\(\int \frac {\cos ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx\) [48]

Optimal result

Integrand size = 38, antiderivative size = 54 \[ \int \frac {\cos ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=-\frac {\cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 3.72 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.93 \[ \int \frac {\cos ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=-\frac {2 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{f \sqrt {a (1+\sin (e+f x))} (c-c \sin (e+f x))^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {2 \, \log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right )}{\sqrt {a} c^{\frac {3}{2}} f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \] Input:

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

Output:

2*log(abs(sin(-1/4*pi + 1/2*f*x + 1/2*e)))/(sqrt(a)*c^(3/2)*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. \[ \int \frac {\cos ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2}{\sqrt {a+a\,\sin \left (e+f\,x\right )}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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