\(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx\) [20]

Optimal result

Integrand size = 38, antiderivative size = 188 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {2 a^3 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{35 c f \sqrt {a+a \sin (e+f x)}}-\frac {4 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{35 c f}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{7 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 c f} \] Output:

-2/35*a^3*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/c/f/(a+a*sin(f*x+e))^(1/2)-4/3 
5*a^2*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)*(c-c*sin(f*x+e))^(7/2)/c/f-1/7*a*c 
os(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2)/c/f-1/7*cos(f*x+e) 
*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(7/2)/c/f
 

Mathematica [A] (verified)

Time = 2.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.43 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a^2 c^2 \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} \left (-35+35 \sin ^2(e+f x)-21 \sin ^4(e+f x)+5 \sin ^6(e+f x)\right ) \tan (e+f x)}{35 f} \] Input:

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

Output:

-1/35*(a^2*c^2*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(-35 + 
35*Sin[e + f*x]^2 - 21*Sin[e + f*x]^4 + 5*Sin[e + f*x]^6)*Tan[e + f*x])/f
 

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 3320, 3042, 3219, 3042, 3219, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

\(\displaystyle \frac {\frac {6}{7} a \int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {6}{7} a \int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}}{a c}\)

\(\Big \downarrow \) 3219

\(\displaystyle \frac {\frac {6}{7} a \left (\frac {2}{3} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {6}{7} a \left (\frac {2}{3} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}}{a c}\)

\(\Big \downarrow \) 3219

\(\displaystyle \frac {\frac {6}{7} a \left (\frac {2}{3} a \left (\frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {6}{7} a \left (\frac {2}{3} a \left (\frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}}{a c}\)

\(\Big \downarrow \) 3217

\(\displaystyle \frac {\frac {6}{7} a \left (\frac {2}{3} a \left (-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{10 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}}{a c}\)

Input:

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

Output:

(-1/7*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2 
))/f + (6*a*(-1/6*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e 
+ f*x])^(7/2))/f + (2*a*(-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)))/3))/7)/(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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.54 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {{\left (5 \, a^{2} c^{2} \cos \left (f x + e\right )^{6} + 6 \, a^{2} c^{2} \cos \left (f x + e\right )^{4} + 8 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} + 16 \, a^{2} c^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{35 \, f \cos \left (f x + e\right )} \] Input:

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

Output:

1/35*(5*a^2*c^2*cos(f*x + e)^6 + 6*a^2*c^2*cos(f*x + e)^4 + 8*a^2*c^2*cos( 
f*x + e)^2 + 16*a^2*c^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c 
)*sin(f*x + e)/(f*cos(f*x + e))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.09 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {32 \, {\left (20 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{14} - 70 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} + 84 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 35 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}\right )} \sqrt {a} \sqrt {c}}{35 \, f} \] Input:

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

Output:

-32/35*(20*a^2*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)^14 - 70*a^2*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)^12 + 84*a^2*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 - 35*a^ 
2*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 = 20.21 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.95 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {\frac {1225\,a^2\,c^2\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{32}+\frac {245\,a^2\,c^2\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{32}+\frac {49\,a^2\,c^2\,\sin \left (5\,e+5\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{32}+\frac {5\,a^2\,c^2\,\sin \left (7\,e+7\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{32}}{70\,f\,\cos \left (e+f\,x\right )} \] Input:

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

Output:

((1225*a^2*c^2*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x) 
)^(1/2))/32 + (245*a^2*c^2*sin(3*e + 3*f*x)*(a + a*sin(e + f*x))^(1/2)*(c 
- c*sin(e + f*x))^(1/2))/32 + (49*a^2*c^2*sin(5*e + 5*f*x)*(a + a*sin(e + 
f*x))^(1/2)*(c - c*sin(e + f*x))^(1/2))/32 + (5*a^2*c^2*sin(7*e + 7*f*x)*( 
a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^(1/2))/32)/(70*f*cos(e + f* 
x))
 

Reduce [F]

\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\sqrt {c}\, \sqrt {a}\, a^{2} c^{2} \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{4}d x -2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2}d x \right )+\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2}d x \right ) \] Input:

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

Output:

sqrt(c)*sqrt(a)*a**2*c**2*(int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) 
 + 1)*cos(e + f*x)**2*sin(e + f*x)**4,x) - 2*int(sqrt(sin(e + f*x) + 1)*sq 
rt( - sin(e + f*x) + 1)*cos(e + f*x)**2*sin(e + f*x)**2,x) + int(sqrt(sin( 
e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*cos(e + f*x)**2,x))