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

Optimal. Leaf size=121 \[ \frac{a^2 c \cos ^5(e+f x)}{5 f}-\frac{a^2 c \cos ^3(e+f x)}{3 f}+\frac{a^2 c \sin ^5(e+f x) \cos (e+f x)}{6 f}-\frac{a^2 c \sin ^3(e+f x) \cos (e+f x)}{24 f}-\frac{a^2 c \sin (e+f x) \cos (e+f x)}{16 f}+\frac{1}{16} a^2 c x \]

[Out]

(a^2*c*x)/16 - (a^2*c*Cos[e + f*x]^3)/(3*f) + (a^2*c*Cos[e + f*x]^5)/(5*f) - (a^2*c*Cos[e + f*x]*Sin[e + f*x])
/(16*f) - (a^2*c*Cos[e + f*x]*Sin[e + f*x]^3)/(24*f) + (a^2*c*Cos[e + f*x]*Sin[e + f*x]^5)/(6*f)

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Rubi [A]  time = 0.17236, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2966, 2633, 2635, 8} \[ \frac{a^2 c \cos ^5(e+f x)}{5 f}-\frac{a^2 c \cos ^3(e+f x)}{3 f}+\frac{a^2 c \sin ^5(e+f x) \cos (e+f x)}{6 f}-\frac{a^2 c \sin ^3(e+f x) \cos (e+f x)}{24 f}-\frac{a^2 c \sin (e+f x) \cos (e+f x)}{16 f}+\frac{1}{16} a^2 c x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c*x)/16 - (a^2*c*Cos[e + f*x]^3)/(3*f) + (a^2*c*Cos[e + f*x]^5)/(5*f) - (a^2*c*Cos[e + f*x]*Sin[e + f*x])
/(16*f) - (a^2*c*Cos[e + f*x]*Sin[e + f*x]^3)/(24*f) + (a^2*c*Cos[e + f*x]*Sin[e + f*x]^5)/(6*f)

Rule 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/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 \sin ^3(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx &=\int \left (a^2 c \sin ^3(e+f x)+a^2 c \sin ^4(e+f x)-a^2 c \sin ^5(e+f x)-a^2 c \sin ^6(e+f x)\right ) \, dx\\ &=\left (a^2 c\right ) \int \sin ^3(e+f x) \, dx+\left (a^2 c\right ) \int \sin ^4(e+f x) \, dx-\left (a^2 c\right ) \int \sin ^5(e+f x) \, dx-\left (a^2 c\right ) \int \sin ^6(e+f x) \, dx\\ &=-\frac{a^2 c \cos (e+f x) \sin ^3(e+f x)}{4 f}+\frac{a^2 c \cos (e+f x) \sin ^5(e+f x)}{6 f}+\frac{1}{4} \left (3 a^2 c\right ) \int \sin ^2(e+f x) \, dx-\frac{1}{6} \left (5 a^2 c\right ) \int \sin ^4(e+f x) \, dx-\frac{\left (a^2 c\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (e+f x)\right )}{f}+\frac{\left (a^2 c\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{a^2 c \cos ^3(e+f x)}{3 f}+\frac{a^2 c \cos ^5(e+f x)}{5 f}-\frac{3 a^2 c \cos (e+f x) \sin (e+f x)}{8 f}-\frac{a^2 c \cos (e+f x) \sin ^3(e+f x)}{24 f}+\frac{a^2 c \cos (e+f x) \sin ^5(e+f x)}{6 f}+\frac{1}{8} \left (3 a^2 c\right ) \int 1 \, dx-\frac{1}{8} \left (5 a^2 c\right ) \int \sin ^2(e+f x) \, dx\\ &=\frac{3}{8} a^2 c x-\frac{a^2 c \cos ^3(e+f x)}{3 f}+\frac{a^2 c \cos ^5(e+f x)}{5 f}-\frac{a^2 c \cos (e+f x) \sin (e+f x)}{16 f}-\frac{a^2 c \cos (e+f x) \sin ^3(e+f x)}{24 f}+\frac{a^2 c \cos (e+f x) \sin ^5(e+f x)}{6 f}-\frac{1}{16} \left (5 a^2 c\right ) \int 1 \, dx\\ &=\frac{1}{16} a^2 c x-\frac{a^2 c \cos ^3(e+f x)}{3 f}+\frac{a^2 c \cos ^5(e+f x)}{5 f}-\frac{a^2 c \cos (e+f x) \sin (e+f x)}{16 f}-\frac{a^2 c \cos (e+f x) \sin ^3(e+f x)}{24 f}+\frac{a^2 c \cos (e+f x) \sin ^5(e+f x)}{6 f}\\ \end{align*}

