3.364 \(\int (4-3 \cos (x)) (1-\frac{\sin (x)}{2})^4 \, dx\)

Optimal. Leaf size=70 \[ \frac{227 x}{32}-\frac{3}{80} \sin ^5(x)+\frac{3 \sin ^4(x)}{8}-\frac{3 \sin ^3(x)}{2}-3 \sin (x)-\frac{2 \cos ^3(x)}{3}-3 \cos ^2(x)+10 \cos (x)-\frac{1}{16} \sin ^3(x) \cos (x)-\frac{99}{32} \sin (x) \cos (x) \]

[Out]

(227*x)/32 + 10*Cos[x] - 3*Cos[x]^2 - (2*Cos[x]^3)/3 - 3*Sin[x] - (99*Cos[x]*Sin[x])/32 - (3*Sin[x]^3)/2 - (Co
s[x]*Sin[x]^3)/16 + (3*Sin[x]^4)/8 - (3*Sin[x]^5)/80

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Rubi [A]  time = 0.153064, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {4401, 2637, 2668, 2669, 2635, 8, 641} \[ \frac{227 x}{32}-\frac{3}{80} \sin ^5(x)+\frac{3 \sin ^4(x)}{8}-\frac{3 \sin ^3(x)}{2}-3 \sin (x)-\frac{2 \cos ^3(x)}{3}-3 \cos ^2(x)+10 \cos (x)-\frac{1}{16} \sin ^3(x) \cos (x)-\frac{99}{32} \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[(4 - 3*Cos[x])*(1 - Sin[x]/2)^4,x]

[Out]

(227*x)/32 + 10*Cos[x] - 3*Cos[x]^2 - (2*Cos[x]^3)/3 - 3*Sin[x] - (99*Cos[x]*Sin[x])/32 - (3*Sin[x]^3)/2 - (Co
s[x]*Sin[x]^3)/16 + (3*Sin[x]^4)/8 - (3*Sin[x]^5)/80

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 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]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rubi steps

\begin{align*} \int (4-3 \cos (x)) \left (1-\frac{\sin (x)}{2}\right )^4 \, dx &=\int \left (4-3 \cos (x)+2 (-4+3 \cos (x)) \sin (x)-\frac{3}{2} (-4+3 \cos (x)) \sin ^2(x)+\frac{1}{2} (-4+3 \cos (x)) \sin ^3(x)-\frac{1}{16} (-4+3 \cos (x)) \sin ^4(x)\right ) \, dx\\ &=4 x-\frac{1}{16} \int (-4+3 \cos (x)) \sin ^4(x) \, dx+\frac{1}{2} \int (-4+3 \cos (x)) \sin ^3(x) \, dx-\frac{3}{2} \int (-4+3 \cos (x)) \sin ^2(x) \, dx+2 \int (-4+3 \cos (x)) \sin (x) \, dx-3 \int \cos (x) \, dx\\ &=4 x-3 \sin (x)-\frac{3 \sin ^3(x)}{2}-\frac{3 \sin ^5(x)}{80}-\frac{1}{54} \operatorname{Subst}\left (\int (-4+x) \left (9-x^2\right ) \, dx,x,3 \cos (x)\right )+\frac{1}{4} \int \sin ^4(x) \, dx-\frac{2}{3} \operatorname{Subst}(\int (-4+x) \, dx,x,3 \cos (x))+6 \int \sin ^2(x) \, dx\\ &=4 x+8 \cos (x)-3 \cos ^2(x)-3 \sin (x)-3 \cos (x) \sin (x)-\frac{3 \sin ^3(x)}{2}-\frac{1}{16} \cos (x) \sin ^3(x)+\frac{3 \sin ^4(x)}{8}-\frac{3 \sin ^5(x)}{80}+\frac{2}{27} \operatorname{Subst}\left (\int \left (9-x^2\right ) \, dx,x,3 \cos (x)\right )+\frac{3}{16} \int \sin ^2(x) \, dx+3 \int 1 \, dx\\ &=7 x+10 \cos (x)-3 \cos ^2(x)-\frac{2 \cos ^3(x)}{3}-3 \sin (x)-\frac{99}{32} \cos (x) \sin (x)-\frac{3 \sin ^3(x)}{2}-\frac{1}{16} \cos (x) \sin ^3(x)+\frac{3 \sin ^4(x)}{8}-\frac{3 \sin ^5(x)}{80}+\frac{3 \int 1 \, dx}{32}\\ &=\frac{227 x}{32}+10 \cos (x)-3 \cos ^2(x)-\frac{2 \cos ^3(x)}{3}-3 \sin (x)-\frac{99}{32} \cos (x) \sin (x)-\frac{3 \sin ^3(x)}{2}-\frac{1}{16} \cos (x) \sin ^3(x)+\frac{3 \sin ^4(x)}{8}-\frac{3 \sin ^5(x)}{80}\\ \end{align*}

