Integrand size = 42, antiderivative size = 57 \[ \int (g \cos (e+f x))^{1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+m} \, dx=-\frac {g (g \cos (e+f x))^{-2 m} \log (1-\sin (e+f x)) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m}{c f} \] Output:
-g*ln(1-sin(f*x+e))*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m/c/f/((g*cos(f*x+ e))^(2*m))
Time = 16.70 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.33 \[ \int (g \cos (e+f x))^{1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+m} \, dx=\frac {g (g \cos (e+f x))^{-2 m} \left (\log \left (\sec ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )\right ) (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^m}{c f} \] Input:
Integrate[(g*Cos[e + f*x])^(1 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 + m),x]
Output:
(g*(Log[Sec[(e + f*x)/2]^2] - 2*Log[1 - Tan[(e + f*x)/2]])*(a*(1 + Sin[e + f*x]))^m*(c - c*Sin[e + f*x])^m)/(c*f*(g*Cos[e + f*x])^(2*m))
Time = 0.47 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3322, 27, 3042, 3146, 16}
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.
\(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{m-1} (g \cos (e+f x))^{1-2 m} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{m-1} (g \cos (e+f x))^{1-2 m}dx\) |
\(\Big \downarrow \) 3322 |
\(\displaystyle (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \frac {g \cos (e+f x)}{c-c \sin (e+f x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle g (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle g (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle -\frac {g (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int \frac {1}{c-c \sin (e+f x)}d(-c \sin (e+f x))}{c f}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle -\frac {g (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m \log (c-c \sin (e+f x)) (g \cos (e+f x))^{-2 m}}{c f}\) |
Input:
Int[(g*Cos[e + f*x])^(1 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x]) ^(-1 + m),x]
Output:
-((g*Log[c - c*Sin[e + f*x]]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^m )/(c*f*(g*Cos[e + f*x])^(2*m)))
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^ IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f* x])^FracPart[m]/(g^(2*IntPart[m])*(g*Cos[e + f*x])^(2*FracPart[m]))) Int[ (g*Cos[e + f*x])^(2*m + p)/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[2 *m + p - 1, 0] && EqQ[m - n - 1, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 5.40 (sec) , antiderivative size = 3232, normalized size of antiderivative = 56.70
Input:
int((g*cos(f*x+e))^(1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+m),x,me thod=_RETURNVERBOSE)
Output:
I/f/c*g*c^m*a^m*g^(-2*m)*exp(1/2*I*Pi*(1+2*m*csgn(I*(exp(I*(f*x+e))-I))*cs gn(I*(exp(I*(f*x+e))-I)^2)^2-csgn(I*g*(exp(I*(f*x+e))-I)*(1+I*exp(-I*(f*x+ e))))^3+2*m*csgn(I*(exp(I*(f*x+e))+I))*csgn(I*(exp(I*(f*x+e))+I)^2)^2+2*m* csgn(I*(exp(I*(f*x+e))+I)*(-1+I*exp(-I*(f*x+e))))*csgn(I*g*(exp(I*(f*x+e)) -I)*(1+I*exp(-I*(f*x+e))))^2+2*m*csgn(I*(exp(I*(f*x+e))-I)*(exp(I*(f*x+e)) +I))*csgn(I*(exp(I*(f*x+e))-I))*csgn(I*(exp(I*(f*x+e))+I))+csgn(c*(exp(I*( f*x+e))-I)^2*exp(-I*(f*x+e)))^2-csgn(c*(exp(I*(f*x+e))-I)^2*exp(-I*(f*x+e) ))^3-csgn(I*(exp(I*(f*x+e))-I)*(exp(I*(f*x+e))+I))^3+2*m*csgn(I*(exp(I*(f* x+e))-I)*(exp(I*(f*x+e))+I))^3+2*m*csgn(I*g*(exp(I*(f*x+e))-I)*(1+I*exp(-I *(f*x+e))))^3+csgn(I*c*(exp(I*(f*x+e))-I)^2*exp(-I*(f*x+e)))^3+csgn(I*(exp (I*(f*x+e))-I)^2*exp(-I*(f*x+e)))^3+csgn(I*(exp(I*(f*x+e))+I)*(-1+I*exp(-I *(f*x+e))))^3+csgn(I*(exp(I*(f*x+e))-I)^2)^3-2*m*csgn(I*g)*csgn(I*(exp(I*( f*x+e))+I)*(-1+I*exp(-I*(f*x+e))))*csgn(I*g*(exp(I*(f*x+e))-I)*(1+I*exp(-I *(f*x+e))))-2*m*csgn(I*(exp(I*(f*x+e))-I)*(exp(I*(f*x+e))+I))*csgn(I*(exp( I*(f*x+e))+I)*(-1+I*exp(-I*(f*x+e))))*csgn(I*exp(-I*(f*x+e)))-m*csgn(I*(ex p(I*(f*x+e))+I)^2)*csgn(I*exp(-I*(f*x+e)))*csgn(I*exp(-I*(f*x+e))*(exp(I*( f*x+e))+I)^2)-m*csgn(I*a)*csgn(I*exp(-I*(f*x+e))*(exp(I*(f*x+e))+I)^2)*csg n(I*a*(exp(I*(f*x+e))+I)^2*exp(-I*(f*x+e)))-m*csgn(I*c)*csgn(I*(exp(I*(f*x +e))-I)^2*exp(-I*(f*x+e)))*csgn(I*c*(exp(I*(f*x+e))-I)^2*exp(-I*(f*x+e)))- m*csgn(I*(exp(I*(f*x+e))-I)^2)*csgn(I*exp(-I*(f*x+e)))*csgn(I*(exp(I*(f...
