3.81 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx\)

Optimal. Leaf size=182 \[ \frac{2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-3}}{a c^2 f \left (4 m^2+24 m+35\right )}+\frac{2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{a c^3 f (2 m+7) \left (4 m^2+16 m+15\right )}+\frac{\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-4}}{a c f (2 m+7)} \]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-4 - m))/(a*c*f*(7 + 2*m)) + (2*Cos[e + f*x]*
(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-3 - m))/(a*c^2*f*(35 + 24*m + 4*m^2)) + (2*Cos[e + f*x]*(a
+ a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-2 - m))/(a*c^3*f*(7 + 2*m)*(15 + 16*m + 4*m^2))

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Rubi [A]  time = 0.448437, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {2841, 2743, 2742} \[ \frac{2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-3}}{a c^2 f \left (4 m^2+24 m+35\right )}+\frac{2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{a c^3 f (2 m+7) \left (4 m^2+16 m+15\right )}+\frac{\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-4}}{a c f (2 m+7)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-5 - m),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-4 - m))/(a*c*f*(7 + 2*m)) + (2*Cos[e + f*x]*
(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-3 - m))/(a*c^2*f*(35 + 24*m + 4*m^2)) + (2*Cos[e + f*x]*(a
+ a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-2 - m))/(a*c^3*f*(7 + 2*m)*(15 + 16*m + 4*m^2))

Rule 2841

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]

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 \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-5-m} \, dx &=\frac{\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-4-m} \, dx}{a c}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-4-m}}{a c f (7+2 m)}+\frac{2 \int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m} \, dx}{a c^2 (7+2 m)}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-4-m}}{a c f (7+2 m)}+\frac{2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m}}{a c^2 f (5+2 m) (7+2 m)}+\frac{2 \int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m} \, dx}{a c^3 (5+2 m) (7+2 m)}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-4-m}}{a c f (7+2 m)}+\frac{2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m}}{a c^2 f (5+2 m) (7+2 m)}+\frac{2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{a c^3 f (3+2 m) (5+2 m) (7+2 m)}\\ \end{align*}

Mathematica [A]  time = 17.2171, size = 176, normalized size = 0.97 \[ \frac{2^{-m-2} \cos ^3\left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) \sin ^{-2 m-7}\left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-5} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{-2 (-m-5)} \left (-2 (2 m+5) \sin (e+f x)+\cos \left (2 \left (-e-f x+\frac{\pi }{2}\right )\right )+4 \left (m^2+5 m+6\right )\right )}{f (2 m+3) (2 m+5) (2 m+7)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-5 - m),x]

[Out]

(2^(-2 - m)*Cos[(-e + Pi/2 - f*x)/2]^3*Sin[(-e + Pi/2 - f*x)/2]^(-7 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e
+ f*x])^(-5 - m)*(4*(6 + 5*m + m^2) + Cos[2*(-e + Pi/2 - f*x)] - 2*(5 + 2*m)*Sin[e + f*x]))/(f*(3 + 2*m)*(5 +
2*m)*(7 + 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(2*(-5 - m)))

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Maple [F]  time = 0.911, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{-5-m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-5-m),x)

[Out]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-5-m),x)

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Maxima [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(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-5-m),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.79568, size = 261, normalized size = 1.43 \begin{align*} -\frac{{\left (2 \, \cos \left (f x + e\right )^{5} + 2 \,{\left (2 \, m + 5\right )} \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) -{\left (4 \, m^{2} + 20 \, m + 25\right )} \cos \left (f x + e\right )^{3}\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 5}}{8 \, f m^{3} + 60 \, f m^{2} + 142 \, f m + 105 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-5-m),x, algorithm="fricas")

[Out]

-(2*cos(f*x + e)^5 + 2*(2*m + 5)*cos(f*x + e)^3*sin(f*x + e) - (4*m^2 + 20*m + 25)*cos(f*x + e)^3)*(a*sin(f*x
+ e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 5)/(8*f*m^3 + 60*f*m^2 + 142*f*m + 105*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(cos(f*x+e)**2*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(-5-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 - 5} \cos \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-5-m),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 5)*cos(f*x + e)^2, x)