\(\int (g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)} \, dx\) [91]

Optimal result

Integrand size = 42, antiderivative size = 178 \[ \int (g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)} \, dx=\frac {2 a c (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {6 a c g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {2 a (g \cos (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{5 f g \sqrt {a+a \sin (e+f x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 3.56 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.96 \[ \int (g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)} \, dx=\frac {(g \cos (e+f x))^{3/2} \sec ^3(e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} \left (3 (3 \cos (e-f x-\arctan (\tan (e)))+\cos (e+f x+\arctan (\tan (e)))) \cot (e) \sqrt {\sec ^2(e)}+2 \cos (e+f x) (-6 \cot (e)+\sin (2 (e+f x)))-6 \cos (e) \csc (f x+\arctan (\tan (e))) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(f x+\arctan (\tan (e)))\right ) \sqrt {\sec ^2(e)} \sqrt {\sin ^2(f x+\arctan (\tan (e)))}\right )}{10 f} \] 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:

((g*Cos[e + f*x])^(3/2)*Sec[e + f*x]^3*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - 
 c*Sin[e + f*x]]*(3*(3*Cos[e - f*x - ArcTan[Tan[e]]] + Cos[e + f*x + ArcTa 
n[Tan[e]]])*Cot[e]*Sqrt[Sec[e]^2] + 2*Cos[e + f*x]*(-6*Cot[e] + Sin[2*(e + 
 f*x)]) - 6*Cos[e]*Csc[f*x + ArcTan[Tan[e]]]*HypergeometricPFQ[{-1/2, -1/4 
}, {3/4}, Cos[f*x + ArcTan[Tan[e]]]^2]*Sqrt[Sec[e]^2]*Sqrt[Sin[f*x + ArcTa 
n[Tan[e]]]^2]))/(10*f)
 

Rubi [A] (verified)

Time = 1.38 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3330, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3330

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3330

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3321

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

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

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. \(993\) vs. \(2(154)=308\).

Time = 14.08 (sec) , antiderivative size = 994, normalized size of antiderivative = 5.58

method	result	size
default	\(\text {Expression too large to display}\)	\(994\)

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:

2*g*(-1/3/f/(1+2^(1/2))*(sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)*((4*cos(1/2 
*f*x+1/2*e)^2-2)*2^(1/2)+4*cos(1/2*f*x+1/2*e)^2-2)+(cos(1/2*f*x+1/2*e)+1)* 
2^(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)*((2^(1/2)*cos(1/2*f*x+1/2*e)-2^(1/2)+2*cos(1/2*f*x+1/2*e)-1)/(co 
s(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))*(-(2*cos(1/2*f*x+1/2*e)*si 
n(1/2*f*x+1/2*e)-1)*c)^(1/2)*((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)/(4*cos(1/2*f*x+1/2*e)^4-4*c 
os(1/2*f*x+1/2*e)^2+1)-1/30/f/(2*2^(1/2)-3)/(1+2^(1/2))*(sin(1/2*f*x+1/2*e 
)*((48*cos(1/2*f*x+1/2*e)^6+48*cos(1/2*f*x+1/2*e)^5-8*cos(1/2*f*x+1/2*e)^4 
-8*cos(1/2*f*x+1/2*e)^3+28*cos(1/2*f*x+1/2*e)^2-8*cos(1/2*f*x+1/2*e)-18)*2 
^(1/2)-48*cos(1/2*f*x+1/2*e)^6-48*cos(1/2*f*x+1/2*e)^5+8*cos(1/2*f*x+1/2*e 
)^4+8*cos(1/2*f*x+1/2*e)^3-28*cos(1/2*f*x+1/2*e)^2+8*cos(1/2*f*x+1/2*e)+18 
)+(9*cos(1/2*f*x+1/2*e)^2+18*cos(1/2*f*x+1/2*e)+9)*2^(1/2)*EllipticE((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)*(-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^(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)+((48*cos(1/2*f*x+1/2*e)^2+96*cos(1/2*f*x+1/2*e)+48 
)*2^(1/2)-76*cos(1/2*f*x+1/2*e)^2-152*cos(1/2*f*x+1/2*e)-76)*((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))...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.65 \[ \int (g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)} \, dx=\frac {2 \, {\left (\sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} g \sin \left (f x + e\right ) - 3 i \, \sqrt {\frac {1}{2}} \sqrt {a c g} g {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 i \, \sqrt {\frac {1}{2}} \sqrt {a c g} g {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )\right )}}{5 \, f} \] 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:

2/5*(sqrt(g*cos(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + 
c)*g*sin(f*x + e) - 3*I*sqrt(1/2)*sqrt(a*c*g)*g*weierstrassZeta(-4, 0, wei 
erstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + 3*I*sqrt(1/2)*sq 
rt(a*c*g)*g*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) 
 - I*sin(f*x + e))))/f
 

Sympy [F(-1)]

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

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

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

Output:

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

Mupad [F(-1)]

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

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