3.251 \(\int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=85 \[ -\frac{a^3 c^2 \cos ^5(e+f x)}{5 f}+\frac{a^3 c^2 \sin (e+f x) \cos ^3(e+f x)}{4 f}+\frac{3 a^3 c^2 \sin (e+f x) \cos (e+f x)}{8 f}+\frac{3}{8} a^3 c^2 x \]

[Out]

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

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Rubi [A]  time = 0.0924749, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2736, 2669, 2635, 8} \[ -\frac{a^3 c^2 \cos ^5(e+f x)}{5 f}+\frac{a^3 c^2 \sin (e+f x) \cos ^3(e+f x)}{4 f}+\frac{3 a^3 c^2 \sin (e+f x) \cos (e+f x)}{8 f}+\frac{3}{8} a^3 c^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2736

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

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2 \, dx &=\left (a^2 c^2\right ) \int \cos ^4(e+f x) (a+a \sin (e+f x)) \, dx\\ &=-\frac{a^3 c^2 \cos ^5(e+f x)}{5 f}+\left (a^3 c^2\right ) \int \cos ^4(e+f x) \, dx\\ &=-\frac{a^3 c^2 \cos ^5(e+f x)}{5 f}+\frac{a^3 c^2 \cos ^3(e+f x) \sin (e+f x)}{4 f}+\frac{1}{4} \left (3 a^3 c^2\right ) \int \cos ^2(e+f x) \, dx\\ &=-\frac{a^3 c^2 \cos ^5(e+f x)}{5 f}+\frac{3 a^3 c^2 \cos (e+f x) \sin (e+f x)}{8 f}+\frac{a^3 c^2 \cos ^3(e+f x) \sin (e+f x)}{4 f}+\frac{1}{8} \left (3 a^3 c^2\right ) \int 1 \, dx\\ &=\frac{3}{8} a^3 c^2 x-\frac{a^3 c^2 \cos ^5(e+f x)}{5 f}+\frac{3 a^3 c^2 \cos (e+f x) \sin (e+f x)}{8 f}+\frac{a^3 c^2 \cos ^3(e+f x) \sin (e+f x)}{4 f}\\ \end{align*}

Mathematica [A]  time = 1.53295, size = 69, normalized size = 0.81 \[ \frac{a^3 c^2 (40 \sin (2 (e+f x))+5 \sin (4 (e+f x))-20 \cos (e+f x)-10 \cos (3 (e+f x))-2 \cos (5 (e+f x))+60 e+60 f x)}{160 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c^2*(60*e + 60*f*x - 20*Cos[e + f*x] - 10*Cos[3*(e + f*x)] - 2*Cos[5*(e + f*x)] + 40*Sin[2*(e + f*x)] + 5
*Sin[4*(e + f*x)]))/(160*f)

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Maple [B]  time = 0.015, size = 160, normalized size = 1.9 \begin{align*}{\frac{1}{f} \left ( -{\frac{{c}^{2}{a}^{3}\cos \left ( fx+e \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) }+{c}^{2}{a}^{3} \left ( -{\frac{\cos \left ( fx+e \right ) }{4} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\sin \left ( fx+e \right ) }{2}} \right ) }+{\frac{3\,fx}{8}}+{\frac{3\,e}{8}} \right ) +{\frac{2\,{c}^{2}{a}^{3} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}-2\,{c}^{2}{a}^{3} \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -{c}^{2}{a}^{3}\cos \left ( fx+e \right ) +{c}^{2}{a}^{3} \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-1/5*c^2*a^3*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)+c^2*a^3*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*c
os(f*x+e)+3/8*f*x+3/8*e)+2/3*c^2*a^3*(2+sin(f*x+e)^2)*cos(f*x+e)-2*c^2*a^3*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x
+1/2*e)-c^2*a^3*cos(f*x+e)+c^2*a^3*(f*x+e))

