\(\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx\) [131]

Optimal result

Integrand size = 42, antiderivative size = 121 \[ \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac {2 g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:

2*(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)-2 
*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 [A] (verified)

Time = 3.97 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.22 \[ \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {2 (g \cos (e+f x))^{3/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \left (-\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\sqrt {\cos (e+f x)} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )}{c f \cos ^{\frac {3}{2}}(e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)}} \] 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*Cos[e + f*x])^(3/2)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(EllipticE 
[(e + f*x)/2, 2]*(-Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + Sqrt[Cos[e + f*x 
]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])))/(c*f*Cos[e + f*x]^(3/2)*Sqrt[a* 
(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 121, 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, 3331, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3331

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3321

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

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:

(2*(g*Cos[e + f*x])^(5/2))/(f*g*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f* 
x])^(3/2)) - (2*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 3331

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*((c + d*Sin[e + f*x])^n/(a* 
f*g*(2*m + p + 1))), x] + Simp[(m + n + p + 1)/(a*(2*m + p + 1))   Int[(g*C 
os[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] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] &&  !LtQ[m, n, -1] && Integers 
Q[2*m, 2*n, 2*p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1041\) vs. \(2(109)=218\).

Time = 25.80 (sec) , antiderivative size = 1042, normalized size of antiderivative = 8.61

method	result	size
default	\(\text {Expression too large to display}\)	\(1042\)
risch	\(\text {Expression too large to display}\)	\(1145\)

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:

-g/c*(2/3/f/(1+2^(1/2))*((cos(1/2*f*x+1/2*e)+1)*(-cos(1/2*f*x+1/2*e)+sin(1 
/2*f*x+1/2*e))*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)+(-cos(1/2*f* 
x+1/2*e)-sin(1/2*f*x+1/2*e))*2^(1/2)-cos(1/2*f*x+1/2*e)-sin(1/2*f*x+1/2*e) 
)*(g*(-1+2*cos(1/2*f*x+1/2*e)^2))^(1/2)/(-cos(1/2*f*x+1/2*e)+sin(1/2*f*x+1 
/2*e))/((2*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)+1)*a)^(1/2)/(-(2*cos(1/2* 
f*x+1/2*e)*sin(1/2*f*x+1/2*e)-1)*c)^(1/2)-1/3/f/(1+2^(1/2))/(2*2^(1/2)-3)* 
(((3*cos(1/2*f*x+1/2*e)^2+6*cos(1/2*f*x+1/2*e)+3)*sin(1/2*f*x+1/2*e)-3*cos 
(1/2*f*x+1/2*e)*(cos(1/2*f*x+1/2*e)^2+2*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)*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)+((-4*cos(1/2*f*x+1/2*e)^2-8*cos(1/2*f* 
x+1/2*e)-4)*sin(1/2*f*x+1/2*e)+4*cos(1/2*f*x+1/2*e)*(cos(1/2*f*x+1/2*e)^2+ 
2*cos(1/2*f*x+1/2*e)+1))*((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)+((12*co...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.24 \[ \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {\frac {1}{2}} \sqrt {a c g} {\left (-i \, g \sin \left (f x + e\right ) + i \, g\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + \sqrt {\frac {1}{2}} \sqrt {a c g} {\left (i \, g \sin \left (f x + e\right ) - i \, g\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} g\right )}}{a c^{2} f \sin \left (f x + e\right ) - a c^{2} f} \] 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*(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))) + sqrt(1/2)*sq 
rt(a*c*g)*(I*g*sin(f*x + e) - I*g)*weierstrassZeta(-4, 0, weierstrassPInve 
rse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) + sqrt(g*cos(f*x + e))*sqrt(a*s 
in(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*g)/(a*c^2*f*sin(f*x + e) - a*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]

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