Integrand size = 42, antiderivative size = 182 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {4 c (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}+\frac {6 c (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}+\frac {6 c g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]
-4/5*c*(g*cos(f*x+e))^(5/2)/f/g/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1 /2)+6/5*c*(g*cos(f*x+e))^(5/2)/a/f/g/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e ))^(1/2)+6/5*c*g*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE (sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(g*cos(f*x+e))^(1/2)/a^2/f/( a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 3.48 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.26 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {4 i g \sqrt {e^{-i (e+f x)} \left (1+e^{2 i (e+f x)}\right ) g} \left (\left (5-4 i e^{i (e+f x)}-3 e^{2 i (e+f x)}\right ) \sqrt {1+e^{2 i (e+f x)}}+e^{i (e+f x)} \left (i+e^{i (e+f x)}\right )^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (e+f x)}\right )\right ) \sqrt {c-c \sin (e+f x)}}{5 a \left (-i+e^{i (e+f x)}\right ) \left (-i a e^{-i (e+f x)} \left (i+e^{i (e+f x)}\right )^2\right )^{3/2} \sqrt {1+e^{2 i (e+f x)}} f} \]
(((4*I)/5)*g*Sqrt[((1 + E^((2*I)*(e + f*x)))*g)/E^(I*(e + f*x))]*((5 - (4* I)*E^(I*(e + f*x)) - 3*E^((2*I)*(e + f*x)))*Sqrt[1 + E^((2*I)*(e + f*x))] + E^(I*(e + f*x))*(I + E^(I*(e + f*x)))^3*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(e + f*x))])*Sqrt[c - c*Sin[e + f*x]])/(a*(-I + E^(I*(e + f*x)) )*(((-I)*a*(I + E^(I*(e + f*x)))^2)/E^(I*(e + f*x)))^(3/2)*Sqrt[1 + E^((2* I)*(e + f*x))]*f)
Time = 1.30 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3329, 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.
| \(\displaystyle \int \frac {\sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{3/2}}{(a \sin (e+f x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{3/2}}{(a \sin (e+f x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3329 |
| \(\displaystyle -\frac {3 c \int \frac {(g \cos (e+f x))^{3/2}}{(\sin (e+f x) a+a)^{3/2} \sqrt {c-c \sin (e+f x)}}dx}{5 a}-\frac {4 c (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 c \int \frac {(g \cos (e+f x))^{3/2}}{(\sin (e+f x) a+a)^{3/2} \sqrt {c-c \sin (e+f x)}}dx}{5 a}-\frac {4 c (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3331 |
| \(\displaystyle -\frac {3 c \left (-\frac {\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}\right )}{5 a}-\frac {4 c (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 c \left (-\frac {\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}\right )}{5 a}-\frac {4 c (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3321 |
| \(\displaystyle -\frac {3 c \left (-\frac {g \cos (e+f x) \int \sqrt {g \cos (e+f x)}dx}{a \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}\right )}{5 a}-\frac {4 c (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 c \left (-\frac {g \cos (e+f x) \int \sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{a \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}\right )}{5 a}-\frac {4 c (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle -\frac {3 c \left (-\frac {g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\cos (e+f x)}dx}{a \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}\right )}{5 a}-\frac {4 c (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 c \left (-\frac {g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{a \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}\right )}{5 a}-\frac {4 c (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {4 c (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}-\frac {3 c \left (-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}-\frac {2 g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{5 a}\) |
(-4*c*(g*Cos[e + f*x])^(5/2))/(5*f*g*(a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c *Sin[e + f*x]]) - (3*c*((-2*(g*Cos[e + f*x])^(5/2))/(f*g*(a + a*Sin[e + f* x])^(3/2)*Sqrt[c - c*Sin[e + f*x]]) - (2*g*Sqrt[Cos[e + f*x]]*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(a*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])))/(5*a)
3.2.46.3.1 Defintions of rubi rules used
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]
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]
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]
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]
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]
Result contains complex when optimal does not.
Time = 0.95 (sec) , antiderivative size = 1492, normalized size of antiderivative = 8.20
int((g*cos(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x,m ethod=_RETURNVERBOSE)
-1/10/f*(-c*(sin(f*x+e)-1))^(1/2)*(g*cos(f*x+e))^(1/2)*g/(a*(1+sin(f*x+e)) )^(1/2)/(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)/(1+cos(f*x+e))^3/a^2*(24*I*(- cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*Ellip ticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(1+cos(f*x+e)))^(1/2)-12*I*(1/(1+cos( f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(csc(f*x+e)-c ot(f*x+e)),I)*cos(f*x+e)*sin(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2 4*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I *(csc(f*x+e)-cot(f*x+e)),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*sin(f*x+e )-12*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*Elliptic F(I*(csc(f*x+e)-cot(f*x+e)),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*tan(f* x+e)+12*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*Ellip ticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*tan (f*x+e)-24*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*El lipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)* sin(f*x+e)+12*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2) *EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*cos(f*x+e)*sin(f*x+e)*(-cos(f*x+e) /(1+cos(f*x+e))^2)^(1/2)-12*I*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*(cos(f* x+e)/(1+cos(f*x+e)))^(1/2)*(1/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(csc(f*x+e )-cot(f*x+e)),I)*sec(f*x+e)+12*I*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*(cos (f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(1...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.15 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {2 \, \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (3 \, g \sin \left (f x + e\right ) + g\right )} - 3 \, {\left (-i \, \sqrt {2} g \cos \left (f x + e\right )^{2} + 2 i \, \sqrt {2} g \sin \left (f x + e\right ) + 2 i \, \sqrt {2} g\right )} \sqrt {a c g} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 3 \, {\left (i \, \sqrt {2} g \cos \left (f x + e\right )^{2} - 2 i \, \sqrt {2} g \sin \left (f x + e\right ) - 2 i \, \sqrt {2} g\right )} \sqrt {a c g} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}{5 \, {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \sin \left (f x + e\right ) - 2 \, a^{3} f\right )}} \]
integrate((g*cos(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/ 2),x, algorithm="fricas")
-1/5*(2*sqrt(g*cos(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*(3*g*sin(f*x + e) + g) - 3*(-I*sqrt(2)*g*cos(f*x + e)^2 + 2*I*sqrt(2 )*g*sin(f*x + e) + 2*I*sqrt(2)*g)*sqrt(a*c*g)*weierstrassZeta(-4, 0, weier strassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) - 3*(I*sqrt(2)*g*cos (f*x + e)^2 - 2*I*sqrt(2)*g*sin(f*x + e) - 2*I*sqrt(2)*g)*sqrt(a*c*g)*weie rstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e ))))/(a^3*f*cos(f*x + e)^2 - 2*a^3*f*sin(f*x + e) - 2*a^3*f)
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} \sqrt {-c \sin \left (f x + e\right ) + c}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integrate((g*cos(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/ 2),x, algorithm="maxima")
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
integrate((g*cos(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/ 2),x, algorithm="giac")
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]