Integrand size = 46, antiderivative size = 37 \[ \int (a-a \sin (e+f x))^m (c+c \sin (e+f x))^n (B (m-n)+B (1+m+n) \sin (e+f x)) \, dx=-\frac {B \cos (e+f x) (a-a \sin (e+f x))^m (c+c \sin (e+f x))^n}{f} \] Output:
-B*cos(f*x+e)*(a-a*sin(f*x+e))^m*(c+c*sin(f*x+e))^n/f
Time = 2.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int (a-a \sin (e+f x))^m (c+c \sin (e+f x))^n (B (m-n)+B (1+m+n) \sin (e+f x)) \, dx=-\frac {B \cos (e+f x) (c (1+\sin (e+f x)))^n (a-a \sin (e+f x))^m}{f} \] Input:
Integrate[(a - a*Sin[e + f*x])^m*(c + c*Sin[e + f*x])^n*(B*(m - n) + B*(1 + m + n)*Sin[e + f*x]),x]
Output:
-((B*Cos[e + f*x]*(c*(1 + Sin[e + f*x]))^n*(a - a*Sin[e + f*x])^m)/f)
Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {3042, 3449}
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.
\(\displaystyle \int (a-a \sin (e+f x))^m (c \sin (e+f x)+c)^n (B (m+n+1) \sin (e+f x)+B (m-n)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a-a \sin (e+f x))^m (c \sin (e+f x)+c)^n (B (m+n+1) \sin (e+f x)+B (m-n))dx\) |
\(\Big \downarrow \) 3449 |
\(\displaystyle -\frac {B \cos (e+f x) (a-a \sin (e+f x))^m (c \sin (e+f x)+c)^n}{f}\) |
Input:
Int[(a - a*Sin[e + f*x])^m*(c + c*Sin[e + f*x])^n*(B*(m - n) + B*(1 + m + n)*Sin[e + f*x]),x]
Output:
-((B*Cos[e + f*x]*(a - a*Sin[e + f*x])^m*(c + c*Sin[e + f*x])^n)/f)
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + n + 1))), x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[A*b*(m + n + 1) + a*B*(m - n), 0] && NeQ[m, -2^(-1)]
Time = 5.55 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.03
method | result | size |
parallelrisch | \(-\frac {\left (-a \left (\sin \left (f x +e \right )-1\right )\right )^{m} \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{n} \cos \left (f x +e \right ) B}{f}\) | \(38\) |
Input:
int((a-a*sin(f*x+e))^m*(c+c*sin(f*x+e))^n*(B*(m-n)+B*(1+m+n)*sin(f*x+e)),x ,method=_RETURNVERBOSE)
Output:
-1/f*(-a*(sin(f*x+e)-1))^m*(c*(1+sin(f*x+e)))^n*cos(f*x+e)*B
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int (a-a \sin (e+f x))^m (c+c \sin (e+f x))^n (B (m-n)+B (1+m+n) \sin (e+f x)) \, dx=-\frac {{\left (-a \sin \left (f x + e\right ) + a\right )}^{m} {\left (c \sin \left (f x + e\right ) + c\right )}^{n} B \cos \left (f x + e\right )}{f} \] Input:
integrate((a-a*sin(f*x+e))^m*(c+c*sin(f*x+e))^n*(B*(m-n)+B*(1+m+n)*sin(f*x +e)),x, algorithm="fricas")
Output:
-(-a*sin(f*x + e) + a)^m*(c*sin(f*x + e) + c)^n*B*cos(f*x + e)/f
