3.9.35 \(\int \frac {\cosh ^3(x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx\) [835]

3.9.35.1 Optimal result

Integrand size = 19, antiderivative size = 299 \[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=-\frac {b x}{c^2}+\frac {2 \left (b^2-a c-\frac {b^3}{\sqrt {b^2-4 a c}}+\frac {3 a b c}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt {b-2 c-\sqrt {b^2-4 a c}} \tanh \left (\frac {x}{2}\right )}{\sqrt {b+2 c-\sqrt {b^2-4 a c}}}\right )}{c^2 \sqrt {b-2 c-\sqrt {b^2-4 a c}} \sqrt {b+2 c-\sqrt {b^2-4 a c}}}+\frac {2 \left (b^2-a c+\frac {b^3}{\sqrt {b^2-4 a c}}-\frac {3 a b c}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt {b-2 c+\sqrt {b^2-4 a c}} \tanh \left (\frac {x}{2}\right )}{\sqrt {b+2 c+\sqrt {b^2-4 a c}}}\right )}{c^2 \sqrt {b-2 c+\sqrt {b^2-4 a c}} \sqrt {b+2 c+\sqrt {b^2-4 a c}}}+\frac {\sinh (x)}{c} \]

output

-b*x/c^2+sinh(x)/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))*(b^2-a*c-b^3/(-4*a*c+b^2)^(1/2)+3*a*b*c/ 
(-4*a*c+b^2)^(1/2))/c^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))*(b^2-a*c+b^3/(-4*a*c+b^2)^(1/2)-3*a*b*c/(-4 
*a*c+b^2)^(1/2))/c^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)
 

3.9.35.2 Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.03 \[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\frac {-b x-\frac {\sqrt {2} \left (b^3-3 a b c+b^2 \sqrt {b^2-4 a c}-a c \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\left (b-2 c+\sqrt {b^2-4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)-2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {-b^2+2 c (a+c)-b \sqrt {b^2-4 a c}}}+\frac {\sqrt {2} \left (-b^3+3 a b c+b^2 \sqrt {b^2-4 a c}-a c \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\left (-b+2 c+\sqrt {b^2-4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)+2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {-b^2+2 c (a+c)+b \sqrt {b^2-4 a c}}}+c \sinh (x)}{c^2} \]

input

Integrate[Cosh[x]^3/(a + b*Cosh[x] + c*Cosh[x]^2),x]
 

output

(-(b*x) - (Sqrt[2]*(b^3 - 3*a*b*c + b^2*Sqrt[b^2 - 4*a*c] - a*c*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 - 4*a*c]*Sqrt[-b^2 + 2*c*(a 
 + c) - b*Sqrt[b^2 - 4*a*c]]) + (Sqrt[2]*(-b^3 + 3*a*b*c + b^2*Sqrt[b^2 - 
4*a*c] - a*c*Sqrt[b^2 - 4*a*c])*ArcTan[((-b + 2*c + Sqrt[b^2 - 4*a*c])*Tan 
h[x/2])/Sqrt[-2*b^2 + 4*c*(a + c) + 2*b*Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4 
*a*c]*Sqrt[-b^2 + 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]]) + c*Sinh[x])/c^2
 

3.9.35.3 Rubi [A] (verified)

Time = 3.53 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3738, 2009}

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.

\(\displaystyle \int \frac {\cosh ^3(x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (i x)^3}{a+b \cos (i x)+c \cos (i x)^2}dx\)

\(\Big \downarrow \) 3738

\(\displaystyle \int \left (\frac {b^2 \cosh (x) \left (1-\frac {a c}{b^2}\right )+a b}{c^2 \left (a+b \cosh (x)+c \cosh ^2(x)\right )}-\frac {b}{c^2}+\frac {\cosh (x)}{c}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (\frac {3 a b c}{\sqrt {b^2-4 a c}}-\frac {b^3}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \text {arctanh}\left (\frac {\tanh \left (\frac {x}{2}\right ) \sqrt {-\sqrt {b^2-4 a c}+b-2 c}}{\sqrt {-\sqrt {b^2-4 a c}+b+2 c}}\right )}{c^2 \sqrt {-\sqrt {b^2-4 a c}+b-2 c} \sqrt {-\sqrt {b^2-4 a c}+b+2 c}}+\frac {2 \left (-\frac {3 a b c}{\sqrt {b^2-4 a c}}+\frac {b^3}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \text {arctanh}\left (\frac {\tanh \left (\frac {x}{2}\right ) \sqrt {\sqrt {b^2-4 a c}+b-2 c}}{\sqrt {\sqrt {b^2-4 a c}+b+2 c}}\right )}{c^2 \sqrt {\sqrt {b^2-4 a c}+b-2 c} \sqrt {\sqrt {b^2-4 a c}+b+2 c}}-\frac {b x}{c^2}+\frac {\sinh (x)}{c}\)

input

Int[Cosh[x]^3/(a + b*Cosh[x] + c*Cosh[x]^2),x]
 

output

-((b*x)/c^2) + (2*(b^2 - a*c - b^3/Sqrt[b^2 - 4*a*c] + (3*a*b*c)/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]]])/(c^2*Sqrt[b - 2*c - Sqrt[b^2 - 4*a*c]]*Sqrt[b + 
2*c - Sqrt[b^2 - 4*a*c]]) + (2*(b^2 - a*c + b^3/Sqrt[b^2 - 4*a*c] - (3*a*b 
*c)/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]]])/(c^2*Sqrt[b - 2*c + Sqrt[b^2 - 4*a* 
c]]*Sqrt[b + 2*c + Sqrt[b^2 - 4*a*c]]) + Sinh[x]/c
 

