\(\int \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)} \, dx\) [263]

Optimal result

Integrand size = 26, antiderivative size = 129 \[ \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)} \, dx=-\frac {3 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{4 d}+\frac {3 a \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d \sqrt {a-a \cos (c+d x)}}-\frac {a \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {a-a \cos (c+d x)}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.20 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.44 \[ \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)} \, dx=-\frac {e^{-\frac {3}{2} i (c+d x)} \left (\sqrt {1+e^{2 i (c+d x)}} \left (1-2 e^{i (c+d x)}-2 e^{2 i (c+d x)}+e^{3 i (c+d x)}\right )+3 e^{2 i (c+d x)} \text {arcsinh}\left (e^{i (c+d x)}\right )+3 e^{2 i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \csc \left (\frac {1}{2} (c+d x)\right )}{8 d \sqrt {1+e^{2 i (c+d x)}}} \] Input:

Integrate[Cos[c + d*x]^(3/2)*Sqrt[a - a*Cos[c + d*x]],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[Cos[c + d*x]]*Sqrt[a - a*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 = 0.57 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3249

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3249

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 220

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

Input:

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

Output:

-1/2*(a*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a - a*Cos[c + d*x]]) - (3 
*((Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[Cos[c + d*x]]*Sqrt[a - a*C 
os[c + d*x]])])/d - (a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a - a*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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(240\) vs. \(2(107)=214\).

Time = 8.27 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.87

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (3 \csc \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right )+\cot \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-10 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-10\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\right ) \sqrt {4}}{16 d \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\)	\(241\)

Input:

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

Output:

-1/16*2^(1/2)/d*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)^2-1)^ 
(1/2)/(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)*(3*csc(1/2*d*x+1/2*c)*2^(1/2)*arctanh(cos(1/2*d*x+1/2*c)/(c 
os(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^(1/2))+cot(1/2*d*x+1/2*c)*(8*cos(1/2*d*x+1/2*c)^3+8*cos(1/2*d*x+1 
/2*c)^2-10*cos(1/2*d*x+1/2*c)-10)*((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x 
+1/2*c)+1)^2)^(1/2))*4^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.20 \[ \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)} \, dx=\frac {3 \, \sqrt {a} \log \left (\frac {4 \, \sqrt {-a \cos \left (d x + c\right ) + a} {\left (2 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \sqrt {\cos \left (d x + c\right )} - {\left (8 \, a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 4 \, \sqrt {-a \cos \left (d x + c\right ) + a} {\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 3\right )} \sqrt {\cos \left (d x + c\right )}}{16 \, d \sin \left (d x + c\right )} \] Input:

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

Output:

1/16*(3*sqrt(a)*log((4*sqrt(-a*cos(d*x + c) + a)*(2*cos(d*x + c)^2 + 3*cos 
(d*x + c) + 1)*sqrt(a)*sqrt(cos(d*x + c)) - (8*a*cos(d*x + c)^2 + 8*a*cos( 
d*x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) - 4*sqrt(-a*cos(d*x 
 + c) + a)*(2*cos(d*x + c)^2 - cos(d*x + c) - 3)*sqrt(cos(d*x + c)))/(d*si 
n(d*x + c))
 

Sympy [F]

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

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

Output:

Integral(sqrt(-a*(cos(c + d*x) - 1))*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. 1063 vs. \(2 (107) = 214\).

Time = 0.35 (sec) , antiderivative size = 1063, normalized size of antiderivative = 8.24 \[ \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/16*(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)))*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))) - sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
 2*c) + 1)) + ((cos(2*d*x + 2*c) - 2)*cos(1/2*arctan2(sin(2*d*x + 2*c), co 
s(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)*sin(1/2*arctan2(sin(2*d*x + 2*c), c 
os(2*d*x + 2*c) + 1)))*sqrt(-a) + 3*sqrt(-a)*(arctan2((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)))*sin(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) + 1)) 
*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (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(si 
n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), c 
os(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1 
))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) + 1) - arctan2((c 
os(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(co 
s(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(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) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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