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

Optimal result

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

[Out]

-1/10*a*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/c/f/(a+a*sin(f*x+e))^(1/2)-1/5*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)*(a+
a*sin(f*x+e))^(1/2)/c/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} \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {\cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 c f}-\frac {a \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{10 c f \sqrt {a \sin (e+f x)+a}} \]

[In]

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

[Out]

-1/10*(a*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(c*f*Sqrt[a + a*Sin[e + f*x]]) - (Cos[e + f*x]*Sqrt[a + a*Si
n[e + f*x]]*(c - c*Sin[e + f*x])^(7/2))/(5*c*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))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx}{a c} \\ & = -\frac {\cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{5 c f}+\frac {2 \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2} \, dx}{5 c} \\ & = -\frac {a \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{10 c f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{5 c f} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.20 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.02 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx=\frac {c^2 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (20 \cos (2 (e+f x))+5 \cos (4 (e+f x))+70 \sin (e+f x)+5 \sin (3 (e+f x))-\sin (5 (e+f x)))}{80 f} \]

[In]

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

[Out]

(c^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)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.99

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

[In]

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

[Out]

-1/10/f*(-c*(sin(f*x+e)-1))^(1/2)*(a*(1+sin(f*x+e)))^(1/2)*c^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 = 0.31 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx=\frac {{\left (5 \, c^{2} \cos \left (f x + e\right )^{4} - 5 \, c^{2} - 2 \, {\left (c^{2} \cos \left (f x + e\right )^{4} - 2 \, c^{2} \cos \left (f x + e\right )^{2} - 4 \, c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{10 \, f \cos \left (f x + e\right )} \]

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/10*(5*c^2*cos(f*x + e)^4 - 5*c^2 - 2*(c^2*cos(f*x + e)^4 - 2*c^2*cos(f*x + e)^2 - 4*c^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. \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.11 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {8 \, {\left (4 \, c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 5 \, c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}\right )} \sqrt {a} \sqrt {c}}{5 \, f} \]

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.19 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.20 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx=\frac {c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (20\,\cos \left (e+f\,x\right )+25\,\cos \left (3\,e+3\,f\,x\right )+5\,\cos \left (5\,e+5\,f\,x\right )+75\,\sin \left (2\,e+2\,f\,x\right )+4\,\sin \left (4\,e+4\,f\,x\right )-\sin \left (6\,e+6\,f\,x\right )\right )}{80\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

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

[Out]

(c^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*cos
(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))