Mathematica [A]  time = 0.0805028, size = 74, normalized size = 1.06 \[ \frac{227 x}{32}-\frac{531 \sin (x)}{128}-\frac{25}{16} \sin (2 x)+\frac{99}{256} \sin (3 x)+\frac{1}{128} \sin (4 x)-\frac{3 \sin (5 x)}{1280}+\frac{19 \cos (x)}{2}-\frac{27}{16} \cos (2 x)-\frac{1}{6} \cos (3 x)+\frac{3}{64} \cos (4 x) \]

Antiderivative was successfully verified.

[In]

Integrate[(4 - 3*Cos[x])*(1 - Sin[x]/2)^4,x]

[Out]

(227*x)/32 + (19*Cos[x])/2 - (27*Cos[2*x])/16 - Cos[3*x]/6 + (3*Cos[4*x])/64 - (531*Sin[x])/128 - (25*Sin[2*x]
)/16 + (99*Sin[3*x])/256 + Sin[4*x]/128 - (3*Sin[5*x])/1280

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Maple [A]  time = 0.039, size = 66, normalized size = 0.9 \begin{align*}{\frac{227\,x}{32}}+8\,\cos \left ( x \right ) -3\,\cos \left ( x \right ) \sin \left ( x \right ) +{\frac{ \left ( 4+2\, \left ( \sin \left ( x \right ) \right ) ^{2} \right ) \cos \left ( x \right ) }{3}}-{\frac{\cos \left ( x \right ) }{16} \left ( \left ( \sin \left ( x \right ) \right ) ^{3}+{\frac{3\,\sin \left ( x \right ) }{2}} \right ) }-3\,\sin \left ( x \right ) -3\, \left ( \cos \left ( x \right ) \right ) ^{2}-{\frac{3\, \left ( \sin \left ( x \right ) \right ) ^{3}}{2}}+{\frac{3\, \left ( \sin \left ( x \right ) \right ) ^{4}}{8}}-{\frac{3\, \left ( \sin \left ( x \right ) \right ) ^{5}}{80}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4-3*cos(x))*(1-1/2*sin(x))^4,x)

[Out]

227/32*x+8*cos(x)-3*cos(x)*sin(x)+2/3*(2+sin(x)^2)*cos(x)-1/16*(sin(x)^3+3/2*sin(x))*cos(x)-3*sin(x)-3*cos(x)^
2-3/2*sin(x)^3+3/8*sin(x)^4-3/80*sin(x)^5

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Maxima [A]  time = 0.942261, size = 73, normalized size = 1.04 \begin{align*} -\frac{3}{80} \, \sin \left (x\right )^{5} + \frac{3}{8} \, \sin \left (x\right )^{4} - \frac{2}{3} \, \cos \left (x\right )^{3} - \frac{3}{2} \, \sin \left (x\right )^{3} - 3 \, \cos \left (x\right )^{2} + \frac{227}{32} \, x + 10 \, \cos \left (x\right ) + \frac{1}{128} \, \sin \left (4 \, x\right ) - \frac{25}{16} \, \sin \left (2 \, x\right ) - 3 \, \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4-3*cos(x))*(1-1/2*sin(x))^4,x, algorithm="maxima")

[Out]

-3/80*sin(x)^5 + 3/8*sin(x)^4 - 2/3*cos(x)^3 - 3/2*sin(x)^3 - 3*cos(x)^2 + 227/32*x + 10*cos(x) + 1/128*sin(4*
x) - 25/16*sin(2*x) - 3*sin(x)

