3.4.39 \(\int \frac {1}{(-\cos (x)+\sec (x))^{5/2}} \, dx\) [339]

3.4.39.1 Optimal result
3.4.39.2 Mathematica [A] (verified)
3.4.39.3 Rubi [A] (verified)
3.4.39.4 Maple [B] (verified)
3.4.39.5 Fricas [B] (verification not implemented)
3.4.39.6 Sympy [F]
3.4.39.7 Maxima [F]
3.4.39.8 Giac [B] (verification not implemented)
3.4.39.9 Mupad [F(-1)]

3.4.39.1 Optimal result

Integrand size = 11, antiderivative size = 91 \[ \int \frac {1}{(-\cos (x)+\sec (x))^{5/2}} \, dx=\frac {3 \cot (x)}{16 \sqrt {\sin (x) \tan (x)}}-\frac {\cot (x) \csc ^2(x)}{4 \sqrt {\sin (x) \tan (x)}}-\frac {3 \arctan \left (\sqrt {\cos (x)}\right ) \sin (x)}{32 \sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}}+\frac {3 \text {arctanh}\left (\sqrt {\cos (x)}\right ) \sin (x)}{32 \sqrt {\cos (x)} \sqrt {\sin (x) \tan (x)}} \]

output
3/16*cot(x)/(sin(x)*tan(x))^(1/2)-1/4*cot(x)*csc(x)^2/(sin(x)*tan(x))^(1/2 
)-3/32*arctan(cos(x)^(1/2))*sin(x)/cos(x)^(1/2)/(sin(x)*tan(x))^(1/2)+3/32 
*arctanh(cos(x)^(1/2))*sin(x)/cos(x)^(1/2)/(sin(x)*tan(x))^(1/2)
 
3.4.39.2 Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(-\cos (x)+\sec (x))^{5/2}} \, dx=-\frac {\cot (x) \left (3 \arctan \left (\sqrt [4]{\cos ^2(x)}\right ) \cos (x)-3 \text {arctanh}\left (\sqrt [4]{\cos ^2(x)}\right ) \cos (x)+\cos ^2(x)^{3/4} (5+3 \cos (2 x)) \cot (x) \csc ^3(x)\right ) \sqrt {\sin (x) \tan (x)}}{32 \cos ^2(x)^{3/4}} \]

input
Integrate[(-Cos[x] + Sec[x])^(-5/2),x]
 
output
-1/32*(Cot[x]*(3*ArcTan[(Cos[x]^2)^(1/4)]*Cos[x] - 3*ArcTanh[(Cos[x]^2)^(1 
/4)]*Cos[x] + (Cos[x]^2)^(3/4)*(5 + 3*Cos[2*x])*Cot[x]*Csc[x]^3)*Sqrt[Sin[ 
x]*Tan[x]])/(Cos[x]^2)^(3/4)
 
3.4.39.3 Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.15, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.455, Rules used = {3042, 4897, 3042, 4900, 3042, 3077, 3042, 3079, 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.

\(\displaystyle \int \frac {1}{(\sec (x)-\cos (x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(\sec (x)-\cos (x))^{5/2}}dx\)

\(\Big \downarrow \) 4897

\(\displaystyle \int \frac {1}{(\sin (x) \tan (x))^{5/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(\sin (x) \tan (x))^{5/2}}dx\)

\(\Big \downarrow \) 4900

\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \int \frac {1}{\sin ^{\frac {5}{2}}(x) \tan ^{\frac {5}{2}}(x)}dx}{\sqrt {\sin (x) \tan (x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \int \frac {1}{\sin (x)^{5/2} \tan (x)^{5/2}}dx}{\sqrt {\sin (x) \tan (x)}}\)

\(\Big \downarrow \) 3077

\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \left (-\frac {3}{8} \int \frac {1}{\sin ^{\frac {5}{2}}(x) \sqrt {\tan (x)}}dx-\frac {1}{4 \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)}\right )}{\sqrt {\sin (x) \tan (x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \left (-\frac {3}{8} \int \frac {1}{\sin (x)^{5/2} \sqrt {\tan (x)}}dx-\frac {1}{4 \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)}\right )}{\sqrt {\sin (x) \tan (x)}}\)

