\(\int \frac {(g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{3/2}} \, dx\) [93]

Optimal result

Integrand size = 42, antiderivative size = 123 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {4 a (g \cos (e+f x))^{5/2}}{f g \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac {6 a g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:

4*a*(g*cos(f*x+e))^(5/2)/f/g/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2) 
-6*a*g*cos(f*x+e)^(1/2)*(g*cos(f*x+e))^(1/2)*EllipticE(sin(1/2*f*x+1/2*e), 
2^(1/2))/c/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 3.47 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.72 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {2 g \sqrt {e^{-i (e+f x)} \left (1+e^{2 i (e+f x)}\right ) g} \left (\left (i-5 e^{i (e+f x)}\right ) \sqrt {1+e^{2 i (e+f x)}}+2 e^{2 i (e+f x)} \left (-i+e^{i (e+f x)}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (e+f x)}\right )\right ) \sqrt {a (1+\sin (e+f x))}}{c \sqrt {i c e^{-i (e+f x)} \left (-i+e^{i (e+f x)}\right )^2} \left (i+e^{i (e+f x)}\right ) \sqrt {1+e^{2 i (e+f x)}} f} \] Input:

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

Output:

(-2*g*Sqrt[((1 + E^((2*I)*(e + f*x)))*g)/E^(I*(e + f*x))]*((I - 5*E^(I*(e 
+ f*x)))*Sqrt[1 + E^((2*I)*(e + f*x))] + 2*E^((2*I)*(e + f*x))*(-I + E^(I* 
(e + f*x)))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(e + f*x))])*Sqrt[a 
*(1 + Sin[e + f*x])])/(c*Sqrt[(I*c*(-I + E^(I*(e + f*x)))^2)/E^(I*(e + f*x 
))]*(I + E^(I*(e + f*x)))*Sqrt[1 + E^((2*I)*(e + f*x))]*f)
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3329, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3329

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3321

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {4 a (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {6 a g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\)

Input:

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

Output:

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

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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 17.48 (sec) , antiderivative size = 1186, normalized size of antiderivative = 9.64

method	result	size
risch	\(\text {Expression too large to display}\)	\(1186\)
default	\(\text {Expression too large to display}\)	\(1279\)

Input:

int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2),x,m 
ethod=_RETURNVERBOSE)
 

Output:

-(-4*I*exp(I*(f*x+e))+exp(I*(f*x+e))^2+3)*(exp(I*(f*x+e))+I)*(exp(I*(f*x+e 
))-I)/f*2^(1/2)*g/c*(g*(exp(I*(f*x+e))^2+1)/exp(I*(f*x+e)))^(1/2)/(exp(I*( 
f*x+e))^2+1)*(-a*(I*exp(I*(f*x+e))^2-I-2*exp(I*(f*x+e)))/exp(I*(f*x+e)))^( 
1/2)/(exp(I*(f*x+e))^2+2*I*exp(I*(f*x+e))-1)/(c*(I*exp(I*(f*x+e))^2-I+2*ex 
p(I*(f*x+e)))/exp(I*(f*x+e)))^(1/2)+1/f*(-exp(I*(f*x+e))+I)*(exp(I*(f*x+e) 
)+I)*(6*(-I*(-exp(I*(f*x+e))+I))^(1/2)*2^(1/2)*(-I*(exp(I*(f*x+e))+I))^(1/ 
2)*(-I*exp(I*(f*x+e)))^(1/2)*(-c*g*a*exp(I*(f*x+e))*(exp(I*(f*x+e))^2+1))^ 
(1/2)*((-exp(I*(f*x+e))+I)*(exp(I*(f*x+e))+I)*exp(I*(f*x+e))*a*c*g)^(1/2)* 
EllipticE((-I*(-exp(I*(f*x+e))+I))^(1/2),1/2*2^(1/2))-3*(-I*(-exp(I*(f*x+e 
))+I))^(1/2)*2^(1/2)*(-I*(exp(I*(f*x+e))+I))^(1/2)*(-I*exp(I*(f*x+e)))^(1/ 
2)*(-c*g*a*exp(I*(f*x+e))*(exp(I*(f*x+e))^2+1))^(1/2)*((-exp(I*(f*x+e))+I) 
*(exp(I*(f*x+e))+I)*exp(I*(f*x+e))*a*c*g)^(1/2)*EllipticF((-I*(-exp(I*(f*x 
+e))+I))^(1/2),1/2*2^(1/2))-2*exp(I*(f*x+e))^2*(-c*g*a*exp(I*(f*x+e))*(exp 
(I*(f*x+e))^2+1))^(1/2)*(-c*g*a*exp(I*(f*x+e))^3-exp(I*(f*x+e))*a*c*g)^(1/ 
2)-4*(-c*g*a*exp(I*(f*x+e))^3-exp(I*(f*x+e))*a*c*g)^(1/2)*((-exp(I*(f*x+e) 
)+I)*(exp(I*(f*x+e))+I)*exp(I*(f*x+e))*a*c*g)^(1/2)*exp(I*(f*x+e))^2-4*(-c 
*g*a*exp(I*(f*x+e))^3-exp(I*(f*x+e))*a*c*g)^(1/2)*((-exp(I*(f*x+e))+I)*(ex 
p(I*(f*x+e))+I)*exp(I*(f*x+e))*a*c*g)^(1/2))/(c*(2*I*exp(I*(f*x+e))-exp(I* 
(f*x+e))^2+1)*exp(I*(f*x+e))*g*(exp(I*(f*x+e))^2+1)*a*(exp(I*(f*x+e))^2+2* 
I*exp(I*(f*x+e))-1))^(1/2)/(-c*g*a*exp(I*(f*x+e))^3-exp(I*(f*x+e))*a*c*...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.23 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {2 \, {\left (3 \, \sqrt {\frac {1}{2}} \sqrt {a c g} {\left (-i \, g \sin \left (f x + e\right ) + i \, g\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {\frac {1}{2}} \sqrt {a c g} {\left (i \, g \sin \left (f x + e\right ) - i \, g\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} g\right )}}{c^{2} f \sin \left (f x + e\right ) - c^{2} f} \] Input:

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/ 
2),x, algorithm="fricas")
 

Output:

-2*(3*sqrt(1/2)*sqrt(a*c*g)*(-I*g*sin(f*x + e) + I*g)*weierstrassZeta(-4, 
0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + 3*sqrt(1/2 
)*sqrt(a*c*g)*(I*g*sin(f*x + e) - I*g)*weierstrassZeta(-4, 0, weierstrassP 
Inverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) + 2*sqrt(g*cos(f*x + e))*sq 
rt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*g)/(c^2*f*sin(f*x + e) - 
c^2*f)
 

Sympy [F(-1)]

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

integrate((g*cos(f*x+e))**(3/2)*(a+a*sin(f*x+e))**(1/2)/(c-c*sin(f*x+e))** 
(3/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/ 
2),x, algorithm="maxima")
 

Output:

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

Giac [F(-1)]

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

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/ 
2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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