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Maxima [B]  time = 1.59996, size = 213, normalized size = 2.51 \begin{align*} -\frac{32 \,{\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} a^{3} c^{2} + 320 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{3} c^{2} - 15 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} c^{2} + 240 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} c^{2} - 480 \,{\left (f x + e\right )} a^{3} c^{2} + 480 \, a^{3} c^{2} \cos \left (f x + e\right )}{480 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480*(32*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*a^3*c^2 + 320*(cos(f*x + e)^3 - 3*cos(f*x
+ e))*a^3*c^2 - 15*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a^3*c^2 + 240*(2*f*x + 2*e - sin(2*
f*x + 2*e))*a^3*c^2 - 480*(f*x + e)*a^3*c^2 + 480*a^3*c^2*cos(f*x + e))/f

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Fricas [A]  time = 1.39263, size = 165, normalized size = 1.94 \begin{align*} -\frac{8 \, a^{3} c^{2} \cos \left (f x + e\right )^{5} - 15 \, a^{3} c^{2} f x - 5 \,{\left (2 \, a^{3} c^{2} \cos \left (f x + e\right )^{3} + 3 \, a^{3} c^{2} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{40 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/40*(8*a^3*c^2*cos(f*x + e)^5 - 15*a^3*c^2*f*x - 5*(2*a^3*c^2*cos(f*x + e)^3 + 3*a^3*c^2*cos(f*x + e))*sin(f
*x + e))/f

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Sympy [A]  time = 5.52163, size = 340, normalized size = 4. \begin{align*} \begin{cases} \frac{3 a^{3} c^{2} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac{3 a^{3} c^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} - a^{3} c^{2} x \sin ^{2}{\left (e + f x \right )} + \frac{3 a^{3} c^{2} x \cos ^{4}{\left (e + f x \right )}}{8} - a^{3} c^{2} x \cos ^{2}{\left (e + f x \right )} + a^{3} c^{2} x - \frac{a^{3} c^{2} \sin ^{4}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{5 a^{3} c^{2} \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} - \frac{4 a^{3} c^{2} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac{2 a^{3} c^{2} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{3 a^{3} c^{2} \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} + \frac{a^{3} c^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{8 a^{3} c^{2} \cos ^{5}{\left (e + f x \right )}}{15 f} + \frac{4 a^{3} c^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{a^{3} c^{2} \cos{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \left (a \sin{\left (e \right )} + a\right )^{3} \left (- c \sin{\left (e \right )} + c\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*a**3*c**2*x*sin(e + f*x)**4/8 + 3*a**3*c**2*x*sin(e + f*x)**2*cos(e + f*x)**2/4 - a**3*c**2*x*sin
(e + f*x)**2 + 3*a**3*c**2*x*cos(e + f*x)**4/8 - a**3*c**2*x*cos(e + f*x)**2 + a**3*c**2*x - a**3*c**2*sin(e +
f*x)**4*cos(e + f*x)/f - 5*a**3*c**2*sin(e + f*x)**3*cos(e + f*x)/(8*f) - 4*a**3*c**2*sin(e + f*x)**2*cos(e +
f*x)**3/(3*f) + 2*a**3*c**2*sin(e + f*x)**2*cos(e + f*x)/f - 3*a**3*c**2*sin(e + f*x)*cos(e + f*x)**3/(8*f) +
a**3*c**2*sin(e + f*x)*cos(e + f*x)/f - 8*a**3*c**2*cos(e + f*x)**5/(15*f) + 4*a**3*c**2*cos(e + f*x)**3/(3*f
) - a**3*c**2*cos(e + f*x)/f, Ne(f, 0)), (x*(a*sin(e) + a)**3*(-c*sin(e) + c)**2, True))

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Giac [A]  time = 2.4822, size = 151, normalized size = 1.78 \begin{align*} \frac{3}{8} \, a^{3} c^{2} x - \frac{a^{3} c^{2} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} - \frac{a^{3} c^{2} \cos \left (3 \, f x + 3 \, e\right )}{16 \, f} - \frac{a^{3} c^{2} \cos \left (f x + e\right )}{8 \, f} + \frac{a^{3} c^{2} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac{a^{3} c^{2} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*a^3*c^2*x - 1/80*a^3*c^2*cos(5*f*x + 5*e)/f - 1/16*a^3*c^2*cos(3*f*x + 3*e)/f - 1/8*a^3*c^2*cos(f*x + e)/f
+ 1/32*a^3*c^2*sin(4*f*x + 4*e)/f + 1/4*a^3*c^2*sin(2*f*x + 2*e)/f