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.53 \[ \int (g \cos (e+f x))^{1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+m} \, dx=-\frac {a \left (\frac {a c}{g^{2}}\right )^{m - 1} \log \left (-\sin \left (f x + e\right ) + 1\right )}{f g} \] Input:
integrate((g*cos(f*x+e))^(1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+m ),x, algorithm="fricas")
Output:
-a*(a*c/g^2)^(m - 1)*log(-sin(f*x + e) + 1)/(f*g)
Timed out. \[ \int (g \cos (e+f x))^{1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+m} \, dx=\text {Timed out} \] Input:
integrate((g*cos(f*x+e))**(1-2*m)*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(- 1+m),x)
Output:
Timed out
\[ \int (g \cos (e+f x))^{1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+m} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m + 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{m - 1} \,d x } \] Input:
integrate((g*cos(f*x+e))^(1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+m ),x, algorithm="maxima")
Output:
integrate((g*cos(f*x + e))^(-2*m + 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(m - 1), x)
Leaf count of result is larger than twice the leaf count of optimal. 934 vs. \(2 (59) = 118\).
Time = 0.47 (sec) , antiderivative size = 934, normalized size of antiderivative = 16.39 \[ \int (g \cos (e+f x))^{1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+m} \, dx=\text {Too large to display} \] Input:
integrate((g*cos(f*x+e))^(1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+m ),x, algorithm="giac")
Output:
1/2*(4*pi*e^(m*log(abs(a)) + m*log(abs(c)) - 2*m*log(abs(g)) - log(abs(c)) + log(abs(g)))*floor(1/4*(pi + 2*f*x - 4*pi*floor(1/2*(pi + f*x + e)/pi) + 2*e)/pi)*tan(1/4*pi + pi*m*floor(-1/4*sgn(a) + 1/2) + pi*m*floor(-1/4*sg n(c) + 1) + 1/4*pi*m*sgn(a) + 1/4*pi*m*sgn(c) - 1/2*pi*m*sgn(g) - pi*floor (-1/4*sgn(c) + 1) - 1/4*pi*sgn(c) + 1/4*pi*sgn(g))^2 + 4*pi*e^(m*log(abs(a )) + m*log(abs(c)) - 2*m*log(abs(g)) - log(abs(c)) + log(abs(g)))*floor(1/ 2*(pi + f*x + e)/pi)*tan(1/4*pi + pi*m*floor(-1/4*sgn(a) + 1/2) + pi*m*flo or(-1/4*sgn(c) + 1) + 1/4*pi*m*sgn(a) + 1/4*pi*m*sgn(c) - 1/2*pi*m*sgn(g) - pi*floor(-1/4*sgn(c) + 1) - 1/4*pi*sgn(c) + 1/4*pi*sgn(g))^2 + 2*pi*e^(m *log(abs(a)) + m*log(abs(c)) - 2*m*log(abs(g)) - log(abs(c)) + log(abs(g)) )*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)*tan(1/4*pi + pi*m*floor(-1/4*sgn(a) + 1/ 2) + pi*m*floor(-1/4*sgn(c) + 1) + 1/4*pi*m*sgn(a) + 1/4*pi*m*sgn(c) - 1/2 *pi*m*sgn(g) - pi*floor(-1/4*sgn(c) + 1) - 1/4*pi*sgn(c) + 1/4*pi*sgn(g))^ 2 + 3*pi*e^(m*log(abs(a)) + m*log(abs(c)) - 2*m*log(abs(g)) - log(abs(c)) + log(abs(g)))*tan(1/4*pi + pi*m*floor(-1/4*sgn(a) + 1/2) + pi*m*floor(-1/ 4*sgn(c) + 1) + 1/4*pi*m*sgn(a) + 1/4*pi*m*sgn(c) - 1/2*pi*m*sgn(g) - pi*f loor(-1/4*sgn(c) + 1) - 1/4*pi*sgn(c) + 1/4*pi*sgn(g))^2 - 2*e*e^(m*log(ab s(a)) + m*log(abs(c)) - 2*m*log(abs(g)) - log(abs(c)) + log(abs(g)))*tan(1 /4*pi + pi*m*floor(-1/4*sgn(a) + 1/2) + pi*m*floor(-1/4*sgn(c) + 1) + 1/4* pi*m*sgn(a) + 1/4*pi*m*sgn(c) - 1/2*pi*m*sgn(g) - pi*floor(-1/4*sgn(c) ...
Timed out. \[ \int (g \cos (e+f x))^{1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+m} \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^{1-2\,m}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m-1} \,d x \] Input:
int((g*cos(e + f*x))^(1 - 2*m)*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x)) ^(m - 1),x)
Output:
int((g*cos(e + f*x))^(1 - 2*m)*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x)) ^(m - 1), x)
\[ \int (g \cos (e+f x))^{1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+m} \, dx=-\frac {\left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (-\sin \left (f x +e \right ) c +c \right )^{m} \cos \left (f x +e \right )}{\cos \left (f x +e \right )^{2 m} \sin \left (f x +e \right )-\cos \left (f x +e \right )^{2 m}}d x \right ) g}{g^{2 m} c} \] Input:
int((g*cos(f*x+e))^(1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+m),x)
Output:
( - int(((sin(e + f*x)*a + a)**m*( - sin(e + f*x)*c + c)**m*cos(e + f*x))/ (cos(e + f*x)**(2*m)*sin(e + f*x) - cos(e + f*x)**(2*m)),x)*g)/(g**(2*m)*c )