\[ \int (a-a \sin (e+f x))^m (c+c \sin (e+f x))^n (B (m-n)+B (1+m+n) \sin (e+f x)) \, dx=B \left (\int m \left (- a \sin {\left (e + f x \right )} + a\right )^{m} \left (c \sin {\left (e + f x \right )} + c\right )^{n}\, dx + \int \left (- n \left (- a \sin {\left (e + f x \right )} + a\right )^{m} \left (c \sin {\left (e + f x \right )} + c\right )^{n}\right )\, dx + \int \left (- a \sin {\left (e + f x \right )} + a\right )^{m} \left (c \sin {\left (e + f x \right )} + c\right )^{n} \sin {\left (e + f x \right )}\, dx + \int m \left (- a \sin {\left (e + f x \right )} + a\right )^{m} \left (c \sin {\left (e + f x \right )} + c\right )^{n} \sin {\left (e + f x \right )}\, dx + \int n \left (- a \sin {\left (e + f x \right )} + a\right )^{m} \left (c \sin {\left (e + f x \right )} + c\right )^{n} \sin {\left (e + f x \right )}\, dx\right ) \] Input:
integrate((a-a*sin(f*x+e))**m*(c+c*sin(f*x+e))**n*(B*(m-n)+B*(1+m+n)*sin(f *x+e)),x)
Output:
B*(Integral(m*(-a*sin(e + f*x) + a)**m*(c*sin(e + f*x) + c)**n, x) + Integ ral(-n*(-a*sin(e + f*x) + a)**m*(c*sin(e + f*x) + c)**n, x) + Integral((-a *sin(e + f*x) + a)**m*(c*sin(e + f*x) + c)**n*sin(e + f*x), x) + Integral( m*(-a*sin(e + f*x) + a)**m*(c*sin(e + f*x) + c)**n*sin(e + f*x), x) + Inte gral(n*(-a*sin(e + f*x) + a)**m*(c*sin(e + f*x) + c)**n*sin(e + f*x), x))
\[ \int (a-a \sin (e+f x))^m (c+c \sin (e+f x))^n (B (m-n)+B (1+m+n) \sin (e+f x)) \, dx=\int { {\left (B {\left (m + n + 1\right )} \sin \left (f x + e\right ) + B {\left (m - n\right )}\right )} {\left (-a \sin \left (f x + e\right ) + a\right )}^{m} {\left (c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \] Input:
integrate((a-a*sin(f*x+e))^m*(c+c*sin(f*x+e))^n*(B*(m-n)+B*(1+m+n)*sin(f*x +e)),x, algorithm="maxima")
Output:
integrate((B*(m + n + 1)*sin(f*x + e) + B*(m - n))*(-a*sin(f*x + e) + a)^m *(c*sin(f*x + e) + c)^n, x)
Leaf count of result is larger than twice the leaf count of optimal. 8948 vs. \(2 (37) = 74\).
Time = 50.24 (sec) , antiderivative size = 8948, normalized size of antiderivative = 241.84 \[ \int (a-a \sin (e+f x))^m (c+c \sin (e+f x))^n (B (m-n)+B (1+m+n) \sin (e+f x)) \, dx=\text {Too large to display} \] Input:
integrate((a-a*sin(f*x+e))^m*(c+c*sin(f*x+e))^n*(B*(m-n)+B*(1+m+n)*sin(f*x +e)),x, algorithm="giac")
Output:
-(B*cos(2*pi*n*floor(-1/8*sgn(4*tan(f*x + e)^2*tan(1/2*f*x + 1/2*e)^2 + 8* tan(f*x + e)^2*tan(1/2*f*x + 1/2*e) + 4*tan(f*x + e)^2 + 2*tan(1/2*f*x + 1 /2*e)^2 + 8*tan(1/2*f*x + 1/2*e) + 2) + 5/8) + 2*pi*m*floor(-1/8*sgn(4*tan (f*x + e)^2*tan(1/2*f*x + 1/2*e)^2 - 8*tan(f*x + e)^2*tan(1/2*f*x + 1/2*e) + 4*tan(f*x + e)^2 + 2*tan(1/2*f*x + 1/2*e)^2 - 8*tan(1/2*f*x + 1/2*e) + 2) + 5/8) + 1/4*pi*n*sgn(4*tan(f*x + e)^2*tan(1/2*f*x + 1/2*e)^2 + 8*tan(f *x + e)^2*tan(1/2*f*x + 1/2*e) + 4*tan(f*x + e)^2 + 2*tan(1/2*f*x + 1/2*e) ^2 + 8*tan(1/2*f*x + 1/2*e) + 2) + 1/4*pi*m*sgn(4*tan(f*x + e)^2*tan(1/2*f *x + 