\(\int \frac {A+B \cosh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx\) [552]

Optimal result

Integrand size = 19, antiderivative size = 121 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\frac {b B x}{b^2-c^2}+\frac {2 \left (a b B-A \left (b^2-c^2\right )\right ) \text {arctanh}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt {a^2-b^2+c^2}}-\frac {B c \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2} \] Output:

b*B*x/(b^2-c^2)+2*(a*b*B-A*(b^2-c^2))*arctanh((c-(a-b)*tanh(1/2*x))/(a^2-b 
^2+c^2)^(1/2))/(b^2-c^2)/(a^2-b^2+c^2)^(1/2)-B*c*ln(a+b*cosh(x)+c*sinh(x)) 
/(b^2-c^2)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.86 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\frac {-\frac {2 \left (a b B+A \left (-b^2+c^2\right )\right ) \arctan \left (\frac {c+(-a+b) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2-c^2}}\right )}{\sqrt {-a^2+b^2-c^2}}+B (b x-c \log (a+b \cosh (x)+c \sinh (x)))}{(b-c) (b+c)} \] Input:

Integrate[(A + B*Cosh[x])/(a + b*Cosh[x] + c*Sinh[x]),x]
 

Output:

((-2*(a*b*B + A*(-b^2 + c^2))*ArcTan[(c + (-a + b)*Tanh[x/2])/Sqrt[-a^2 + 
b^2 - c^2]])/Sqrt[-a^2 + b^2 - c^2] + B*(b*x - c*Log[a + b*Cosh[x] + c*Sin 
h[x]]))/((b - c)*(b + c))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3617, 3042, 3603, 1083, 219}

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.

\(\displaystyle \int \frac {A+B \cosh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \cos (i x)}{a+b \cos (i x)-i c \sin (i x)}dx\)

\(\Big \downarrow \) 3617

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

\(\Big \downarrow \) 3042

\(\displaystyle \left (A-\frac {a b B}{b^2-c^2}\right ) \int \frac {1}{a+b \cos (i x)-i c \sin (i x)}dx-\frac {B c \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {b B x}{b^2-c^2}\)

\(\Big \downarrow \) 3603

\(\displaystyle 2 \left (A-\frac {a b B}{b^2-c^2}\right ) \int \frac {1}{-\left ((a-b) \tanh ^2\left (\frac {x}{2}\right )\right )+2 c \tanh \left (\frac {x}{2}\right )+a+b}d\tanh \left (\frac {x}{2}\right )-\frac {B c \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {b B x}{b^2-c^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle -4 \left (A-\frac {a b B}{b^2-c^2}\right ) \int \frac {1}{4 \left (a^2-b^2+c^2\right )-\left (2 c-2 (a-b) \tanh \left (\frac {x}{2}\right )\right )^2}d\left (2 c-2 (a-b) \tanh \left (\frac {x}{2}\right )\right )-\frac {B c \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {b B x}{b^2-c^2}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {2 \left (A-\frac {a b B}{b^2-c^2}\right ) \text {arctanh}\left (\frac {2 c-2 (a-b) \tanh \left (\frac {x}{2}\right )}{2 \sqrt {a^2-b^2+c^2}}\right )}{\sqrt {a^2-b^2+c^2}}-\frac {B c \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {b B x}{b^2-c^2}\)

Input:

Int[(A + B*Cosh[x])/(a + b*Cosh[x] + c*Sinh[x]),x]
 

Output:

(b*B*x)/(b^2 - c^2) - (2*(A - (a*b*B)/(b^2 - c^2))*ArcTanh[(2*c - 2*(a - b 
)*Tanh[x/2])/(2*Sqrt[a^2 - b^2 + c^2])])/Sqrt[a^2 - b^2 + c^2] - (B*c*Log[ 
a + b*Cosh[x] + c*Sinh[x]])/(b^2 - c^2)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1083

