Integrand size = 23, antiderivative size = 137 \[ \int \frac {A+B \cosh (x)+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\frac {(b B-c C) x}{b^2-c^2}-\frac {2 \left (A b^2-a b B-A c^2+a c C\right ) \text {arctanh}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt {a^2-b^2+c^2}}-\frac {(B c-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2} \] Output:
(B*b-C*c)*x/(b^2-c^2)-2*(A*b^2-A*c^2-B*a*b+C*a*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-C*b)*ln(a+b* cosh(x)+c*sinh(x))/(b^2-c^2)
Time = 0.79 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.87 \[ \int \frac {A+B \cosh (x)+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\frac {(b B-c C) x+\frac {2 \left (A b^2-a b B-A c^2+a c C\right ) \arctan \left (\frac {c+(-a+b) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2-c^2}}\right )}{\sqrt {-a^2+b^2-c^2}}+(-B c+b C) \log (a+b \cosh (x)+c \sinh (x))}{(b-c) (b+c)} \] Input:
Integrate[(A + B*Cosh[x] + C*Sinh[x])/(a + b*Cosh[x] + c*Sinh[x]),x]
Output:
((b*B - c*C)*x + (2*(A*b^2 - a*b*B - A*c^2 + a*c*C)*ArcTan[(c + (-a + b)*T anh[x/2])/Sqrt[-a^2 + b^2 - c^2]])/Sqrt[-a^2 + b^2 - c^2] + (-(B*c) + b*C) *Log[a + b*Cosh[x] + c*Sinh[x]])/((b - c)*(b + c))
Time = 0.44 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3615, 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.
\(\displaystyle \int \frac {A+B \cosh (x)+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \cos (i x)-i C \sin (i x)}{a+b \cos (i x)-i c \sin (i x)}dx\) |
\(\Big \downarrow \) 3615 |
\(\displaystyle \frac {\left (-a b B+a c C+A b^2-A c^2\right ) \int \frac {1}{a+b \cosh (x)+c \sinh (x)}dx}{b^2-c^2}-\frac {(B c-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {x (b B-c C)}{b^2-c^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\left (-a b B+a c C+A b^2-A c^2\right ) \int \frac {1}{a+b \cos (i x)-i c \sin (i x)}dx}{b^2-c^2}-\frac {(B c-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {x (b B-c C)}{b^2-c^2}\) |
\(\Big \downarrow \) 3603 |
\(\displaystyle \frac {2 \left (-a b B+a c C+A b^2-A c^2\right ) \int \frac {1}{-\left ((a-b) \tanh ^2\left (\frac {x}{2}\right )\right )+2 c \tanh \left (\frac {x}{2}\right )+a+b}d\tanh \left (\frac {x}{2}\right )}{b^2-c^2}-\frac {(B c-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {x (b B-c C)}{b^2-c^2}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle -\frac {4 \left (-a b B+a c C+A b^2-A c^2\right ) \int \frac {1}{4 \left (a^2-b^2+c^2\right )-\left (2 c-2 (a-b) \tanh \left (\frac {x}{2}\right )\right )^2}d\left (2 c-2 (a-b) \tanh \left (\frac {x}{2}\right )\right )}{b^2-c^2}-\frac {(B c-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {x (b B-c C)}{b^2-c^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2 \left (-a b B+a c C+A b^2-A c^2\right ) \text {arctanh}\left (\frac {2 c-2 (a-b) \tanh \left (\frac {x}{2}\right )}{2 \sqrt {a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt {a^2-b^2+c^2}}-\frac {(B c-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {x (b B-c C)}{b^2-c^2}\) |
Input:
Int[(A + B*Cosh[x] + C*Sinh[x])/(a + b*Cosh[x] + c*Sinh[x]),x]
Output:
((b*B - c*C)*x)/(b^2 - c^2) - (2*(A*b^2 - a*b*B - A*c^2 + a*c*C)*ArcTanh[( 2*c - 2*(a - b)*Tanh[x/2])/(2*Sqrt[a^2 - b^2 + c^2])])/((b^2 - c^2)*Sqrt[a ^2 - b^2 + c^2]) - ((B*c - b*C)*Log[a + b*Cosh[x] + c*Sinh[x]])/(b^2 - c^2 )
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])
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]
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]
Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]) /((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x _Symbol] :> Simp[(b*B + c*C)*(x/(b^2 + c^2)), x] + (Simp[(c*B - 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*(b*B + 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, B, C}, x] && NeQ[b^2 + c^2, 0] && NeQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]
Time = 0.40 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.73
method | result | size |
default | \(\frac {\frac {2 \left (-a B c +b B c +a b C -C \,b^{2}\right ) \ln \left (a \tanh \left (\frac {x}{2}\right )^{2}-b \tanh \left (\frac {x}{2}\right )^{2}-2 c \tanh \left (\frac {x}{2}\right )-a -b \right )}{2 a -2 b}+\frac {2 \left (-A \,b^{2}+A \,c^{2}+a b B +B \,c^{2}-a c C -C b c +\frac {\left (-a B c +b B c +a b C -C \,b^{2}\right ) c}{a -b}\right ) \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right )}{\sqrt {-a^{2}+b^{2}-c^{2}}}}{\left (b -c \right ) \left (b +c \right )}+\frac {2 \left (B -C \right ) \ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{2 b -2 c}+\frac {2 \left (-B -C \right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b +2 c}\) | \(237\) |
