3.91 \(\int \cos ^2(a+b x) \sin ^4(a+b x) \, dx\)

Optimal. Leaf size=69 \[ -\frac{\sin ^3(a+b x) \cos ^3(a+b x)}{6 b}-\frac{\sin (a+b x) \cos ^3(a+b x)}{8 b}+\frac{\sin (a+b x) \cos (a+b x)}{16 b}+\frac{x}{16} \]

[Out]

x/16 + (Cos[a + b*x]*Sin[a + b*x])/(16*b) - (Cos[a + b*x]^3*Sin[a + b*x])/(8*b) - (Cos[a + b*x]^3*Sin[a + b*x]
^3)/(6*b)

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Rubi [A]  time = 0.0683697, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2568, 2635, 8} \[ -\frac{\sin ^3(a+b x) \cos ^3(a+b x)}{6 b}-\frac{\sin (a+b x) \cos ^3(a+b x)}{8 b}+\frac{\sin (a+b x) \cos (a+b x)}{16 b}+\frac{x}{16} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Sin[a + b*x]^4,x]

[Out]

x/16 + (Cos[a + b*x]*Sin[a + b*x])/(16*b) - (Cos[a + b*x]^3*Sin[a + b*x])/(8*b) - (Cos[a + b*x]^3*Sin[a + b*x]
^3)/(6*b)

Rule 2568

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

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 \cos ^2(a+b x) \sin ^4(a+b x) \, dx &=-\frac{\cos ^3(a+b x) \sin ^3(a+b x)}{6 b}+\frac{1}{2} \int \cos ^2(a+b x) \sin ^2(a+b x) \, dx\\ &=-\frac{\cos ^3(a+b x) \sin (a+b x)}{8 b}-\frac{\cos ^3(a+b x) \sin ^3(a+b x)}{6 b}+\frac{1}{8} \int \cos ^2(a+b x) \, dx\\ &=\frac{\cos (a+b x) \sin (a+b x)}{16 b}-\frac{\cos ^3(a+b x) \sin (a+b x)}{8 b}-\frac{\cos ^3(a+b x) \sin ^3(a+b x)}{6 b}+\frac{\int 1 \, dx}{16}\\ &=\frac{x}{16}+\frac{\cos (a+b x) \sin (a+b x)}{16 b}-\frac{\cos ^3(a+b x) \sin (a+b x)}{8 b}-\frac{\cos ^3(a+b x) \sin ^3(a+b x)}{6 b}\\ \end{align*}

Mathematica [A]  time = 0.0672379, size = 40, normalized size = 0.58 \[ \frac{-3 \sin (2 (a+b x))-3 \sin (4 (a+b x))+\sin (6 (a+b x))+12 b x}{192 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2*Sin[a + b*x]^4,x]

[Out]

(12*b*x - 3*Sin[2*(a + b*x)] - 3*Sin[4*(a + b*x)] + Sin[6*(a + b*x)])/(192*b)

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Maple [A]  time = 0.01, size = 61, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{3} \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{6}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{3}\sin \left ( bx+a \right ) }{8}}+{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{16}}+{\frac{bx}{16}}+{\frac{a}{16}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2*sin(b*x+a)^4,x)

[Out]

1/b*(-1/6*cos(b*x+a)^3*sin(b*x+a)^3-1/8*cos(b*x+a)^3*sin(b*x+a)+1/16*cos(b*x+a)*sin(b*x+a)+1/16*b*x+1/16*a)

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Maxima [A]  time = 1.10965, size = 50, normalized size = 0.72 \begin{align*} -\frac{4 \, \sin \left (2 \, b x + 2 \, a\right )^{3} - 12 \, b x - 12 \, a + 3 \, \sin \left (4 \, b x + 4 \, a\right )}{192 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

-1/192*(4*sin(2*b*x + 2*a)^3 - 12*b*x - 12*a + 3*sin(4*b*x + 4*a))/b

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Fricas [A]  time = 1.66524, size = 117, normalized size = 1.7 \begin{align*} \frac{3 \, b x +{\left (8 \, \cos \left (b x + a\right )^{5} - 14 \, \cos \left (b x + a\right )^{3} + 3 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{48 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

1/48*(3*b*x + (8*cos(b*x + a)^5 - 14*cos(b*x + a)^3 + 3*cos(b*x + a))*sin(b*x + a))/b

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Sympy [A]  time = 3.8449, size = 136, normalized size = 1.97 \begin{align*} \begin{cases} \frac{x \sin ^{6}{\left (a + b x \right )}}{16} + \frac{3 x \sin ^{4}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{16} + \frac{3 x \sin ^{2}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{16} + \frac{x \cos ^{6}{\left (a + b x \right )}}{16} + \frac{\sin ^{5}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{16 b} - \frac{\sin ^{3}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{6 b} - \frac{\sin{\left (a + b x \right )} \cos ^{5}{\left (a + b x \right )}}{16 b} & \text{for}\: b \neq 0 \\x \sin ^{4}{\left (a \right )} \cos ^{2}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**2*sin(b*x+a)**4,x)

[Out]

Piecewise((x*sin(a + b*x)**6/16 + 3*x*sin(a + b*x)**4*cos(a + b*x)**2/16 + 3*x*sin(a + b*x)**2*cos(a + b*x)**4
/16 + x*cos(a + b*x)**6/16 + sin(a + b*x)**5*cos(a + b*x)/(16*b) - sin(a + b*x)**3*cos(a + b*x)**3/(6*b) - sin
(a + b*x)*cos(a + b*x)**5/(16*b), Ne(b, 0)), (x*sin(a)**4*cos(a)**2, True))

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Giac [A]  time = 1.17517, size = 62, normalized size = 0.9 \begin{align*} \frac{1}{16} \, x + \frac{\sin \left (6 \, b x + 6 \, a\right )}{192 \, b} - \frac{\sin \left (4 \, b x + 4 \, a\right )}{64 \, b} - \frac{\sin \left (2 \, b x + 2 \, a\right )}{64 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(b*x+a)^4,x, algorithm="giac")

[Out]

1/16*x + 1/192*sin(6*b*x + 6*a)/b - 1/64*sin(4*b*x + 4*a)/b - 1/64*sin(2*b*x + 2*a)/b