\(\int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx\) [62]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 38, antiderivative size = 43 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\cos (e+f x)}{a f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2920, 2817} \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\cos (e+f x)}{a f (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}} \]

[In]

Int[Cos[e + f*x]^2/((a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]]),x]

[Out]

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

Rule 2817

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)]

Rule 2920

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{3/2}} \, dx}{a c} \\ & = -\frac {\cos (e+f x)}{a f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.76 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.86 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{f (a (1+\sin (e+f x)))^{5/2} \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[Cos[e + f*x]^2/((a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]]),x]

[Out]

-(((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]]))

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.65

method result size
default \(\frac {\left (-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )\right ) \left (1+\cos \left (f x +e \right )\right )}{f \left (\cos \left (f x +e \right )+\sin \left (f x +e \right )+1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}\) \(71\)

[In]

int(cos(f*x+e)^2/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.40 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{3} c f \cos \left (f x + e\right )} \]

[In]

integrate(cos(f*x+e)^2/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-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))

Sympy [F]

\[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {\cos ^{2}{\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \]

[In]

integrate(cos(f*x+e)**2/(a+a*sin(f*x+e))**(5/2)/(c-c*sin(f*x+e))**(1/2),x)

[Out]

Integral(cos(e + f*x)**2/((a*(sin(e + f*x) + 1))**(5/2)*sqrt(-c*(sin(e + f*x) - 1))), x)

Maxima [F]

\[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \]

[In]

integrate(cos(f*x+e)^2/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(cos(f*x + e)^2/((a*sin(f*x + e) + a)^(5/2)*sqrt(-c*sin(f*x + e) + c)), x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=\frac {1}{2 \, a^{\frac {5}{2}} \sqrt {c} f \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]

[In]

integrate(cos(f*x+e)^2/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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)))

Mupad [B] (verification not implemented)

Time = 9.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {2\,\cos \left (e+f\,x\right )\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}}{a^2\,c\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}} \]

[In]

int(cos(e + f*x)^2/((a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(1/2)),x)

[Out]

-(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))