\(\int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [144]

Optimal result

Integrand size = 42, antiderivative size = 243 \[ \int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {154 c^3 (g \cos (e+f x))^{5/2}}{15 a^2 f g \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {154 c^3 g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {44 c^2 (g \cos (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{5 a f g (a+a \sin (e+f x))^{3/2}}-\frac {4 c (g \cos (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}}{5 f g (a+a \sin (e+f x))^{5/2}} \] Output:

154/15*c^3*(g*cos(f*x+e))^(5/2)/a^2/f/g/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f* 
x+e))^(1/2)+154/5*c^3*g*cos(f*x+e)^(1/2)*(g*cos(f*x+e))^(1/2)*EllipticE(si 
n(1/2*f*x+1/2*e),2^(1/2))/a^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1 
/2)+44/5*c^2*(g*cos(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(1/2)/a/f/g/(a+a*sin(f* 
x+e))^(3/2)-4/5*c*(g*cos(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(3/2)/f/g/(a+a*sin 
(f*x+e))^(5/2)
 

Mathematica [A] (verified)

Time = 5.22 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.95 \[ \int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {c^2 (g \cos (e+f x))^{3/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sqrt {c-c \sin (e+f x)} \left (924 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+\sqrt {\cos (e+f x)} \left (226 \cos \left (\frac {1}{2} (e+f x)\right )+327 \cos \left (\frac {3}{2} (e+f x)\right )-5 \cos \left (\frac {5}{2} (e+f x)\right )-226 \sin \left (\frac {1}{2} (e+f x)\right )+327 \sin \left (\frac {3}{2} (e+f x)\right )+5 \sin \left (\frac {5}{2} (e+f x)\right )\right )\right )}{30 f \cos ^{\frac {3}{2}}(e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{5/2}} \] Input:

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

Output:

(c^2*(g*Cos[e + f*x])^(3/2)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*Sqrt[c 
 - c*Sin[e + f*x]]*(924*EllipticE[(e + f*x)/2, 2]*(Cos[(e + f*x)/2] + Sin[ 
(e + f*x)/2])^3 + Sqrt[Cos[e + f*x]]*(226*Cos[(e + f*x)/2] + 327*Cos[(3*(e 
 + f*x))/2] - 5*Cos[(5*(e + f*x))/2] - 226*Sin[(e + f*x)/2] + 327*Sin[(3*( 
e + f*x))/2] + 5*Sin[(5*(e + f*x))/2])))/(30*f*Cos[e + f*x]^(3/2)*(Cos[(e 
+ f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(5/2))
 

Rubi [A] (verified)

Time = 1.89 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3329, 3042, 3329, 3042, 3330, 3042, 3321, 3042, 3121, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3329

\(\displaystyle -\frac {11 c \int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{(\sin (e+f x) a+a)^{3/2}}dx}{5 a}-\frac {4 c (c-c \sin (e+f x))^{3/2} (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {11 c \int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{(\sin (e+f x) a+a)^{3/2}}dx}{5 a}-\frac {4 c (c-c \sin (e+f x))^{3/2} (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3329

\(\displaystyle -\frac {11 c \left (-\frac {7 c \int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx}{a}-\frac {4 c \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2}}\right )}{5 a}-\frac {4 c (c-c \sin (e+f x))^{3/2} (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {11 c \left (-\frac {7 c \int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx}{a}-\frac {4 c \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2}}\right )}{5 a}-\frac {4 c (c-c \sin (e+f x))^{3/2} (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3330

\(\displaystyle -\frac {11 c \left (-\frac {7 c \left (c \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx+\frac {2 c (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{a}-\frac {4 c \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2}}\right )}{5 a}-\frac {4 c (c-c \sin (e+f x))^{3/2} (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {11 c \left (-\frac {7 c \left (c \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx+\frac {2 c (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{a}-\frac {4 c \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2}}\right )}{5 a}-\frac {4 c (c-c \sin (e+f x))^{3/2} (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3321

