3.6 \(\int \sin ^6(a+b x) \, dx\)

Optimal. Leaf size=67 \[ -\frac{\sin ^5(a+b x) \cos (a+b x)}{6 b}-\frac{5 \sin ^3(a+b x) \cos (a+b x)}{24 b}-\frac{5 \sin (a+b x) \cos (a+b x)}{16 b}+\frac{5 x}{16} \]

[Out]

(5*x)/16 - (5*Cos[a + b*x]*Sin[a + b*x])/(16*b) - (5*Cos[a + b*x]*Sin[a + b*x]^3)/(24*b) - (Cos[a + b*x]*Sin[a
 + b*x]^5)/(6*b)

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Rubi [A]  time = 0.0327676, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2635, 8} \[ -\frac{\sin ^5(a+b x) \cos (a+b x)}{6 b}-\frac{5 \sin ^3(a+b x) \cos (a+b x)}{24 b}-\frac{5 \sin (a+b x) \cos (a+b x)}{16 b}+\frac{5 x}{16} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^6,x]

[Out]

(5*x)/16 - (5*Cos[a + b*x]*Sin[a + b*x])/(16*b) - (5*Cos[a + b*x]*Sin[a + b*x]^3)/(24*b) - (Cos[a + b*x]*Sin[a
 + b*x]^5)/(6*b)

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 ^6(a+b x) \, dx &=-\frac{\cos (a+b x) \sin ^5(a+b x)}{6 b}+\frac{5}{6} \int \sin ^4(a+b x) \, dx\\ &=-\frac{5 \cos (a+b x) \sin ^3(a+b x)}{24 b}-\frac{\cos (a+b x) \sin ^5(a+b x)}{6 b}+\frac{5}{8} \int \sin ^2(a+b x) \, dx\\ &=-\frac{5 \cos (a+b x) \sin (a+b x)}{16 b}-\frac{5 \cos (a+b x) \sin ^3(a+b x)}{24 b}-\frac{\cos (a+b x) \sin ^5(a+b x)}{6 b}+\frac{5 \int 1 \, dx}{16}\\ &=\frac{5 x}{16}-\frac{5 \cos (a+b x) \sin (a+b x)}{16 b}-\frac{5 \cos (a+b x) \sin ^3(a+b x)}{24 b}-\frac{\cos (a+b x) \sin ^5(a+b x)}{6 b}\\ \end{align*}

Mathematica [A]  time = 0.0419155, size = 45, normalized size = 0.67 \[ \frac{-45 \sin (2 (a+b x))+9 \sin (4 (a+b x))-\sin (6 (a+b x))+60 a+60 b x}{192 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^6,x]

[Out]

(60*a + 60*b*x - 45*Sin[2*(a + b*x)] + 9*Sin[4*(a + b*x)] - Sin[6*(a + b*x)])/(192*b)

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Maple [A]  time = 0.036, size = 48, normalized size = 0.7 \begin{align*}{\frac{1}{b} \left ( -{\frac{\cos \left ( bx+a \right ) }{6} \left ( \left ( \sin \left ( bx+a \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( bx+a \right ) }{8}} \right ) }+{\frac{5\,bx}{16}}+{\frac{5\,a}{16}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^6,x)

[Out]

1/b*(-1/6*(sin(b*x+a)^5+5/4*sin(b*x+a)^3+15/8*sin(b*x+a))*cos(b*x+a)+5/16*b*x+5/16*a)

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Maxima [A]  time = 1.00742, size = 65, normalized size = 0.97 \begin{align*} \frac{4 \, \sin \left (2 \, b x + 2 \, a\right )^{3} + 60 \, b x + 60 \, a + 9 \, \sin \left (4 \, b x + 4 \, a\right ) - 48 \, \sin \left (2 \, b x + 2 \, a\right )}{192 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^6,x, algorithm="maxima")

[Out]

1/192*(4*sin(2*b*x + 2*a)^3 + 60*b*x + 60*a + 9*sin(4*b*x + 4*a) - 48*sin(2*b*x + 2*a))/b

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Fricas [A]  time = 2.20246, size = 120, normalized size = 1.79 \begin{align*} \frac{15 \, b x -{\left (8 \, \cos \left (b x + a\right )^{5} - 26 \, \cos \left (b x + a\right )^{3} + 33 \, \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(sin(b*x+a)^6,x, algorithm="fricas")

[Out]

1/48*(15*b*x - (8*cos(b*x + a)^5 - 26*cos(b*x + a)^3 + 33*cos(b*x + a))*sin(b*x + a))/b

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Sympy [A]  time = 3.54554, size = 139, normalized size = 2.07 \begin{align*} \begin{cases} \frac{5 x \sin ^{6}{\left (a + b x \right )}}{16} + \frac{15 x \sin ^{4}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{16} + \frac{15 x \sin ^{2}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{16} + \frac{5 x \cos ^{6}{\left (a + b x \right )}}{16} - \frac{11 \sin ^{5}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{16 b} - \frac{5 \sin ^{3}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{6 b} - \frac{5 \sin{\left (a + b x \right )} \cos ^{5}{\left (a + b x \right )}}{16 b} & \text{for}\: b \neq 0 \\x \sin ^{6}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)**6,x)

[Out]

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

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Giac [A]  time = 1.10528, size = 62, normalized size = 0.93 \begin{align*} \frac{5}{16} \, x - \frac{\sin \left (6 \, b x + 6 \, a\right )}{192 \, b} + \frac{3 \, \sin \left (4 \, b x + 4 \, a\right )}{64 \, b} - \frac{15 \, \sin \left (2 \, b x + 2 \, a\right )}{64 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^6,x, algorithm="giac")

[Out]

5/16*x - 1/192*sin(6*b*x + 6*a)/b + 3/64*sin(4*b*x + 4*a)/b - 15/64*sin(2*b*x + 2*a)/b