Mathematica [A]  time = 0.124349, size = 77, normalized size = 0.64 \[ \frac{a^2 c (-15 \sin (2 (e+f x))-15 \sin (4 (e+f x))+5 \sin (6 (e+f x))-120 \cos (e+f x)-20 \cos (3 (e+f x))+12 \cos (5 (e+f x))+60 e+60 f x)}{960 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c*(60*e + 60*f*x - 120*Cos[e + f*x] - 20*Cos[3*(e + f*x)] + 12*Cos[5*(e + f*x)] - 15*Sin[2*(e + f*x)] - 1
5*Sin[4*(e + f*x)] + 5*Sin[6*(e + f*x)]))/(960*f)

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Maple [A]  time = 0.026, size = 147, normalized size = 1.2 \begin{align*}{\frac{1}{f} \left ( -{a}^{2}c \left ( -{\frac{\cos \left ( fx+e \right ) }{6} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( fx+e \right ) }{8}} \right ) }+{\frac{5\,fx}{16}}+{\frac{5\,e}{16}} \right ) +{\frac{{a}^{2}c\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 ) }+{a}^{2}c \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{{a}^{2}c \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-a^2*c*(-1/6*(sin(f*x+e)^5+5/4*sin(f*x+e)^3+15/8*sin(f*x+e))*cos(f*x+e)+5/16*f*x+5/16*e)+1/5*a^2*c*(8/3+s
in(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)+a^2*c*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)-1
/3*a^2*c*(2+sin(f*x+e)^2)*cos(f*x+e))

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Maxima [A]  time = 0.976141, size = 198, normalized size = 1.64 \begin{align*} \frac{64 \,{\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^{2} c + 320 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} c - 5 \,{\left (4 \, \sin \left (2 \, f x + 2 \, e\right )^{3} + 60 \, f x + 60 \, e + 9 \, \sin \left (4 \, f x + 4 \, e\right ) - 48 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c + 30 \,{\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^{2} c}{960 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(64*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*a^2*c + 320*(cos(f*x + e)^3 - 3*cos(f*x + e
))*a^2*c - 5*(4*sin(2*f*x + 2*e)^3 + 60*f*x + 60*e + 9*sin(4*f*x + 4*e) - 48*sin(2*f*x + 2*e))*a^2*c + 30*(12*
f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a^2*c)/f

