\(\int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx\) [337]

Optimal result

Integrand size = 11, antiderivative size = 52 \[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=\frac {\arctan \left (\sqrt {\cos (x)}\right ) \sin (x)}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}}-\frac {\text {arctanh}\left (\sqrt {\cos (x)}\right ) \sin (x)}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}} \]

[Out]

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)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {4482, 4485, 2681, 2645, 335, 304, 209, 212} \[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=\frac {\sin (x) \arctan \left (\sqrt {\cos (x)}\right )}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}}-\frac {\sin (x) \text {arctanh}\left (\sqrt {\cos (x)}\right )}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}} \]

[In]

Int[1/Sqrt[-Cos[x] + Sec[x]],x]

[Out]

(ArcTan[Sqrt[Cos[x]]]*Sin[x])/(Sqrt[Cos[x]]*Sqrt[Sin[x]*Tan[x]]) - (ArcTanh[Sqrt[Cos[x]]]*Sin[x])/(Sqrt[Cos[x]
]*Sqrt[Sin[x]*Tan[x]])

Rule 209

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(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])

Rule 2681

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[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])

Rule 4482

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 4485

Int[(u_.)*((v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> With[{uu = ActivateTrig[u], vv = ActivateTrig[v], ww = Ac
tivateTrig[w]}, Dist[(vv^m*ww^n)^FracPart[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])

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\sqrt {\sin (x) \tan (x)}} \, dx \\ & = \frac {\left (\sqrt {\sin (x)} \sqrt {\tan (x)}\right ) \int \frac {1}{\sqrt {\sin (x)} \sqrt {\tan (x)}} \, dx}{\sqrt {\sin (x) \tan (x)}} \\ & = \frac {\sin (x) \int \sqrt {\cos (x)} \csc (x) \, dx}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}} \\ & = -\frac {\sin (x) \text {Subst}\left (\int \frac {\sqrt {x}}{1-x^2} \, dx,x,\cos (x)\right )}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}} \\ & = -\frac {(2 \sin (x)) \text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\sqrt {\cos (x)}\right )}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}} \\ & = -\frac {\sin (x) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {\cos (x)}\right )}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}}+\frac {\sin (x) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {\cos (x)}\right )}{\sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}} \\ & = \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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=\frac {\left (\arctan \left (\sqrt [4]{\cos ^2(x)}\right )-\text {arctanh}\left (\sqrt [4]{\cos ^2(x)}\right )\right ) \cos (x) \cot (x) \sqrt {\sin (x) \tan (x)}}{\cos ^2(x)^{3/4}} \]

[In]

Integrate[1/Sqrt[-Cos[x] + Sec[x]],x]

[Out]

((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)

Maple [B] (verified)

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	\(\frac {\sin \left (x \right ) \left (\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}}\right )+\ln \left (\frac {2 \cos \left (x \right ) \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}-\cos \left (x \right )+1}{\cos \left (x \right )+1}\right )\right )}{2 \left (\cos \left (x \right )+1\right ) \sqrt {\sin \left (x \right ) \tan \left (x \right )}\, \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}}\)	\(90\)

[In]

int(1/(-cos(x)+sec(x))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=-\frac {1}{2} \, \arctan \left (\frac {2 \, \sqrt {-\frac {\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}} \cos \left (x\right )}{{\left (\cos \left (x\right ) - 1\right )} \sin \left (x\right )}\right ) + \frac {1}{2} \, \log \left (\frac {{\left (\cos \left (x\right ) + 1\right )} \sin \left (x\right ) - 2 \, \sqrt {-\frac {\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}} \cos \left (x\right )}{{\left (\cos \left (x\right ) - 1\right )} \sin \left (x\right )}\right ) \]

[In]

integrate(1/(-cos(x)+sec(x))^(1/2),x, algorithm="fricas")

[Out]

-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*sq
rt(-(cos(x)^2 - 1)/cos(x))*cos(x))/((cos(x) - 1)*sin(x)))

Sympy [F]

\[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=\int \frac {1}{\sqrt {- \cos {\left (x \right )} + \sec {\left (x \right )}}}\, dx \]

[In]

integrate(1/(-cos(x)+sec(x))**(1/2),x)

[Out]

Integral(1/sqrt(-cos(x) + sec(x)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=\int { \frac {1}{\sqrt {-\cos \left (x\right ) + \sec \left (x\right )}} \,d x } \]

[In]

integrate(1/(-cos(x)+sec(x))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(-cos(x) + sec(x)), x)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=\frac {1}{2} \, \arcsin \left (\tan \left (\frac {1}{2} \, x\right )^{2}\right ) - \frac {1}{2} \, \log \left (-\frac {\sqrt {-\tan \left (\frac {1}{2} \, x\right )^{4} + 1} - 1}{\tan \left (\frac {1}{2} \, x\right )^{2}}\right ) \]

[In]

integrate(1/(-cos(x)+sec(x))^(1/2),x, algorithm="giac")

[Out]

1/2*arcsin(tan(1/2*x)^2) - 1/2*log(-(sqrt(-tan(1/2*x)^4 + 1) - 1)/tan(1/2*x)^2)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {-\cos (x)+\sec (x)}} \, dx=\int \frac {1}{\sqrt {\frac {1}{\cos \left (x\right )}-\cos \left (x\right )}} \,d x \]

[In]

int(1/(1/cos(x) - cos(x))^(1/2),x)

[Out]

int(1/(1/cos(x) - cos(x))^(1/2), x)