Integrand size = 38, antiderivative size = 182 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-4-m}}{a c f (7+2 m)}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m}}{a c^2 f \left (35+24 m+4 m^2\right )}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{a c^3 f (7+2 m) \left (15+16 m+4 m^2\right )} \]
cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)*(c-c*sin(f*x+e))^(-4-m)/a/c/f/(7+2*m)+2* cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)*(c-c*sin(f*x+e))^(-3-m)/a/c^2/f/(4*m^2+2 4*m+35)+2*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)*(c-c*sin(f*x+e))^(-2-m)/a/c^3/ f/(8*m^3+60*m^2+142*m+105)
Time = 6.61 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.58 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=\frac {\cos ^3(e+f x) (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^{-m} \left (-4 \left (6+5 m+m^2\right )+\cos (2 (e+f x))+2 (5+2 m) \sin (e+f x)\right )}{c^5 f (3+2 m) (5+2 m) (7+2 m) (-1+\sin (e+f x))^5} \]
(Cos[e + f*x]^3*(a*(1 + Sin[e + f*x]))^m*(-4*(6 + 5*m + m^2) + Cos[2*(e + f*x)] + 2*(5 + 2*m)*Sin[e + f*x]))/(c^5*f*(3 + 2*m)*(5 + 2*m)*(7 + 2*m)*(- 1 + Sin[e + f*x])^5*(c - c*Sin[e + f*x])^m)
Time = 0.85 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.96, 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, 3222, 3042, 3222, 3042, 3221}
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.
| \(\displaystyle \int \cos ^2(e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-5} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (e+f x)^2 (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-5}dx\) |
| \(\Big \downarrow \) 3320 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-4}dx}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-4}dx}{a c}\) |
| \(\Big \downarrow \) 3222 |
| \(\displaystyle \frac {\frac {2 \int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-3}dx}{c (2 m+7)}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-4}}{f (2 m+7)}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 \int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-3}dx}{c (2 m+7)}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-4}}{f (2 m+7)}}{a c}\) |
| \(\Big \downarrow \) 3222 |
| \(\displaystyle \frac {\frac {2 \left (\frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-2}dx}{c (2 m+5)}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-3}}{f (2 m+5)}\right )}{c (2 m+7)}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-4}}{f (2 m+7)}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 \left (\frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-2}dx}{c (2 m+5)}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-3}}{f (2 m+5)}\right )}{c (2 m+7)}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-4}}{f (2 m+7)}}{a c}\) |
| \(\Big \downarrow \) 3221 |
| \(\displaystyle \frac {\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-4}}{f (2 m+7)}+\frac {2 \left (\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-3}}{f (2 m+5)}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{c f (2 m+3) (2 m+5)}\right )}{c (2 m+7)}}{a c}\) |
((Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-4 - m)) /(f*(7 + 2*m)) + (2*((Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin [e + f*x])^(-3 - m))/(f*(5 + 2*m)) + (Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-2 - m))/(c*f*(3 + 2*m)*(5 + 2*m))))/(c*(7 + 2 *m)))/(a*c)
3.1.81.3.1 Defintions of rubi rules used
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*( (c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && Ne Q[m, -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*( (c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp[(m + n + 1)/(a*(2*m + 1) ) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; Free Q[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m, 1] || !SumSimplerQ[n, 1])
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]
\[\int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-5-m}d x\]
Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.58 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=-\frac {{\left (2 \, \cos \left (f x + e\right )^{5} + 2 \, {\left (2 \, m + 5\right )} \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - {\left (4 \, m^{2} + 20 \, m + 25\right )} \cos \left (f x + e\right )^{3}\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 5}}{8 \, f m^{3} + 60 \, f m^{2} + 142 \, f m + 105 \, f} \]
-(2*cos(f*x + e)^5 + 2*(2*m + 5)*cos(f*x + e)^3*sin(f*x + e) - (4*m^2 + 20 *m + 25)*cos(f*x + e)^3)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 5)/(8*f*m^3 + 60*f*m^2 + 142*f*m + 105*f)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m - 5} \cos ^{2}{\left (e + f x \right )}\, dx \]
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 5} \cos \left (f x + e\right )^{2} \,d x } \]
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 5} \cos \left (f x + e\right )^{2} \,d x } \]
Time = 17.22 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.84 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx=\frac {\cos \left (e+f\,x\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (24\,m^2+120\,m+140\right )}{8\,f\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+5}\,\left (8\,m^3+60\,m^2+142\,m+105\right )}-\frac {\cos \left (5\,e+5\,f\,x\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{8\,f\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+5}\,\left (8\,m^3+60\,m^2+142\,m+105\right )}+\frac {\cos \left (3\,e+3\,f\,x\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (8\,m^2+40\,m+45\right )}{8\,f\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+5}\,\left (8\,m^3+60\,m^2+142\,m+105\right )}+\frac {\sin \left (4\,e+4\,f\,x\right )\,\left (m\,4{}\mathrm {i}+10{}\mathrm {i}\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,1{}\mathrm {i}}{8\,f\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+5}\,\left (8\,m^3+60\,m^2+142\,m+105\right )}+\frac {\sin \left (2\,e+2\,f\,x\right )\,\left (m\,8{}\mathrm {i}+20{}\mathrm {i}\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,1{}\mathrm {i}}{8\,f\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+5}\,\left (8\,m^3+60\,m^2+142\,m+105\right )} \]
(cos(e + f*x)*(a + a*sin(e + f*x))^m*(120*m + 24*m^2 + 140))/(8*f*(c - c*s in(e + f*x))^(m + 5)*(142*m + 60*m^2 + 8*m^3 + 105)) - (cos(5*e + 5*f*x)*( a + a*sin(e + f*x))^m)/(8*f*(c - c*sin(e + f*x))^(m + 5)*(142*m + 60*m^2 + 8*m^3 + 105)) + (sin(4*e + 4*f*x)*(m*4i + 10i)*(a + a*sin(e + f*x))^m*1i) /(8*f*(c - c*sin(e + f*x))^(m + 5)*(142*m + 60*m^2 + 8*m^3 + 105)) + (sin( 2*e + 2*f*x)*(m*8i + 20i)*(a + a*sin(e + f*x))^m*1i)/(8*f*(c - c*sin(e + f *x))^(m + 5)*(142*m + 60*m^2 + 8*m^3 + 105)) + (cos(3*e + 3*f*x)*(a + a*si n(e + f*x))^m*(40*m + 8*m^2 + 45))/(8*f*(c - c*sin(e + f*x))^(m + 5)*(142* m + 60*m^2 + 8*m^3 + 105))