3.420 \(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx\)

Optimal. Leaf size=101 \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{c f \left (4 m^2+8 m+3\right )}+\frac{\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)} \]

[Out]

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

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Rubi [A]  time = 0.134702, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {2743, 2742} \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{c f \left (4 m^2+8 m+3\right )}+\frac{\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2743

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])

Rule 2742

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)]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx &=\frac{\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)}+\frac{\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m} \, dx}{c (3+2 m)}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)}+\frac{\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c f \left (3+8 m+4 m^2\right )}\\ \end{align*}

Mathematica [A]  time = 3.12659, size = 136, normalized size = 1.35 \[ -\frac{2^{-m} \cos \left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) \sin ^{-2 m-3}\left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) (\sin (e+f x)-2 (m+1)) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{-2 (-m-2)}}{f \left (8 m^2+16 m+6\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[(-e + Pi/2 - f*x)/2]*Sin[(-e + Pi/2 - f*x)/2]^(-3 - 2*m)*(-2*(1 + m) + Sin[e + f*x])*(a + a*Sin[e + f*x
])^m*(c - c*Sin[e + f*x])^(-2 - m))/(2^m*f*(6 + 16*m + 8*m^2)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(2*(-2 - m
))))

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Maple [F]  time = 0.298, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{-2-m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.14477, size = 178, normalized size = 1.76 \begin{align*} \frac{{\left (2 \,{\left (m + 1\right )} \cos \left (f x + e\right ) - \cos \left (f x + e\right ) \sin \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 - 2}}{4 \, f m^{2} + 8 \, f m + 3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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