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

Optimal result

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

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

Mathematica [A] (verified)

Time = 2.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.02 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx=\frac {c^2 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (20 \cos (2 (e+f x))+5 \cos (4 (e+f x))+70 \sin (e+f x)+5 \sin (3 (e+f x))-\sin (5 (e+f x)))}{80 f} \] Input:

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

Output:

(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)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3320, 3042, 3219, 3042, 3217}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{7/2}dx}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{7/2}dx}{a c}\)

\(\Big \downarrow \) 3219

\(\displaystyle \frac {\frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}}{a c}\)

\(\Big \downarrow \) 3217

\(\displaystyle \frac {-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{10 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}}{a c}\)

Input:

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

Output:

(-1/10*(a^2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(f*Sqrt[a + a*Sin[e + 
 f*x]]) - (a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7 
/2))/(5*f))/(a*c)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3217

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 3219

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] + Simp[a*((2*m - 1)/(m + n 
))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; Fre 
eQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I 
GtQ[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 3320

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx=\frac {{\left (5 \, c^{2} \cos \left (f x + e\right )^{4} - 5 \, c^{2} - 2 \, {\left (c^{2} \cos \left (f x + e\right )^{4} - 2 \, c^{2} \cos \left (f x + e\right )^{2} - 4 \, c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{10 \, f \cos \left (f x + e\right )} \] Input:

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

Output:

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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)

Time = 0.36 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.11 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {8 \, {\left (4 \, c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 5 \, c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}\right )} \sqrt {a} \sqrt {c}}{5 \, f} \] Input:

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

Output:

-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 = 18.88 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.20 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx=\frac {c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (20\,\cos \left (e+f\,x\right )+25\,\cos \left (3\,e+3\,f\,x\right )+5\,\cos \left (5\,e+5\,f\,x\right )+75\,\sin \left (2\,e+2\,f\,x\right )+4\,\sin \left (4\,e+4\,f\,x\right )-\sin \left (6\,e+6\,f\,x\right )\right )}{80\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:

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

Output:

(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))
 

Reduce [F]

\[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx=\sqrt {c}\, \sqrt {a}\, c^{2} \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2}d x -2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )d x \right )+\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2}d x \right ) \] Input:

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

Output:

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