Integrand size = 11, antiderivative size = 52 \[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=\frac {\arctan \left (\sqrt {\cos (x)}\right ) \sin (x)}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}}-\frac {\text {arctanh}\left (\sqrt {\cos (x)}\right ) \sin (x)}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}} \]
arctan(cos(x)^(1/2))*sin(x)/cos(x)^(1/2)/(sin(x)*tan(x))^(1/2)-arctanh(cos (x)^(1/2))*sin(x)/cos(x)^(1/2)/(sin(x)*tan(x))^(1/2)
Time = 0.11 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=\frac {\left (\arctan \left (\sqrt [4]{\cos ^2(x)}\right )-\text {arctanh}\left (\sqrt [4]{\cos ^2(x)}\right )\right ) \cos (x) \cot (x) \sqrt {\sin (x) \tan (x)}}{\cos ^2(x)^{3/4}} \]
((ArcTan[(Cos[x]^2)^(1/4)] - ArcTanh[(Cos[x]^2)^(1/4)])*Cos[x]*Cot[x]*Sqrt [Sin[x]*Tan[x]])/(Cos[x]^2)^(3/4)
Time = 0.36 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.091, Rules used = {3042, 4897, 3042, 4900, 3042, 3081, 3042, 3045, 266, 827, 216, 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.
\(\displaystyle \int \frac {1}{\sqrt {\sec (x)-\cos (x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {\sec (x)-\cos (x)}}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \frac {1}{\sqrt {\sin (x) \tan (x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {\sin (x) \tan (x)}}dx\) |
\(\Big \downarrow \) 4900 |
\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \int \frac {1}{\sqrt {\sin (x)} \sqrt {\tan (x)}}dx}{\sqrt {\sin (x) \tan (x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \int \frac {1}{\sqrt {\sin (x)} \sqrt {\tan (x)}}dx}{\sqrt {\sin (x) \tan (x)}}\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle \frac {\sin (x) \int \sqrt {\cos (x)} \csc (x)dx}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (x) \int \frac {\sqrt {\cos (x)}}{\sin (x)}dx}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}}\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\sin (x) \int \frac {\sqrt {\cos (x)}}{1-\cos ^2(x)}d\cos (x)}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {2 \sin (x) \int \frac {\cos (x)}{1-\cos ^2(x)}d\sqrt {\cos (x)}}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle -\frac {2 \sin (x) \left (\frac {1}{2} \int \frac {1}{1-\cos (x)}d\sqrt {\cos (x)}-\frac {1}{2} \int \frac {1}{\cos (x)+1}d\sqrt {\cos (x)}\right )}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {2 \sin (x) \left (\frac {1}{2} \int \frac {1}{1-\cos (x)}d\sqrt {\cos (x)}-\frac {1}{2} \arctan \left (\sqrt {\cos (x)}\right )\right )}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2 \sin (x) \left (\frac {1}{2} \text {arctanh}\left (\sqrt {\cos (x)}\right )-\frac {1}{2} \arctan \left (\sqrt {\cos (x)}\right )\right )}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}}\) |
(-2*(-1/2*ArcTan[Sqrt[Cos[x]]] + ArcTanh[Sqrt[Cos[x]]]/2)*Sin[x])/(Sqrt[Co s[x]]*Sqrt[Sin[x]*Tan[x]])
3.4.37.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[Cos[e + f*x]^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^ n) Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(- 1)]) || IntegersQ[m - 1/2, n - 1/2])
Int[(u_.)*((v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> With[{uu = ActivateTri g[u], vv = ActivateTrig[v], ww = ActivateTrig[w]}, Simp[(vv^m*ww^n)^FracPar t[p]/(vv^(m*FracPart[p])*ww^(n*FracPart[p])) Int[uu*vv^(m*p)*ww^(n*p), x] , x]] /; FreeQ[{m, n, p}, x] && !IntegerQ[p] && ( !InertTrigFreeQ[v] || ! InertTrigFreeQ[w])
Leaf count of result is larger than twice the leaf count of optimal. \(89\) vs. \(2(40)=80\).
Time = 1.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.73
method | result | size |
default | \(\frac {\sin \left (x \right ) \left (\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}}\right )+\ln \left (\frac {2 \cos \left (x \right ) \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}-\cos \left (x \right )+1}{\cos \left (x \right )+1}\right )\right )}{2 \left (\cos \left (x \right )+1\right ) \sqrt {\sin \left (x \right ) \tan \left (x \right )}\, \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}}\) | \(90\) |
1/2*sin(x)*(arctan(1/2/(-cos(x)/(cos(x)+1)^2)^(1/2))+ln((2*cos(x)*(-cos(x) /(cos(x)+1)^2)^(1/2)+2*(-cos(x)/(cos(x)+1)^2)^(1/2)-cos(x)+1)/(cos(x)+1))) /(cos(x)+1)/(sin(x)*tan(x))^(1/2)/(-cos(x)/(cos(x)+1)^2)^(1/2)
Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=-\frac {1}{2} \, \arctan \left (\frac {2 \, \sqrt {-\frac {\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}} \cos \left (x\right )}{{\left (\cos \left (x\right ) - 1\right )} \sin \left (x\right )}\right ) + \frac {1}{2} \, \log \left (\frac {{\left (\cos \left (x\right ) + 1\right )} \sin \left (x\right ) - 2 \, \sqrt {-\frac {\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}} \cos \left (x\right )}{{\left (\cos \left (x\right ) - 1\right )} \sin \left (x\right )}\right ) \]
-1/2*arctan(2*sqrt(-(cos(x)^2 - 1)/cos(x))*cos(x)/((cos(x) - 1)*sin(x))) + 1/2*log(((cos(x) + 1)*sin(x) - 2*sqrt(-(cos(x)^2 - 1)/cos(x))*cos(x))/((c os(x) - 1)*sin(x)))
\[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=\int \frac {1}{\sqrt {- \cos {\left (x \right )} + \sec {\left (x \right )}}}\, dx \]
\[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=\int { \frac {1}{\sqrt {-\cos \left (x\right ) + \sec \left (x\right )}} \,d x } \]
Time = 0.34 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=\frac {1}{2} \, \arcsin \left (\tan \left (\frac {1}{2} \, x\right )^{2}\right ) - \frac {1}{2} \, \log \left (-\frac {\sqrt {-\tan \left (\frac {1}{2} \, x\right )^{4} + 1} - 1}{\tan \left (\frac {1}{2} \, x\right )^{2}}\right ) \]
Timed out. \[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=\int \frac {1}{\sqrt {\frac {1}{\cos \left (x\right )}-\cos \left (x\right )}} \,d x \]