3.227 \(\int (a+a \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx\)

Optimal. Leaf size=116 \[ \frac{7 a c^4 \cos ^3(e+f x)}{12 f}+\frac{a \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{5 f}+\frac{7 a \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{20 f}+\frac{7 a c^4 \sin (e+f x) \cos (e+f x)}{8 f}+\frac{7}{8} a c^4 x \]

[Out]

(7*a*c^4*x)/8 + (7*a*c^4*Cos[e + f*x]^3)/(12*f) + (7*a*c^4*Cos[e + f*x]*Sin[e + f*x])/(8*f) + (a*Cos[e + f*x]^
3*(c^2 - c^2*Sin[e + f*x])^2)/(5*f) + (7*a*Cos[e + f*x]^3*(c^4 - c^4*Sin[e + f*x]))/(20*f)

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Rubi [A]  time = 0.162034, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {2736, 2678, 2669, 2635, 8} \[ \frac{7 a c^4 \cos ^3(e+f x)}{12 f}+\frac{a \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{5 f}+\frac{7 a \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{20 f}+\frac{7 a c^4 \sin (e+f x) \cos (e+f x)}{8 f}+\frac{7}{8} a c^4 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a*c^4*x)/8 + (7*a*c^4*Cos[e + f*x]^3)/(12*f) + (7*a*c^4*Cos[e + f*x]*Sin[e + f*x])/(8*f) + (a*Cos[e + f*x]^
3*(c^2 - c^2*Sin[e + f*x])^2)/(5*f) + (7*a*Cos[e + f*x]^3*(c^4 - c^4*Sin[e + f*x]))/(20*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 2678

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

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)) (c-c \sin (e+f x))^4 \, dx &=(a c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^3 \, dx\\ &=\frac{a \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{5 f}+\frac{1}{5} \left (7 a c^2\right ) \int \cos ^2(e+f x) (c-c \sin (e+f x))^2 \, dx\\ &=\frac{a \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{5 f}+\frac{7 a \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{20 f}+\frac{1}{4} \left (7 a c^3\right ) \int \cos ^2(e+f x) (c-c \sin (e+f x)) \, dx\\ &=\frac{7 a c^4 \cos ^3(e+f x)}{12 f}+\frac{a \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{5 f}+\frac{7 a \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{20 f}+\frac{1}{4} \left (7 a c^4\right ) \int \cos ^2(e+f x) \, dx\\ &=\frac{7 a c^4 \cos ^3(e+f x)}{12 f}+\frac{7 a c^4 \cos (e+f x) \sin (e+f x)}{8 f}+\frac{a \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{5 f}+\frac{7 a \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{20 f}+\frac{1}{8} \left (7 a c^4\right ) \int 1 \, dx\\ &=\frac{7}{8} a c^4 x+\frac{7 a c^4 \cos ^3(e+f x)}{12 f}+\frac{7 a c^4 \cos (e+f x) \sin (e+f x)}{8 f}+\frac{a \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{5 f}+\frac{7 a \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{20 f}\\ \end{align*}

Mathematica [A]  time = 0.548112, size = 64, normalized size = 0.55 \[ \frac{a c^4 (120 \sin (2 (e+f x))-45 \sin (4 (e+f x))+420 \cos (e+f x)+130 \cos (3 (e+f x))-6 \cos (5 (e+f x))+420 f x)}{480 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^4*(420*f*x + 420*Cos[e + f*x] + 130*Cos[3*(e + f*x)] - 6*Cos[5*(e + f*x)] + 120*Sin[2*(e + f*x)] - 45*Sin
[4*(e + f*x)]))/(480*f)

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Maple [A]  time = 0.021, size = 149, normalized size = 1.3 \begin{align*}{\frac{1}{f} \left ( -{\frac{a{c}^{4}\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 ) }-3\,a{c}^{4} \left ( -1/4\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+3/2\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -{\frac{2\,a{c}^{4} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+2\,a{c}^{4} \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) +3\,a{c}^{4}\cos \left ( fx+e \right ) +a{c}^{4} \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))*(c-c*sin(f*x+e))^4,x)

[Out]

