\(\int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx\) [54]

Optimal result

Integrand size = 38, antiderivative size = 51 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) \log (1+\sin (e+f x))}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:

cos(f*x+e)*ln(1+sin(f*x+e))/a/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1 
/2)
 

Mathematica [A] (verified)

Time = 2.63 (sec) , antiderivative size = 102, normalized size of antiderivative = 2.00 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx=\frac {2 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{f (a (1+\sin (e+f x)))^{3/2} \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[Cos[e + f*x]^2/((a + a*Sin[e + f*x])^(3/2)*Sqrt[c - c*Sin[e + f* 
x]]),x]
 

Output:

(2*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*(Cos[(e + f*x)/2] - Sin[(e + f 
*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/(f*(a*(1 + Sin[e + f*x])) 
^(3/2)*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {3042, 3320, 3042, 3216, 3042, 3146, 16}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

\(\displaystyle \frac {\int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx}{a c}\)

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

\(\displaystyle \frac {\cos (e+f x) \log (a \sin (e+f x)+a)}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\)

Input:

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

Output:

(Cos[e + f*x]*Log[a + a*Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[ 
c - c*Sin[e + f*x]])
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 3042

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

rule 3146

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x 
)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I 
ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/ 
2])
 

rule 3216

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

rule 3320

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(a^(p/ 
2)*c^(p/2))   Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(n + 
p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && 
EqQ[a^2 - b^2, 0] && IntegerQ[p/2]
 

Maple [F]

\[\int \frac {\cos \left (f x +e \right )^{2}}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c -c \sin \left (f x +e \right )}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(cos(f*x+e)^2/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="fricas")
 

Output:

integral(sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(a^2*c*sin(f*x 
 + e) + a^2*c), x)
 

Sympy [F]

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

integrate(cos(f*x+e)**2/(a+a*sin(f*x+e))**(3/2)/(c-c*sin(f*x+e))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(cos(f*x+e)^2/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {2 \, \log \left ({\left | \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right )}{a^{\frac {3}{2}} \sqrt {c} f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \] Input:

integrate(cos(f*x+e)^2/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="giac")
 

Output:

-2*log(abs(cos(-1/4*pi + 1/2*f*x + 1/2*e)))/(a^(3/2)*sqrt(c)*f*sgn(cos(-1/ 
4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - sqrt(c)*sqrt(a)*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)* 
cos(e + f*x)**2)/(sin(e + f*x)**3 + sin(e + f*x)**2 - sin(e + f*x) - 1),x) 
)/(a**2*c)