Integrand size = 38, antiderivative size = 140 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2} \, dx=\frac {c^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{15 a f \sqrt {c-c \sin (e+f x)}}+\frac {2 c \cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)}}{15 a f}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2}}{6 a f} \]
1/6*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(3/2)/a/f+1/15*c^2* cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/f/(c-c*sin(f*x+e))^(1/2)+2/15*c*cos(f* x+e)*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(1/2)/a/f
Time = 7.71 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.09 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {c (-1+\sin (e+f x)) (a (1+\sin (e+f x)))^{5/2} \sqrt {c-c \sin (e+f x)} (-75 \cos (2 (e+f x))-30 \cos (4 (e+f x))-5 \cos (6 (e+f x))+600 \sin (e+f x)+100 \sin (3 (e+f x))+12 \sin (5 (e+f x)))}{960 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]
-1/960*(c*(-1 + Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c - c*Sin[ e + f*x]]*(-75*Cos[2*(e + f*x)] - 30*Cos[4*(e + f*x)] - 5*Cos[6*(e + f*x)] + 600*Sin[e + f*x] + 100*Sin[3*(e + f*x)] + 12*Sin[5*(e + f*x)]))/(f*(Cos [(e + f*x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^ 5)
Time = 0.87 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3320, 3042, 3219, 3042, 3219, 3042, 3217}
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.
| \(\displaystyle \int \cos ^2(e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (e+f x)^2 (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}dx\) |
| \(\Big \downarrow \) 3320 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^{7/2} (c-c \sin (e+f x))^{5/2}dx}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^{7/2} (c-c \sin (e+f x))^{5/2}dx}{a c}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {\frac {2}{3} c \int (\sin (e+f x) a+a)^{7/2} (c-c \sin (e+f x))^{3/2}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2}{3} c \int (\sin (e+f x) a+a)^{7/2} (c-c \sin (e+f x))^{3/2}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f}}{a c}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {\frac {2}{3} c \left (\frac {2}{5} c \int (\sin (e+f x) a+a)^{7/2} \sqrt {c-c \sin (e+f x)}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt {c-c \sin (e+f x)}}{5 f}\right )+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2}{3} c \left (\frac {2}{5} c \int (\sin (e+f x) a+a)^{7/2} \sqrt {c-c \sin (e+f x)}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt {c-c \sin (e+f x)}}{5 f}\right )+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f}}{a c}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle \frac {\frac {2}{3} c \left (\frac {c^2 \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 f \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt {c-c \sin (e+f x)}}{5 f}\right )+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f}}{a c}\) |
((c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(3/2))/(6 *f) + (2*c*((c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(10*f*Sqrt[c - c *Sin[e + f*x]]) + (c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*Sqrt[c - c*Si n[e + f*x]])/(5*f)))/3)/(a*c)
3.1.21.3.1 Defintions of rubi rules used
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f _.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^ n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^ (m - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Simp[a*((2*m - 1)/(m + n )) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; Fre eQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I GtQ[m - 1/2, 0] && !LtQ[n, -1] && !(IGtQ[n - 1/2, 0] && LtQ[n, m]) && !( ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(a^(p/ 2)*c^(p/2)) Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p/2]
Time = 0.17 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.66
| method | result | size |
| default | \(\frac {\sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, c \,a^{2} \left (-5 \left (\cos ^{5}\left (f x +e \right )\right )+6 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )+8 \cos \left (f x +e \right ) \sin \left (f x +e \right )+16 \tan \left (f x +e \right )+5 \sec \left (f x +e \right )\right )}{30 f}\) | \(92\) |
1/30/f*(-c*(sin(f*x+e)-1))^(1/2)*(a*(1+sin(f*x+e)))^(1/2)*c*a^2*(-5*cos(f* x+e)^5+6*cos(f*x+e)^3*sin(f*x+e)+8*cos(f*x+e)*sin(f*x+e)+16*tan(f*x+e)+5*s ec(f*x+e))
Time = 0.27 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.73 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {{\left (5 \, a^{2} c \cos \left (f x + e\right )^{6} - 5 \, a^{2} c - 2 \, {\left (3 \, a^{2} c \cos \left (f x + e\right )^{4} + 4 \, a^{2} c \cos \left (f x + e\right )^{2} + 8 \, a^{2} c\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{30 \, f \cos \left (f x + e\right )} \]
-1/30*(5*a^2*c*cos(f*x + e)^6 - 5*a^2*c - 2*(3*a^2*c*cos(f*x + e)^4 + 4*a^ 2*c*cos(f*x + e)^2 + 8*a^2*c)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt( -c*sin(f*x + e) + c)/(f*cos(f*x + e))
Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2} \, dx=\text {Timed out} \]
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{2} \,d x } \]
Time = 0.33 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.40 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {16 \, {\left (10 \, a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} - 36 \, a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} + 45 \, a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 20 \, a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6}\right )} \sqrt {a} \sqrt {c}}{15 \, f} \]
-16/15*(10*a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2 *f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^12 - 36*a^2*c*sgn(cos(-1/4*p i + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/ 2*f*x + 1/2*e)^10 + 45*a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(- 1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^8 - 20*a^2*c*sgn (cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(- 1/4*pi + 1/2*f*x + 1/2*e)^6)*sqrt(a)*sqrt(c)/f
Time = 12.40 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.87 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {a^2\,c\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (75\,\cos \left (e+f\,x\right )+105\,\cos \left (3\,e+3\,f\,x\right )+35\,\cos \left (5\,e+5\,f\,x\right )+5\,\cos \left (7\,e+7\,f\,x\right )-700\,\sin \left (2\,e+2\,f\,x\right )-112\,\sin \left (4\,e+4\,f\,x\right )-12\,\sin \left (6\,e+6\,f\,x\right )\right )}{960\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]