1/2*e)^2 - 8*tan(f*x + e)^2*tan(1/2*f*x + 1/2*e) + 4*tan(f*x + e)^2 + 2*tan(1/2*f*x + 1/2*e)^2 - 8*tan(1/2*f*x + 1/2*e) + 2) - 1/4*pi*m - 1/4*p i*n)*e^(-m*log(2) - n*log(2) + m*log(sqrt(2)*sqrt(abs(4*tan(f*x + e)^2*tan (1/2*f*x + 1/2*e)^2 - 8*tan(f*x + e)^2*tan(1/2*f*x + 1/2*e) + 4*tan(f*x + e)^2 + 2*tan(1/2*f*x + 1/2*e)^2 - 8*tan(1/2*f*x + 1/2*e) + 2)*tan(f*x + e) ^2*tan(1/2*f*x + 1/2*e)^2 + abs(4*tan(f*x + e)^2*tan(1/2*f*x + 1/2*e)^2 - 8*tan(f*x + e)^2*tan(1/2*f*x + 1/2*e) + 4*tan(f*x + e)^2 + 2*tan(1/2*f*x + 1/2*e)^2 - 8*tan(1/2*f*x + 1/2*e) + 2)*tan(f*x + e)^2 + abs(4*tan(f*x + e )^2*tan(1/2*f*x + 1/2*e)^2 - 8*tan(f*x + e)^2*tan(1/2*f*x + 1/2*e) + 4*tan (f*x + e)^2 + 2*tan(1/2*f*x + 1/2*e)^2 - 8*tan(1/2*f*x + 1/2*e) + 2)*tan(1 /2*f*x + 1/2*e)^2 + abs(4*tan(f*x + e)^2*tan(1/2*f*x + 1/2*e)^2 - 8*tan(f* x + e)^2*tan(1/2*f*x + 1/2*e) + 4*tan(f*x + e)^2 + 2*tan(1/2*f*x + 1/2*...
Time = 35.48 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int (a-a \sin (e+f x))^m (c+c \sin (e+f x))^n (B (m-n)+B (1+m+n) \sin (e+f x)) \, dx=-\frac {B\,\cos \left (e+f\,x\right )\,{\left (-a\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,{\left (c\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^n}{f} \] Input:
int((B*(m - n) + B*sin(e + f*x)*(m + n + 1))*(a - a*sin(e + f*x))^m*(c + c *sin(e + f*x))^n,x)
Output:
-(B*cos(e + f*x)*(-a*(sin(e + f*x) - 1))^m*(c*(sin(e + f*x) + 1))^n)/f
\[ \int (a-a \sin (e+f x))^m (c+c \sin (e+f x))^n (B (m-n)+B (1+m+n) \sin (e+f x)) \, dx=b \left (\left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n} \left (-a \sin \left (f x +e \right )+a \right )^{m} \sin \left (f x +e \right )d x \right ) m +\left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n} \left (-a \sin \left (f x +e \right )+a \right )^{m} \sin \left (f x +e \right )d x \right ) n +\int \left (\sin \left (f x +e \right ) c +c \right )^{n} \left (-a \sin \left (f x +e \right )+a \right )^{m} \sin \left (f x +e \right )d x +\left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n} \left (-a \sin \left (f x +e \right )+a \right )^{m}d x \right ) m -\left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n} \left (-a \sin \left (f x +e \right )+a \right )^{m}d x \right ) n \right ) \] Input:
int((a-a*sin(f*x+e))^m*(c+c*sin(f*x+e))^n*(B*(m-n)+B*(1+m+n)*sin(f*x+e)),x )
Output:
b*(int((sin(e + f*x)*c + c)**n*( - sin(e + f*x)*a + a)**m*sin(e + f*x),x)* m + int((sin(e + f*x)*c + c)**n*( - sin(e + f*x)*a + a)**m*sin(e + f*x),x) *n + int((sin(e + f*x)*c + c)**n*( - sin(e + f*x)*a + a)**m*sin(e + f*x),x ) + int((sin(e + f*x)*c + c)**n*( - sin(e + f*x)*a + a)**m,x)*m - int((sin (e + f*x)*c + c)**n*( - sin(e + f*x)*a + a)**m,x)*n)