\(\int \frac {\cos ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx\) [47]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.22 (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} \[ \int \frac {\cos ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\cos (e+f x) \sqrt {c-c \sin (e+f x)}}{c f \sqrt {a \sin (e+f x)+a}} \]

[In]

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

[Out]

-((Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(c*f*Sqrt[a + a*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 \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)} \, dx}{a c} \\ & = -\frac {\cos (e+f x) \sqrt {c-c \sin (e+f x)}}{c f \sqrt {a+a \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\sin (2 (e+f x))}{2 f \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{f \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(42\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.14 \[ \int \frac {\cos ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{a c f \cos \left (f x + e\right )} \]

[In]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {\cos ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx=\frac {2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{\sqrt {a} \sqrt {c} f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]

[In]

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

[Out]

2*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2/(sqrt(a)*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 [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.21 \[ \int \frac {\cos ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\sin \left (2\,e+2\,f\,x\right )\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}}{2\,c\,f\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (\sin \left (e+f\,x\right )-1\right )} \]

[In]

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

[Out]

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