Integrand size = 42, antiderivative size = 298 \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {4 a (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {12 a^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{c f g (c-c \sin (e+f x))^{3/2}}-\frac {154 a^4 (g \cos (e+f x))^{5/2}}{5 c^2 f g \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {462 a^4 g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {66 a^3 (g \cos (e+f x))^{5/2} \sqrt {a+a \sin (e+f x)}}{5 c^2 f g \sqrt {c-c \sin (e+f x)}} \]
4/5*a*(g*cos(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(5/2)/f/g/(c-c*sin(f*x+e))^(5/ 2)-12*a^2*(g*cos(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(3/2)/c/f/g/(c-c*sin(f*x+e ))^(3/2)-154/5*a^4*(g*cos(f*x+e))^(5/2)/c^2/f/g/(a+a*sin(f*x+e))^(1/2)/(c- c*sin(f*x+e))^(1/2)+462/5*a^4*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)/c^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)-66/5*a^3*(g*co s(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2)/c^2/f/g/(c-c*sin(f*x+e))^(1/2)
Time = 11.30 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.90 \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {a^3 (g \cos (e+f x))^{3/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sqrt {a (1+\sin (e+f x))} \left (-1848 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+\sqrt {\cos (e+f x)} \left (487 \cos \left (\frac {1}{2} (e+f x)\right )+633 \cos \left (\frac {3}{2} (e+f x)\right )-17 \cos \left (\frac {5}{2} (e+f x)\right )+\cos \left (\frac {7}{2} (e+f x)\right )+487 \sin \left (\frac {1}{2} (e+f x)\right )-633 \sin \left (\frac {3}{2} (e+f x)\right )-17 \sin \left (\frac {5}{2} (e+f x)\right )-\sin \left (\frac {7}{2} (e+f x)\right )\right )\right )}{20 c^2 f \cos ^{\frac {3}{2}}(e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \]
-1/20*(a^3*(g*Cos[e + f*x])^(3/2)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2* Sqrt[a*(1 + Sin[e + f*x])]*(-1848*EllipticE[(e + f*x)/2, 2]*(Cos[(e + f*x) /2] - Sin[(e + f*x)/2])^3 + Sqrt[Cos[e + f*x]]*(487*Cos[(e + f*x)/2] + 633 *Cos[(3*(e + f*x))/2] - 17*Cos[(5*(e + f*x))/2] + Cos[(7*(e + f*x))/2] + 4 87*Sin[(e + f*x)/2] - 633*Sin[(3*(e + f*x))/2] - 17*Sin[(5*(e + f*x))/2] - Sin[(7*(e + f*x))/2])))/(c^2*f*Cos[e + f*x]^(3/2)*(Cos[(e + f*x)/2] + Sin [(e + f*x)/2])*(-1 + Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])
Time = 2.21 (sec) , antiderivative size = 296, normalized size of antiderivative = 0.99, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3329, 3042, 3329, 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.
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{7/2} (g \cos (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{7/2} (g \cos (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 3329 |
\(\displaystyle \frac {4 a (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {3 a \int \frac {(g \cos (e+f x))^{3/2} (\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{3/2}}dx}{c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 a (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {3 a \int \frac {(g \cos (e+f x))^{3/2} (\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{3/2}}dx}{c}\) |
\(\Big \downarrow \) 3329 |
\(\displaystyle \frac {4 a (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {3 a \left (\frac {4 a (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{f g (c-c \sin (e+f x))^{3/2}}-\frac {11 a \int \frac {(g \cos (e+f x))^{3/2} (\sin (e+f x) a+a)^{3/2}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\right )}{c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 a (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {3 a \left (\frac {4 a (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{f g (c-c \sin (e+f x))^{3/2}}-\frac {11 a \int \frac {(g \cos (e+f x))^{3/2} (\sin (e+f x) a+a)^{3/2}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\right )}{c}\) |
\(\Big \downarrow \) 3330 |
\(\displaystyle \frac {4 a (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {3 a \left (\frac {4 a (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{f g (c-c \sin (e+f x))^{3/2}}-\frac {11 a \left (\frac {7}{5} a \int \frac {(g \cos (e+f x))^{3/2} \sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {2 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g \sqrt {c-c \sin (e+f x)}}\right )}{c}\right )}{c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 a (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {3 a \left (\frac {4 a (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{f g (c-c \sin (e+f x))^{3/2}}-\frac {11 a \left (\frac {7}{5} a \int \frac {(g \cos (e+f x))^{3/2} \sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {2 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g \sqrt {c-c \sin (e+f x)}}\right )}{c}\right )}{c}\) |
\(\Big \downarrow \) 3330 |
\(\displaystyle \frac {4 a (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {3 a \left (\frac {4 a (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{f g (c-c \sin (e+f x))^{3/2}}-\frac {11 a \left (\frac {7}{5} a \left (a \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx-\frac {2 a (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )-\frac {2 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g \sqrt {c-c \sin (e+f x)}}\right )}{c}\right )}{c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 a (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {3 a \left (\frac {4 a (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{f g (c-c \sin (e+f x))^{3/2}}-\frac {11 a \left (\frac {7}{5} a \left (a \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx-\frac {2 a (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )-\frac {2 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g \sqrt {c-c \sin (e+f x)}}\right )}{c}\right )}{c}\) |
\(\Big \downarrow \) 3321 |
