\(\int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {a-a \cos (c+d x)}} \, dx\) [276]

Optimal result

Integrand size = 26, antiderivative size = 141 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {a-a \cos (c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {a-a \cos (c+d x)}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.50 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.62 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {a-a \cos (c+d x)}} \, dx=-\frac {i e^{-i (c+d x)} \left (-1+e^{i (c+d x)}\right ) \left (\sqrt {2} e^{i (c+d x)} \text {arcsinh}\left (e^{i (c+d x)}\right )-4 e^{i (c+d x)} \text {arctanh}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\sqrt {2} \left (\left (1+e^{i (c+d x)}\right ) \sqrt {1+e^{2 i (c+d x)}}+e^{i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )\right ) \sqrt {\cos (c+d x)}}{2 \sqrt {2} d \sqrt {1+e^{2 i (c+d x)}} \sqrt {a-a \cos (c+d x)}} \] Input:

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

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3257, 3042, 3461, 3042, 3254, 220, 3261, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3257

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3461

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 220

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 221

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

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(375\) vs. \(2(118)=236\).

Time = 6.67 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.67

method	result	size
default	\(\frac {\sqrt {2}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-\operatorname {arctanh}\left (\frac {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right )+\ln \left (\frac {2 \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}-4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1}\right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right )+\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {4}}{4 d \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\)	\(376\)

Input:

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

Output:

1/4*2^(1/2)/d*sin(1/2*d*x+1/2*c)*(-arctanh((-1+2*cos(1/2*d*x+1/2*c))/(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))+ln(2*(((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*cos 
(1/2*d*x+1/2*c)+((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2 
)-2*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/2*c)+1))+2^(1/2)*arctanh(cos(1/2* 
d*x+1/2*c)/(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)*2^(1/2))+cos(1/2*d*x+1/2*c)*(2*cos(1/2*d*x+1/2*c)+2)*( 
(2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^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)/(a*sin(1/2*d*x+1/2*c)^2)^(1/2)/( 
(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 = 0.09 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.50 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {a-a \cos (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {a} \log \left (-\frac {\frac {2 \, \sqrt {2} \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + \sqrt {a} \log \left (-\frac {2 \, \sqrt {-a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )} + {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 2 \, \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{2 \, a d \sin \left (d x + c\right )} \] Input:

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

Output:

1/2*(sqrt(2)*sqrt(a)*log(-(2*sqrt(2)*sqrt(-a*cos(d*x + c) + a)*(cos(d*x + 
c) + 1)*sqrt(cos(d*x + c))/sqrt(a) - (3*cos(d*x + c) + 1)*sin(d*x + c))/(( 
cos(d*x + c) - 1)*sin(d*x + c)))*sin(d*x + c) + sqrt(a)*log(-(2*sqrt(-a*co 
s(d*x + c) + a)*sqrt(a)*(cos(d*x + c) + 1)*sqrt(cos(d*x + c)) + (2*a*cos(d 
*x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) + 2*sqrt(-a*cos(d*x 
+ c) + a)*(cos(d*x + c) + 1)*sqrt(cos(d*x + c)))/(a*d*sin(d*x + c))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 581 vs. \(2 (118) = 236\).

Time = 5.37 (sec) , antiderivative size = 581, normalized size of antiderivative = 4.12 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {a-a \cos (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(3/2)/(a-a*cos(d*x+c))^(1/2),x, algorithm="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)/(sqrt(a)*sgn(sin(1/2*d*x + 1/2*c))) - sqr 
t(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) + 3))/(sqrt(a)*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))/(sqrt(a)*sgn(sin(1/2*d*x + 1/2*c))) + 2*log(1 
/2*abs(2*tan(1/4*d*x + 1/4*c)^2 - 4*sqrt(2) - 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 + 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))/(sqrt 
(a)*sgn(sin(1/2*d*x + 1/2*c))) - 8*sqrt(2)*(3*(tan(1/4*d*x + 1/4*c)^2 - sq 
rt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1))^3*sqrt(a) - 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*sqrt(a) + (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))*sqrt(a) + 11*sqrt(a))/(((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*a*sgn(sin(1/2*d*x + 1/2*c))))/d
 

Mupad [F(-1)]

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

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