3.221 \(\int (a-a \sin (e+f x))^m (c+c \sin (e+f x))^3 (B (-3+m)+B (4+m) \sin (e+f x)) \, dx\)

Optimal. Leaf size=35 \[ -\frac{a^3 B c^3 \cos ^7(e+f x) (a-a \sin (e+f x))^{m-3}}{f} \]

[Out]

-((a^3*B*c^3*Cos[e + f*x]^7*(a - a*Sin[e + f*x])^(-3 + m))/f)

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Rubi [A]  time = 0.238397, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.047, Rules used = {2967, 2854} \[ -\frac{a^3 B c^3 \cos ^7(e+f x) (a-a \sin (e+f x))^{m-3}}{f} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sin[e + f*x])^m*(c + c*Sin[e + f*x])^3*(B*(-3 + m) + B*(4 + m)*Sin[e + f*x]),x]

[Out]

-((a^3*B*c^3*Cos[e + f*x]^7*(a - a*Sin[e + f*x])^(-3 + m))/f)

Rule 2967

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rule 2854

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

Rubi steps

\begin{align*} \int (a-a \sin (e+f x))^m (c+c \sin (e+f x))^3 (B (-3+m)+B (4+m) \sin (e+f x)) \, dx &=\left (a^3 c^3\right ) \int \cos ^6(e+f x) (a-a \sin (e+f x))^{-3+m} (B (-3+m)+B (4+m) \sin (e+f x)) \, dx\\ &=-\frac{a^3 B c^3 \cos ^7(e+f x) (a-a \sin (e+f x))^{-3+m}}{f}\\ \end{align*}

Mathematica [A]  time = 0.544851, size = 61, normalized size = 1.74 \[ \frac{B c^3 (-14 \sin (2 (e+f x))+\sin (4 (e+f x))-14 \cos (e+f x)+6 \cos (3 (e+f x))) (a-a \sin (e+f x))^m}{8 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sin[e + f*x])^m*(c + c*Sin[e + f*x])^3*(B*(-3 + m) + B*(4 + m)*Sin[e + f*x]),x]

[Out]

(B*c^3*(a - a*Sin[e + f*x])^m*(-14*Cos[e + f*x] + 6*Cos[3*(e + f*x)] - 14*Sin[2*(e + f*x)] + Sin[4*(e + f*x)])
)/(8*f)

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Maple [F]  time = 2.238, 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 ) ^{3} \left ( B \left ( m-3 \right ) +B \left ( 4+m \right ) \sin \left ( fx+e \right ) \right ) \, 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))^3*(B*(m-3)+B*(4+m)*sin(f*x+e)),x)

[Out]

int((a-a*sin(f*x+e))^m*(c+c*sin(f*x+e))^3*(B*(m-3)+B*(4+m)*sin(f*x+e)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B{\left (m + 4\right )} \sin \left (f x + e\right ) + B{\left (m - 3\right )}\right )}{\left (c \sin \left (f x + e\right ) + c\right )}^{3}{\left (-a \sin \left (f x + e\right ) + a\right )}^{m}\,{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))^3*(B*(-3+m)+B*(4+m)*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((B*(m + 4)*sin(f*x + e) + B*(m - 3))*(c*sin(f*x + e) + c)^3*(-a*sin(f*x + e) + a)^m, x)

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Fricas [B]  time = 2.03624, size = 184, normalized size = 5.26 \begin{align*} \frac{{\left (3 \, B c^{3} \cos \left (f x + e\right )^{3} - 4 \, B c^{3} \cos \left (f x + e\right ) +{\left (B c^{3} \cos \left (f x + e\right )^{3} - 4 \, B c^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}{\left (-a \sin \left (f x + e\right ) + a\right )}^{m}}{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))^3*(B*(-3+m)+B*(4+m)*sin(f*x+e)),x, algorithm="fricas")

[Out]

(3*B*c^3*cos(f*x + e)^3 - 4*B*c^3*cos(f*x + e) + (B*c^3*cos(f*x + e)^3 - 4*B*c^3*cos(f*x + e))*sin(f*x + e))*(
-a*sin(f*x + e) + a)^m/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))**3*(B*(-3+m)+B*(4+m)*sin(f*x+e)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B{\left (m + 4\right )} \sin \left (f x + e\right ) + B{\left (m - 3\right )}\right )}{\left (c \sin \left (f x + e\right ) + c\right )}^{3}{\left (-a \sin \left (f x + e\right ) + a\right )}^{m}\,{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))^3*(B*(-3+m)+B*(4+m)*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate((B*(m + 4)*sin(f*x + e) + B*(m - 3))*(c*sin(f*x + e) + c)^3*(-a*sin(f*x + e) + a)^m, x)