Integrand size = 8, antiderivative size = 67 \[ \int \sin ^6(a+b x) \, dx=\frac {5 x}{16}-\frac {5 \cos (a+b x) \sin (a+b x)}{16 b}-\frac {5 \cos (a+b x) \sin ^3(a+b x)}{24 b}-\frac {\cos (a+b x) \sin ^5(a+b x)}{6 b} \] Output:
5/16*x-5/16*cos(b*x+a)*sin(b*x+a)/b-5/24*cos(b*x+a)*sin(b*x+a)^3/b-1/6*cos (b*x+a)*sin(b*x+a)^5/b
Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.67 \[ \int \sin ^6(a+b x) \, dx=\frac {60 a+60 b x-45 \sin (2 (a+b x))+9 \sin (4 (a+b x))-\sin (6 (a+b x))}{192 b} \] Input:
Integrate[Sin[a + b*x]^6,x]
Output:
(60*a + 60*b*x - 45*Sin[2*(a + b*x)] + 9*Sin[4*(a + b*x)] - Sin[6*(a + b*x )])/(192*b)
Time = 0.52 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {3042, 3115, 3042, 3115, 3042, 3115, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sin ^6(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^6dx\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {5}{6} \int \sin ^4(a+b x)dx-\frac {\sin ^5(a+b x) \cos (a+b x)}{6 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \int \sin (a+b x)^4dx-\frac {\sin ^5(a+b x) \cos (a+b x)}{6 b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \int \sin ^2(a+b x)dx-\frac {\sin ^3(a+b x) \cos (a+b x)}{4 b}\right )-\frac {\sin ^5(a+b x) \cos (a+b x)}{6 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \int \sin (a+b x)^2dx-\frac {\sin ^3(a+b x) \cos (a+b x)}{4 b}\right )-\frac {\sin ^5(a+b x) \cos (a+b x)}{6 b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}-\frac {\sin (a+b x) \cos (a+b x)}{2 b}\right )-\frac {\sin ^3(a+b x) \cos (a+b x)}{4 b}\right )-\frac {\sin ^5(a+b x) \cos (a+b x)}{6 b}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {x}{2}-\frac {\sin (a+b x) \cos (a+b x)}{2 b}\right )-\frac {\sin ^3(a+b x) \cos (a+b x)}{4 b}\right )-\frac {\sin ^5(a+b x) \cos (a+b x)}{6 b}\) |
Input:
Int[Sin[a + b*x]^6,x]
Output:
-1/6*(Cos[a + b*x]*Sin[a + b*x]^5)/b + (5*(-1/4*(Cos[a + b*x]*Sin[a + b*x] ^3)/b + (3*(x/2 - (Cos[a + b*x]*Sin[a + b*x])/(2*b)))/4))/6
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Time = 9.67 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.66
method | result | size |
parallelrisch | \(\frac {60 b x -\sin \left (6 b x +6 a \right )+9 \sin \left (4 b x +4 a \right )-45 \sin \left (2 b x +2 a \right )}{192 b}\) | \(44\) |
risch | \(\frac {5 x}{16}-\frac {\sin \left (6 b x +6 a \right )}{192 b}+\frac {3 \sin \left (4 b x +4 a \right )}{64 b}-\frac {15 \sin \left (2 b x +2 a \right )}{64 b}\) | \(47\) |
derivativedivides | \(\frac {-\frac {\left (\sin \left (b x +a \right )^{5}+\frac {5 \sin \left (b x +a \right )^{3}}{4}+\frac {15 \sin \left (b x +a \right )}{8}\right ) \cos \left (b x +a \right )}{6}+\frac {5 b x}{16}+\frac {5 a}{16}}{b}\) | \(48\) |
default | \(\frac {-\frac {\left (\sin \left (b x +a \right )^{5}+\frac {5 \sin \left (b x +a \right )^{3}}{4}+\frac {15 \sin \left (b x +a \right )}{8}\right ) \cos \left (b x +a \right )}{6}+\frac {5 b x}{16}+\frac {5 a}{16}}{b}\) | \(48\) |
norman | \(\frac {\frac {5 x}{16}-\frac {5 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}-\frac {85 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}}{24 b}-\frac {33 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{5}}{4 b}+\frac {33 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{7}}{4 b}+\frac {85 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{9}}{24 b}+\frac {5 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{11}}{8 b}+\frac {15 x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{8}+\frac {75 x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}{16}+\frac {25 x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}}{4}+\frac {75 x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{8}}{16}+\frac {15 x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{10}}{8}+\frac {5 x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{12}}{16}}{\left (1+\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )^{6}}\) | \(199\) |