\(\Big \downarrow \) 3079

\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \left (-\frac {3}{8} \left (\frac {1}{4} \int \frac {1}{\sqrt {\sin (x)} \sqrt {\tan (x)}}dx-\frac {1}{2 \sqrt {\sin (x)} \tan ^{\frac {3}{2}}(x)}\right )-\frac {1}{4 \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)}\right )}{\sqrt {\sin (x) \tan (x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \left (-\frac {3}{8} \left (\frac {1}{4} \int \frac {1}{\sqrt {\sin (x)} \sqrt {\tan (x)}}dx-\frac {1}{2 \sqrt {\sin (x)} \tan ^{\frac {3}{2}}(x)}\right )-\frac {1}{4 \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)}\right )}{\sqrt {\sin (x) \tan (x)}}\)

\(\Big \downarrow \) 3081

\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \left (-\frac {3}{8} \left (\frac {\sqrt {\sin (x)} \int \sqrt {\cos (x)} \csc (x)dx}{4 \sqrt {\cos (x)} \sqrt {\tan (x)}}-\frac {1}{2 \sqrt {\sin (x)} \tan ^{\frac {3}{2}}(x)}\right )-\frac {1}{4 \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)}\right )}{\sqrt {\sin (x) \tan (x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \left (-\frac {3}{8} \left (\frac {\sqrt {\sin (x)} \int \frac {\sqrt {\cos (x)}}{\sin (x)}dx}{4 \sqrt {\cos (x)} \sqrt {\tan (x)}}-\frac {1}{2 \sqrt {\sin (x)} \tan ^{\frac {3}{2}}(x)}\right )-\frac {1}{4 \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)}\right )}{\sqrt {\sin (x) \tan (x)}}\)

\(\Big \downarrow \) 3045

\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \left (-\frac {3}{8} \left (-\frac {\sqrt {\sin (x)} \int \frac {\sqrt {\cos (x)}}{1-\cos ^2(x)}d\cos (x)}{4 \sqrt {\cos (x)} \sqrt {\tan (x)}}-\frac {1}{2 \sqrt {\sin (x)} \tan ^{\frac {3}{2}}(x)}\right )-\frac {1}{4 \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)}\right )}{\sqrt {\sin (x) \tan (x)}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \left (-\frac {3}{8} \left (-\frac {\sqrt {\sin (x)} \int \frac {\cos (x)}{1-\cos ^2(x)}d\sqrt {\cos (x)}}{2 \sqrt {\cos (x)} \sqrt {\tan (x)}}-\frac {1}{2 \sqrt {\sin (x)} \tan ^{\frac {3}{2}}(x)}\right )-\frac {1}{4 \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)}\right )}{\sqrt {\sin (x) \tan (x)}}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \left (-\frac {3}{8} \left (-\frac {\sqrt {\sin (x)} \left (\frac {1}{2} \int \frac {1}{1-\cos (x)}d\sqrt {\cos (x)}-\frac {1}{2} \int \frac {1}{\cos (x)+1}d\sqrt {\cos (x)}\right )}{2 \sqrt {\cos (x)} \sqrt {\tan (x)}}-\frac {1}{2 \sqrt {\sin (x)} \tan ^{\frac {3}{2}}(x)}\right )-\frac {1}{4 \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)}\right )}{\sqrt {\sin (x) \tan (x)}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \left (-\frac {3}{8} \left (-\frac {\sqrt {\sin (x)} \left (\frac {1}{2} \int \frac {1}{1-\cos (x)}d\sqrt {\cos (x)}-\frac {1}{2} \arctan \left (\sqrt {\cos (x)}\right )\right )}{2 \sqrt {\cos (x)} \sqrt {\tan (x)}}-\frac {1}{2 \sqrt {\sin (x)} \tan ^{\frac {3}{2}}(x)}\right )-\frac {1}{4 \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)}\right )}{\sqrt {\sin (x) \tan (x)}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {\sin (x)} \sqrt {\tan (x)} \left (-\frac {3}{8} \left (-\frac {\sqrt {\sin (x)} \left (\frac {1}{2} \text {arctanh}\left (\sqrt {\cos (x)}\right )-\frac {1}{2} \arctan \left (\sqrt {\cos (x)}\right )\right )}{2 \sqrt {\cos (x)} \sqrt {\tan (x)}}-\frac {1}{2 \sqrt {\sin (x)} \tan ^{\frac {3}{2}}(x)}\right )-\frac {1}{4 \sin ^{\frac {5}{2}}(x) \tan ^{\frac {3}{2}}(x)}\right )}{\sqrt {\sin (x) \tan (x)}}\)

input
Int[(-Cos[x] + Sec[x])^(-5/2),x]
 
output
(Sqrt[Sin[x]]*((-3*(-1/2*1/(Sqrt[Sin[x]]*Tan[x]^(3/2)) - ((-1/2*ArcTan[Sqr 
t[Cos[x]]] + ArcTanh[Sqrt[Cos[x]]]/2)*Sqrt[Sin[x]])/(2*Sqrt[Cos[x]]*Sqrt[T 
an[x]])))/8 - 1/(4*Sin[x]^(5/2)*Tan[x]^(3/2)))*Sqrt[Tan[x]])/Sqrt[Sin[x]*T 
an[x]]
 

