3.4.56 \(\int \sin ^6(2 x) \, dx\) [356]

Optimal. Leaf size=46 \[ \frac {5 x}{16}-\frac {5}{32} \cos (2 x) \sin (2 x)-\frac {5}{48} \cos (2 x) \sin ^3(2 x)-\frac {1}{12} \cos (2 x) \sin ^5(2 x) \]

[Out]

5/16*x-5/32*cos(2*x)*sin(2*x)-5/48*cos(2*x)*sin(2*x)^3-1/12*cos(2*x)*sin(2*x)^5

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Rubi [A]
time = 0.01, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2715, 8} \begin {gather*} \frac {5 x}{16}-\frac {1}{12} \sin ^5(2 x) \cos (2 x)-\frac {5}{48} \sin ^3(2 x) \cos (2 x)-\frac {5}{32} \sin (2 x) \cos (2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[2*x]^6,x]

[Out]

(5*x)/16 - (5*Cos[2*x]*Sin[2*x])/32 - (5*Cos[2*x]*Sin[2*x]^3)/48 - (Cos[2*x]*Sin[2*x]^5)/12

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

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] && Integ
erQ[2*n]

Rubi steps

\begin {align*} \int \sin ^6(2 x) \, dx &=-\frac {1}{12} \cos (2 x) \sin ^5(2 x)+\frac {5}{6} \int \sin ^4(2 x) \, dx\\ &=-\frac {5}{48} \cos (2 x) \sin ^3(2 x)-\frac {1}{12} \cos (2 x) \sin ^5(2 x)+\frac {5}{8} \int \sin ^2(2 x) \, dx\\ &=-\frac {5}{32} \cos (2 x) \sin (2 x)-\frac {5}{48} \cos (2 x) \sin ^3(2 x)-\frac {1}{12} \cos (2 x) \sin ^5(2 x)+\frac {5 \int 1 \, dx}{16}\\ &=\frac {5 x}{16}-\frac {5}{32} \cos (2 x) \sin (2 x)-\frac {5}{48} \cos (2 x) \sin ^3(2 x)-\frac {1}{12} \cos (2 x) \sin ^5(2 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 30, normalized size = 0.65 \begin {gather*} \frac {5 x}{16}-\frac {15}{128} \sin (4 x)+\frac {3}{128} \sin (8 x)-\frac {1}{384} \sin (12 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[2*x]^6,x]

[Out]

(5*x)/16 - (15*Sin[4*x])/128 + (3*Sin[8*x])/128 - Sin[12*x]/384

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Maple [A]
time = 0.03, size = 32, normalized size = 0.70

method	result	size
risch	\(\frac {5 x}{16}-\frac {\sin \left (12 x \right )}{384}+\frac {3 \sin \left (8 x \right )}{128}-\frac {15 \sin \left (4 x \right )}{128}\)	\(23\)
derivativedivides	\(-\frac {\left (\sin ^{5}\left (2 x \right )+\frac {5 \left (\sin ^{3}\left (2 x \right )\right )}{4}+\frac {15 \sin \left (2 x \right )}{8}\right ) \cos \left (2 x \right )}{12}+\frac {5 x}{16}\)	\(32\)
default	\(-\frac {\left (\sin ^{5}\left (2 x \right )+\frac {5 \left (\sin ^{3}\left (2 x \right )\right )}{4}+\frac {15 \sin \left (2 x \right )}{8}\right ) \cos \left (2 x \right )}{12}+\frac {5 x}{16}\)	\(32\)
norman	\(\frac {\frac {5 x}{16}-\frac {85 \left (\tan ^{3}\left (x \right )\right )}{48}-\frac {33 \left (\tan ^{5}\left (x \right )\right )}{8}+\frac {33 \left (\tan ^{7}\left (x \right )\right )}{8}+\frac {85 \left (\tan ^{9}\left (x \right )\right )}{48}+\frac {5 \left (\tan ^{11}\left (x \right )\right )}{16}+\frac {15 x \left (\tan ^{2}\left (x \right )\right )}{8}+\frac {75 x \left (\tan ^{4}\left (x \right )\right )}{16}+\frac {25 x \left (\tan ^{6}\left (x \right )\right )}{4}+\frac {75 x \left (\tan ^{8}\left (x \right )\right )}{16}+\frac {15 x \left (\tan ^{10}\left (x \right )\right )}{8}+\frac {5 x \left (\tan ^{12}\left (x \right )\right )}{16}-\frac {5 \tan \left (x \right )}{16}}{\left (1+\tan ^{2}\left (x \right )\right )^{6}}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*x)^6,x,method=_RETURNVERBOSE)

[Out]

-1/12*(sin(2*x)^5+5/4*sin(2*x)^3+15/8*sin(2*x))*cos(2*x)+5/16*x

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Maxima [A]
time = 1.70, size = 24, normalized size = 0.52 \begin {gather*} \frac {1}{96} \, \sin \left (4 \, x\right )^{3} + \frac {5}{16} \, x + \frac {3}{128} \, \sin \left (8 \, x\right ) - \frac {1}{8} \, \sin \left (4 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)^6,x, algorithm="maxima")

[Out]

1/96*sin(4*x)^3 + 5/16*x + 3/128*sin(8*x) - 1/8*sin(4*x)

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Fricas [A]
time = 0.54, size = 33, normalized size = 0.72 \begin {gather*} -\frac {1}{96} \, {\left (8 \, \cos \left (2 \, x\right )^{5} - 26 \, \cos \left (2 \, x\right )^{3} + 33 \, \cos \left (2 \, x\right )\right )} \sin \left (2 \, x\right ) + \frac {5}{16} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)^6,x, algorithm="fricas")

[Out]

-1/96*(8*cos(2*x)^5 - 26*cos(2*x)^3 + 33*cos(2*x))*sin(2*x) + 5/16*x

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Sympy [A]
time = 0.01, size = 46, normalized size = 1.00 \begin {gather*} \frac {5 x}{16} - \frac {\sin ^{5}{\left (2 x \right )} \cos {\left (2 x \right )}}{12} - \frac {5 \sin ^{3}{\left (2 x \right )} \cos {\left (2 x \right )}}{48} - \frac {5 \sin {\left (2 x \right )} \cos {\left (2 x \right )}}{32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)**6,x)

[Out]

5*x/16 - sin(2*x)**5*cos(2*x)/12 - 5*sin(2*x)**3*cos(2*x)/48 - 5*sin(2*x)*cos(2*x)/32

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Giac [A]
time = 1.25, size = 22, normalized size = 0.48 \begin {gather*} \frac {5}{16} \, x - \frac {1}{384} \, \sin \left (12 \, x\right ) + \frac {3}{128} \, \sin \left (8 \, x\right ) - \frac {15}{128} \, \sin \left (4 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)^6,x, algorithm="giac")

[Out]

5/16*x - 1/384*sin(12*x) + 3/128*sin(8*x) - 15/128*sin(4*x)

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Mupad [B]
time = 0.21, size = 22, normalized size = 0.48 \begin {gather*} \frac {5\,x}{16}-\frac {15\,\sin \left (4\,x\right )}{128}+\frac {3\,\sin \left (8\,x\right )}{128}-\frac {\sin \left (12\,x\right )}{384} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*x)^6,x)

[Out]

(5*x)/16 - (15*sin(4*x))/128 + (3*sin(8*x))/128 - sin(12*x)/384

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