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

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Maxima [A]  time = 1.20619, size = 197, normalized size = 1.7 \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 c^{4} - 320 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a c^{4} + 45 \,{\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 c^{4} - 240 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a c^{4} - 480 \,{\left (f x + e\right )} a c^{4} - 1440 \, a c^{4} \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))*(c-c*sin(f*x+e))^4,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*c^4 - 320*(cos(f*x + e)^3 - 3*cos(f*x +
e))*a*c^4 + 45*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a*c^4 - 240*(2*f*x + 2*e - sin(2*f*x +
2*e))*a*c^4 - 480*(f*x + e)*a*c^4 - 1440*a*c^4*cos(f*x + e))/f

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Fricas [A]  time = 1.64101, size = 196, normalized size = 1.69 \begin{align*} -\frac{24 \, a c^{4} \cos \left (f x + e\right )^{5} - 160 \, a c^{4} \cos \left (f x + e\right )^{3} - 105 \, a c^{4} f x + 15 \,{\left (6 \, a c^{4} \cos \left (f x + e\right )^{3} - 7 \, a c^{4} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{120 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(c-c*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

-1/120*(24*a*c^4*cos(f*x + e)^5 - 160*a*c^4*cos(f*x + e)^3 - 105*a*c^4*f*x + 15*(6*a*c^4*cos(f*x + e)^3 - 7*a*
c^4*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 4.43833, size = 314, normalized size = 2.71 \begin{align*} \begin{cases} - \frac{9 a c^{4} x \sin ^{4}{\left (e + f x \right )}}{8} - \frac{9 a c^{4} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + a c^{4} x \sin ^{2}{\left (e + f x \right )} - \frac{9 a c^{4} x \cos ^{4}{\left (e + f x \right )}}{8} + a c^{4} x \cos ^{2}{\left (e + f x \right )} + a c^{4} x - \frac{a c^{4} \sin ^{4}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{15 a c^{4} \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} - \frac{4 a c^{4} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{2 a c^{4} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{9 a c^{4} \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac{a c^{4} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{8 a c^{4} \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac{4 a c^{4} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac{3 a c^{4} \cos{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \left (a \sin{\left (e \right )} + a\right ) \left (- c \sin{\left (e \right )} + c\right )^{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(c-c*sin(f*x+e))**4,x)

[Out]

Piecewise((-9*a*c**4*x*sin(e + f*x)**4/8 - 9*a*c**4*x*sin(e + f*x)**2*cos(e + f*x)**2/4 + a*c**4*x*sin(e + f*x
)**2 - 9*a*c**4*x*cos(e + f*x)**4/8 + a*c**4*x*cos(e + f*x)**2 + a*c**4*x - a*c**4*sin(e + f*x)**4*cos(e + f*x
)/f + 15*a*c**4*sin(e + f*x)**3*cos(e + f*x)/(8*f) - 4*a*c**4*sin(e + f*x)**2*cos(e + f*x)**3/(3*f) - 2*a*c**4
*sin(e + f*x)**2*cos(e + f*x)/f + 9*a*c**4*sin(e + f*x)*cos(e + f*x)**3/(8*f) - a*c**4*sin(e + f*x)*cos(e + f*
x)/f - 8*a*c**4*cos(e + f*x)**5/(15*f) - 4*a*c**4*cos(e + f*x)**3/(3*f) + 3*a*c**4*cos(e + f*x)/f, Ne(f, 0)),
(x*(a*sin(e) + a)*(-c*sin(e) + c)**4, True))

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Giac [A]  time = 1.71468, size = 135, normalized size = 1.16 \begin{align*} \frac{7}{8} \, a c^{4} x - \frac{a c^{4} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac{13 \, a c^{4} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} + \frac{7 \, a c^{4} \cos \left (f x + e\right )}{8 \, f} - \frac{3 \, a c^{4} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac{a c^{4} \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))*(c-c*sin(f*x+e))^4,x, algorithm="giac")

[Out]

7/8*a*c^4*x - 1/80*a*c^4*cos(5*f*x + 5*e)/f + 13/48*a*c^4*cos(3*f*x + 3*e)/f + 7/8*a*c^4*cos(f*x + e)/f - 3/32
*a*c^4*sin(4*f*x + 4*e)/f + 1/4*a*c^4*sin(2*f*x + 2*e)/f