\(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx\) [22]

Optimal result

Integrand size = 38, antiderivative size = 92 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=\frac {c \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{10 a f \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)}}{5 a f} \]

[Out]

1/10*c*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/f/(c-c*sin(f*x+e))^(1/2)+1/5*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)*(c-c
*sin(f*x+e))^(1/2)/a/f

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2920, 2819, 2817} \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt {c-c \sin (e+f x)}}{5 a f}+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 a f \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(10*a*f*Sqrt[c - c*Sin[e + f*x]]) + (Cos[e + f*x]*(a + a*Sin[e + f
*x])^(7/2)*Sqrt[c - c*Sin[e + f*x]])/(5*a*f)

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 2819

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 - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Dist[a*((2*m - 1)/(
m + n)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m
]) &&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

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 (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx}{a c} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)}}{5 a f}+\frac {2 \int (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)} \, dx}{5 a} \\ & = \frac {c \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{10 a f \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)}}{5 a f} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.94 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a^2 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (20 \cos (2 (e+f x))+5 \cos (4 (e+f x))-70 \sin (e+f x)-5 \sin (3 (e+f x))+\sin (5 (e+f x)))}{80 f} \]

[In]

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

[Out]

-1/80*(a^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(20*Cos[2*(e + f*x)] + 5*Cos[4*(e
+ f*x)] - 70*Sin[e + f*x] - 5*Sin[3*(e + f*x)] + Sin[5*(e + f*x)]))/f

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.99

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, a^{2} \left (2 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )+5 \left (\cos ^{3}\left (f x +e \right )\right )-4 \cos \left (f x +e \right ) \sin \left (f x +e \right )-8 \tan \left (f x +e \right )-5 \sec \left (f x +e \right )\right )}{10 f}\)	\(91\)

[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]

-1/10/f*(a*(1+sin(f*x+e)))^(1/2)*(-c*(sin(f*x+e)-1))^(1/2)*a^2*(2*cos(f*x+e)^3*sin(f*x+e)+5*cos(f*x+e)^3-4*cos
(f*x+e)*sin(f*x+e)-8*tan(f*x+e)-5*sec(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=-\frac {{\left (5 \, a^{2} \cos \left (f x + e\right )^{4} - 5 \, a^{2} + 2 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} - 4 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{10 \, f \cos \left (f x + e\right )} \]

[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]

-1/10*(5*a^2*cos(f*x + e)^4 - 5*a^2 + 2*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 - 4*a^2)*sin(f*x + e))*sqrt
(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=\text {Timed out} \]

[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

Maxima [F]

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

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.11 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=\frac {8 \, {\left (4 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a} \sqrt {c}}{5 \, f} \]

[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]

8/5*(4*a^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^10*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1
/2*e)) - 5*a^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^8*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x
+ 1/2*e)))*sqrt(a)*sqrt(c)/f

Mupad [B] (verification not implemented)

Time = 2.64 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.17 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (20\,\cos \left (e+f\,x\right )+25\,\cos \left (3\,e+3\,f\,x\right )+5\,\cos \left (5\,e+5\,f\,x\right )-75\,\sin \left (2\,e+2\,f\,x\right )-4\,\sin \left (4\,e+4\,f\,x\right )+\sin \left (6\,e+6\,f\,x\right )\right )}{80\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[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]

-(a^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(20*cos(e + f*x) + 25*cos(3*e + 3*f*x) + 5*co
s(5*e + 5*f*x) - 75*sin(2*e + 2*f*x) - 4*sin(4*e + 4*f*x) + sin(6*e + 6*f*x)))/(80*f*(cos(2*e + 2*f*x) + 1))