\(\displaystyle -\frac {11 c \left (-\frac {7 c \left (\frac {c g \cos (e+f x) \int \sqrt {g \cos (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {2 c (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{a}-\frac {4 c \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2}}\right )}{5 a}-\frac {4 c (c-c \sin (e+f x))^{3/2} (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {11 c \left (-\frac {7 c \left (\frac {c g \cos (e+f x) \int \sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {2 c (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{a}-\frac {4 c \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2}}\right )}{5 a}-\frac {4 c (c-c \sin (e+f x))^{3/2} (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3121

\(\displaystyle -\frac {11 c \left (-\frac {7 c \left (\frac {c g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\cos (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {2 c (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{a}-\frac {4 c \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2}}\right )}{5 a}-\frac {4 c (c-c \sin (e+f x))^{3/2} (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {11 c \left (-\frac {7 c \left (\frac {c g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {2 c (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{a}-\frac {4 c \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2}}\right )}{5 a}-\frac {4 c (c-c \sin (e+f x))^{3/2} (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {4 c (c-c \sin (e+f x))^{3/2} (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2}}-\frac {11 c \left (-\frac {4 c \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2}}-\frac {7 c \left (\frac {2 c (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {2 c g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{a}\right )}{5 a}\)

Input:

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

Output:

(-4*c*(g*Cos[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(3/2))/(5*f*g*(a + a*Sin 
[e + f*x])^(5/2)) - (11*c*((-4*c*(g*Cos[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + 
 f*x]])/(f*g*(a + a*Sin[e + f*x])^(3/2)) - (7*c*((2*c*(g*Cos[e + f*x])^(5/ 
2))/(3*f*g*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) + (2*c*g*Sqr 
t[Cos[e + f*x]]*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(f*Sqrt[a 
+ a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])))/a))/(5*a)
 

Defintions of rubi rules used

rule 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3321

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

rule 3329

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

rule 3330

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1927\) vs. \(2(211)=422\).

Time = 18.85 (sec) , antiderivative size = 1928, normalized size of antiderivative = 7.93

method	result	size
default	\(\text {Expression too large to display}\)	\(1928\)

Input:

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

Output:

-1/15/f/(2*2^(1/2)-3)/(1+2^(1/2))*c^2/g/a^2*(150*(1+(-1+2*cos(1/2*f*x+1/2* 
e)*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e)+2*cos(1/2*f*x+1/2*e)^2)* 
2^(1/2)-2*cos(1/2*f*x+1/2*e)*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e 
)-2*cos(1/2*f*x+1/2*e)^2)*g^(5/2)*ln(2/g^(1/2)*(g^(1/2)*(g*(-1+2*cos(1/2*f 
*x+1/2*e)^2)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)*cos(1/2*f*x+1/2*e)+g^(1/2)*(g 
*(-1+2*cos(1/2*f*x+1/2*e)^2)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)-2*cos(1/2*f*x 
+1/2*e)*g-g)/(cos(1/2*f*x+1/2*e)+1))+150*(-1+(1-2*cos(1/2*f*x+1/2*e)*(-1+2 
*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e)-2*cos(1/2*f*x+1/2*e)^2)*2^(1/2)+ 
2*cos(1/2*f*x+1/2*e)*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e)+2*cos( 
1/2*f*x+1/2*e)^2)*g^(5/2)*ln(4/g^(1/2)*(g^(1/2)*(g*(-1+2*cos(1/2*f*x+1/2*e 
)^2)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)*cos(1/2*f*x+1/2*e)+g^(1/2)*(g*(-1+2*c 
os(1/2*f*x+1/2*e)^2)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)-2*cos(1/2*f*x+1/2*e)* 
g-g)/(cos(1/2*f*x+1/2*e)+1))+120*(-2+(1-2*cos(1/2*f*x+1/2*e)*(-1+2*cos(1/2 
*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e)-2*cos(1/2*f*x+1/2*e)^2)*2^(1/2)+4*cos(1/ 
2*f*x+1/2*e)*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e)+4*cos(1/2*f*x+ 
1/2*e)^2)*g^(5/2)*ln(4*g^(1/2)*(g*(-1+2*cos(1/2*f*x+1/2*e)^2)/(cos(1/2*f*x 
+1/2*e)+1)^2)^(1/2)*2^(1/2)*cos(1/2*f*x+1/2*e)+4*2^(1/2)*g^(1/2)*(g*(-1+2* 
cos(1/2*f*x+1/2*e)^2)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)+8*cos(1/2*f*x+1/2*e) 
*g)+120*(2+(-1+2*cos(1/2*f*x+1/2*e)*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f* 
x+1/2*e)+2*cos(1/2*f*x+1/2*e)^2)*2^(1/2)-4*cos(1/2*f*x+1/2*e)*(-1+2*cos...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.97 \[ \int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (231 \, \sqrt {\frac {1}{2}} {\left (-i \, c^{2} g \cos \left (f x + e\right )^{2} + 2 i \, c^{2} g \sin \left (f x + e\right ) + 2 i \, c^{2} g\right )} \sqrt {a c g} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 231 \, \sqrt {\frac {1}{2}} {\left (i \, c^{2} g \cos \left (f x + e\right )^{2} - 2 i \, c^{2} g \sin \left (f x + e\right ) - 2 i \, c^{2} g\right )} \sqrt {a c g} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + {\left (5 \, c^{2} g \cos \left (f x + e\right )^{2} - 166 \, c^{2} g \sin \left (f x + e\right ) - 142 \, c^{2} g\right )} \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}\right )}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \sin \left (f x + e\right ) - 2 \, a^{3} f\right )}} \] Input:

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

Output:

2/15*(231*sqrt(1/2)*(-I*c^2*g*cos(f*x + e)^2 + 2*I*c^2*g*sin(f*x + e) + 2* 
I*c^2*g)*sqrt(a*c*g)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos 
(f*x + e) + I*sin(f*x + e))) + 231*sqrt(1/2)*(I*c^2*g*cos(f*x + e)^2 - 2*I 
*c^2*g*sin(f*x + e) - 2*I*c^2*g)*sqrt(a*c*g)*weierstrassZeta(-4, 0, weiers 
trassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) + (5*c^2*g*cos(f*x + 
e)^2 - 166*c^2*g*sin(f*x + e) - 142*c^2*g)*sqrt(g*cos(f*x + e))*sqrt(a*sin 
(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(a^3*f*cos(f*x + e)^2 - 2*a^3*f* 
sin(f*x + e) - 2*a^3*f)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {too large to display} \] Input:

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

Output:

(sqrt(g)*sqrt(c)*sqrt(a)*c**2*g*( - 18*sqrt(sin(e + f*x) + 1)*sqrt( - sin( 
e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x)**2 - 4*sqrt(sin(e + f*x) + 1 
)*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x) - 8*sqrt(sin(e 
 + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x)) + 30*int((sqrt(s 
in(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*cos(e + f*x) 
*sin(e + f*x)**3)/(sin(e + f*x)**4 + 2*sin(e + f*x)**3 - 2*sin(e + f*x) - 
1),x)*sin(e + f*x)**2*f + 60*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f 
*x) + 1)*sqrt(cos(e + f*x))*cos(e + f*x)*sin(e + f*x)**3)/(sin(e + f*x)**4 
 + 2*sin(e + f*x)**3 - 2*sin(e + f*x) - 1),x)*sin(e + f*x)*f + 30*int((sqr 
t(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*cos(e + f 
*x)*sin(e + f*x)**3)/(sin(e + f*x)**4 + 2*sin(e + f*x)**3 - 2*sin(e + f*x) 
 - 1),x)*f - 9*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sqrt( 
cos(e + f*x))*sin(e + f*x)**5)/(cos(e + f*x)*sin(e + f*x)**4 + 2*cos(e + f 
*x)*sin(e + f*x)**3 - 2*cos(e + f*x)*sin(e + f*x) - cos(e + f*x)),x)*sin(e 
 + f*x)**2*f - 18*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sq 
rt(cos(e + f*x))*sin(e + f*x)**5)/(cos(e + f*x)*sin(e + f*x)**4 + 2*cos(e 
+ f*x)*sin(e + f*x)**3 - 2*cos(e + f*x)*sin(e + f*x) - cos(e + f*x)),x)*si 
n(e + f*x)*f - 9*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sqr 
t(cos(e + f*x))*sin(e + f*x)**5)/(cos(e + f*x)*sin(e + f*x)**4 + 2*cos(e + 
 f*x)*sin(e + f*x)**3 - 2*cos(e + f*x)*sin(e + f*x) - cos(e + f*x)),x)*...