3.9.35.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3738

Int[cos[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + cos[(d_.) + (e_.)*(x_)]^(n_.)*(b 
_.) + cos[(d_.) + (e_.)*(x_)]^(n2_.)*(c_.))^(p_), x_Symbol] :> Int[ExpandTr 
ig[cos[d + e*x]^m*(a + b*cos[d + e*x]^n + c*cos[d + e*x]^(2*n))^p, x], x] / 
; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && Integ 
ersQ[m, n, p]
 

3.9.35.4 Maple [A] (verified)

Time = 1.50 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.18

method	result	size
default	\(-\frac {1}{c \left (1+\tanh \left (\frac {x}{2}\right )\right )}-\frac {b \ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{c^{2}}-\frac {1}{c \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{c^{2}}+\frac {2 \left (a -b +c \right ) \left (\frac {\left (-a b \sqrt {-4 a c +b^{2}}-a c \sqrt {-4 a c +b^{2}}+b^{2} \sqrt {-4 a c +b^{2}}-2 a^{2} c +a \,b^{2}+3 b c a -b^{3}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b -c \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \left (a -b +c \right ) \sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}+\frac {\left (-a b \sqrt {-4 a c +b^{2}}-a c \sqrt {-4 a c +b^{2}}+b^{2} \sqrt {-4 a c +b^{2}}+2 a^{2} c -a \,b^{2}-3 b c a +b^{3}\right ) \arctan \left (\frac {\left (a -b +c \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \left (a -b +c \right ) \sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}\right )}{c^{2}}\)	\(354\)
risch	\(\text {Expression too large to display}\)	\(2096\)

input

int(cosh(x)^3/(a+b*cosh(x)+c*cosh(x)^2),x,method=_RETURNVERBOSE)
 

output

-1/c/(1+tanh(1/2*x))-b/c^2*ln(1+tanh(1/2*x))-1/c/(tanh(1/2*x)-1)+b/c^2*ln( 
tanh(1/2*x)-1)+2/c^2*(a-b+c)*(1/2*(-a*b*(-4*a*c+b^2)^(1/2)-a*c*(-4*a*c+b^2 
)^(1/2)+b^2*(-4*a*c+b^2)^(1/2)-2*a^2*c+a*b^2+3*b*c*a-b^3)/(-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))+1/2*(-a*b*(-4*a*c+b^2)^(1 
/2)-a*c*(-4*a*c+b^2)^(1/2)+b^2*(-4*a*c+b^2)^(1/2)+2*a^2*c-a*b^2-3*b*c*a+b^ 
3)/(-4*a*c+b^2)^(1/2)/(a-b+c)/(((-4*a*c+b^2)^(1/2)-a+c)*(a-b+c))^(1/2)*arc 
tan((a-b+c)*tanh(1/2*x)/(((-4*a*c+b^2)^(1/2)-a+c)*(a-b+c))^(1/2)))
 

3.9.35.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6794 vs. \(2 (255) = 510\).

Time = 1.08 (sec) , antiderivative size = 6794, normalized size of antiderivative = 22.72 \[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\text {Too large to display} \]

input

integrate(cosh(x)^3/(a+b*cosh(x)+c*cosh(x)^2),x, algorithm="fricas")
 

output

Too large to include
 

3.9.35.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\text {Timed out} \]

input

integrate(cosh(x)**3/(a+b*cosh(x)+c*cosh(x)**2),x)
 

output

Timed out
 

3.9.35.7 Maxima [F]

\[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\int { \frac {\cosh \left (x\right )^{3}}{c \cosh \left (x\right )^{2} + b \cosh \left (x\right ) + a} \,d x } \]

input

integrate(cosh(x)^3/(a+b*cosh(x)+c*cosh(x)^2),x, algorithm="maxima")
 

output

-1/2*(2*b*x*e^x - c*e^(2*x) + c)*e^(-x)/c^2 - 1/8*integrate(-16*(2*a*b*e^( 
2*x) + (b^2 - a*c)*e^(3*x) + (b^2 - a*c)*e^x)/(c^3*e^(4*x) + 2*b*c^2*e^(3* 
x) + 2*b*c^2*e^x + c^3 + 2*(2*a*c^2 + c^3)*e^(2*x)), x)
 

3.9.35.8 Giac [A] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.08 \[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=-\frac {b x}{c^{2}} - \frac {e^{\left (-x\right )}}{2 \, c} + \frac {e^{x}}{2 \, c} \]

input

integrate(cosh(x)^3/(a+b*cosh(x)+c*cosh(x)^2),x, algorithm="giac")
 

output

-b*x/c^2 - 1/2*e^(-x)/c + 1/2*e^x/c
 

3.9.35.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)+c \cosh ^2(x)} \, dx=\text {Hanged} \]

input

int(cosh(x)^3/(a + b*cosh(x) + c*cosh(x)^2),x)
 

output

\text{Hanged}