\(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx\) [282]

Optimal result

Integrand size = 25, antiderivative size = 161 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=\frac {7 \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{4 d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {2} \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}+\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{4 d \sqrt {1-\cos (c+d x)}}+\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {1-\cos (c+d x)}} \] Output:

7/4*arctanh(sin(d*x+c)/(1-cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2))/d-2^(1/2)*ar 
ctanh(1/2*sin(d*x+c)*2^(1/2)/(1-cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2))/d+1/4* 
cos(d*x+c)^(1/2)*sin(d*x+c)/d/(1-cos(d*x+c))^(1/2)+1/2*cos(d*x+c)^(3/2)*si 
n(d*x+c)/d/(1-cos(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.76 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.58 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=-\frac {i e^{-2 i (c+d x)} \left (-1+e^{i (c+d x)}\right ) \left (7 \sqrt {2} e^{2 i (c+d x)} \text {arcsinh}\left (e^{i (c+d x)}\right )-16 e^{2 i (c+d x)} \text {arctanh}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\sqrt {2} \left (\sqrt {1+e^{2 i (c+d x)}} \left (1+2 e^{i (c+d x)}+2 e^{2 i (c+d x)}+e^{3 i (c+d x)}\right )+7 e^{2 i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )\right ) \sqrt {\cos (c+d x)}}{8 \sqrt {2} d \sqrt {1+e^{2 i (c+d x)}} \sqrt {1-\cos (c+d x)}} \] Input:

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

Output:

((-1/8*I)*(-1 + E^(I*(c + d*x)))*(7*Sqrt[2]*E^((2*I)*(c + d*x))*ArcSinh[E^ 
(I*(c + d*x))] - 16*E^((2*I)*(c + d*x))*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqr 
t[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + Sqrt[2]*(Sqrt[1 + E^((2*I)*(c + d*x 
))]*(1 + 2*E^(I*(c + d*x)) + 2*E^((2*I)*(c + d*x)) + E^((3*I)*(c + d*x))) 
+ 7*E^((2*I)*(c + d*x))*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))*Sqrt[Cos[ 
c + d*x]])/(Sqrt[2]*d*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Sq 
rt[1 - Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3257, 3042, 3462, 27, 3042, 3461, 3042, 3254, 220, 3261, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3257

\(\displaystyle \frac {1}{4} \int \frac {\sqrt {\cos (c+d x)} (\cos (c+d x)+3)}{\sqrt {1-\cos (c+d x)}}dx+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {1-\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right )+3\right )}{\sqrt {1-\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {1-\cos (c+d x)}}\)

\(\Big \downarrow \) 3462

\(\displaystyle \frac {1}{4} \left (\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {1-\cos (c+d x)}}-\int -\frac {7 \cos (c+d x)+1}{2 \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}dx\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {1-\cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {7 \cos (c+d x)+1}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}dx+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {1-\cos (c+d x)}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {1-\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {7 \sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sqrt {1-\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {1-\cos (c+d x)}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {1-\cos (c+d x)}}\)

\(\Big \downarrow \) 3461

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (8 \int \frac {1}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}dx-7 \int \frac {\sqrt {1-\cos (c+d x)}}{\sqrt {\cos (c+d x)}}dx\right )+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {1-\cos (c+d x)}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {1-\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (8 \int \frac {1}{\sqrt {1-\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-7 \int \frac {\sqrt {1-\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {1-\cos (c+d x)}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {1-\cos (c+d x)}}\)

\(\Big \downarrow \) 3254

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (8 \int \frac {1}{\sqrt {1-\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {14 \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x)}{1-\cos (c+d x)}-1}d\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}}{d}\right )+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {1-\cos (c+d x)}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {1-\cos (c+d x)}}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (8 \int \frac {1}{\sqrt {1-\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {14 \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}\right )+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {1-\cos (c+d x)}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {1-\cos (c+d x)}}\)

\(\Big \downarrow \) 3261

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {14 \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}-\frac {16 \int \frac {1}{2-\frac {\sin (c+d x) \tan (c+d x)}{1-\cos (c+d x)}}d\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}}{d}\right )+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {1-\cos (c+d x)}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {1-\cos (c+d x)}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {14 \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}-\frac {8 \sqrt {2} \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {2} \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}\right )+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {1-\cos (c+d x)}}\right )+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {1-\cos (c+d x)}}\)

Input:

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

Output:

(Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[1 - Cos[c + d*x]]) + (((14*Arc 
Tanh[Sin[c + d*x]/(Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])])/d - (8*Sqr 
t[2]*ArcTanh[Sin[c + d*x]/(Sqrt[2]*Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x 
]])])/d)/2 + (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[1 - Cos[c + d*x]])) 
/4
 

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 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 220

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

rule 3042

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

rule 3254

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b + d*x^2), x], 
x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[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 3257

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. 
) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*d*Cos[e + f*x]*((c + d*Sin[e + f*x]) 
^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(b*(2*n - 1)) 
   Int[((c + d*Sin[e + f*x])^(n - 2)/Sqrt[a + b*Sin[e + f*x]])*Simp[a*c*d - 
 b*(2*d^2*(n - 1) + c^2*(2*n - 1)) + d*(a*d - b*c*(4*n - 3))*Sin[e + f*x], 
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] && GtQ[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 3461

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

rule 3462

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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + 
n + 1))), x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Sin[e + f*x])^m*(c + d*S 
in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m 
 + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
- d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 0])
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 8.56 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\frac {\sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \left (4 \cos \left (d x +c \right )+2\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{2}+7 \sqrt {2}\, \operatorname {arctanh}\left (\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )-8 \,\operatorname {arctanh}\left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \operatorname {csgn}\left (\sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(147\)

Input:

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

Output:

1/8/d*sec(1/2*d*x+1/2*c)*(1/2*2^(1/2)*(cos(d*x+c)+1)*(4*cos(d*x+c)+2)*(cos 
(d*x+c)/(cos(d*x+c)+1))^(1/2)+7*2^(1/2)*arctanh((cos(d*x+c)/(cos(d*x+c)+1) 
)^(1/2))-8*arctanh(1/2*2^(1/2)/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)))*cos(d*x 
+c)^(1/2)*csgn(sin(1/2*d*x+1/2*c))/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=\frac {4 \, \sqrt {2} \log \left (-\frac {2 \, {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 2 \, {\left (2 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} + 7 \, \log \left (\frac {2 \, {\left (\sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 7 \, \log \left (\frac {2 \, {\left (\sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right )}{8 \, d \sin \left (d x + c\right )} \] Input:

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

Output:

1/8*(4*sqrt(2)*log(-(2*(sqrt(2)*cos(d*x + c) + sqrt(2))*sqrt(-cos(d*x + c) 
 + 1)*sqrt(cos(d*x + c)) - (3*cos(d*x + c) + 1)*sin(d*x + c))/((cos(d*x + 
c) - 1)*sin(d*x + c)))*sin(d*x + c) + 2*(2*cos(d*x + c)^2 + 3*cos(d*x + c) 
 + 1)*sqrt(-cos(d*x + c) + 1)*sqrt(cos(d*x + c)) + 7*log(2*(sqrt(-cos(d*x 
+ c) + 1)*sqrt(cos(d*x + c)) + sin(d*x + c))/sin(d*x + c))*sin(d*x + c) - 
7*log(2*(sqrt(-cos(d*x + c) + 1)*sqrt(cos(d*x + c)) - sin(d*x + c))/sin(d* 
x + c))*sin(d*x + c))/(d*sin(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 747 vs. \(2 (136) = 272\).

Time = 6.90 (sec) , antiderivative size = 747, normalized size of antiderivative = 4.64 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(2*sqrt(2)*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)/sgn(sin(1/2*d*x + 1/2*c)) - 2*sqrt(2)*l 
og(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))/sgn(sin(1/2*d*x + 1/2*c)) - 2*sqrt(2)*log(abs(-ta 
n(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))/sgn(sin(1/2*d*x + 1/2*c)) - 7*log(tan(1/4*d*x + 1/4*c)^2 + 
2*sqrt(2) - sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 
1)/sgn(sin(1/2*d*x + 1/2*c)) + 7*log(abs(-tan(1/4*d*x + 1/4*c)^2 + 2*sqrt( 
2) + sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) - 1))/sgn 
(sin(1/2*d*x + 1/2*c)) - 4*sqrt(2)*(17*(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))^7 - 73*(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))^6 
+ 157*(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))^5 - 597*(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))^4 + 1603*(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 - 875*(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))^2 - 1585*tan(1/4*d*x + 1/4*c)^2 + 1585*sqrt(tan(1/4*d*x + 1/4*c)^ 
4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 1737)/(((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))^2 + 2*tan(1/4*...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

 - int((sqrt( - cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**2)/(cos 
(c + d*x) - 1),x)