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]
 

rule 3042

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

rule 3603

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ 
(-1), x_Symbol] :> Module[{f = FreeFactors[Tan[(d + e*x)/2], x]}, Simp[2*(f 
/e)   Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d + e*x) 
/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]
 

rule 3617

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.))/((a_.) + cos[(d_.) + (e_.)*(x_) 
]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[b*B*((d + e*x)/ 
(e*(b^2 + c^2))), x] + (Simp[c*B*(Log[a + b*Cos[d + e*x] + c*Sin[d + e*x]]/ 
(e*(b^2 + c^2))), x] + Simp[(A*(b^2 + c^2) - a*b*B)/(b^2 + c^2)   Int[1/(a 
+ b*Cos[d + e*x] + c*Sin[d + e*x]), x], x]) /; FreeQ[{a, b, c, d, e, A, B}, 
 x] && NeQ[b^2 + c^2, 0] && NeQ[A*(b^2 + c^2) - a*b*B, 0]
 

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.63

method	result	size
default	\(-\frac {2 B \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b +2 c}+\frac {\frac {2 \left (-a B c +b B c \right ) \ln \left (a \tanh \left (\frac {x}{2}\right )^{2}-b \tanh \left (\frac {x}{2}\right )^{2}-2 c \tanh \left (\frac {x}{2}\right )-a -b \right )}{2 a -2 b}+\frac {2 \left (-A \,b^{2}+A \,c^{2}+a b B +B \,c^{2}+\frac {\left (-a B c +b B c \right ) c}{a -b}\right ) \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right )}{\sqrt {-a^{2}+b^{2}-c^{2}}}}{\left (b -c \right ) \left (b +c \right )}+\frac {2 B \ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{2 b -2 c}\)	\(197\)
risch	\(\text {Expression too large to display}\)	\(2436\)

Input:

int((A+B*cosh(x))/(a+b*cosh(x)+c*sinh(x)),x,method=_RETURNVERBOSE)
 

Output:

-2*B/(2*b+2*c)*ln(tanh(1/2*x)-1)+2/(b-c)/(b+c)*(1/2*(-B*a*c+B*b*c)/(a-b)*l 
n(a*tanh(1/2*x)^2-b*tanh(1/2*x)^2-2*c*tanh(1/2*x)-a-b)+(-A*b^2+A*c^2+a*b*B 
+B*c^2+(-B*a*c+B*b*c)*c/(a-b))/(-a^2+b^2-c^2)^(1/2)*arctan(1/2*(2*(a-b)*ta 
nh(1/2*x)-2*c)/(-a^2+b^2-c^2)^(1/2)))+2*B/(2*b-2*c)*ln(1+tanh(1/2*x))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 508, normalized size of antiderivative = 4.20 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\left [-\frac {{\left (B a b - A b^{2} + A c^{2}\right )} \sqrt {a^{2} - b^{2} + c^{2}} \log \left (\frac {{\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \left (x\right )^{2} + {\left (b^{2} + 2 \, b c + c^{2}\right )} \sinh \left (x\right )^{2} + 2 \, a^{2} - b^{2} + c^{2} + 2 \, {\left (a b + a c\right )} \cosh \left (x\right ) + 2 \, {\left (a b + a c + {\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) - 2 \, \sqrt {a^{2} - b^{2} + c^{2}} {\left ({\left (b + c\right )} \cosh \left (x\right ) + {\left (b + c\right )} \sinh \left (x\right ) + a\right )}}{{\left (b + c\right )} \cosh \left (x\right )^{2} + {\left (b + c\right )} \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left ({\left (b + c\right )} \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) + b - c}\right ) - {\left (B a^{2} b - B b^{3} + B b c^{2} + B c^{3} + {\left (B a^{2} - B b^{2}\right )} c\right )} x + {\left (B c^{3} + {\left (B a^{2} - B b^{2}\right )} c\right )} \log \left (\frac {2 \, {\left (b \cosh \left (x\right ) + c \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} b^{2} - b^{4} - c^{4} - {\left (a^{2} - 2 \, b^{2}\right )} c^{2}}, -\frac {2 \, {\left (B a b - A b^{2} + A c^{2}\right )} \sqrt {-a^{2} + b^{2} - c^{2}} \arctan \left (\frac {\sqrt {-a^{2} + b^{2} - c^{2}} {\left ({\left (b + c\right )} \cosh \left (x\right ) + {\left (b + c\right )} \sinh \left (x\right ) + a\right )}}{a^{2} - b^{2} + c^{2}}\right ) - {\left (B a^{2} b - B b^{3} + B b c^{2} + B c^{3} + {\left (B a^{2} - B b^{2}\right )} c\right )} x + {\left (B c^{3} + {\left (B a^{2} - B b^{2}\right )} c\right )} \log \left (\frac {2 \, {\left (b \cosh \left (x\right ) + c \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} b^{2} - b^{4} - c^{4} - {\left (a^{2} - 2 \, b^{2}\right )} c^{2}}\right ] \] Input:

integrate((A+B*cosh(x))/(a+b*cosh(x)+c*sinh(x)),x, algorithm="fricas")
 

Output:

[-((B*a*b - A*b^2 + A*c^2)*sqrt(a^2 - b^2 + c^2)*log(((b^2 + 2*b*c + c^2)* 
cosh(x)^2 + (b^2 + 2*b*c + c^2)*sinh(x)^2 + 2*a^2 - b^2 + c^2 + 2*(a*b + a 
*c)*cosh(x) + 2*(a*b + a*c + (b^2 + 2*b*c + c^2)*cosh(x))*sinh(x) - 2*sqrt 
(a^2 - b^2 + c^2)*((b + c)*cosh(x) + (b + c)*sinh(x) + a))/((b + c)*cosh(x 
)^2 + (b + c)*sinh(x)^2 + 2*a*cosh(x) + 2*((b + c)*cosh(x) + a)*sinh(x) + 
b - c)) - (B*a^2*b - B*b^3 + B*b*c^2 + B*c^3 + (B*a^2 - B*b^2)*c)*x + (B*c 
^3 + (B*a^2 - B*b^2)*c)*log(2*(b*cosh(x) + c*sinh(x) + a)/(cosh(x) - sinh( 
x))))/(a^2*b^2 - b^4 - c^4 - (a^2 - 2*b^2)*c^2), -(2*(B*a*b - A*b^2 + A*c^ 
2)*sqrt(-a^2 + b^2 - c^2)*arctan(sqrt(-a^2 + b^2 - c^2)*((b + c)*cosh(x) + 
 (b + c)*sinh(x) + a)/(a^2 - b^2 + c^2)) - (B*a^2*b - B*b^3 + B*b*c^2 + B* 
c^3 + (B*a^2 - B*b^2)*c)*x + (B*c^3 + (B*a^2 - B*b^2)*c)*log(2*(b*cosh(x) 
+ c*sinh(x) + a)/(cosh(x) - sinh(x))))/(a^2*b^2 - b^4 - c^4 - (a^2 - 2*b^2 
)*c^2)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\text {Timed out} \] Input:

integrate((A+B*cosh(x))/(a+b*cosh(x)+c*sinh(x)),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((A+B*cosh(x))/(a+b*cosh(x)+c*sinh(x)),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(c^2-b^2+a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.01 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=-\frac {B c \log \left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + 2 \, a e^{x} + b - c\right )}{b^{2} - c^{2}} + \frac {B x}{b - c} - \frac {2 \, {\left (B a b - A b^{2} + A c^{2}\right )} \arctan \left (\frac {b e^{x} + c e^{x} + a}{\sqrt {-a^{2} + b^{2} - c^{2}}}\right )}{\sqrt {-a^{2} + b^{2} - c^{2}} {\left (b^{2} - c^{2}\right )}} \] Input:

integrate((A+B*cosh(x))/(a+b*cosh(x)+c*sinh(x)),x, algorithm="giac")
 

Output:

-B*c*log(b*e^(2*x) + c*e^(2*x) + 2*a*e^x + b - c)/(b^2 - c^2) + B*x/(b - c 
) - 2*(B*a*b - A*b^2 + A*c^2)*arctan((b*e^x + c*e^x + a)/sqrt(-a^2 + b^2 - 
 c^2))/(sqrt(-a^2 + b^2 - c^2)*(b^2 - c^2))
 