3.4.39.3.1 Defintions of rubi rules used

rule 216
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])
 

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 266
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]
 

rule 827
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]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3045
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])
 

rule 3077
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n + 1)/(b*f*(m 
 + n + 1))), x] - Simp[(n + 1)/(b^2*(m + n + 1))   Int[(a*Sin[e + f*x])^m*( 
b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] 
&& NeQ[m + n + 1, 0] && IntegersQ[2*m, 2*n] &&  !(EqQ[n, -3/2] && EqQ[m, 1] 
)
 

rule 3079
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_.), x_Symbol] :> Simp[b*(a*Sin[e + f*x])^(m + 2)*((b*Tan[e + f*x])^(n - 1) 
/(a^2*f*(m + n + 1))), x] + Simp[(m + 2)/(a^2*(m + n + 1))   Int[(a*Sin[e + 
 f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && L 
tQ[m, -1] && NeQ[m + n + 1, 0] && IntegersQ[2*m, 2*n]
 

rule 3081
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])
 

rule 4897
Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]
 

rule 4900
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])
 
3.4.39.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(194\) vs. \(2(67)=134\).

Time = 1.23 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.14

method result size
default \(-\frac {12 \cot \left (x \right )^{3} \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}-3 \cot \left (x \right ) \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}}\right )-3 \cot \left (x \right ) \ln \left (\frac {2 \cos \left (x \right ) \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}-\cos \left (x \right )+1}{\cos \left (x \right )+1}\right )+3 \csc \left (x \right ) \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}}\right )+3 \csc \left (x \right ) \ln \left (\frac {2 \cos \left (x \right ) \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}-\cos \left (x \right )+1}{\cos \left (x \right )+1}\right )+4 \csc \left (x \right )^{2} \cot \left (x \right ) \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}}{64 \sqrt {\sin \left (x \right ) \tan \left (x \right )}\, \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}}\) \(195\)

input
int(1/(-cos(x)+sec(x))^(5/2),x,method=_RETURNVERBOSE)
 
output
-1/64/(sin(x)*tan(x))^(1/2)/(-cos(x)/(cos(x)+1)^2)^(1/2)*(12*cot(x)^3*(-co 
s(x)/(cos(x)+1)^2)^(1/2)-3*cot(x)*arctan(1/2/(-cos(x)/(cos(x)+1)^2)^(1/2)) 
-3*cot(x)*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))+3*csc(x)*arctan(1/2/(-cos(x)/(cos(x)+1)^2)^ 
(1/2))+3*csc(x)*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))+4*csc(x)^2*cot(x)*(-cos(x)/(cos(x)+1) 
^2)^(1/2))
 
3.4.39.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (67) = 134\).

Time = 0.27 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.62 \[ \int \frac {1}{(-\cos (x)+\sec (x))^{5/2}} \, dx=\frac {3 \, {\left (\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1\right )} \arctan \left (\frac {2 \, \sqrt {-\frac {\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}} \cos \left (x\right )}{{\left (\cos \left (x\right ) - 1\right )} \sin \left (x\right )}\right ) \sin \left (x\right ) + 3 \, {\left (\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1\right )} \log \left (\frac {{\left (\cos \left (x\right ) + 1\right )} \sin \left (x\right ) + 2 \, \sqrt {-\frac {\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}} \cos \left (x\right )}{{\left (\cos \left (x\right ) - 1\right )} \sin \left (x\right )}\right ) \sin \left (x\right ) - 4 \, {\left (3 \, \cos \left (x\right )^{4} + \cos \left (x\right )^{2}\right )} \sqrt {-\frac {\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}}}{64 \, {\left (\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1\right )} \sin \left (x\right )} \]

input
integrate(1/(-cos(x)+sec(x))^(5/2),x, algorithm="fricas")
 
