\(\int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx\) [240]

Optimal result

Integrand size = 25, antiderivative size = 97 \[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.34 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {1+\cos (c+d x)} \left (\arcsin \left (\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \sqrt {1+\cos (c+d x)}+2 \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d (a (1+\cos (c+d x)))^{3/2}} \] Input:

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

Output:

(Cos[(c + d*x)/2]*Sqrt[1 + Cos[c + d*x]]*(ArcSin[Sin[(c + d*x)/2]/Sqrt[Cos 
[(c + d*x)/2]^2]]*Sqrt[1 + Cos[c + d*x]] + 2*Sqrt[Cos[c + d*x]/(1 + Cos[c 
+ d*x])]*Sin[(c + d*x)/2]))/(2*d*(a*(1 + Cos[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3243, 27, 3042, 3261, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3243

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {\arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}\)

Input:

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

Output:

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

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 218

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3243

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

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]
 

Maple [A] (verified)

Time = 3.96 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.35

method	result	size
default	\(\frac {\sqrt {\cos \left (d x +c \right )}\, \left (\csc \left (d x +c \right )^{2} \left (1-\cos \left (d x +c \right )\right )^{2}+1\right ) \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \left (\sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right )}{8 d \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{2}}\)	\(131\)

Input:

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

Output:

1/8/d*cos(d*x+c)^(1/2)*(csc(d*x+c)^2*(1-cos(d*x+c))^2+1)*2^(1/2)*(a*(cos(d 
*x+c)+1))^(1/2)*(2^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(csc(d*x+c)-cot 
(d*x+c))-arcsin(cot(d*x+c)-csc(d*x+c)))/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/ 
a^2
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.49 \[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) + 2 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

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

Output:

1/4*(sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*arctan(1/2*sqrt 
(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*co 
s(d*x + c)^2 + a*cos(d*x + c))) + 2*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x 
+ c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d*x)**2 
+ 2*cos(c + d*x) + 1),x))/a**2