\(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx\) [83]

Optimal result

Integrand size = 38, antiderivative size = 54 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{a c f (3+2 m)} \]

[Out]

cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)*(c-c*sin(f*x+e))^(-2-m)/a/c/f/(3+2*m)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2920, 2821} \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{a c f (2 m+3)} \]

[In]

Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-3 - m),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-2 - m))/(a*c*f*(3 + 2*m))

Rule 2821

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] /; FreeQ[{a, b, c, d, e, f
, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && NeQ[m, -2^(-1)]

Rule 2920

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[1/(a^(p/2)*c^(p/2)), Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(
n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p
/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m} \, dx}{a c} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{a c f (3+2 m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.84 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.11 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=-\frac {\cos ^3(e+f x) (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^{-m}}{c^3 f (3+2 m) (-1+\sin (e+f x))^3} \]

[In]

Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-3 - m),x]

[Out]

-((Cos[e + f*x]^3*(a*(1 + Sin[e + f*x]))^m)/(c^3*f*(3 + 2*m)*(-1 + Sin[e + f*x])^3*(c - c*Sin[e + f*x])^m))

Maple [F]

\[\int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-3-m}d x\]

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m),x)

[Out]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m),x)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3} \cos \left (f x + e\right )^{3}}{2 \, f m + 3 \, f} \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m),x, algorithm="fricas")

[Out]

(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 3)*cos(f*x + e)^3/(2*f*m + 3*f)

Sympy [F]

\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m - 3} \cos ^{2}{\left (e + f x \right )}\, dx \]

[In]

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(-3-m),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(-c*(sin(e + f*x) - 1))**(-m - 3)*cos(e + f*x)**2, x)

Maxima [F]

\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3} \cos \left (f x + e\right )^{2} \,d x } \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 3)*cos(f*x + e)^2, x)

Giac [F]

\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3} \cos \left (f x + e\right )^{2} \,d x } \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 3)*cos(f*x + e)^2, x)

Mupad [B] (verification not implemented)

Time = 1.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.87 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=-\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (6\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2-4\right )}{c^3\,f\,\left (2\,m+3\right )\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,\left (12\,{\sin \left (e+f\,x\right )}^2-15\,\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )+4\right )} \]

[In]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(m + 3),x)

[Out]

-((a*(sin(e + f*x) + 1))^m*(6*sin(e/2 + (f*x)/2)^2 + 2*sin((3*e)/2 + (3*f*x)/2)^2 - 4))/(c^3*f*(2*m + 3)*(-c*(
sin(e + f*x) - 1))^m*(sin(3*e + 3*f*x) - 15*sin(e + f*x) + 12*sin(e + f*x)^2 + 4))