\(\int \frac {A+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx\) [549]

Optimal result

Integrand size = 19, antiderivative size = 120 \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=-\frac {c C x}{b^2-c^2}-\frac {2 \left (A \left (b^2-c^2\right )+a c C\right ) \text {arctanh}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt {a^2-b^2+c^2}}+\frac {b C \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2} \] Output:

-c*C*x/(b^2-c^2)-2*(A*(b^2-c^2)+a*c*C)*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 = 0.87 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87 \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\frac {\frac {2 \left (A \left (b^2-c^2\right )+a c C\right ) \arctan \left (\frac {c+(-a+b) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2-c^2}}\right )}{\sqrt {-a^2+b^2-c^2}}+C (-c x+b \log (a+b \cosh (x)+c \sinh (x)))}{(b-c) (b+c)} \] Input:

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

Output:

((2*(A*(b^2 - c^2) + a*c*C)*ArcTan[(c + (-a + b)*Tanh[x/2])/Sqrt[-a^2 + b^ 
2 - c^2]])/Sqrt[-a^2 + b^2 - c^2] + C*(-(c*x) + b*Log[a + b*Cosh[x] + c*Si 
nh[x]]))/((b - c)*(b + c))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 115, 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, 3616, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3616

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3603

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-((c*C*x)/(b^2 - c^2)) - (2*(A + (a*c*C)/(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*L 
og[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 3616

Int[((A_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_) 
]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[c*C*((d + e*x)/ 
(e*(b^2 + c^2))), x] + (-Simp[b*C*(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*c*C)/(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, C} 
, x] && NeQ[b^2 + c^2, 0] && NeQ[A*(b^2 + c^2) - a*c*C, 0]
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.67

method	result	size
default	\(-\frac {2 C \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b +2 c}+\frac {\frac {2 \left (a b C -C \,b^{2}\right ) \ln \left (a \tanh \left (\frac {x}{2}\right )^{2}-b \tanh \left (\frac {x}{2}\right )^{2}-2 c \tanh \left (\frac {x}{2}\right )-a -b \right )}{2 a -2 b}+\frac {2 \left (-A \,b^{2}+A \,c^{2}-a c C -C b c +\frac {\left (a b C -C \,b^{2}\right ) c}{a -b}\right ) \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right )}{\sqrt {-a^{2}+b^{2}-c^{2}}}}{\left (b -c \right ) \left (b +c \right )}-\frac {2 C \ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{2 b -2 c}\)	\(200\)
risch	\(\text {Expression too large to display}\)	\(2418\)

Input:

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

Output:

-2*C/(2*b+2*c)*ln(tanh(1/2*x)-1)+2/(b-c)/(b+c)*(1/2*(C*a*b-C*b^2)/(a-b)*ln 
(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*c*C- 
C*b*c+(C*a*b-C*b^2)*c/(a-b))/(-a^2+b^2-c^2)^(1/2)*arctan(1/2*(2*(a-b)*tanh 
(1/2*x)-2*c)/(-a^2+b^2-c^2)^(1/2)))-2*C/(2*b-2*c)*ln(1+tanh(1/2*x))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 504, normalized size of antiderivative = 4.20 \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\left [\frac {{\left (A b^{2} + C a c - A c^{2}\right )} \sqrt {a^{2} - b^{2} + c^{2}} \log \left (\frac {{\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \left (x\right )^{2} + {\left (b^{2} + 2 \, b c + c^{2}\right )} \sinh \left (x\right )^{2} + 2 \, a^{2} - b^{2} + c^{2} + 2 \, {\left (a b + a c\right )} \cosh \left (x\right ) + 2 \, {\left (a b + a c + {\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) - 2 \, \sqrt {a^{2} - b^{2} + c^{2}} {\left ({\left (b + c\right )} \cosh \left (x\right ) + {\left (b + c\right )} \sinh \left (x\right ) + a\right )}}{{\left (b + c\right )} \cosh \left (x\right )^{2} + {\left (b + c\right )} \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left ({\left (b + c\right )} \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) + b - c}\right ) - {\left (C a^{2} b - C b^{3} + C b c^{2} + C c^{3} + {\left (C a^{2} - C b^{2}\right )} c\right )} x + {\left (C a^{2} b - C b^{3} + C b c^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \left (x\right ) + c \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} b^{2} - b^{4} - c^{4} - {\left (a^{2} - 2 \, b^{2}\right )} c^{2}}, \frac {2 \, {\left (A b^{2} + C a c - A c^{2}\right )} \sqrt {-a^{2} + b^{2} - c^{2}} \arctan \left (\frac {\sqrt {-a^{2} + b^{2} - c^{2}} {\left ({\left (b + c\right )} \cosh \left (x\right ) + {\left (b + c\right )} \sinh \left (x\right ) + a\right )}}{a^{2} - b^{2} + c^{2}}\right ) - {\left (C a^{2} b - C b^{3} + C b c^{2} + C c^{3} + {\left (C a^{2} - C b^{2}\right )} c\right )} x + {\left (C a^{2} b - C b^{3} + C b c^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \left (x\right ) + c \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} b^{2} - b^{4} - c^{4} - {\left (a^{2} - 2 \, b^{2}\right )} c^{2}}\right ] \] Input:

