\(\int \frac {1}{\sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)} \, dx\) [287]

Optimal result

Integrand size = 25, antiderivative size = 122 \[ \int \frac {1}{\sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {2} \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}+\frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.39 \[ \int \frac {1}{\sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \left (-\frac {3 e^{-\frac {3}{2} i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^2 \text {arctanh}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )}{2 \sqrt {2}}+2 \sqrt {1+e^{2 i (c+d x)}} \cos \left (\frac {1}{2} (c+d x)\right ) (1+\cos (c+d x))\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{3 d \sqrt {1+e^{2 i (c+d x)}} \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)} \] Input:

Integrate[1/(Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(5/2)),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))]*Sqrt[1 - Cos[c + d*x]]*Cos[c 
 + d*x]^(3/2))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3258, 3042, 3463, 27, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(2*Sin[c + d*x])/(3*d*Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(3/2)) + ((-3*Sq 
rt[2]*ArcTanh[Sin[c + d*x]/(Sqrt[2]*Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d* 
x]])])/d + (2*Sin[c + d*x])/(d*Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])) 
/3
 

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 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 [C] (warning: unable to verify)

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

Time = 6.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {4 \left (-3 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+\sqrt {2}\, \left (\cos \left (d x +c \right )+1\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \operatorname {csgn}\left (\sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \cos \left (d x +c \right )^{\frac {3}{2}} \left (\cos \left (d x +c \right )+1\right )}\)	\(110\)

Input:

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

Output:

4/3/d*(-3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)/(cos(d*x+c 
)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+2^(1/2)*(cos(d*x+c)+1))*cos(1/2*d*x+1/ 
2*c)^3*csgn(sin(1/2*d*x+1/2*c))/cos(d*x+c)^(3/2)/(cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.29 \[ \int \frac {1}{\sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {3 \, \sqrt {2} \cos \left (d x + c\right )^{2} \log \left (-\frac {2 \, {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 4 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{6 \, d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 563, normalized size of antiderivative = 4.61 \[ \int \frac {1}{\sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

1/3*(3*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)* 
log(4*(abs(I*e^(I*d*x + I*c) - I)^2 + 2*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d* 
x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x + 2*c), co 
s(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c 
) + 1))^2) - 2*(sqrt(2)*abs(I*e^(I*d*x + I*c) - I)*sin(1/2*arctan2(sin(2*d 
*x + 2*c), cos(2*d*x + 2*c) + 1)) - 2*sqrt(2)*cos(1/2*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)^(1/4) + 4)/abs(I*e^(I*d*x + I*c) - I)^2) - 2*(sqrt( 
2)*sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 
 (sqrt(2)*cos(d*x + c) + 3*sqrt(2))*cos(1/2*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)^(3/4) - 4*(sqrt(2)*sin(d*x + c)*sin(3/2*arctan2(sin(2*d*x + 2 
*c), cos(2*d*x + 2*c) + 1)) + (sqrt(2)*cos(d*x + c) - sqrt(2))*cos(3/2*arc 
tan2(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)^(1/4))/((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))*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (103) = 206\).

Time = 0.48 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.90 \[ \int \frac {1}{\sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {\sqrt {2} {\left (\frac {8 \, {\left ({\left ({\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 3\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 3\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 1\right )}}{{\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1\right )}^{\frac {3}{2}}} - 3 \, \log \left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1\right ) + 3 \, \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 3 \right |}\right ) + 3 \, \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1 \right |}\right )\right )}}{6 \, d \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \] Input:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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