\(\int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx\) [419]

Optimal result

Integrand size = 30, antiderivative size = 164 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\frac {\cos (e+f x) (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac {2 \cos (e+f x) (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{c f \left (15+16 m+4 m^2\right )}+\frac {2 \cos (e+f x) (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c^2 f (5+2 m) \left (3+8 m+4 m^2\right )} \]

[Out]

cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m)/f/(5+2*m)+2*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e
))^(-2-m)/c/f/(4*m^2+16*m+15)+2*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1-m)/c^2/f/(8*m^3+36*m^2+46*m
+15)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2822, 2821} \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\frac {2 \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{c^2 f (2 m+5) \left (4 m^2+8 m+3\right )}+\frac {2 \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}}{c f \left (4 m^2+16 m+15\right )}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-3}}{f (2 m+5)} \]

[In]

Int[(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])^m*(c - c*Sin[e + f*x])^(-3 - m))/(f*(5 + 2*m)) + (2*Cos[e + f*x]*(a + a*Sin
[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m))/(c*f*(15 + 16*m + 4*m^2)) + (2*Cos[e + f*x]*(a + a*Sin[e + f*x])^m
*(c - c*Sin[e + f*x])^(-1 - m))/(c^2*f*(5 + 2*m)*(3 + 8*m + 4*m^2))

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 2822

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] + Dist[(m + n + 1)/(a*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m
, 1] ||  !SumSimplerQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \frac {\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac {2 \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx}{c (5+2 m)} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{c f \left (15+16 m+4 m^2\right )}+\frac {2 \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m} \, dx}{c^2 (3+2 m) (5+2 m)} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{c f \left (15+16 m+4 m^2\right )}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c^2 f (1+2 m) (3+2 m) (5+2 m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.65 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=\frac {3^m \sec (e+f x) (1+\sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-m} \left (7+12 m+4 m^2-2 (3+2 m) \sin (e+f x)+2 \sin ^2(e+f x)\right )}{c^3 f (1+2 m) (3+2 m) (5+2 m) (-1+\sin (e+f x))^2} \]

[In]

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

[Out]

(3^m*Sec[e + f*x]*(1 + Sin[e + f*x])^(1 + m)*(7 + 12*m + 4*m^2 - 2*(3 + 2*m)*Sin[e + f*x] + 2*Sin[e + f*x]^2))
/(c^3*f*(1 + 2*m)*(3 + 2*m)*(5 + 2*m)*(-1 + Sin[e + f*x])^2*(c - c*Sin[e + f*x])^m)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.62 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=-\frac {{\left (2 \, \cos \left (f x + e\right )^{3} + 2 \, {\left (2 \, m + 3\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (4 \, m^{2} + 12 \, m + 9\right )} \cos \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f} \]

[In]

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

[Out]

-(2*cos(f*x + e)^3 + 2*(2*m + 3)*cos(f*x + e)*sin(f*x + e) - (4*m^2 + 12*m + 9)*cos(f*x + e))*(a*sin(f*x + e)
+ a)^m*(-c*sin(f*x + e) + c)^(-m - 3)/(8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)

Sympy [F]

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

[In]

integrate((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), x)

Maxima [F]

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

[In]

integrate((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), x)

Giac [F]

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

[In]

integrate((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), x)

Mupad [B] (verification not implemented)

Time = 8.36 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.91 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx=-\frac {2\,{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (15\,\cos \left (e+f\,x\right )-\cos \left (3\,e+3\,f\,x\right )-6\,\sin \left (2\,e+2\,f\,x\right )+24\,m\,\cos \left (e+f\,x\right )+8\,m^2\,\cos \left (e+f\,x\right )-4\,m\,\sin \left (2\,e+2\,f\,x\right )\right )}{c^3\,f\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,\left (8\,m^3+36\,m^2+46\,m+15\right )\,\left (15\,\sin \left (e+f\,x\right )+6\,\cos \left (2\,e+2\,f\,x\right )-\sin \left (3\,e+3\,f\,x\right )-10\right )} \]

[In]

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

[Out]

-(2*(a*(sin(e + f*x) + 1))^m*(15*cos(e + f*x) - cos(3*e + 3*f*x) - 6*sin(2*e + 2*f*x) + 24*m*cos(e + f*x) + 8*
m^2*cos(e + f*x) - 4*m*sin(2*e + 2*f*x)))/(c^3*f*(-c*(sin(e + f*x) - 1))^m*(46*m + 36*m^2 + 8*m^3 + 15)*(15*si
n(e + f*x) + 6*cos(2*e + 2*f*x) - sin(3*e + 3*f*x) - 10))