Integrand size = 40, antiderivative size = 34 \[ \int (g \cos (e+f x))^p (a-a \sin (e+f x))^m (A m+A (1+m+p) \sin (e+f x)) \, dx=-\frac {A (g \cos (e+f x))^{1+p} (a-a \sin (e+f x))^m}{f g} \] Output:
-A*(g*cos(f*x+e))^(p+1)*(a-a*sin(f*x+e))^m/f/g
Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int (g \cos (e+f x))^p (a-a \sin (e+f x))^m (A m+A (1+m+p) \sin (e+f x)) \, dx=-\frac {A \cos (e+f x) (g \cos (e+f x))^p (a-a \sin (e+f x))^m}{f} \] Input:
Integrate[(g*Cos[e + f*x])^p*(a - a*Sin[e + f*x])^m*(A*m + A*(1 + m + p)*S in[e + f*x]),x]
Output:
-((A*Cos[e + f*x]*(g*Cos[e + f*x])^p*(a - a*Sin[e + f*x])^m)/f)
Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {3042, 3333}
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.
\(\displaystyle \int (a-a \sin (e+f x))^m (g \cos (e+f x))^p (A (m+p+1) \sin (e+f x)+A m) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a-a \sin (e+f x))^m (g \cos (e+f x))^p (A (m+p+1) \sin (e+f x)+A m)dx\) |
\(\Big \downarrow \) 3333 |
\(\displaystyle -\frac {A (a-a \sin (e+f x))^m (g \cos (e+f x))^{p+1}}{f g}\) |
Input:
Int[(g*Cos[e + f*x])^p*(a - a*Sin[e + f*x])^m*(A*m + A*(1 + m + p)*Sin[e + f*x]),x]
Output:
-((A*(g*Cos[e + f*x])^(1 + p)*(a - a*Sin[e + f*x])^m)/(f*g))
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[a*d*m + b *c*(m + p + 1), 0]
Time = 5.76 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06
method | result | size |
parallelrisch | \(-\frac {\left (g \cos \left (f x +e \right )\right )^{p} \left (-a \left (\sin \left (f x +e \right )-1\right )\right )^{m} \cos \left (f x +e \right ) A}{f}\) | \(36\) |
Input:
int((g*cos(f*x+e))^p*(a-a*sin(f*x+e))^m*(A*m+A*(1+m+p)*sin(f*x+e)),x,metho d=_RETURNVERBOSE)
Output:
-1/f*(g*cos(f*x+e))^p*(-a*(sin(f*x+e)-1))^m*cos(f*x+e)*A
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int (g \cos (e+f x))^p (a-a \sin (e+f x))^m (A m+A (1+m+p) \sin (e+f x)) \, dx=-\frac {\left (g \cos \left (f x + e\right )\right )^{p} {\left (-a \sin \left (f x + e\right ) + a\right )}^{m} A \cos \left (f x + e\right )}{f} \] Input:
integrate((g*cos(f*x+e))^p*(a-a*sin(f*x+e))^m*(A*m+A*(1+m+p)*sin(f*x+e)),x , algorithm="fricas")
Output:
-(g*cos(f*x + e))^p*(-a*sin(f*x + e) + a)^m*A*cos(f*x + e)/f
\[ \int (g \cos (e+f x))^p (a-a \sin (e+f x))^m (A m+A (1+m+p) \sin (e+f x)) \, dx=A \left (\int m \left (g \cos {\left (e + f x \right )}\right )^{p} \left (- a \sin {\left (e + f x \right )} + a\right )^{m}\, dx + \int \left (g \cos {\left (e + f x \right )}\right )^{p} \left (- a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\, dx + \int m \left (g \cos {\left (e + f x \right )}\right )^{p} \left (- a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\, dx + \int p \left (g \cos {\left (e + f x \right )}\right )^{p} \left (- a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\, dx\right ) \] Input:
integrate((g*cos(f*x+e))**p*(a-a*sin(f*x+e))**m*(A*m+A*(1+m+p)*sin(f*x+e)) ,x)
Output:
A*(Integral(m*(g*cos(e + f*x))**p*(-a*sin(e + f*x) + a)**m, x) + Integral( (g*cos(e + f*x))**p*(-a*sin(e + f*x) + a)**m*sin(e + f*x), x) + Integral(m *(g*cos(e + f*x))**p*(-a*sin(e + f*x) + a)**m*sin(e + f*x), x) + Integral( p*(g*cos(e + f*x))**p*(-a*sin(e + f*x) + a)**m*sin(e + f*x), x))
\[ \int (g \cos (e+f x))^p (a-a \sin (e+f x))^m (A m+A (1+m+p) \sin (e+f x)) \, dx=\int { {\left (A {\left (m + p + 1\right )} \sin \left (f x + e\right ) + A m\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (-a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:
integrate((g*cos(f*x+e))^p*(a-a*sin(f*x+e))^m*(A*m+A*(1+m+p)*sin(f*x+e)),x , algorithm="maxima")
Output:
integrate((A*(m + p + 1)*sin(f*x + e) + A*m)*(g*cos(f*x + e))^p*(-a*sin(f* x + e) + a)^m, x)
Leaf count of result is larger than twice the leaf count of optimal. 1775 vs. \(2 (34) = 68\).