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

Output:

[((A*b^2 + C*a*c - A*c^2)*sqrt(a^2 - b^2 + c^2)*log(((b^2 + 2*b*c + c^2)*c 
osh(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)) - (C*a^2*b - C*b^3 + C*b*c^2 + C*c^3 + (C*a^2 - C*b^2)*c)*x + (C*a^ 
2*b - C*b^3 + C*b*c^2)*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*(A*b^2 + C*a*c - 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)) - (C*a^2*b - C*b^3 + C*b*c^2 + C*c^ 
3 + (C*a^2 - C*b^2)*c)*x + (C*a^2*b - C*b^3 + C*b*c^2)*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. \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((A+C*sinh(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 = 0.13 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.02 \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\frac {C b \log \left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + 2 \, a e^{x} + b - c\right )}{b^{2} - c^{2}} - \frac {C x}{b - c} + \frac {2 \, {\left (A b^{2} + C a c - A c^{2}\right )} \arctan \left (\frac {b e^{x} + c e^{x} + a}{\sqrt {-a^{2} + b^{2} - c^{2}}}\right )}{\sqrt {-a^{2} + b^{2} - c^{2}} {\left (b^{2} - c^{2}\right )}} \] Input:

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

Output:

C*b*log(b*e^(2*x) + c*e^(2*x) + 2*a*e^x + b - c)/(b^2 - c^2) - C*x/(b - c) 
 + 2*(A*b^2 + C*a*c - 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 = 0.81 (sec) , antiderivative size = 377, normalized size of antiderivative = 3.14 \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\frac {\ln \left (b\,\sqrt {a^2-b^2+c^2}-c\,\sqrt {a^2-b^2+c^2}-a^2\,{\mathrm {e}}^x+b^2\,{\mathrm {e}}^x-c^2\,{\mathrm {e}}^x+a\,{\mathrm {e}}^x\,\sqrt {a^2-b^2+c^2}\right )\,\left (C\,b^3+A\,b^2\,\sqrt {a^2-b^2+c^2}-C\,a^2\,b-A\,c^2\,\sqrt {a^2-b^2+c^2}-C\,b\,c^2+C\,a\,c\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4}-\frac {\ln \left (b\,\sqrt {a^2-b^2+c^2}-c\,\sqrt {a^2-b^2+c^2}+a^2\,{\mathrm {e}}^x-b^2\,{\mathrm {e}}^x+c^2\,{\mathrm {e}}^x+a\,{\mathrm {e}}^x\,\sqrt {a^2-b^2+c^2}\right )\,\left (A\,b^2\,\sqrt {a^2-b^2+c^2}-C\,b^3+C\,a^2\,b-A\,c^2\,\sqrt {a^2-b^2+c^2}+C\,b\,c^2+C\,a\,c\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4}-\frac {C\,x}{b-c} \] Input:

int((A + C*sinh(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))*(C*b^3 + A*b^2 
*(a^2 - b^2 + c^2)^(1/2) - C*a^2*b - A*c^2*(a^2 - b^2 + c^2)^(1/2) - C*b*c 
^2 + C*a*c*(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))*(A*b^ 
2*(a^2 - b^2 + c^2)^(1/2) - C*b^3 + C*a^2*b - A*c^2*(a^2 - b^2 + c^2)^(1/2 
) + C*b*c^2 + C*a*c*(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) - (C*x)/(b - c)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.36 \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\frac {-2 \sqrt {-a^{2}+b^{2}-c^{2}}\, \mathit {atan} \left (\frac {e^{x} b +e^{x} c +a}{\sqrt {-a^{2}+b^{2}-c^{2}}}\right ) a \,b^{2}+\mathrm {log}\left (a +\cosh \left (x \right ) b +\sinh \left (x \right ) c \right ) a^{2} b c -\mathrm {log}\left (a +\cosh \left (x \right ) b +\sinh \left (x \right ) c \right ) b^{3} c +\mathrm {log}\left (a +\cosh \left (x \right ) b +\sinh \left (x \right ) c \right ) b \,c^{3}-a^{2} c^{2} x +b^{2} c^{2} x -c^{4} x}{a^{2} b^{2}-a^{2} c^{2}-b^{4}+2 b^{2} c^{2}-c^{4}} \] Input:

int((A+C*sinh(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*b**2 + log(cosh(x)*b + sinh(x)*c + a)*a**2*b*c - log(cos 
h(x)*b + sinh(x)*c + a)*b**3*c + log(cosh(x)*b + sinh(x)*c + a)*b*c**3 - a 
**2*c**2*x + b**2*c**2*x - c**4*x)/(a**2*b**2 - a**2*c**2 - b**4 + 2*b**2* 
c**2 - c**4)