\(\int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx\) [269]

Optimal result

Integrand size = 25, antiderivative size = 114 \[ \int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx=-\frac {3 \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{4 d}+\frac {3 \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d \sqrt {1-\cos (c+d x)}}-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {1-\cos (c+d x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.44 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.59 \[ \int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx=-\frac {e^{-\frac {3}{2} i (c+d x)} \left (\sqrt {1+e^{2 i (c+d x)}} \left (1-2 e^{i (c+d x)}-2 e^{2 i (c+d x)}+e^{3 i (c+d x)}\right )+3 e^{2 i (c+d x)} \text {arcsinh}\left (e^{i (c+d x)}\right )+3 e^{2 i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))} \csc \left (\frac {1}{2} (c+d x)\right )}{8 d \sqrt {1+e^{2 i (c+d x)}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3249, 3042, 3249, 3042, 3254, 220}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3249

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3249

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 220

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

Input:

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

Output:

-1/2*(Cos[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[1 - Cos[c + d*x]]) - (3*(Ar 
cTanh[Sin[c + d*x]/(Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])]/d - (Sqrt[ 
Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[1 - Cos[c + d*x]])))/4
 

Defintions of rubi rules used

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 3249

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

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]
 

Maple [C] (warning: unable to verify)

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

Time = 8.99 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.08

method	result	size
default	\(-\frac {\operatorname {csgn}\left (\sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \left (\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-10 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \operatorname {arctanh}\left (\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )\right ) \sqrt {2}}{8 d \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(123\)

Input:

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

Output:

-1/8/d*csgn(sin(1/2*d*x+1/2*c))*cos(d*x+c)^(1/2)/(cos(d*x+c)/(cos(d*x+c)+1 
))^(1/2)*((cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(8*cos(1/2*d*x+1/2*c)^3-10*cos 
(1/2*d*x+1/2*c))+3*sec(1/2*d*x+1/2*c)*arctanh((cos(d*x+c)/(cos(d*x+c)+1))^ 
(1/2)))*2^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.09 \[ \int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx=-\frac {2 \, {\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 3\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - 3 \, \log \left (-\frac {2 \, {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - {\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right )}{8 \, d \sin \left (d x + c\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1305 vs. \(2 (96) = 192\).

Time = 0.29 (sec) , antiderivative size = 1305, normalized size of antiderivative = 11.45 \[ \int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/32*(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(2*d*x + 2*c) - 2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d* 
x + 2*c))) + sin(2*d*x + 2*c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x 
+ 2*c))) + cos(2*d*x + 2*c) - 2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d 
*x + 2*c) + 1)) - (cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*si 
n(2*d*x + 2*c) - (cos(2*d*x + 2*c) - 2)*sin(1/2*arctan2(sin(2*d*x + 2*c), 
cos(2*d*x + 2*c))) - sin(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), c 
os(2*d*x + 2*c) + 1))) + 3*log(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), cos(2*d*x + 
2*c) + 1))^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x 
+ 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + 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)* 
cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - 3*log(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), cos(2*d*x + 2*c) + 1))^2 + sqrt(cos(2*d*x + 2*c 
)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d 
*x + 2*c), cos(2*d*x + 2*c) + 1))^2 - 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)*cos(1/2*arctan2(sin(2*d*x + 2*c), c 
os(2*d*x + 2*c) + 1)) + 1) + 3*log(((cos(1/2*arctan2(sin(2*d*x + 2*c), cos 
(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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