risch | \(\text {Expression too large to display}\) | \(6027\) |
Input:
int((A+B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x)),x,method=_RETURNVERBOS E)
Output:
2/(b-c)/(b+c)*(1/2*(-B*a*c+B*b*c+C*a*b-C*b^2)/(a-b)*ln(a*tanh(1/2*x)^2-b*t anh(1/2*x)^2-2*c*tanh(1/2*x)-a-b)+(-A*b^2+A*c^2+a*b*B+B*c^2-a*c*C-C*b*c+(- B*a*c+B*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*(B-C)/(2*b-2*c)*ln(1+tanh(1/2*x ))+2*(-B-C)/(2*b+2*c)*ln(tanh(1/2*x)-1)
Time = 0.11 (sec) , antiderivative size = 608, normalized size of antiderivative = 4.44 \[ \int \frac {A+B \cosh (x)+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx =\text {Too large to display} \] Input:
integrate((A+B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x)),x, algorithm="fr icas")
Output:
[-((B*a*b - A*b^2 - C*a*c + 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)*si nh(x) + b - c)) - ((B - C)*a^2*b - (B - C)*b^3 + (B - C)*b*c^2 + (B - C)*c ^3 + ((B - C)*a^2 - (B - C)*b^2)*c)*x - (C*a^2*b - C*b^3 + C*b*c^2 - 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 - C*a*c + A*c^2)*sqrt(-a^2 + b^2 - c^2)*arctan(sqrt(-a^2 + b^2 - c^2)*((b + c)*cos h(x) + (b + c)*sinh(x) + a)/(a^2 - b^2 + c^2)) - ((B - C)*a^2*b - (B - C)* b^3 + (B - C)*b*c^2 + (B - C)*c^3 + ((B - C)*a^2 - (B - C)*b^2)*c)*x - (C* a^2*b - C*b^3 + C*b*c^2 - 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)]
Timed out. \[ \int \frac {A+B \cosh (x)+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\text {Timed out} \] Input:
integrate((A+B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x)),x)
Output:
Timed out
Exception generated. \[ \int \frac {A+B \cosh (x)+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((A+B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x)),x, algorithm="ma xima")
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
Time = 0.13 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.99 \[ \int \frac {A+B \cosh (x)+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\frac {{\left (B - C\right )} x}{b - c} + \frac {{\left (C b - B c\right )} \log \left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + 2 \, a e^{x} + b - c\right )}{b^{2} - c^{2}} - \frac {2 \, {\left (B a b - A b^{2} - C a c + A c^{2}\right )} \arctan \left (\frac {b e^{x} + c e^{x} + a}{\sqrt {-a^{2} + b^{2} - c^{2}}}\right )}{\sqrt {-a^{2} + b^{2} - c^{2}} {\left (b^{2} - c^{2}\right )}} \] Input:
integrate((A+B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x)),x, algorithm="gi ac")
Output:
(B - C)*x/(b - c) + (C*b - B*c)*log(b*e^(2*x) + c*e^(2*x) + 2*a*e^x + b - c)/(b^2 - c^2) - 2*(B*a*b - 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))
Time = 1.83 (sec) , antiderivative size = 454, normalized size of antiderivative = 3.31 \[ \int \frac {A+B \cosh (x)+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=\frac {\ln \left (b\,\sqrt {a^2-b^2+c^2}-c\,\sqrt {a^2-b^2+c^2}+a^2\,{\mathrm {e}}^x-b^2\,{\mathrm {e}}^x+c^2\,{\mathrm {e}}^x+a\,{\mathrm {e}}^x\,\sqrt {a^2-b^2+c^2}\right )\,\left (B\,c^3+C\,b^3-A\,b^2\,\sqrt {a^2-b^2+c^2}+B\,a^2\,c-C\,a^2\,b+A\,c^2\,\sqrt {a^2-b^2+c^2}-B\,b^2\,c-C\,b\,c^2+B\,a\,b\,\sqrt {a^2-b^2+c^2}-C\,a\,c\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4}+\frac {\ln \left (b\,\sqrt {a^2-b^2+c^2}-c\,\sqrt {a^2-b^2+c^2}-a^2\,{\mathrm {e}}^x+b^2\,{\mathrm {e}}^x-c^2\,{\mathrm {e}}^x+a\,{\mathrm {e}}^x\,\sqrt {a^2-b^2+c^2}\right )\,\left (B\,c^3+C\,b^3+A\,b^2\,\sqrt {a^2-b^2+c^2}+B\,a^2\,c-C\,a^2\,b-A\,c^2\,\sqrt {a^2-b^2+c^2}-B\,b^2\,c-C\,b\,c^2-B\,a\,b\,\sqrt {a^2-b^2+c^2}+C\,a\,c\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4}+\frac {x\,\left (B-C\right )}{b-c} \] Input:
int((A + B*cosh(x) + 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))*(B*c^3 + C*b^3 - A*b^2*(a^2 - b^2 + c^2)^(1/2) + B*a^2*c - C*a^2*b + A*c^2*(a^2 - b^2 + c^2)^(1/2) - B*b^2*c - C*b*c^2 + B*a*b*(a^2 - b^2 + c^2)^(1/2) - 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*e xp(x) - c^2*exp(x) + a*exp(x)*(a^2 - b^2 + c^2)^(1/2))*(B*c^3 + C*b^3 + A* b^2*(a^2 - b^2 + c^2)^(1/2) + B*a^2*c - C*a^2*b - A*c^2*(a^2 - b^2 + c^2)^ (1/2) - B*b^2*c - C*b*c^2 - B*a*b*(a^2 - b^2 + c^2)^(1/2) + 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) + (x*(B - C) )/(b - c)
Time = 0.24 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.01 \[ \int \frac {A+B \cosh (x)+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx=x \] Input:
int((A+B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x)),x)
Output:
x