Integrand size = 42, antiderivative size = 186 \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {4 a (g \cos (e+f x))^{5/2} \sqrt {a+a \sin (e+f x)}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {28 a^2 (g \cos (e+f x))^{5/2}}{5 c f g \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {42 a^2 g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]
-28/5*a^2*(g*cos(f*x+e))^(5/2)/c/f/g/(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e ))^(1/2)+4/5*a*(g*cos(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2)/f/g/(c-c*sin(f* x+e))^(5/2)+42/5*a^2*g*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Ell ipticE(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(g*cos(f*x+e))^(1/2)/c ^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)
Time = 6.28 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.03 \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {a \sqrt {\cos (e+f x)} (g \cos (e+f x))^{3/2} \sqrt {a (1+\sin (e+f x))} \left (-42 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+8 \sqrt {\cos (e+f x)} \left (\cos \left (\frac {1}{2} (e+f x)\right )+2 \cos \left (\frac {3}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )-2 \sin \left (\frac {3}{2} (e+f x)\right )\right )\right )}{5 c^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \]
-1/5*(a*Sqrt[Cos[e + f*x]]*(g*Cos[e + f*x])^(3/2)*Sqrt[a*(1 + Sin[e + f*x] )]*(-42*EllipticE[(e + f*x)/2, 2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 + 8*Sqrt[Cos[e + f*x]]*(Cos[(e + f*x)/2] + 2*Cos[(3*(e + f*x))/2] + Sin[(e + f*x)/2] - 2*Sin[(3*(e + f*x))/2])))/(c^2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(-1 + Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])
Time = 1.34 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.99, 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, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3329 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {7 a \int \frac {(g \cos (e+f x))^{3/2} \sqrt {\sin (e+f x) a+a}}{(c-c \sin (e+f x))^{3/2}}dx}{5 c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {7 a \int \frac {(g \cos (e+f x))^{3/2} \sqrt {\sin (e+f x) a+a}}{(c-c \sin (e+f x))^{3/2}}dx}{5 c}\) |
| \(\Big \downarrow \) 3329 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {7 a \left (\frac {4 a (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {3 a \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{c}\right )}{5 c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {7 a \left (\frac {4 a (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {3 a \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{c}\right )}{5 c}\) |
| \(\Big \downarrow \) 3321 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {7 a \left (\frac {4 a (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {3 a g \cos (e+f x) \int \sqrt {g \cos (e+f x)}dx}{c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{5 c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {7 a \left (\frac {4 a (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {3 a g \cos (e+f x) \int \sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{5 c}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {7 a \left (\frac {4 a (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {3 a g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\cos (e+f x)}dx}{c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{5 c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {7 a \left (\frac {4 a (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {3 a g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{5 c}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {4 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {7 a \left (\frac {4 a (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {6 a g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{5 c}\) |
(4*a*(g*Cos[e + f*x])^(5/2)*Sqrt[a + a*Sin[e + f*x]])/(5*f*g*(c - c*Sin[e + f*x])^(5/2)) - (7*a*((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]])))/(5*c)
3.2.2.3.1 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]
Result contains complex when optimal does not.
Time = 1.50 (sec) , antiderivative size = 2547, normalized size of antiderivative = 13.69
int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2),x,m ethod=_RETURNVERBOSE)
2/5/f*(g*cos(f*x+e))^(1/2)*(a*(1+sin(f*x+e)))^(1/2)*g*a/(cos(f*x+e)+sin(f* x+e)+1)/c^2/(-c*(sin(f*x+e)-1))^(1/2)*(-41-21*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)-cot(f*x+e)),I)*se c(f*x+e)+21*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*E llipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*sec(f*x+e)+21*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)-cot(f*x+e)) ,I)*tan(f*x+e)-21*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)*tan(f*x+e)+21*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)-cot(f *x+e)),I)*sin(f*x+e)-21*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)*sin(f*x+e)-21*I*(1/(1+co s(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(csc(f*x+e) -cot(f*x+e)),I)*cos(f*x+e)+21*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+co s(f*x+e)))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*cos(f*x+e)+5*ln(2* (2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+2*(-cos(f*x+e)/(1+cos(f *x+e))^2)^(1/2)-cos(f*x+e)+1)/(1+cos(f*x+e)))*(-cos(f*x+e)/(1+cos(f*x+e))^ 2)^(3/2)*cos(f*x+e)*sin(f*x+e)-5*ln((2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2 )*cos(f*x+e)+2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)-cos(f*x+e)+1)/(1+cos(f *x+e)))*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*cos(f*x+e)*sin(f*x+e)+5*ln(2* (2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+2*(-cos(f*x+e)/(1+co...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.18 \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {8 \, {\left (4 \, a g \sin \left (f x + e\right ) - 3 \, a g\right )} \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} + 21 \, {\left (i \, \sqrt {2} a g \cos \left (f x + e\right )^{2} + 2 i \, \sqrt {2} a g \sin \left (f x + e\right ) - 2 i \, \sqrt {2} a g\right )} \sqrt {a c g} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 21 \, {\left (-i \, \sqrt {2} a g \cos \left (f x + e\right )^{2} - 2 i \, \sqrt {2} a g \sin \left (f x + e\right ) + 2 i \, \sqrt {2} a g\right )} \sqrt {a c g} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}{5 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} + 2 \, c^{3} f \sin \left (f x + e\right ) - 2 \, c^{3} f\right )}} \]
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/ 2),x, algorithm="fricas")
-1/5*(8*(4*a*g*sin(f*x + e) - 3*a*g)*sqrt(g*cos(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c) + 21*(I*sqrt(2)*a*g*cos(f*x + e)^2 + 2* I*sqrt(2)*a*g*sin(f*x + e) - 2*I*sqrt(2)*a*g)*sqrt(a*c*g)*weierstrassZeta( -4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + 21*(-I *sqrt(2)*a*g*cos(f*x + e)^2 - 2*I*sqrt(2)*a*g*sin(f*x + e) + 2*I*sqrt(2)*a *g)*sqrt(a*c*g)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))))/(c^3*f*cos(f*x + e)^2 + 2*c^3*f*sin(f*x + e) - 2* c^3*f)
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/ 2),x, algorithm="maxima")
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/ 2),x, algorithm="giac")
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]