\(\int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx\) [342]

Optimal result

Integrand size = 28, antiderivative size = 104 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {64 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^3 f}+\frac {16 \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 c f}-\frac {2 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 c^2 f} \] Output:

-64/15*sec(f*x+e)^5*(c-c*sin(f*x+e))^(5/2)/a^3/f+16/3*sec(f*x+e)^5*(c-c*si 
n(f*x+e))^(7/2)/a^3/c/f-2*sec(f*x+e)^5*(c-c*sin(f*x+e))^(9/2)/a^3/c^2/f
 

Mathematica [A] (verified)

Time = 3.98 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-29+15 \cos (2 (e+f x))-20 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{15 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3} \] Input:

Integrate[(c - c*Sin[e + f*x])^(5/2)/(a + a*Sin[e + f*x])^3,x]
 

Output:

(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-29 + 15*Cos[2*(e + f*x)] - 20 
*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(15*a^3*f*(Cos[(e + f*x)/2] - Sin 
[(e + f*x)/2])*(1 + Sin[e + f*x])^3)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 105, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

\(\displaystyle \frac {\int \sec ^6(e+f x) (c-c \sin (e+f x))^{11/2}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(c-c \sin (e+f x))^{11/2}}{\cos (e+f x)^6}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-8 c \int \sec ^6(e+f x) (c-c \sin (e+f x))^{9/2}dx-\frac {2 c \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-8 c \int \frac {(c-c \sin (e+f x))^{9/2}}{\cos (e+f x)^6}dx-\frac {2 c \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{f}}{a^3 c^3}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-8 c \left (-\frac {4}{3} c \int \sec ^6(e+f x) (c-c \sin (e+f x))^{7/2}dx-\frac {2 c \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 f}\right )-\frac {2 c \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-8 c \left (-\frac {4}{3} c \int \frac {(c-c \sin (e+f x))^{7/2}}{\cos (e+f x)^6}dx-\frac {2 c \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 f}\right )-\frac {2 c \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{f}}{a^3 c^3}\)

\(\Big \downarrow \) 3152

\(\displaystyle \frac {-8 c \left (\frac {8 c^2 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{15 f}-\frac {2 c \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 f}\right )-\frac {2 c \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{f}}{a^3 c^3}\)

Input:

Int[(c - c*Sin[e + f*x])^(5/2)/(a + a*Sin[e + f*x])^3,x]
 

Output:

((-2*c*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(9/2))/f - 8*c*((8*c^2*Sec[e + 
f*x]^5*(c - c*Sin[e + f*x])^(5/2))/(15*f) - (2*c*Sec[e + f*x]^5*(c - c*Sin 
[e + f*x])^(7/2))/(3*f)))/(a^3*c^3)
 

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 = 133.96 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.68

method	result	size
default	\(\frac {2 c^{3} \left (-1+\sin \left (f x +e \right )\right ) \left (15 \sin \left (f x +e \right )^{2}+10 \sin \left (f x +e \right )+7\right )}{15 a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(71\)

Input:

int((c-c*sin(f*x+e))^(5/2)/(a+sin(f*x+e)*a)^3,x,method=_RETURNVERBOSE)
 

Output:

2/15*c^3/a^3*(-1+sin(f*x+e))/(1+sin(f*x+e))^2*(15*sin(f*x+e)^2+10*sin(f*x+ 
e)+7)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, c^{2} \cos \left (f x + e\right )^{2} - 10 \, c^{2} \sin \left (f x + e\right ) - 22 \, c^{2}\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \] Input:

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Timed out} \] Input:

integrate((c-c*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e))**3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (94) = 188\).

Time = 0.13 (sec) , antiderivative size = 380, normalized size of antiderivative = 3.65 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=\frac {2 \, {\left (7 \, c^{\frac {5}{2}} + \frac {20 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {95 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {80 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {250 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {120 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {250 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {80 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac {95 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac {20 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}} + \frac {7 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}}\right )}}{15 \, {\left (a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} f {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} \] Input:

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")
 

Output:

2/15*(7*c^(5/2) + 20*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 95*c^(5/2)* 
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 80*c^(5/2)*sin(f*x + e)^3/(cos(f*x + 
 e) + 1)^3 + 250*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 120*c^(5/2) 
*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 250*c^(5/2)*sin(f*x + e)^6/(cos(f*x 
 + e) + 1)^6 + 80*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 95*c^(5/2) 
*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 20*c^(5/2)*sin(f*x + e)^9/(cos(f*x 
+ e) + 1)^9 + 7*c^(5/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)/((a^3 + 5*a 
^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 
 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/ 
(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*f*(sin(f*x 
 + e)^2/(cos(f*x + e) + 1)^2 + 1)^(5/2))
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.59 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=\frac {16 \, \sqrt {2} {\left (c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \frac {5 \, c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {10 \, c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )} \sqrt {c}}{15 \, a^{3} f {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )}^{5}} \] Input:

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")
 

Output:

16/15*sqrt(2)*(c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*c^2*(cos(-1/4*p 
i + 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) + 10*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*s 
gn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2) 
*sqrt(c)/(a^3*f*((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)
 

Mupad [B] (verification not implemented)

Time = 21.54 (sec) , antiderivative size = 453, normalized size of antiderivative = 4.36 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

int((c - c*sin(e + f*x))^(5/2)/(a + a*sin(e + f*x))^3,x)
 

Output:

(c^2*exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + 
f*x*1i)*1i)/2))^(1/2)*32i)/(3*a^3*f*(exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + 
f*x*1i) + 1i)^2) - (4*c^2*exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i) 
*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2))/(a^3*f*(exp(e*1i + f*x*1i) - 1 
i)*(exp(e*1i + f*x*1i) + 1i)) + (352*c^2*exp(e*1i + f*x*1i)*(c - c*((exp(- 
 e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2))/(15*a^3*f*(exp( 
e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^3) - (c^2*exp(e*1i + f*x*1i 
)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)* 
128i)/(5*a^3*f*(exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^4) - (6 
4*c^2*exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + 
 f*x*1i)*1i)/2))^(1/2))/(5*a^3*f*(exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x 
*1i) + 1i)^5)
 

Reduce [F]

\[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=\frac {\sqrt {c}\, c^{2} \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x +\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x -2 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right )\right )}{a^{3}} \] Input:

int((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x)
 

Output:

(sqrt(c)*c**2*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**3 + 3*sin(e + 
f*x)**2 + 3*sin(e + f*x) + 1),x) + int((sqrt( - sin(e + f*x) + 1)*sin(e + 
f*x)**2)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x) - 2 
*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**3 + 3*sin(e + 
 f*x)**2 + 3*sin(e + f*x) + 1),x)))/a**3