Integrand size = 18, antiderivative size = 57 \[ \int \frac {A+B \cosh (x)}{\sqrt {a-a \cosh (x)}} \, dx=-\frac {\sqrt {2} (A+B) \arctan \left (\frac {\sqrt {a} \sinh (x)}{\sqrt {2} \sqrt {a-a \cosh (x)}}\right )}{\sqrt {a}}+\frac {2 B \sinh (x)}{\sqrt {a-a \cosh (x)}} \] Output:
-2^(1/2)*(A+B)*arctan(1/2*a^(1/2)*sinh(x)*2^(1/2)/(a-a*cosh(x))^(1/2))/a^( 1/2)+2*B*sinh(x)/(a-a*cosh(x))^(1/2)
Time = 0.18 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89 \[ \int \frac {A+B \cosh (x)}{\sqrt {a-a \cosh (x)}} \, dx=\frac {2 \left (2 B \cosh \left (\frac {x}{2}\right )-(A+B) \left (\log \left (\cosh \left (\frac {x}{4}\right )\right )-\log \left (\sinh \left (\frac {x}{4}\right )\right )\right )\right ) \sinh \left (\frac {x}{2}\right )}{\sqrt {a-a \cosh (x)}} \] Input:
Integrate[(A + B*Cosh[x])/Sqrt[a - a*Cosh[x]],x]
Output:
(2*(2*B*Cosh[x/2] - (A + B)*(Log[Cosh[x/4]] - Log[Sinh[x/4]]))*Sinh[x/2])/ Sqrt[a - a*Cosh[x]]
Time = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3042, 3230, 3042, 3128, 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.
\(\displaystyle \int \frac {A+B \cosh (x)}{\sqrt {a-a \cosh (x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (\frac {\pi }{2}+i x\right )}{\sqrt {a-a \sin \left (\frac {\pi }{2}+i x\right )}}dx\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle (A+B) \int \frac {1}{\sqrt {a-a \cosh (x)}}dx+\frac {2 B \sinh (x)}{\sqrt {a-a \cosh (x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 B \sinh (x)}{\sqrt {a-a \cosh (x)}}+(A+B) \int \frac {1}{\sqrt {a-a \sin \left (i x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle \frac {2 B \sinh (x)}{\sqrt {a-a \cosh (x)}}+2 i (A+B) \int \frac {1}{\frac {a^2 \sinh ^2(x)}{a-a \cosh (x)}+2 a}d\frac {i a \sinh (x)}{\sqrt {a-a \cosh (x)}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 B \sinh (x)}{\sqrt {a-a \cosh (x)}}-\frac {\sqrt {2} (A+B) \arctan \left (\frac {\sqrt {a} \sinh (x)}{\sqrt {2} \sqrt {a-a \cosh (x)}}\right )}{\sqrt {a}}\) |
Input:
Int[(A + B*Cosh[x])/Sqrt[a - a*Cosh[x]],x]
Output:
-((Sqrt[2]*(A + B)*ArcTan[(Sqrt[a]*Sinh[x])/(Sqrt[2]*Sqrt[a - a*Cosh[x]])] )/Sqrt[a]) + (2*B*Sinh[x])/Sqrt[a - a*Cosh[x]]
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[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Time = 0.86 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.11
method | result | size |
default | \(\frac {\sinh \left (\frac {x}{2}\right ) \left (\ln \left (\cosh \left (\frac {x}{2}\right )-1\right ) A -\ln \left (\cosh \left (\frac {x}{2}\right )+1\right ) A +\ln \left (\cosh \left (\frac {x}{2}\right )-1\right ) B -\ln \left (\cosh \left (\frac {x}{2}\right )+1\right ) B +4 B \cosh \left (\frac {x}{2}\right )\right )}{\sqrt {-2 \sinh \left (\frac {x}{2}\right )^{2} a}}\) | \(63\) |
parts | \(-\frac {2 A \sinh \left (\frac {x}{2}\right ) \operatorname {arctanh}\left (\cosh \left (\frac {x}{2}\right )\right )}{\sqrt {-2 \sinh \left (\frac {x}{2}\right )^{2} a}}+\frac {B \sinh \left (\frac {x}{2}\right ) \left (4 \cosh \left (\frac {x}{2}\right )+\ln \left (\cosh \left (\frac {x}{2}\right )-1\right )-\ln \left (\cosh \left (\frac {x}{2}\right )+1\right )\right )}{\sqrt {-2 \sinh \left (\frac {x}{2}\right )^{2} a}}\) | \(65\) |
Input:
int((A+B*cosh(x))/(a-cosh(x)*a)^(1/2),x,method=_RETURNVERBOSE)
Output:
sinh(1/2*x)*(ln(cosh(1/2*x)-1)*A-ln(cosh(1/2*x)+1)*A+ln(cosh(1/2*x)-1)*B-l n(cosh(1/2*x)+1)*B+4*B*cosh(1/2*x))/(-2*sinh(1/2*x)^2*a)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (46) = 92\).
Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.74 \[ \int \frac {A+B \cosh (x)}{\sqrt {a-a \cosh (x)}} \, dx=\frac {\sqrt {2} {\left (A + B\right )} a \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a}{\cosh \left (x\right ) + \sinh \left (x\right )}} \sqrt {-\frac {1}{a}} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} - \cosh \left (x\right ) - \sinh \left (x\right ) - 1}{\cosh \left (x\right ) + \sinh \left (x\right ) - 1}\right ) - 2 \, \sqrt {\frac {1}{2}} {\left (B \cosh \left (x\right ) + B \sinh \left (x\right ) + B\right )} \sqrt {-\frac {a}{\cosh \left (x\right ) + \sinh \left (x\right )}}}{a} \] Input:
integrate((A+B*cosh(x))/(a-a*cosh(x))^(1/2),x, algorithm="fricas")
Output:
(sqrt(2)*(A + B)*a*sqrt(-1/a)*log((2*sqrt(2)*sqrt(1/2)*sqrt(-a/(cosh(x) + sinh(x)))*sqrt(-1/a)*(cosh(x) + sinh(x)) - cosh(x) - sinh(x) - 1)/(cosh(x) + sinh(x) - 1)) - 2*sqrt(1/2)*(B*cosh(x) + B*sinh(x) + B)*sqrt(-a/(cosh(x ) + sinh(x))))/a
\[ \int \frac {A+B \cosh (x)}{\sqrt {a-a \cosh (x)}} \, dx=\int \frac {A + B \cosh {\left (x \right )}}{\sqrt {- a \left (\cosh {\left (x \right )} - 1\right )}}\, dx \] Input:
integrate((A+B*cosh(x))/(a-a*cosh(x))**(1/2),x)
Output:
Integral((A + B*cosh(x))/sqrt(-a*(cosh(x) - 1)), x)
\[ \int \frac {A+B \cosh (x)}{\sqrt {a-a \cosh (x)}} \, dx=\int { \frac {B \cosh \left (x\right ) + A}{\sqrt {-a \cosh \left (x\right ) + a}} \,d x } \] Input:
integrate((A+B*cosh(x))/(a-a*cosh(x))^(1/2),x, algorithm="maxima")
Output:
integrate((B*cosh(x) + A)/sqrt(-a*cosh(x) + a), x)
Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.47 \[ \int \frac {A+B \cosh (x)}{\sqrt {a-a \cosh (x)}} \, dx=-\frac {2 \, {\left (\sqrt {2} A + \sqrt {2} B\right )} \arctan \left (\frac {\sqrt {-a e^{x}}}{\sqrt {a}}\right )}{\sqrt {a} \mathrm {sgn}\left (-e^{x} + 1\right )} - \frac {\sqrt {2} B}{\sqrt {-a e^{x}} \mathrm {sgn}\left (-e^{x} + 1\right )} + \frac {\sqrt {2} \sqrt {-a e^{x}} B}{a \mathrm {sgn}\left (-e^{x} + 1\right )} \] Input:
integrate((A+B*cosh(x))/(a-a*cosh(x))^(1/2),x, algorithm="giac")
Output:
-2*(sqrt(2)*A + sqrt(2)*B)*arctan(sqrt(-a*e^x)/sqrt(a))/(sqrt(a)*sgn(-e^x + 1)) - sqrt(2)*B/(sqrt(-a*e^x)*sgn(-e^x + 1)) + sqrt(2)*sqrt(-a*e^x)*B/(a *sgn(-e^x + 1))
Timed out. \[ \int \frac {A+B \cosh (x)}{\sqrt {a-a \cosh (x)}} \, dx=\int \frac {A+B\,\mathrm {cosh}\left (x\right )}{\sqrt {a-a\,\mathrm {cosh}\left (x\right )}} \,d x \] Input:
int((A + B*cosh(x))/(a - a*cosh(x))^(1/2),x)
Output:
int((A + B*cosh(x))/(a - a*cosh(x))^(1/2), x)
\[ \int \frac {A+B \cosh (x)}{\sqrt {a-a \cosh (x)}} \, dx=-\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {-\cosh \left (x \right )+1}}{\cosh \left (x \right )-1}d x \right ) a +\left (\int \frac {\sqrt {-\cosh \left (x \right )+1}\, \cosh \left (x \right )}{\cosh \left (x \right )-1}d x \right ) b \right )}{a} \] Input:
int((A+B*cosh(x))/(a-a*cosh(x))^(1/2),x)
Output:
( - sqrt(a)*(int(sqrt( - cosh(x) + 1)/(cosh(x) - 1),x)*a + int((sqrt( - co sh(x) + 1)*cosh(x))/(cosh(x) - 1),x)*b))/a