\(\displaystyle \frac {4 a (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {3 a \left (\frac {4 a (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{f g (c-c \sin (e+f x))^{3/2}}-\frac {11 a \left (\frac {7}{5} a \left (\frac {a g \cos (e+f x) \int \sqrt {g \cos (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 a (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )-\frac {2 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g \sqrt {c-c \sin (e+f x)}}\right )}{c}\right )}{c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 a (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {3 a \left (\frac {4 a (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{f g (c-c \sin (e+f x))^{3/2}}-\frac {11 a \left (\frac {7}{5} a \left (\frac {a g \cos (e+f x) \int \sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 a (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )-\frac {2 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g \sqrt {c-c \sin (e+f x)}}\right )}{c}\right )}{c}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {4 a (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {3 a \left (\frac {4 a (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{f g (c-c \sin (e+f x))^{3/2}}-\frac {11 a \left (\frac {7}{5} a \left (\frac {a g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\cos (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 a (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )-\frac {2 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g \sqrt {c-c \sin (e+f x)}}\right )}{c}\right )}{c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 a (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {3 a \left (\frac {4 a (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{f g (c-c \sin (e+f x))^{3/2}}-\frac {11 a \left (\frac {7}{5} a \left (\frac {a g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 a (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )-\frac {2 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g \sqrt {c-c \sin (e+f x)}}\right )}{c}\right )}{c}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {4 a (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{5 f g (c-c \sin (e+f x))^{5/2}}-\frac {3 a \left (\frac {4 a (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{f g (c-c \sin (e+f x))^{3/2}}-\frac {11 a \left (\frac {7}{5} a \left (\frac {2 a g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 a (g \cos (e+f x))^{5/2}}{3 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )-\frac {2 a \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{5 f g \sqrt {c-c \sin (e+f x)}}\right )}{c}\right )}{c}\) |
(4*a*(g*Cos[e + f*x])^(5/2)*(a + a*Sin[e + f*x])^(5/2))/(5*f*g*(c - c*Sin[ e + f*x])^(5/2)) - (3*a*((4*a*(g*Cos[e + f*x])^(5/2)*(a + a*Sin[e + f*x])^ (3/2))/(f*g*(c - c*Sin[e + f*x])^(3/2)) - (11*a*((-2*a*(g*Cos[e + f*x])^(5 /2)*Sqrt[a + a*Sin[e + f*x]])/(5*f*g*Sqrt[c - c*Sin[e + f*x]]) + (7*a*((-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]]) + (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]])))/5))/c ))/c
3.2.21.3.1 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 - 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]
Result contains complex when optimal does not.
Time = 3.26 (sec) , antiderivative size = 2596, normalized size of antiderivative = 8.71
int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(7/2)/(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^3/(cos(f*x+e)+sin (f*x+e)+1)/(-c*(sin(f*x+e)-1))^(1/2)/c^2*(391-40*ln(2*(2*(-cos(f*x+e)/(1+c os(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)+40*ln((2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+2*(-co s(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)-40*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)-co s(f*x+e)+1)/(1+cos(f*x+e)))*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*sec(f*x+e )*tan(f*x+e)+40*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)*sec(f*x+e)*tan(f*x+e)+10*cos(f*x+e)^2+462*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)-462*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)^2*sin(f*x+e )-16*tan(f*x+e)-32*sec(f*x+e)*tan(f*x+e)-16*sec(f*x+e)-78*cos(f*x+e)-87*si n(f*x+e)+cos(f*x+e)^3+40*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)^2-40*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/...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.14 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.89 \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {2 \, {\left (8 \, a^{3} g \cos \left (f x + e\right )^{2} - 146 \, a^{3} g + {\left (a^{3} g \cos \left (f x + e\right )^{2} + 162 \, a^{3} g\right )} \sin \left (f x + e\right )\right )} \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} + 231 \, {\left (i \, \sqrt {2} a^{3} g \cos \left (f x + e\right )^{2} + 2 i \, \sqrt {2} a^{3} g \sin \left (f x + e\right ) - 2 i \, \sqrt {2} a^{3} g\right )} \sqrt {a c g} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 231 \, {\left (-i \, \sqrt {2} a^{3} g \cos \left (f x + e\right )^{2} - 2 i \, \sqrt {2} a^{3} g \sin \left (f x + e\right ) + 2 i \, \sqrt {2} a^{3} g\right )} \sqrt {a c g} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}{5 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} + 2 \, c^{3} f \sin \left (f x + e\right ) - 2 \, c^{3} f\right )}} \]
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(5/ 2),x, algorithm="fricas")
-1/5*(2*(8*a^3*g*cos(f*x + e)^2 - 146*a^3*g + (a^3*g*cos(f*x + e)^2 + 162* a^3*g)*sin(f*x + e))*sqrt(g*cos(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c *sin(f*x + e) + c) + 231*(I*sqrt(2)*a^3*g*cos(f*x + e)^2 + 2*I*sqrt(2)*a^3 *g*sin(f*x + e) - 2*I*sqrt(2)*a^3*g)*sqrt(a*c*g)*weierstrassZeta(-4, 0, we ierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + 231*(-I*sqrt(2) *a^3*g*cos(f*x + e)^2 - 2*I*sqrt(2)*a^3*g*sin(f*x + e) + 2*I*sqrt(2)*a^3*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. \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(5/ 2),x, algorithm="maxima")
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(5/ 2),x, algorithm="giac")
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]