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

Optimal. Leaf size=43 \[ -\frac {\cos (e+f x)}{a f (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}} \]

[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]  time = 0.32, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2841, 2738} \[ -\frac {\cos (e+f x)}{a f (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[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 2738

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 2841

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*} \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx &=\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]  time = 0.52, size = 80, normalized size = 1.86 \[ -\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3}{f (a (\sin (e+f x)+1))^{5/2} \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[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]]))

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 60, normalized size = 1.40 \[ -\frac {\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{3} c f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (f x + e\right )^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {-c \sin \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

________________________________________________________________________________________

maple [A]  time = 0.36, size = 50, normalized size = 1.16 \[ \frac {\left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sin \left (f x +e \right )}{f \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (f x + e\right )^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {-c \sin \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 9.26, size = 55, normalized size = 1.28 \[ -\frac {2\,\cos \left (e+f\,x\right )\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}}{a^2\,c\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

Timed out

________________________________________________________________________________________