Integrand size = 40, antiderivative size = 32 \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (A m-A (1+m+p) \sin (e+f x)) \, dx=\frac {A (g \cos (e+f x))^{1+p} (a+a \sin (e+f x))^m}{f g} \] Output:
A*(g*cos(f*x+e))^(p+1)*(a+a*sin(f*x+e))^m/f/g
Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (A m-A (1+m+p) \sin (e+f x)) \, dx=\frac {A \cos (e+f x) (g \cos (e+f x))^p (a (1+\sin (e+f x)))^m}{f} \] 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*(1 + Sin[e + f*x]))^m)/f
Time = 0.29 (sec) , antiderivative size = 32, 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.
\(\displaystyle \int (a \sin (e+f x)+a)^m (g \cos (e+f x))^p (A m-A (m+p+1) \sin (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^m (g \cos (e+f x))^p (A m-A (m+p+1) \sin (e+f x))dx\) |
\(\Big \downarrow \) 3333 |
\(\displaystyle \frac {A (a \sin (e+f x)+a)^m (g \cos (e+f x))^{p+1}}{f g}\) |
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 = 5.34 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06
method | result | size |
parallelrisch | \(\frac {\left (g \cos \left (f x +e \right )\right )^{p} \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{m} \cos \left (f x +e \right ) A}{f}\) | \(34\) |
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*(1+sin(f*x+e)))^m*cos(f*x+e)*A
Time = 0.09 (sec) , antiderivative size = 33, 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
\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (A m-A (1+m+p) \sin (e+f x)) \, dx=- A \left (\int \left (- m \left (g \cos {\left (e + f x \right )}\right )^{p} \left (a \sin {\left (e + f x \right )} + a\right )^{m}\right )\, dx + \int \left (g \cos {\left (e + f x \right )}\right )^{p} \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\, dx + \int m \left (g \cos {\left (e + f x \right )}\right )^{p} \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\, dx + \int p \left (g \cos {\left (e + f x \right )}\right )^{p} \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\, dx\right ) \] 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. 1790 vs. \(2 (32) = 64\).
Time = 2.41 (sec) , antiderivative size = 1790, normalized size of antiderivative = 55.94 \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (A m-A (1+m+p) \sin (e+f x)) \, dx=\text {Too large to display} \] 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(4*abs(g)*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)) + 2*m*log(2*abs(tan(1/8*pi - 1 /4*f*x - 1/4*e)^2 - 1)/(tan(1/8*pi - 1/4*f*x - 1/4*e)^2 + 1)) + p*log(2*ab s(tan(1/8*pi - 1/4*f*x - 1/4*e)^2 - 1)/(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) + pi*m*floor(1 /2*f*x/pi + 1/2*e/pi - floor(1/2*f*x/pi + 1/2*e/pi + 1/2) + 3/4) + 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) + 3/4) + 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) + 1/2) + 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*sg n(a) - 3/4*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(4*abs(g)*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)) + 2*m*log(2*abs(tan(1/8*pi - 1/4*f*x - 1/4*e)^2 - 1)/(tan(1/8*pi - 1/4*f*x - 1/4*e)^2 + 1)) + p*log(2*abs(tan(1/8*pi - 1/4*f *x - 1/4*e)^2 - 1)/(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) + pi*m*floor(1/2*f*x/pi + 1/2*e...
Time = 33.64 (sec) , antiderivative size = 33, 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
\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (A m-A (1+m+p) \sin (e+f x)) \, dx=g^{p} a \left (-\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \cos \left (f x +e \right )^{p} \sin \left (f x +e \right )d x \right ) m -\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \cos \left (f x +e \right )^{p} \sin \left (f x +e \right )d x \right ) p -\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \cos \left (f x +e \right )^{p} \sin \left (f x +e \right )d x \right )+\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \cos \left (f x +e \right )^{p}d x \right ) m \right ) \] 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)