\(\int \frac {(c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx\) [326]

Optimal result

Integrand size = 28, antiderivative size = 98 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=-\frac {64 c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a f}+\frac {16 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a f}+\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a f} \] Output:

-64/3*c^2*sec(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f+16/3*c*sec(f*x+e)*(c-c*sin 
(f*x+e))^(3/2)/a/f+2/3*sec(f*x+e)*(c-c*sin(f*x+e))^(5/2)/a/f
 

Mathematica [A] (verified)

Time = 8.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.04 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=-\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (45+\cos (2 (e+f x))+20 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{3 a f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))} \] Input:

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

Output:

-1/3*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(45 + Cos[2*(e + f*x)] + 2 
0*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(a*f*(Cos[(e + f*x)/2] - Sin[(e 
+ f*x)/2])*(1 + Sin[e + f*x]))
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3215, 3042, 3153, 3042, 3153, 3042, 3152}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3153

\(\displaystyle \frac {\frac {8}{3} c \int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {8}{3} c \int \frac {(c-c \sin (e+f x))^{5/2}}{\cos (e+f x)^2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}}{a c}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {\frac {8}{3} c \left (4 c \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {8}{3} c \left (4 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\cos (e+f x)^2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}}{a c}\)

\(\Big \downarrow \) 3152

\(\displaystyle \frac {\frac {8}{3} c \left (\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}-\frac {8 c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}}{a c}\)

Input:

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

Output:

((2*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(3*f) + (8*c*((-8*c^2*Sec[e 
 + f*x]*Sqrt[c - c*Sin[e + f*x]])/f + (2*c*Sec[e + f*x]*(c - c*Sin[e + f*x 
])^(3/2))/f))/3)/(a*c)
 

Defintions of rubi rules used

rule 3042

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

rule 3152

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x 
])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 
 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
 

rule 3153

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + 
f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p))   Int[(g* 
Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, 
g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && 
NeQ[m + p, 0]
 

rule 3215

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m   Int[Cos[e + f*x]^(2*m)*(c + 
 d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ 
b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((Lt 
Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 

Maple [A] (verified)

Time = 1.51 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.60

method	result	size
default	\(-\frac {2 c^{3} \left (-1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )^{2}-10 \sin \left (f x +e \right )-23\right )}{3 a \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(59\)

Input:

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

Output:

-2/3*c^3/a*(-1+sin(f*x+e))*(sin(f*x+e)^2-10*sin(f*x+e)-23)/cos(f*x+e)/(c-c 
*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=-\frac {2 \, {\left (c^{2} \cos \left (f x + e\right )^{2} + 10 \, c^{2} \sin \left (f x + e\right ) + 22 \, c^{2}\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, a f \cos \left (f x + e\right )} \] Input:

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

Output:

-2/3*(c^2*cos(f*x + e)^2 + 10*c^2*sin(f*x + e) + 22*c^2)*sqrt(-c*sin(f*x + 
 e) + c)/(a*f*cos(f*x + e))
 

Sympy [F]

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

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

Output:

(Integral(c**2*sqrt(-c*sin(e + f*x) + c)/(sin(e + f*x) + 1), x) + Integral 
(-2*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)/(sin(e + f*x) + 1), x) + I 
ntegral(c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2/(sin(e + f*x) + 1), 
 x))/a
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (86) = 172\).

Time = 0.13 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.96 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=\frac {2 \, {\left (23 \, c^{\frac {5}{2}} + \frac {20 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {65 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {40 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {65 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {20 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {23 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )}}{3 \, {\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} f {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} \] Input:

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

Output:

2/3*(23*c^(5/2) + 20*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 65*c^(5/2)* 
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 40*c^(5/2)*sin(f*x + e)^3/(cos(f*x + 
 e) + 1)^3 + 65*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 20*c^(5/2)*s 
in(f*x + e)^5/(cos(f*x + e) + 1)^5 + 23*c^(5/2)*sin(f*x + e)^6/(cos(f*x + 
e) + 1)^6)/((a + a*sin(f*x + e)/(cos(f*x + e) + 1))*f*(sin(f*x + e)^2/(cos 
(f*x + e) + 1)^2 + 1)^(5/2))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (86) = 172\).

Time = 0.18 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.30 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=\frac {8 \, \sqrt {2} \sqrt {c} {\left (\frac {3 \, c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{a {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )}} - \frac {5 \, c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \frac {12 \, c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {3 \, c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}}{a {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 1\right )}^{3}}\right )}}{3 \, f} \] Input:

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

Output:

8/3*sqrt(2)*sqrt(c)*(3*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(a*((cos(-1 
/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 1)) - 
 (5*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 12*c^2*(cos(-1/4*pi + 1/2*f* 
x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x 
 + 1/2*e) + 1) + 3*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4 
*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)/(a*((cos(- 
1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 1)^3 
))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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