\(\int \frac {\cos ^2(e+f x) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx\) [80]

Optimal result

Integrand size = 36, antiderivative size = 50 \[ \int \frac {\cos ^2(e+f x) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=\frac {2 \cos (e+f x) (c+c \sin (e+f x))^{1+m}}{c f (3+2 m) \sqrt {a-a \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2920, 2817} \[ \int \frac {\cos ^2(e+f x) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=\frac {2 \cos (e+f x) (c \sin (e+f x)+c)^{m+1}}{c f (2 m+3) \sqrt {a-a \sin (e+f x)}} \]

[In]

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

[Out]

(2*Cos[e + f*x]*(c + c*Sin[e + f*x])^(1 + m))/(c*f*(3 + 2*m)*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)} (c+c \sin (e+f x))^{1+m} \, dx}{a c} \\ & = \frac {2 \cos (e+f x) (c+c \sin (e+f x))^{1+m}}{c f (3+2 m) \sqrt {a-a \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.70 \[ \int \frac {\cos ^2(e+f x) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (c (1+\sin (e+f x)))^m}{f (3+2 m) \sqrt {a-a \sin (e+f x)}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (48) = 96\).

Time = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.16 \[ \int \frac {\cos ^2(e+f x) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=-\frac {2 \, {\left (\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2\right )} \sqrt {-a \sin \left (f x + e\right ) + a} {\left (c \sin \left (f x + e\right ) + c\right )}^{m}}{2 \, a f m + 3 \, a f + {\left (2 \, a f m + 3 \, a f\right )} \cos \left (f x + e\right ) - {\left (2 \, a f m + 3 \, a f\right )} \sin \left (f x + e\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 10.37 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.36 \[ \int \frac {\cos ^2(e+f x) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=-\frac {\sqrt {-a\,\left (\sin \left (e+f\,x\right )-1\right )}\,{\left (c\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (2\,\cos \left (e+f\,x\right )+\sin \left (2\,e+2\,f\,x\right )\right )}{a\,f\,\left (2\,m+3\right )\,\left (\sin \left (e+f\,x\right )-1\right )} \]

[In]

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

[Out]

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