Integrand size = 27, antiderivative size = 89 \[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=-\frac {(e \cos (c+d x))^{-1-m} (a+a \sin (c+d x))^m}{d e (1-m)}+\frac {(e \cos (c+d x))^{-1-m} (a+a \sin (c+d x))^{1+m}}{a d e \left (1-m^2\right )} \] Output:
-(e*cos(d*x+c))^(-1-m)*(a+a*sin(d*x+c))^m/d/e/(1-m)+(e*cos(d*x+c))^(-1-m)* (a+a*sin(d*x+c))^(1+m)/a/d/e/(-m^2+1)
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=\frac {(e \cos (c+d x))^{-1-m} (m-\sin (c+d x)) (a (1+\sin (c+d x)))^m}{d e (-1+m) (1+m)} \] Input:
Integrate[(e*Cos[c + d*x])^(-2 - m)*(a + a*Sin[c + d*x])^m,x]
Output:
((e*Cos[c + d*x])^(-1 - m)*(m - Sin[c + d*x])*(a*(1 + Sin[c + d*x]))^m)/(d *e*(-1 + m)*(1 + m))
Time = 0.42 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3042, 3151, 3042, 3150}
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.
\(\displaystyle \int (a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-2}dx\) |
\(\Big \downarrow \) 3151 |
\(\displaystyle \frac {\int (e \cos (c+d x))^{-m-2} (\sin (c+d x) a+a)^{m+1}dx}{a (1-m)}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-1}}{d e (1-m)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (e \cos (c+d x))^{-m-2} (\sin (c+d x) a+a)^{m+1}dx}{a (1-m)}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-1}}{d e (1-m)}\) |
\(\Big \downarrow \) 3150 |
\(\displaystyle \frac {(a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-1}}{a d e (1-m) (m+1)}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-1}}{d e (1-m)}\) |
Input:
Int[(e*Cos[c + d*x])^(-2 - m)*(a + a*Sin[c + d*x])^m,x]
Output:
-(((e*Cos[c + d*x])^(-1 - m)*(a + a*Sin[c + d*x])^m)/(d*e*(1 - m))) + ((e* Cos[c + d*x])^(-1 - m)*(a + a*Sin[c + d*x])^(1 + m))/(a*d*e*(1 - m)*(1 + m ))
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^m/(a*f*g*m)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[Simplify[m + p + 1], 0] && !ILtQ[p, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^m/(a*f*g*Simplify[2*m + p + 1])), x] + Simp[Simplify[m + p + 1]/(a*Simpl ify[2*m + p + 1]) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x] , x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simpli fy[m + p + 1], 0] && NeQ[2*m + p + 1, 0] && !IGtQ[m, 0]
\[\int \left (e \cos \left (d x +c \right )\right )^{-2-m} \left (a +a \sin \left (d x +c \right )\right )^{m}d x\]
Input:
int((e*cos(d*x+c))^(-2-m)*(a+a*sin(d*x+c))^m,x)
Output:
int((e*cos(d*x+c))^(-2-m)*(a+a*sin(d*x+c))^m,x)
Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.69 \[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=\frac {{\left (m \cos \left (d x + c\right ) - \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \left (e \cos \left (d x + c\right )\right )^{-m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{2} - d} \] Input:
integrate((e*cos(d*x+c))^(-2-m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")
Output:
(m*cos(d*x + c) - cos(d*x + c)*sin(d*x + c))*(e*cos(d*x + c))^(-m - 2)*(a* sin(d*x + c) + a)^m/(d*m^2 - d)
\[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos {\left (c + d x \right )}\right )^{- m - 2}\, dx \] Input:
integrate((e*cos(d*x+c))**(-2-m)*(a+a*sin(d*x+c))**m,x)
Output:
Integral((a*(sin(c + d*x) + 1))**m*(e*cos(c + d*x))**(-m - 2), x)
\[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \] Input:
integrate((e*cos(d*x+c))^(-2-m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")
Output:
integrate((e*cos(d*x + c))^(-m - 2)*(a*sin(d*x + c) + a)^m, x)
\[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \] Input:
integrate((e*cos(d*x+c))^(-2-m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")
Output:
integrate((e*cos(d*x + c))^(-m - 2)*(a*sin(d*x + c) + a)^m, x)
Time = 16.14 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.80 \[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=-\frac {\left (\sin \left (2\,c+2\,d\,x\right )-2\,m\,\cos \left (c+d\,x\right )\right )\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m}{d\,e^2\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )\,{\left (e\,\cos \left (c+d\,x\right )\right )}^m\,\left (m^2-1\right )} \] Input:
int((a + a*sin(c + d*x))^m/(e*cos(c + d*x))^(m + 2),x)
Output:
-((sin(2*c + 2*d*x) - 2*m*cos(c + d*x))*(a*(sin(c + d*x) + 1))^m)/(d*e^2*( cos(2*c + 2*d*x) + 1)*(e*cos(c + d*x))^m*(m^2 - 1))
\[ \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx=\frac {\int \frac {\left (\sin \left (d x +c \right ) a +a \right )^{m}}{\cos \left (d x +c \right )^{m} \cos \left (d x +c \right )^{2}}d x}{e^{m} e^{2}} \] Input:
int((e*cos(d*x+c))^(-2-m)*(a+a*sin(d*x+c))^m,x)
Output:
int((sin(c + d*x)*a + a)**m/(cos(c + d*x)**m*cos(c + d*x)**2),x)/(e**m*e** 2)