\(\int \sin ^6(x) \, dx\) [32]

Optimal result

Integrand size = 4, antiderivative size = 34 \[ \int \sin ^6(x) \, dx=\frac {5 x}{16}-\frac {5}{16} \cos (x) \sin (x)-\frac {5}{24} \cos (x) \sin ^3(x)-\frac {1}{6} \cos (x) \sin ^5(x) \]

[Out]

5/16*x-5/16*cos(x)*sin(x)-5/24*cos(x)*sin(x)^3-1/6*cos(x)*sin(x)^5

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2715, 8} \[ \int \sin ^6(x) \, dx=\frac {5 x}{16}-\frac {1}{6} \sin ^5(x) \cos (x)-\frac {5}{24} \sin ^3(x) \cos (x)-\frac {5}{16} \sin (x) \cos (x) \]

[In]

Int[Sin[x]^6,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \sin ^6(x) \, dx=\frac {5 x}{16}-\frac {15}{64} \sin (2 x)+\frac {3}{64} \sin (4 x)-\frac {1}{192} \sin (6 x) \]

[In]

Integrate[Sin[x]^6,x]

[Out]

(5*x)/16 - (15*Sin[2*x])/64 + (3*Sin[4*x])/64 - Sin[6*x]/192

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.68

method	result	size
risch	\(\frac {5 x}{16}-\frac {\sin \left (6 x \right )}{192}+\frac {3 \sin \left (4 x \right )}{64}-\frac {15 \sin \left (2 x \right )}{64}\)	\(23\)
parallelrisch	\(\frac {5 x}{16}-\frac {\sin \left (6 x \right )}{192}+\frac {3 \sin \left (4 x \right )}{64}-\frac {15 \sin \left (2 x \right )}{64}\)	\(23\)
default	\(-\frac {\left (\sin ^{5}\left (x \right )+\frac {5 \left (\sin ^{3}\left (x \right )\right )}{4}+\frac {15 \sin \left (x \right )}{8}\right ) \cos \left (x \right )}{6}+\frac {5 x}{16}\)	\(24\)
norman	\(\frac {\frac {5 x}{16}-\frac {85 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{24}-\frac {33 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{4}+\frac {33 \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{4}+\frac {85 \left (\tan ^{9}\left (\frac {x}{2}\right )\right )}{24}+\frac {5 \left (\tan ^{11}\left (\frac {x}{2}\right )\right )}{8}+\frac {15 x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{8}+\frac {75 x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{16}+\frac {25 x \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{4}+\frac {75 x \left (\tan ^{8}\left (\frac {x}{2}\right )\right )}{16}+\frac {15 x \left (\tan ^{10}\left (\frac {x}{2}\right )\right )}{8}+\frac {5 x \left (\tan ^{12}\left (\frac {x}{2}\right )\right )}{16}-\frac {5 \tan \left (\frac {x}{2}\right )}{8}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{6}}\)	\(116\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int \sin ^6(x) \, dx=-\frac {1}{48} \, {\left (8 \, \cos \left (x\right )^{5} - 26 \, \cos \left (x\right )^{3} + 33 \, \cos \left (x\right )\right )} \sin \left (x\right ) + \frac {5}{16} \, x \]

[In]

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

[Out]

-1/48*(8*cos(x)^5 - 26*cos(x)^3 + 33*cos(x))*sin(x) + 5/16*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \sin ^6(x) \, dx=\frac {5 x}{16} - \frac {\sin ^{5}{\left (x \right )} \cos {\left (x \right )}}{6} - \frac {5 \sin ^{3}{\left (x \right )} \cos {\left (x \right )}}{24} - \frac {5 \sin {\left (x \right )} \cos {\left (x \right )}}{16} \]

[In]

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

[Out]

5*x/16 - sin(x)**5*cos(x)/6 - 5*sin(x)**3*cos(x)/24 - 5*sin(x)*cos(x)/16

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \sin ^6(x) \, dx=\frac {1}{48} \, \sin \left (2 \, x\right )^{3} + \frac {5}{16} \, x + \frac {3}{64} \, \sin \left (4 \, x\right ) - \frac {1}{4} \, \sin \left (2 \, x\right ) \]

[In]

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

[Out]

1/48*sin(2*x)^3 + 5/16*x + 3/64*sin(4*x) - 1/4*sin(2*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \sin ^6(x) \, dx=\frac {5}{16} \, x - \frac {1}{192} \, \sin \left (6 \, x\right ) + \frac {3}{64} \, \sin \left (4 \, x\right ) - \frac {15}{64} \, \sin \left (2 \, x\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \sin ^6(x) \, dx=\frac {5\,x}{16}-\frac {15\,\sin \left (2\,x\right )}{64}+\frac {3\,\sin \left (4\,x\right )}{64}-\frac {\sin \left (6\,x\right )}{192} \]

[In]

int(sin(x)^6,x)

[Out]

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