Mupad [B] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 375, normalized size of antiderivative = 3.10 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\frac {\ln \left (b\,\sqrt {a^2-b^2+c^2}-c\,\sqrt {a^2-b^2+c^2}+a^2\,{\mathrm {e}}^x-b^2\,{\mathrm {e}}^x+c^2\,{\mathrm {e}}^x+a\,{\mathrm {e}}^x\,\sqrt {a^2-b^2+c^2}\right )\,\left (B\,c^3-A\,b^2\,\sqrt {a^2-b^2+c^2}+B\,a^2\,c+A\,c^2\,\sqrt {a^2-b^2+c^2}-B\,b^2\,c+B\,a\,b\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4}+\frac {\ln \left (b\,\sqrt {a^2-b^2+c^2}-c\,\sqrt {a^2-b^2+c^2}-a^2\,{\mathrm {e}}^x+b^2\,{\mathrm {e}}^x-c^2\,{\mathrm {e}}^x+a\,{\mathrm {e}}^x\,\sqrt {a^2-b^2+c^2}\right )\,\left (B\,c^3+A\,b^2\,\sqrt {a^2-b^2+c^2}+B\,a^2\,c-A\,c^2\,\sqrt {a^2-b^2+c^2}-B\,b^2\,c-B\,a\,b\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4}+\frac {B\,x}{b-c} \] Input:

int((A + B*cosh(x))/(a + b*cosh(x) + c*sinh(x)),x)
 

Output:

(log(b*(a^2 - b^2 + c^2)^(1/2) - c*(a^2 - b^2 + c^2)^(1/2) + a^2*exp(x) - 
b^2*exp(x) + c^2*exp(x) + a*exp(x)*(a^2 - b^2 + c^2)^(1/2))*(B*c^3 - A*b^2 
*(a^2 - b^2 + c^2)^(1/2) + B*a^2*c + A*c^2*(a^2 - b^2 + c^2)^(1/2) - B*b^2 
*c + B*a*b*(a^2 - b^2 + c^2)^(1/2)))/(b^4 + c^4 - a^2*b^2 + a^2*c^2 - 2*b^ 
2*c^2) + (log(b*(a^2 - b^2 + c^2)^(1/2) - c*(a^2 - b^2 + c^2)^(1/2) - a^2* 
exp(x) + b^2*exp(x) - c^2*exp(x) + a*exp(x)*(a^2 - b^2 + c^2)^(1/2))*(B*c^ 
3 + A*b^2*(a^2 - b^2 + c^2)^(1/2) + B*a^2*c - A*c^2*(a^2 - b^2 + c^2)^(1/2 
) - B*b^2*c - B*a*b*(a^2 - b^2 + c^2)^(1/2)))/(b^4 + c^4 - a^2*b^2 + a^2*c 
^2 - 2*b^2*c^2) + (B*x)/(b - c)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.35 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\frac {2 \sqrt {-a^{2}+b^{2}-c^{2}}\, \mathit {atan} \left (\frac {e^{x} b +e^{x} c +a}{\sqrt {-a^{2}+b^{2}-c^{2}}}\right ) a \,c^{2}-\mathrm {log}\left (a +\cosh \left (x \right ) b +\sinh \left (x \right ) c \right ) a^{2} b c +\mathrm {log}\left (a +\cosh \left (x \right ) b +\sinh \left (x \right ) c \right ) b^{3} c -\mathrm {log}\left (a +\cosh \left (x \right ) b +\sinh \left (x \right ) c \right ) b \,c^{3}+a^{2} b^{2} x -b^{4} x +b^{2} c^{2} x}{a^{2} b^{2}-a^{2} c^{2}-b^{4}+2 b^{2} c^{2}-c^{4}} \] Input:

int((A+B*cosh(x))/(a+b*cosh(x)+c*sinh(x)),x)
 

Output:

(2*sqrt( - a**2 + b**2 - c**2)*atan((e**x*b + e**x*c + a)/sqrt( - a**2 + b 
**2 - c**2))*a*c**2 - log(cosh(x)*b + sinh(x)*c + a)*a**2*b*c + log(cosh(x 
)*b + sinh(x)*c + a)*b**3*c - log(cosh(x)*b + sinh(x)*c + a)*b*c**3 + a**2 
*b**2*x - b**4*x + b**2*c**2*x)/(a**2*b**2 - a**2*c**2 - b**4 + 2*b**2*c** 
2 - c**4)