\(\int \frac {(g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx\) [92]

Optimal result

Integrand size = 42, antiderivative size = 122 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {2 a (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {2 a g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:

-2/3*a*(g*cos(f*x+e))^(5/2)/f/g/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1 
/2)+2*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))/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 = 3.32 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.61 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {i \sqrt {i c e^{-i (e+f x)} \left (-i+e^{i (e+f x)}\right )^2} g \sqrt {e^{-i (e+f x)} \left (1+e^{2 i (e+f x)}\right ) g} \left (-i \sqrt {1+e^{2 i (e+f x)}} \left (1-6 i e^{i (e+f x)}+e^{2 i (e+f x)}\right )+4 e^{3 i (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (e+f x)}\right )\right ) \sqrt {a (1+\sin (e+f x))}}{3 c \left (1+e^{2 i (e+f x)}\right )^{3/2} f} \] Input:

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

Output:

((-1/3*I)*Sqrt[(I*c*(-I + E^(I*(e + f*x)))^2)/E^(I*(e + f*x))]*g*Sqrt[((1 
+ E^((2*I)*(e + f*x)))*g)/E^(I*(e + f*x))]*((-I)*Sqrt[1 + E^((2*I)*(e + f* 
x))]*(1 - (6*I)*E^(I*(e + f*x)) + E^((2*I)*(e + f*x))) + 4*E^((3*I)*(e + f 
*x))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(e + f*x))])*Sqrt[a*(1 + S 
in[e + f*x])])/(c*(1 + E^((2*I)*(e + f*x)))^(3/2)*f)
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 122, 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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3330

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3321

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

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

Output:

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

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 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] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2082\) vs. \(2(108)=216\).

Time = 11.43 (sec) , antiderivative size = 2083, normalized size of antiderivative = 17.07

method	result	size
default	\(\text {Expression too large to display}\)	\(2083\)

Input:

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

Output:

g*(2/f/g^2/(1+2^(1/2))*(((cos(1/2*f*x+1/2*e)+1)*2^(1/2)+cos(1/2*f*x+1/2*e) 
+1)*g^(3/2)*(g*(-1+2*cos(1/2*f*x+1/2*e)^2)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2) 
*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))+((-cos(1/2*f*x+1/2*e)-1)*2^(1/2)-cos(1/2*f*x+1/2*e)-1)*g^(3/2)*(g*(-1+ 
2*cos(1/2*f*x+1/2*e)^2)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)*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*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))+(-cos(1/2*f*x+1 
/2*e)-1)*2^(1/2)*((2^(1/2)*cos(1/2*f*x+1/2*e)-2^(1/2)+2*cos(1/2*f*x+1/2*e) 
-1)/(cos(1/2*f*x+1/2*e)+1))^(1/2)*(-2*(2^(1/2)*cos(1/2*f*x+1/2*e)-2^(1/2)- 
2*cos(1/2*f*x+1/2*e)+1)/(cos(1/2*f*x+1/2*e)+1))^(1/2)*EllipticF((1+2^(1/2) 
)*(csc(1/2*f*x+1/2*e)-cot(1/2*f*x+1/2*e)),-2*2^(1/2)+3)*g^2+((-1+2*cos(1/2 
*f*x+1/2*e)^2)*2^(1/2)+2*cos(1/2*f*x+1/2*e)^2-1)*g^2)*((2*cos(1/2*f*x+1/2* 
e)*sin(1/2*f*x+1/2*e)+1)*a)^(1/2)*(g*(-1+2*cos(1/2*f*x+1/2*e)^2))^(1/2)/(2 
*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)+1)/(-(2*cos(1/2*f*x+1/2*e)*sin(1/2* 
f*x+1/2*e)-1)*c)^(1/2)+1/3/f/g^2/(1+2^(1/2))/(2*2^(1/2)-3)*(((12*cos(1/2*f 
*x+1/2*e)^2+24*cos(1/2*f*x+1/2*e)+12)*2^(1/2)-12*cos(1/2*f*x+1/2*e)^2-24*c 
os(1/2*f*x+1/2*e)-12)*(g*(-1+2*cos(1/2*f*x+1/2*e)^2)/(cos(1/2*f*x+1/2*e...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.92 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {2 \, {\left (3 i \, \sqrt {\frac {1}{2}} \sqrt {a c g} g {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 3 i \, \sqrt {\frac {1}{2}} \sqrt {a c g} g {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} g\right )}}{3 \, c f} \] Input:

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

Output:

-2/3*(3*I*sqrt(1/2)*sqrt(a*c*g)*g*weierstrassZeta(-4, 0, weierstrassPInver 
se(-4, 0, cos(f*x + e) + I*sin(f*x + e))) - 3*I*sqrt(1/2)*sqrt(a*c*g)*g*we 
ierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + 
 e))) + sqrt(g*cos(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) 
 + c)*g)/(c*f)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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