\(\int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx\) [367]

Optimal result

Integrand size = 30, antiderivative size = 134 \[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {2 a^3 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f} \] Output:

-2/15*a^3*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/f/(a+a*sin(f*x+e))^(1/2)-1/5*a 
^2*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)*(c-c*sin(f*x+e))^(5/2)/f-1/5*a*cos(f* 
x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2)/f
 

Mathematica [A] (verified)

Time = 1.71 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.52 \[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {a^2 c^2 \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} \left (15-10 \sin ^2(e+f x)+3 \sin ^4(e+f x)\right ) \tan (e+f x)}{15 f} \] Input:

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

Output:

(a^2*c^2*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(15 - 10*Sin[ 
e + f*x]^2 + 3*Sin[e + f*x]^4)*Tan[e + f*x])/(15*f)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

\(\displaystyle \frac {4}{5} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{5} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}\)

\(\Big \downarrow \) 3219

\(\displaystyle \frac {4}{5} a \left (\frac {1}{2} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{5} a \left (\frac {1}{2} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}\)

\(\Big \downarrow \) 3217

\(\displaystyle \frac {4}{5} a \left (-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}\)

Input:

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

Output:

-1/5*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(5/2) 
)/f + (4*a*(-1/6*(a^2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(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])^(5/2))/(4*f)))/5
 

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])
 

Maple [A] (verified)

Time = 3.24 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {8 \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (6 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+3 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) a^{2} c^{2} \sqrt {4}}{15 f}\)	\(111\)

Input:

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

Output:

8/15/f*sin(1/4*Pi+1/2*f*x+1/2*e)^4*tan(1/4*Pi+1/2*f*x+1/2*e)*(a*sin(1/4*Pi 
+1/2*f*x+1/2*e)^2)^(1/2)*(c*cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*(6*cos(1/4* 
Pi+1/2*f*x+1/2*e)^4+3*cos(1/4*Pi+1/2*f*x+1/2*e)^2+1)*a^2*c^2*4^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.63 \[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {{\left (3 \, a^{2} c^{2} \cos \left (f x + e\right )^{4} + 4 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} + 8 \, a^{2} c^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )} \] Input:

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

Output:

1/15*(3*a^2*c^2*cos(f*x + e)^4 + 4*a^2*c^2*cos(f*x + e)^2 + 8*a^2*c^2)*sqr 
t(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. \[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.16 \[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {16 \, {\left (6 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 15 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} + 10 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6}\right )} \sqrt {a} \sqrt {c}}{15 \, f} \] Input:

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

Output:

16/15*(6*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 - 15*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 + 10*a^2*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(si 
n(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^6)*sqrt(a)*sq 
rt(c)/f
 

Mupad [B] (verification not implemented)

Time = 1.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.62 \[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {a^2\,c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (175\,\sin \left (2\,e+2\,f\,x\right )+28\,\sin \left (4\,e+4\,f\,x\right )+3\,\sin \left (6\,e+6\,f\,x\right )\right )}{240\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:

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

Output:

(a^2*c^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(175*s 
in(2*e + 2*f*x) + 28*sin(4*e + 4*f*x) + 3*sin(6*e + 6*f*x)))/(240*f*(cos(2 
*e + 2*f*x) + 1))
 

Reduce [F]

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

int((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)*sin(e + f*x)**4,x) - 2*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f* 
x) + 1)*sin(e + f*x)**2,x) + int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f* 
x) + 1),x))