Integrand size = 38, antiderivative size = 43 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\cos (e+f x)}{a f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \] Output:
-cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2)
Time = 2.94 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.86 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{f (a (1+\sin (e+f x)))^{5/2} \sqrt {c-c \sin (e+f x)}} \] Input:
Integrate[Cos[e + f*x]^2/((a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f* x]]),x]
Output:
-(((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]))^(5/2)*Sqrt[c - c*Sin[e + f*x]]))
Time = 0.56 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3042, 3320, 3042, 3217}
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.
\(\displaystyle \int \frac {\cos ^2(e+f x)}{(a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (e+f x)^2}{(a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}dx\) |
\(\Big \downarrow \) 3320 |
\(\displaystyle \frac {\int \frac {\sqrt {c-c \sin (e+f x)}}{(\sin (e+f x) a+a)^{3/2}}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {c-c \sin (e+f x)}}{(\sin (e+f x) a+a)^{3/2}}dx}{a c}\) |
\(\Big \downarrow \) 3217 |
\(\displaystyle -\frac {\cos (e+f x)}{a f (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}\) |
Input:
Int[Cos[e + f*x]^2/((a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]]),x ]
Output:
-(Cos[e + f*x]/(a*f*(a + a*Sin[e + f*x])^(3/2)*Sqrt[c - c*Sin[e + f*x]]))
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] /; FreeQ[{a, b, c, d, e, f, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]
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]
\[\int \frac {\cos \left (f x +e \right )^{2}}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \sqrt {c -c \sin \left (f x +e \right )}}d x\]
Input:
int(cos(f*x+e)^2/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x)
Output:
int(cos(f*x+e)^2/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x)
Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.40 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{3} c f \cos \left (f x + e\right )} \] Input:
integrate(cos(f*x+e)^2/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x, al gorithm="fricas")
Output:
-sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(a^3*c*f*cos(f*x + e)* sin(f*x + e) + a^3*c*f*cos(f*x + e))
\[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {\cos ^{2}{\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \] Input:
integrate(cos(f*x+e)**2/(a+a*sin(f*x+e))**(5/2)/(c-c*sin(f*x+e))**(1/2),x)
Output:
Integral(cos(e + f*x)**2/((a*(sin(e + f*x) + 1))**(5/2)*sqrt(-c*(sin(e + f *x) - 1))), x)
\[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \] Input:
integrate(cos(f*x+e)^2/(a+a*sin(f*x+e))^(5/2)/(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)^(5/2)*sqrt(-c*sin(f*x + e) + c)), x)
Time = 0.38 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=\frac {1}{2 \, a^{\frac {5}{2}} \sqrt {c} f \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \] Input:
integrate(cos(f*x+e)^2/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x, al gorithm="giac")
Output:
1/2/(a^(5/2)*sqrt(c)*f*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))
Time = 17.93 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {2\,\cos \left (e+f\,x\right )\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}}{a^2\,c\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}} \] Input:
int(cos(e + f*x)^2/((a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(1/2)) ,x)
Output:
-(2*cos(e + f*x)*(-c*(sin(e + f*x) - 1))^(1/2))/(a^2*c*f*(cos(2*e + 2*f*x) + 1)*(a*(sin(e + f*x) + 1))^(1/2))
\[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\sqrt {c}\, \sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{4}+2 \sin \left (f x +e \right )^{3}-2 \sin \left (f x +e \right )-1}d x \right )}{a^{3} c} \] Input:
int(cos(f*x+e)^2/(a+a*sin(f*x+e))^(5/2)/(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)**4 + 2*sin(e + f*x)**3 - 2*sin(e + f*x) - 1 ),x))/(a**3*c)