\(\int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx\) [334]

Optimal result

Integrand size = 28, antiderivative size = 100 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {64 c \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}-\frac {16 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 f}+\frac {2 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 c f} \] Output:

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

Mathematica [A] (verified)

Time = 3.56 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.04 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (25-3 \cos (2 (e+f x))+36 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{3 a^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2} \] Input:

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

Output:

(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(25 - 3*Cos[2*(e + f*x)] + 36*S 
in[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(3*a^2*f*(Cos[(e + f*x)/2] - Sin[(e 
 + f*x)/2])*(1 + Sin[e + f*x])^2)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.03, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

\(\displaystyle \frac {\int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(c-c \sin (e+f x))^{9/2}}{\cos (e+f x)^4}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3153

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3153

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3152

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

Input:

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

Output:

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

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 = 20.68 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.71

method	result	size
default	\(-\frac {2 c^{3} \left (-1+\sin \left (f x +e \right )\right ) \left (3 \sin \left (f x +e \right )^{2}+18 \sin \left (f x +e \right )+11\right )}{3 a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(71\)

Input:

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

Output:

-2/3*c^3/a^2*(-1+sin(f*x+e))/(1+sin(f*x+e))*(3*sin(f*x+e)^2+18*sin(f*x+e)+ 
11)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.76 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (3 \, c^{2} \cos \left (f x + e\right )^{2} - 18 \, c^{2} \sin \left (f x + e\right ) - 14 \, c^{2}\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, {\left (a^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} f \cos \left (f x + e\right )\right )}} \] Input:

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

Output:

-2/3*(3*c^2*cos(f*x + e)^2 - 18*c^2*sin(f*x + e) - 14*c^2)*sqrt(-c*sin(f*x 
 + e) + c)/(a^2*f*cos(f*x + e)*sin(f*x + e) + a^2*f*cos(f*x + e))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 288 vs. \(2 (92) = 184\).

Time = 0.13 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.88 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (11 \, c^{\frac {5}{2}} + \frac {36 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {56 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {108 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {90 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {108 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {56 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {36 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac {11 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\right )}}{3 \, {\left (a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} f {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} \] Input:

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

Output:

-2/3*(11*c^(5/2) + 36*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 56*c^(5/2) 
*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 108*c^(5/2)*sin(f*x + e)^3/(cos(f*x 
 + e) + 1)^3 + 90*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 108*c^(5/2 
)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 56*c^(5/2)*sin(f*x + e)^6/(cos(f*x 
 + e) + 1)^6 + 36*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 11*c^(5/2) 
*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)/((a^2 + 3*a^2*sin(f*x + e)/(cos(f*x 
+ e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3 
/(cos(f*x + e) + 1)^3)*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. \(2 (92) = 184\).

Time = 0.18 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.25 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {4 \, \sqrt {2} \sqrt {c} {\left (\frac {3 \, c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{a^{2} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 1\right )}} - \frac {5 \, c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \frac {12 \, c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {3 \, c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}}{a^{2} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )}^{3}}\right )}}{3 \, f} \] Input:

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

Output:

4/3*sqrt(2)*sqrt(c)*(3*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(a^2*((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^2*((c 
os(-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 [B] (verification not implemented)

Time = 23.05 (sec) , antiderivative size = 360, normalized size of antiderivative = 3.60 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {\sqrt {c-c\,\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}\,\left (\frac {2\,c^2}{a^2\,f}-\frac {c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a^2\,f}\right )}{{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-\mathrm {i}}+\frac {16\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\sqrt {c-c\,\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}}{a^2\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}-\frac {c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\sqrt {c-c\,\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}\,32{}\mathrm {i}}{3\,a^2\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-\mathrm {i}\right )\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}^2}-\frac {32\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\sqrt {c-c\,\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}}{3\,a^2\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-\mathrm {i}\right )\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}^3} \] Input:

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

Output:

((c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*( 
(2*c^2)/(a^2*f) - (c^2*exp(e*1i + f*x*1i)*2i)/(a^2*f)))/(exp(e*1i + f*x*1i 
) - 1i) + (16*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^2*f*(exp(e*1i + f*x*1i) - 1i)*(exp( 
e*1i + f*x*1i) + 1i)) - (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^2*f*(exp(e*1i + f* 
x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^2) - (32*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))/(3*a^ 
2*f*(exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^3)
 

Reduce [F]

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

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

Output:

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