orering | \(x \sin \left (b x +a \right )^{6}-\frac {49 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{5}}{24 b}+\frac {49 x \left (30 \sin \left (b x +a \right )^{4} b^{2} \cos \left (b x +a \right )^{2}-6 \sin \left (b x +a \right )^{6} b^{2}\right )}{144 b^{2}}-\frac {7 \left (120 \sin \left (b x +a \right )^{3} b^{3} \cos \left (b x +a \right )^{3}-96 \sin \left (b x +a \right )^{5} b^{3} \cos \left (b x +a \right )\right )}{288 b^{4}}+\frac {7 x \left (360 \cos \left (b x +a \right )^{4} \sin \left (b x +a \right )^{2} b^{4}-840 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{4} b^{4}+96 \sin \left (b x +a \right )^{6} b^{4}\right )}{288 b^{4}}-\frac {-4800 \cos \left (b x +a \right )^{3} \sin \left (b x +a \right )^{3} b^{5}+720 \cos \left (b x +a \right )^{5} \sin \left (b x +a \right ) b^{5}+2256 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{5} b^{5}}{2304 b^{6}}+\frac {x \left (25680 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{4} b^{6}-18000 \cos \left (b x +a \right )^{4} \sin \left (b x +a \right )^{2} b^{6}+720 \cos \left (b x +a \right )^{6} b^{6}-2256 b^{6} \sin \left (b x +a \right )^{6}\right )}{2304 b^{6}}\) | \(320\) |
Input:
int(sin(b*x+a)^6,x,method=_RETURNVERBOSE)
Output:
1/192*(60*b*x-sin(6*b*x+6*a)+9*sin(4*b*x+4*a)-45*sin(2*b*x+2*a))/b
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.70 \[ \int \sin ^6(a+b x) \, dx=\frac {15 \, b x - {\left (8 \, \cos \left (b x + a\right )^{5} - 26 \, \cos \left (b x + a\right )^{3} + 33 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{48 \, b} \] Input:
integrate(sin(b*x+a)^6,x, algorithm="fricas")
Output:
1/48*(15*b*x - (8*cos(b*x + a)^5 - 26*cos(b*x + a)^3 + 33*cos(b*x + a))*si n(b*x + a))/b
Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (61) = 122\).
Time = 0.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.07 \[ \int \sin ^6(a+b x) \, dx=\begin {cases} \frac {5 x \sin ^{6}{\left (a + b x \right )}}{16} + \frac {15 x \sin ^{4}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{16} + \frac {15 x \sin ^{2}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{16} + \frac {5 x \cos ^{6}{\left (a + b x \right )}}{16} - \frac {11 \sin ^{5}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{16 b} - \frac {5 \sin ^{3}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{6 b} - \frac {5 \sin {\left (a + b x \right )} \cos ^{5}{\left (a + b x \right )}}{16 b} & \text {for}\: b \neq 0 \\x \sin ^{6}{\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate(sin(b*x+a)**6,x)
Output:
Piecewise((5*x*sin(a + b*x)**6/16 + 15*x*sin(a + b*x)**4*cos(a + b*x)**2/1 6 + 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 ))
Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.72 \[ \int \sin ^6(a+b x) \, dx=\frac {4 \, \sin \left (2 \, b x + 2 \, a\right )^{3} + 60 \, b x + 60 \, a + 9 \, \sin \left (4 \, b x + 4 \, a\right ) - 48 \, \sin \left (2 \, b x + 2 \, a\right )}{192 \, b} \] Input:
integrate(sin(b*x+a)^6,x, algorithm="maxima")
Output:
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
Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.69 \[ \int \sin ^6(a+b x) \, dx=\frac {5}{16} \, x - \frac {\sin \left (6 \, b x + 6 \, a\right )}{192 \, b} + \frac {3 \, \sin \left (4 \, b x + 4 \, a\right )}{64 \, b} - \frac {15 \, \sin \left (2 \, b x + 2 \, a\right )}{64 \, b} \] Input:
integrate(sin(b*x+a)^6,x, algorithm="giac")
Output:
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
Time = 25.49 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.64 \[ \int \sin ^6(a+b x) \, dx=\frac {5\,x}{16}-\frac {\frac {15\,\sin \left (2\,a+2\,b\,x\right )}{64}-\frac {3\,\sin \left (4\,a+4\,b\,x\right )}{64}+\frac {\sin \left (6\,a+6\,b\,x\right )}{192}}{b} \] Input:
int(sin(a + b*x)^6,x)
Output:
(5*x)/16 - ((15*sin(2*a + 2*b*x))/64 - (3*sin(4*a + 4*b*x))/64 + sin(6*a + 6*b*x)/192)/b
Time = 0.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.84 \[ \int \sin ^6(a+b x) \, dx=\frac {-8 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{5}-10 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{3}-15 \cos \left (b x +a \right ) \sin \left (b x +a \right )+15 b x}{48 b} \] Input:
int(sin(b*x+a)^6,x)
Output:
( - 8*cos(a + b*x)*sin(a + b*x)**5 - 10*cos(a + b*x)*sin(a + b*x)**3 - 15* cos(a + b*x)*sin(a + b*x) + 15*b*x)/(48*b)