output
1/64*(3*(cos(x)^4 - 2*cos(x)^2 + 1)*arctan(2*sqrt(-(cos(x)^2 - 1)/cos(x))* 
cos(x)/((cos(x) - 1)*sin(x)))*sin(x) + 3*(cos(x)^4 - 2*cos(x)^2 + 1)*log(( 
(cos(x) + 1)*sin(x) + 2*sqrt(-(cos(x)^2 - 1)/cos(x))*cos(x))/((cos(x) - 1) 
*sin(x)))*sin(x) - 4*(3*cos(x)^4 + cos(x)^2)*sqrt(-(cos(x)^2 - 1)/cos(x))) 
/((cos(x)^4 - 2*cos(x)^2 + 1)*sin(x))
 
3.4.39.6 Sympy [F]

\[ \int \frac {1}{(-\cos (x)+\sec (x))^{5/2}} \, dx=\int \frac {1}{\left (- \cos {\left (x \right )} + \sec {\left (x \right )}\right )^{\frac {5}{2}}}\, dx \]

input
integrate(1/(-cos(x)+sec(x))**(5/2),x)
 
output
Integral((-cos(x) + sec(x))**(-5/2), x)
 
3.4.39.7 Maxima [F]

\[ \int \frac {1}{(-\cos (x)+\sec (x))^{5/2}} \, dx=\int { \frac {1}{{\left (-\cos \left (x\right ) + \sec \left (x\right )\right )}^{\frac {5}{2}}} \,d x } \]

input
integrate(1/(-cos(x)+sec(x))^(5/2),x, algorithm="maxima")
 
output
integrate((-cos(x) + sec(x))^(-5/2), x)
 
3.4.39.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (67) = 134\).

Time = 0.38 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.66 \[ \int \frac {1}{(-\cos (x)+\sec (x))^{5/2}} \, dx=\frac {{\left (\frac {4 \, {\left (\sqrt {-\tan \left (\frac {1}{2} \, x\right )^{4} + 1} - 1\right )}}{\tan \left (\frac {1}{2} \, x\right )^{2}} + 1\right )} \tan \left (\frac {1}{2} \, x\right )^{4}}{256 \, {\left (\sqrt {-\tan \left (\frac {1}{2} \, x\right )^{4} + 1} - 1\right )}^{2}} - \frac {1}{64} \, \sqrt {-\tan \left (\frac {1}{2} \, x\right )^{4} + 1} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} - 2\right )} - \frac {\sqrt {-\tan \left (\frac {1}{2} \, x\right )^{4} + 1} - 1}{64 \, \tan \left (\frac {1}{2} \, x\right )^{2}} - \frac {{\left (\sqrt {-\tan \left (\frac {1}{2} \, x\right )^{4} + 1} - 1\right )}^{2}}{256 \, \tan \left (\frac {1}{2} \, x\right )^{4}} - \frac {3}{64} \, \arcsin \left (\tan \left (\frac {1}{2} \, x\right )^{2}\right ) + \frac {3}{64} \, \log \left (-\frac {\sqrt {-\tan \left (\frac {1}{2} \, x\right )^{4} + 1} - 1}{\tan \left (\frac {1}{2} \, x\right )^{2}}\right ) \]

input
integrate(1/(-cos(x)+sec(x))^(5/2),x, algorithm="giac")
 
output
1/256*(4*(sqrt(-tan(1/2*x)^4 + 1) - 1)/tan(1/2*x)^2 + 1)*tan(1/2*x)^4/(sqr 
t(-tan(1/2*x)^4 + 1) - 1)^2 - 1/64*sqrt(-tan(1/2*x)^4 + 1)*(tan(1/2*x)^2 - 
 2) - 1/64*(sqrt(-tan(1/2*x)^4 + 1) - 1)/tan(1/2*x)^2 - 1/256*(sqrt(-tan(1 
/2*x)^4 + 1) - 1)^2/tan(1/2*x)^4 - 3/64*arcsin(tan(1/2*x)^2) + 3/64*log(-( 
sqrt(-tan(1/2*x)^4 + 1) - 1)/tan(1/2*x)^2)
 
3.4.39.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(-\cos (x)+\sec (x))^{5/2}} \, dx=\int \frac {1}{{\left (\frac {1}{\cos \left (x\right )}-\cos \left (x\right )\right )}^{5/2}} \,d x \]

input
int(1/(1/cos(x) - cos(x))^(5/2),x)
 
output
int(1/(1/cos(x) - cos(x))^(5/2), x)