\(\int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx\) [283]

Optimal result

Integrand size = 25, antiderivative size = 118 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {2} \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}+\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {1-\cos (c+d x)}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.50 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.92 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=-\frac {i e^{-i (c+d x)} \left (-1+e^{i (c+d x)}\right ) \left (\sqrt {2} e^{i (c+d x)} \text {arcsinh}\left (e^{i (c+d x)}\right )-4 e^{i (c+d x)} \text {arctanh}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\sqrt {2} \left (\left (1+e^{i (c+d x)}\right ) \sqrt {1+e^{2 i (c+d x)}}+e^{i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )\right ) \sqrt {\cos (c+d x)}}{2 \sqrt {2} d \sqrt {1+e^{2 i (c+d x)}} \sqrt {1-\cos (c+d x)}} \] Input:

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

Output:

((-1/2*I)*(-1 + E^(I*(c + d*x)))*(Sqrt[2]*E^(I*(c + d*x))*ArcSinh[E^(I*(c 
+ d*x))] - 4*E^(I*(c + d*x))*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 
 + E^((2*I)*(c + d*x))])] + Sqrt[2]*((1 + E^(I*(c + d*x)))*Sqrt[1 + E^((2* 
I)*(c + d*x))] + E^(I*(c + d*x))*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))* 
Sqrt[Cos[c + d*x]])/(Sqrt[2]*d*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x) 
)]*Sqrt[1 - Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3257, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3257

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3461

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 220

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

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]
 

Maple [C] (warning: unable to verify)

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

Time = 8.62 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.14

method	result	size
default	\(\frac {\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\sqrt {2}\, \operatorname {arctanh}\left (\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )-2 \,\operatorname {arctanh}\left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \operatorname {csgn}\left (\sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(135\)

Input:

int(cos(d*x+c)^(3/2)/(1-cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/2/d*sec(1/2*d*x+1/2*c)*(2^(1/2)*(cos(d*x+c)+1)*(cos(d*x+c)/(cos(d*x+c)+1 
))^(1/2)+2^(1/2)*arctanh((cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-2*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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (103) = 206\).

Time = 0.09 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.91 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=\frac {\sqrt {2} \log \left (-\frac {2 \, {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 2 \, {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} + \log \left (\frac {2 \, {\left (\sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - \log \left (\frac {2 \, {\left (\sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right )}{2 \, d \sin \left (d x + c\right )} \] Input:

integrate(cos(d*x+c)^(3/2)/(1-cos(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

1/2*(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*(cos(d*x + c) + 1)*sqrt(-cos(d*x + c 
) + 1)*sqrt(cos(d*x + c)) + 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) - 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*si 
n(d*x + c))
 

Sympy [F]

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

integrate(cos(d*x+c)**(3/2)/(1-cos(d*x+c))**(1/2),x)
 

Output:

Integral(cos(c + d*x)**(3/2)/sqrt(1 - cos(c + d*x)), x)
 

Maxima [F]

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

integrate(cos(d*x+c)^(3/2)/(1-cos(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(cos(d*x + c)^(3/2)/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. 560 vs. \(2 (103) = 206\).

Time = 5.19 (sec) , antiderivative size = 560, normalized size of antiderivative = 4.75 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(3/2)/(1-cos(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

1/2*(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)) - sqrt(2)*log(a 
bs(-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)) - sqrt(2)*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))/sgn(sin(1/2*d*x + 1/2*c)) - 2*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)) + 2*log(abs(-tan(1/4*d*x + 1/4*c)^2 + 2*sqrt(2) + s 
qrt(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)) - 8*sqrt(2)*(3*(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 - 7*(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 + 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) + 11)/(((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*d*x + 1/4*c)^2 - 2*sqrt(tan(1/4*d* 
x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) - 7)^2*sgn(sin(1/2*d*x + 1/2* 
c))))/d
 

Mupad [F(-1)]

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

int(cos(c + d*x)^(3/2)/(1 - cos(c + d*x))^(1/2),x)
 

Output:

int(cos(c + d*x)^(3/2)/(1 - cos(c + d*x))^(1/2), x)
 

Reduce [F]

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

int(cos(d*x+c)^(3/2)/(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))/(cos(c 
+ d*x) - 1),x)