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Fricas [A]  time = 2.1721, size = 194, normalized size = 2.77 \begin{align*} \frac{3}{8} \, \cos \left (x\right )^{4} - \frac{2}{3} \, \cos \left (x\right )^{3} - \frac{15}{4} \, \cos \left (x\right )^{2} - \frac{1}{160} \,{\left (6 \, \cos \left (x\right )^{4} - 10 \, \cos \left (x\right )^{3} - 252 \, \cos \left (x\right )^{2} + 505 \, \cos \left (x\right ) + 726\right )} \sin \left (x\right ) + \frac{227}{32} \, x + 10 \, \cos \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4-3*cos(x))*(1-1/2*sin(x))^4,x, algorithm="fricas")

[Out]

3/8*cos(x)^4 - 2/3*cos(x)^3 - 15/4*cos(x)^2 - 1/160*(6*cos(x)^4 - 10*cos(x)^3 - 252*cos(x)^2 + 505*cos(x) + 72
6)*sin(x) + 227/32*x + 10*cos(x)

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Sympy [B]  time = 1.67097, size = 162, normalized size = 2.31 \begin{align*} \frac{3 x \sin ^{4}{\left (x \right )}}{32} + \frac{3 x \sin ^{2}{\left (x \right )} \cos ^{2}{\left (x \right )}}{16} + 3 x \sin ^{2}{\left (x \right )} + \frac{3 x \cos ^{4}{\left (x \right )}}{32} + 3 x \cos ^{2}{\left (x \right )} + 4 x - \frac{3 \sin ^{5}{\left (x \right )}}{80} - \frac{5 \sin ^{3}{\left (x \right )} \cos{\left (x \right )}}{32} - \frac{3 \sin ^{3}{\left (x \right )}}{2} - \frac{3 \sin ^{2}{\left (x \right )} \cos ^{2}{\left (x \right )}}{4} + 2 \sin ^{2}{\left (x \right )} \cos{\left (x \right )} + 3 \sin ^{2}{\left (x \right )} - \frac{3 \sin{\left (x \right )} \cos ^{3}{\left (x \right )}}{32} - 3 \sin{\left (x \right )} \cos{\left (x \right )} - 3 \sin{\left (x \right )} - \frac{3 \cos ^{4}{\left (x \right )}}{8} + \frac{4 \cos ^{3}{\left (x \right )}}{3} + 8 \cos{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4-3*cos(x))*(1-1/2*sin(x))**4,x)

[Out]

3*x*sin(x)**4/32 + 3*x*sin(x)**2*cos(x)**2/16 + 3*x*sin(x)**2 + 3*x*cos(x)**4/32 + 3*x*cos(x)**2 + 4*x - 3*sin
(x)**5/80 - 5*sin(x)**3*cos(x)/32 - 3*sin(x)**3/2 - 3*sin(x)**2*cos(x)**2/4 + 2*sin(x)**2*cos(x) + 3*sin(x)**2
- 3*sin(x)*cos(x)**3/32 - 3*sin(x)*cos(x) - 3*sin(x) - 3*cos(x)**4/8 + 4*cos(x)**3/3 + 8*cos(x)

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Giac [A]  time = 1.09989, size = 73, normalized size = 1.04 \begin{align*} \frac{227}{32} \, x + \frac{3}{64} \, \cos \left (4 \, x\right ) - \frac{1}{6} \, \cos \left (3 \, x\right ) - \frac{27}{16} \, \cos \left (2 \, x\right ) + \frac{19}{2} \, \cos \left (x\right ) - \frac{3}{1280} \, \sin \left (5 \, x\right ) + \frac{1}{128} \, \sin \left (4 \, x\right ) + \frac{99}{256} \, \sin \left (3 \, x\right ) - \frac{25}{16} \, \sin \left (2 \, x\right ) - \frac{531}{128} \, \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4-3*cos(x))*(1-1/2*sin(x))^4,x, algorithm="giac")

[Out]

227/32*x + 3/64*cos(4*x) - 1/6*cos(3*x) - 27/16*cos(2*x) + 19/2*cos(x) - 3/1280*sin(5*x) + 1/128*sin(4*x) + 99
/256*sin(3*x) - 25/16*sin(2*x) - 531/128*sin(x)