Time = 2.53 (sec) , antiderivative size = 1775, normalized size of antiderivative = 52.21 \[ \int (g \cos (e+f x))^p (a-a \sin (e+f x))^m (A m+A (1+m+p) \sin (e+f x)) \, dx=\text {Too large to display} \] Input:
integrate((g*cos(f*x+e))^p*(a-a*sin(f*x+e))^m*(A*m+A*(1+m+p)*sin(f*x+e)),x , algorithm="giac")
Output:
-(A*e^(-m*log(2) - p*log(2) + p*log(2*abs(tan(1/8*pi - 1/4*f*x - 1/4*e)^2 - 1)*abs(g)/(tan(1/8*pi - 1/4*f*x - 1/4*e)^2 + 1)) + 2*m*log(4*abs(tan(1/8 *pi - 1/4*f*x - 1/4*e))/(tan(1/8*pi - 1/4*f*x - 1/4*e)^2 + 1)) + p*log(4*a bs(tan(1/8*pi - 1/4*f*x - 1/4*e))/(tan(1/8*pi - 1/4*f*x - 1/4*e)^2 + 1)) + m*log(abs(a)))*tan(-1/4*pi*p*sgn(g*tan(1/2*f*x + 1/2*e)^2 + 2*g*tan(1/2*f *x + 1/2*e) + g)*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)*sgn(g) + 1/2*pi*p*floor(1 /2*f*x/pi + 1/2*e/pi - floor(1/2*f*x/pi + 1/2*e/pi + 1/2) + 3/4) + pi*m*fl oor(1/2*f*x/pi + 1/2*e/pi - floor(1/2*f*x/pi + 1/2*e/pi + 1/2) + 1/4) + 1/ 2*pi*p*floor(1/2*f*x/pi + 1/2*e/pi - floor(1/2*f*x/pi + 1/2*e/pi + 1/2) + 1/4) + pi*m*floor(1/2*f*x/pi + 1/2*e/pi + 1/2) + pi*p*floor(1/2*f*x/pi + 1 /2*e/pi + 1/2) + pi*m*floor(-1/4*sgn(a) + 1) - 1/4*pi*p*sgn(g*tan(1/2*f*x + 1/2*e)^2 + 2*g*tan(1/2*f*x + 1/2*e) + g) + 1/2*pi*m*sgn(tan(1/2*f*x + 1/ 2*e)^2 - 1) + 1/4*pi*p*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 1/4*pi*m*sgn(a) + 1/4*pi*m + 1/4*pi*p)^2*tan(1/2*f*x + 1/2*e)^2 - A*e^(-m*log(2) - p*log(2) + p*log(2*abs(tan(1/8*pi - 1/4*f*x - 1/4*e)^2 - 1)*abs(g)/(tan(1/8*pi - 1 /4*f*x - 1/4*e)^2 + 1)) + 2*m*log(4*abs(tan(1/8*pi - 1/4*f*x - 1/4*e))/(ta n(1/8*pi - 1/4*f*x - 1/4*e)^2 + 1)) + p*log(4*abs(tan(1/8*pi - 1/4*f*x - 1 /4*e))/(tan(1/8*pi - 1/4*f*x - 1/4*e)^2 + 1)) + m*log(abs(a)))*tan(-1/4*pi *p*sgn(g*tan(1/2*f*x + 1/2*e)^2 + 2*g*tan(1/2*f*x + 1/2*e) + g)*sgn(tan(1/ 2*f*x + 1/2*e)^2 - 1)*sgn(g) + 1/2*pi*p*floor(1/2*f*x/pi + 1/2*e/pi - f...
Time = 34.53 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int (g \cos (e+f x))^p (a-a \sin (e+f x))^m (A m+A (1+m+p) \sin (e+f x)) \, dx=-\frac {A\,\cos \left (e+f\,x\right )\,{\left (g\,\cos \left (e+f\,x\right )\right )}^p\,{\left (-a\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m}{f} \] Input:
int((g*cos(e + f*x))^p*(A*m + A*sin(e + f*x)*(m + p + 1))*(a - a*sin(e + f *x))^m,x)
Output:
-(A*cos(e + f*x)*(g*cos(e + f*x))^p*(-a*(sin(e + f*x) - 1))^m)/f
\[ \int (g \cos (e+f x))^p (a-a \sin (e+f x))^m (A m+A (1+m+p) \sin (e+f x)) \, dx=g^{p} a \left (\left (\int \left (-a \sin \left (f x +e \right )+a \right )^{m} \cos \left (f x +e \right )^{p} \sin \left (f x +e \right )d x \right ) m +\left (\int \left (-a \sin \left (f x +e \right )+a \right )^{m} \cos \left (f x +e \right )^{p} \sin \left (f x +e \right )d x \right ) p +\int \left (-a \sin \left (f x +e \right )+a \right )^{m} \cos \left (f x +e \right )^{p} \sin \left (f x +e \right )d x +\left (\int \left (-a \sin \left (f x +e \right )+a \right )^{m} \cos \left (f x +e \right )^{p}d x \right ) m \right ) \] Input:
int((g*cos(f*x+e))^p*(a-a*sin(f*x+e))^m*(A*m+A*(1+m+p)*sin(f*x+e)),x)
Output:
g**p*a*(int(( - sin(e + f*x)*a + a)**m*cos(e + f*x)**p*sin(e + f*x),x)*m + int(( - sin(e + f*x)*a + a)**m*cos(e + f*x)**p*sin(e + f*x),x)*p + int(( - sin(e + f*x)*a + a)**m*cos(e + f*x)**p*sin(e + f*x),x) + int(( - sin(e + f*x)*a + a)**m*cos(e + f*x)**p,x)*m)