Timed out. \[ \int \frac {d+e \cosh (x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\text {Hanged} \]