Integrand size = 38, antiderivative size = 54 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{a c f (3+2 m)} \] Output:
cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)*(c-c*sin(f*x+e))^(-2-m)/a/c/f/(3+2*m)
Time = 4.96 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.11 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=-\frac {\cos ^3(e+f x) (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^{-m}}{c^3 f (3+2 m) (-1+\sin (e+f x))^3} \] Input:
Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-3 - m),x]
Output:
-((Cos[e + f*x]^3*(a*(1 + Sin[e + f*x]))^m)/(c^3*f*(3 + 2*m)*(-1 + Sin[e + f*x])^3*(c - c*Sin[e + f*x])^m))
Time = 0.49 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3042, 3320, 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.
\(\displaystyle \int \cos ^2(e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (e+f x)^2 (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-3}dx\) |
\(\Big \downarrow \) 3320 |
\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-2}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-2}dx}{a c}\) |
\(\Big \downarrow \) 3221 |
\(\displaystyle \frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{a c f (2 m+3)}\) |
Input:
Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-3 - m),x]
Output:
(Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-2 - m))/ (a*c*f*(3 + 2*m))
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[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]
\[\int \cos \left (f x +e \right )^{2} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-3-m}d x\]
Input:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m),x)
Output:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m),x)
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3} \cos \left (f x + e\right )^{3}}{2 \, f m + 3 \, f} \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m),x, algor ithm="fricas")
Output:
(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 3)*cos(f*x + e)^3/(2*f* m + 3*f)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m - 3} \cos ^{2}{\left (e + f x \right )}\, dx \] Input:
integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(-3-m),x)
Output:
Integral((a*(sin(e + f*x) + 1))**m*(-c*(sin(e + f*x) - 1))**(-m - 3)*cos(e + f*x)**2, x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m),x, algor ithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 3)*cos(f*x + e)^2, x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m),x, algor ithm="giac")
Output:
integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 3)*cos(f*x + e)^2, x)
Time = 0.61 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.87 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=-\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (6\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2-4\right )}{c^3\,f\,\left (2\,m+3\right )\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,\left (12\,{\sin \left (e+f\,x\right )}^2-15\,\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )+4\right )} \] Input:
int((cos(e + f*x)^2*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(m + 3),x )
Output:
-((a*(sin(e + f*x) + 1))^m*(6*sin(e/2 + (f*x)/2)^2 + 2*sin((3*e)/2 + (3*f* x)/2)^2 - 4))/(c^3*f*(2*m + 3)*(-c*(sin(e + f*x) - 1))^m*(sin(3*e + 3*f*x) - 15*sin(e + f*x) + 12*sin(e + f*x)^2 + 4))
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=-\frac {\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \cos \left (f x +e \right )^{2}}{\left (-\sin \left (f x +e \right ) c +c \right )^{m} \sin \left (f x +e \right )^{3}-3 \left (-\sin \left (f x +e \right ) c +c \right )^{m} \sin \left (f x +e \right )^{2}+3 \left (-\sin \left (f x +e \right ) c +c \right )^{m} \sin \left (f x +e \right )-\left (-\sin \left (f x +e \right ) c +c \right )^{m}}d x}{c^{3}} \] Input:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m),x)
Output:
( - int(((sin(e + f*x)*a + a)**m*cos(e + f*x)**2)/(( - sin(e + f*x)*c + c) **m*sin(e + f*x)**3 - 3*( - sin(e + f*x)*c + c)**m*sin(e + f*x)**2 + 3*( - sin(e + f*x)*c + c)**m*sin(e + f*x) - ( - sin(e + f*x)*c + c)**m),x))/c** 3