\(\int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx\) [280]

Optimal result

Integrand size = 26, antiderivative size = 135 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}} \] Output:

-2^(1/2)*arctanh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/cos(d*x+c)^(1/2)/(a-a*cos( 
d*x+c))^(1/2))/a^(1/2)/d+2/3*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a-a*cos(d*x+c) 
)^(1/2)+2/3*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a-a*cos(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.77 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.27 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=\frac {2 \left (-\frac {3 e^{-\frac {3}{2} i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^2 \text {arctanh}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )}{2 \sqrt {2}}+2 \sqrt {1+e^{2 i (c+d x)}} \cos \left (\frac {1}{2} (c+d x)\right ) (1+\cos (c+d x))\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{3 d \sqrt {1+e^{2 i (c+d x)}} \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \] Input:

Integrate[1/(Cos[c + d*x]^(5/2)*Sqrt[a - a*Cos[c + d*x]]),x]
 

Output:

(2*((-3*(1 + E^((2*I)*(c + d*x)))^2*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2] 
*Sqrt[1 + E^((2*I)*(c + d*x))])])/(2*Sqrt[2]*E^(((3*I)/2)*(c + d*x))) + 2* 
Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[(c + d*x)/2]*(1 + Cos[c + d*x]))*Sin[(c 
+ d*x)/2])/(3*d*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^(3/2)*Sqrt[a - 
a*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 3258, 3042, 3463, 27, 3042, 3261, 221}

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.

\(\displaystyle \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3258

\(\displaystyle \frac {\int \frac {2 \cos (c+d x) a+a}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}dx}{3 a}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {2 \sin \left (c+d x+\frac {\pi }{2}\right ) a+a}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}-\frac {2 \int -\frac {3 a^2}{2 \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}dx}{a}}{3 a}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 a \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}dx+\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}}{3 a}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}}{3 a}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\)

\(\Big \downarrow \) 3261

\(\displaystyle \frac {\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}-\frac {6 a^2 \int \frac {1}{2 a^2-\frac {a^3 \sin (c+d x) \tan (c+d x)}{a-a \cos (c+d x)}}d\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}}{d}}{3 a}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}-\frac {3 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{d}}{3 a}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\)

Input:

Int[1/(Cos[c + d*x]^(5/2)*Sqrt[a - a*Cos[c + d*x]]),x]
 

Output:

(2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a - a*Cos[c + d*x]]) + ((-3* 
Sqrt[2]*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]] 
*Sqrt[a - a*Cos[c + d*x]])])/d + (2*a*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]* 
Sqrt[a - a*Cos[c + d*x]]))/(3*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 221

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

rule 3042

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

rule 3258

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. 
) + (f_.)*(x_)]], x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((c + d*Sin[e + f*x]) 
^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(2* 
b*(n + 1)*(c^2 - d^2))   Int[(c + d*Sin[e + f*x])^(n + 1)*(Simp[a*d - 2*b*c 
*(n + 1) + b*d*(2*n + 3)*Sin[e + f*x], x]/Sqrt[a + b*Sin[e + f*x]]), x], x] 
 /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] 
&& NeQ[c^2 - d^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
 

rule 3261

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f)   Subst[Int[1/(2*b^2 - (a*c 
 - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S 
in[e + f*x]]))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 
 EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3463

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n 
+ 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2))   Int[(a 
 + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 
1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], 
 x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq 
Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m 
 + 1/2, 0])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(386\) vs. \(2(112)=224\).

Time = 13.65 (sec) , antiderivative size = 387, normalized size of antiderivative = 2.87

method	result	size
default	\(-\frac {\sqrt {2}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3\right ) \operatorname {arctanh}\left (\frac {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}+\left (-6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right ) \ln \left (\frac {2 \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}-4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1}\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}\right ) \sqrt {4}}{12 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{\frac {3}{2}} \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sqrt {2}-3\right )^{2} \left (3+2 \sqrt {2}\right )^{2}}\)	\(387\)

Input:

int(1/cos(d*x+c)^(5/2)/(a-a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/12*2^(1/2)/d*sin(1/2*d*x+1/2*c)*((6*cos(1/2*d*x+1/2*c)^3+6*cos(1/2*d*x+ 
1/2*c)^2-3*cos(1/2*d*x+1/2*c)-3)*arctanh((-1+2*cos(1/2*d*x+1/2*c))/(cos(1/ 
2*d*x+1/2*c)+1)/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2 
))*((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)+(-6*cos(1/2 
*d*x+1/2*c)^3-6*cos(1/2*d*x+1/2*c)^2+3*cos(1/2*d*x+1/2*c)+3)*ln(2*(((2*cos 
(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*cos(1/2*d*x+1/2*c)+(( 
2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)-2*cos(1/2*d*x+1/ 
2*c)-1)/(cos(1/2*d*x+1/2*c)+1))*((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1 
/2*c)+1)^2)^(1/2)-8*cos(1/2*d*x+1/2*c)^3)/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2) 
/(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*4^(1/2)/(2*2^(1/2)-3)^2/(3+2*2^(1/2))^2
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.22 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=\frac {3 \, \sqrt {2} \sqrt {a} \cos \left (d x + c\right )^{2} \log \left (-\frac {\frac {2 \, \sqrt {2} \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 4 \, \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{6 \, a d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )} \] Input:

integrate(1/cos(d*x+c)^(5/2)/(a-a*cos(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

1/6*(3*sqrt(2)*sqrt(a)*cos(d*x + c)^2*log(-(2*sqrt(2)*sqrt(-a*cos(d*x + c) 
 + a)*(cos(d*x + c) + 1)*sqrt(cos(d*x + c))/sqrt(a) - (3*cos(d*x + c) + 1) 
*sin(d*x + c))/((cos(d*x + c) - 1)*sin(d*x + c)))*sin(d*x + c) + 4*sqrt(-a 
*cos(d*x + c) + a)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(cos(d*x + c) 
))/(a*d*cos(d*x + c)^2*sin(d*x + c))
 

Sympy [F]

\[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=\int \frac {1}{\sqrt {- a \left (\cos {\left (c + d x \right )} - 1\right )} \cos ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \] Input:

integrate(1/cos(d*x+c)**(5/2)/(a-a*cos(d*x+c))**(1/2),x)
 

Output:

Integral(1/(sqrt(-a*(cos(c + d*x) - 1))*cos(c + d*x)**(5/2)), x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 504, normalized size of antiderivative = 3.73 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(1/cos(d*x+c)^(5/2)/(a-a*cos(d*x+c))^(1/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

-1/3*(3*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt( 
2)*cos(2*d*x + 2*c) + sqrt(2))*arctan2(2*sqrt(2)*(cos(2*d*x + 2*c)^2 + sin 
(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x 
+ 2*c), cos(2*d*x + 2*c) + 1))/(sqrt(a)*abs(e^(I*d*x + I*c) - 1)), 2*(sqrt 
(2)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/ 
4)*sqrt(a)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - sqrt 
(-a)*abs(e^(I*d*x + I*c) - 1) + 2*sqrt(a))/(a*abs(e^(I*d*x + I*c) - 1))) - 
 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4 
)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - 
 (cos(d*x + c) + 3)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1 
))) - 4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) 
^(1/4)*(cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + 
 c) - (cos(d*x + c) - 1)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c 
) + 1))))/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 
 1)*sqrt(-a)*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (112) = 224\).

Time = 0.76 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.74 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} {\left (\frac {8 \, {\left ({\left ({\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 3\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 3\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 1\right )}}{{\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1\right )}^{\frac {3}{2}}} - 3 \, \log \left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1\right ) + 3 \, \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 3 \right |}\right ) + 3 \, \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1 \right |}\right )\right )}}{6 \, \sqrt {a} d \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \] Input:

integrate(1/cos(d*x+c)^(5/2)/(a-a*cos(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

-1/6*sqrt(2)*(8*(((tan(1/4*d*x + 1/4*c)^2 - 3)*tan(1/4*d*x + 1/4*c)^2 + 3) 
*tan(1/4*d*x + 1/4*c)^2 - 1)/(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4 
*c)^2 + 1)^(3/2) - 3*log(tan(1/4*d*x + 1/4*c)^2 - sqrt(tan(1/4*d*x + 1/4*c 
)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 1) + 3*log(abs(-tan(1/4*d*x + 1/4*c) 
^2 + sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 3)) + 3 
*log(abs(-tan(1/4*d*x + 1/4*c)^2 + sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4 
*d*x + 1/4*c)^2 + 1) + 1)))/(sqrt(a)*d*sgn(sin(1/2*d*x + 1/2*c)))
 

Mupad [F(-1)]

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

int(1/(cos(c + d*x)^(5/2)*(a - a*cos(c + d*x))^(1/2)),x)
 

Output:

int(1/(cos(c + d*x)^(5/2)*(a - a*cos(c + d*x))^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=-\frac {\sqrt {a}\, \left (\int \frac {\sqrt {-\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}-\cos \left (d x +c \right )^{3}}d x \right )}{a} \] Input:

int(1/cos(d*x+c)^(5/2)/(a-a*cos(d*x+c))^(1/2),x)
 

Output:

( - sqrt(a)*int((sqrt( - cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d* 
x)**4 - cos(c + d*x)**3),x))/a