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Fricas [A]  time = 1.94822, size = 225, normalized size = 1.86 \begin{align*} \frac{48 \, a^{2} c \cos \left (f x + e\right )^{5} - 80 \, a^{2} c \cos \left (f x + e\right )^{3} + 15 \, a^{2} c f x + 5 \,{\left (8 \, a^{2} c \cos \left (f x + e\right )^{5} - 14 \, a^{2} c \cos \left (f x + e\right )^{3} + 3 \, a^{2} c \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{240 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(48*a^2*c*cos(f*x + e)^5 - 80*a^2*c*cos(f*x + e)^3 + 15*a^2*c*f*x + 5*(8*a^2*c*cos(f*x + e)^5 - 14*a^2*c
*cos(f*x + e)^3 + 3*a^2*c*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 4.78227, size = 415, normalized size = 3.43 \begin{align*} \begin{cases} - \frac{5 a^{2} c x \sin ^{6}{\left (e + f x \right )}}{16} - \frac{15 a^{2} c x \sin ^{4}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{16} + \frac{3 a^{2} c x \sin ^{4}{\left (e + f x \right )}}{8} - \frac{15 a^{2} c x \sin ^{2}{\left (e + f x \right )} \cos ^{4}{\left (e + f x \right )}}{16} + \frac{3 a^{2} c x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} - \frac{5 a^{2} c x \cos ^{6}{\left (e + f x \right )}}{16} + \frac{3 a^{2} c x \cos ^{4}{\left (e + f x \right )}}{8} + \frac{11 a^{2} c \sin ^{5}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{16 f} + \frac{a^{2} c \sin ^{4}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{5 a^{2} c \sin ^{3}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{6 f} - \frac{5 a^{2} c \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} + \frac{4 a^{2} c \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{a^{2} c \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{5 a^{2} c \sin{\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{16 f} - \frac{3 a^{2} c \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} + \frac{8 a^{2} c \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac{2 a^{2} c \cos ^{3}{\left (e + f x \right )}}{3 f} & \text{for}\: f \neq 0 \\x \left (a \sin{\left (e \right )} + a\right )^{2} \left (- c \sin{\left (e \right )} + c\right ) \sin ^{3}{\left (e \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-5*a**2*c*x*sin(e + f*x)**6/16 - 15*a**2*c*x*sin(e + f*x)**4*cos(e + f*x)**2/16 + 3*a**2*c*x*sin(e
+ f*x)**4/8 - 15*a**2*c*x*sin(e + f*x)**2*cos(e + f*x)**4/16 + 3*a**2*c*x*sin(e + f*x)**2*cos(e + f*x)**2/4 -
5*a**2*c*x*cos(e + f*x)**6/16 + 3*a**2*c*x*cos(e + f*x)**4/8 + 11*a**2*c*sin(e + f*x)**5*cos(e + f*x)/(16*f) +
a**2*c*sin(e + f*x)**4*cos(e + f*x)/f + 5*a**2*c*sin(e + f*x)**3*cos(e + f*x)**3/(6*f) - 5*a**2*c*sin(e + f*x
)**3*cos(e + f*x)/(8*f) + 4*a**2*c*sin(e + f*x)**2*cos(e + f*x)**3/(3*f) - a**2*c*sin(e + f*x)**2*cos(e + f*x)
/f + 5*a**2*c*sin(e + f*x)*cos(e + f*x)**5/(16*f) - 3*a**2*c*sin(e + f*x)*cos(e + f*x)**3/(8*f) + 8*a**2*c*cos
(e + f*x)**5/(15*f) - 2*a**2*c*cos(e + f*x)**3/(3*f), Ne(f, 0)), (x*(a*sin(e) + a)**2*(-c*sin(e) + c)*sin(e)**
3, True))

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Giac [A]  time = 1.27495, size = 161, normalized size = 1.33 \begin{align*} \frac{1}{16} \, a^{2} c x + \frac{a^{2} c \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} - \frac{a^{2} c \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac{a^{2} c \cos \left (f x + e\right )}{8 \, f} + \frac{a^{2} c \sin \left (6 \, f x + 6 \, e\right )}{192 \, f} - \frac{a^{2} c \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} - \frac{a^{2} c \sin \left (2 \, f x + 2 \, e\right )}{64 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*a^2*c*x + 1/80*a^2*c*cos(5*f*x + 5*e)/f - 1/48*a^2*c*cos(3*f*x + 3*e)/f - 1/8*a^2*c*cos(f*x + e)/f + 1/19
2*a^2*c*sin(6*f*x + 6*e)/f - 1/64*a^2*c*sin(4*f*x + 4*e)/f - 1/64*a^2*c*sin(2*f*x + 2*e)/f