\(\int \frac {\cos ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{7/2}} \, dx\) [8]

Optimal result

Integrand size = 38, antiderivative size = 48 \[ \int \frac {\cos ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 a c f (c-c \sin (e+f x))^{5/2}} \]

[Out]

1/4*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/a/c/f/(c-c*sin(f*x+e))^(5/2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 48, 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, 2821} \[ \int \frac {\cos ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 a c f (c-c \sin (e+f x))^{5/2}} \]

[In]

Int[(Cos[e + f*x]^2*Sqrt[a + a*Sin[e + f*x]])/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(4*a*c*f*(c - c*Sin[e + f*x])^(5/2))

Rule 2821

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] /; FreeQ[{a, b, c, d, e, f
, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && NeQ[m, -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 {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx}{a c} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 a c f (c-c \sin (e+f x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.88 \[ \int \frac {\cos ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {\sin (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)}}{c^4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

Integrate[(Cos[e + f*x]^2*Sqrt[a + a*Sin[e + f*x]])/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(Sin[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]])/(c^4*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]
)^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \tan \left (f x +e \right )}{f \left (\sin \left (f x +e \right )-1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3}}\)	\(50\)

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.65 \[ \int \frac {\cos ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{c^{4} f \cos \left (f x + e\right )^{3} + 2 \, c^{4} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, c^{4} f \cos \left (f x + e\right )} \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e)/(c^4*f*cos(f*x + e)^3 + 2*c^4*f*cos(f*x + e)*
sin(f*x + e) - 2*c^4*f*cos(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**(1/2)/(c-c*sin(f*x+e))**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{7/2}} \, dx=\int { \frac {\sqrt {a \sin \left (f x + e\right ) + a} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(f*x + e) + a)*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(7/2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (42) = 84\).

Time = 0.32 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.90 \[ \int \frac {\cos ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {{\left (2 \, \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{4 \, c^{4} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}} \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

1/4*(2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - sqrt(c)*sgn(cos(-1/4*pi
+ 1/2*f*x + 1/2*e)))*sqrt(a)/(c^4*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^4)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{7/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \]

[In]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^(1/2))/(c - c*sin(e + f*x))^(7/2),x)

[Out]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^(1/2))/(c - c*sin(e + f*x))^(7/2), x)