Integrand size = 10, antiderivative size = 85 \[ \int x^2 \sin ^3(k x) \, dx=\frac {14 \cos (k x)}{9 k^3}-\frac {2 x^2 \cos (k x)}{3 k}-\frac {2 \cos ^3(k x)}{27 k^3}+\frac {4 x \sin (k x)}{3 k^2}-\frac {x^2 \cos (k x) \sin ^2(k x)}{3 k}+\frac {2 x \sin ^3(k x)}{9 k^2} \]
14/9*cos(k*x)/k^3-2/3*x^2*cos(k*x)/k-2/27*cos(k*x)^3/k^3+4/3*x*sin(k*x)/k^ 2-1/3*x^2*cos(k*x)*sin(k*x)^2/k+2/9*x*sin(k*x)^3/k^2
Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.65 \[ \int x^2 \sin ^3(k x) \, dx=\frac {-81 \left (-2+k^2 x^2\right ) \cos (k x)+\left (-2+9 k^2 x^2\right ) \cos (3 k x)-6 k x (-27 \sin (k x)+\sin (3 k x))}{108 k^3} \]
(-81*(-2 + k^2*x^2)*Cos[k*x] + (-2 + 9*k^2*x^2)*Cos[3*k*x] - 6*k*x*(-27*Si n[k*x] + Sin[3*k*x]))/(108*k^3)
Time = 0.47 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.14, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.100, Rules used = {3042, 3792, 3042, 3113, 2009, 3777, 3042, 3777, 25, 3042, 3118}
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.
\(\displaystyle \int x^2 \sin ^3(k x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int x^2 \sin (k x)^3dx\) |
\(\Big \downarrow \) 3792 |
\(\displaystyle -\frac {2 \int \sin ^3(k x)dx}{9 k^2}+\frac {2}{3} \int x^2 \sin (k x)dx+\frac {2 x \sin ^3(k x)}{9 k^2}-\frac {x^2 \sin ^2(k x) \cos (k x)}{3 k}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 \int \sin (k x)^3dx}{9 k^2}+\frac {2}{3} \int x^2 \sin (k x)dx+\frac {2 x \sin ^3(k x)}{9 k^2}-\frac {x^2 \sin ^2(k x) \cos (k x)}{3 k}\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle \frac {2 \int \left (1-\cos ^2(k x)\right )d\cos (k x)}{9 k^3}+\frac {2}{3} \int x^2 \sin (k x)dx+\frac {2 x \sin ^3(k x)}{9 k^2}-\frac {x^2 \sin ^2(k x) \cos (k x)}{3 k}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{3} \int x^2 \sin (k x)dx+\frac {2 \left (\cos (k x)-\frac {1}{3} \cos ^3(k x)\right )}{9 k^3}+\frac {2 x \sin ^3(k x)}{9 k^2}-\frac {x^2 \sin ^2(k x) \cos (k x)}{3 k}\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {2}{3} \left (\frac {2 \int x \cos (k x)dx}{k}-\frac {x^2 \cos (k x)}{k}\right )+\frac {2 \left (\cos (k x)-\frac {1}{3} \cos ^3(k x)\right )}{9 k^3}+\frac {2 x \sin ^3(k x)}{9 k^2}-\frac {x^2 \sin ^2(k x) \cos (k x)}{3 k}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} \left (\frac {2 \int x \sin \left (k x+\frac {\pi }{2}\right )dx}{k}-\frac {x^2 \cos (k x)}{k}\right )+\frac {2 \left (\cos (k x)-\frac {1}{3} \cos ^3(k x)\right )}{9 k^3}+\frac {2 x \sin ^3(k x)}{9 k^2}-\frac {x^2 \sin ^2(k x) \cos (k x)}{3 k}\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {2}{3} \left (\frac {2 \left (\frac {\int -\sin (k x)dx}{k}+\frac {x \sin (k x)}{k}\right )}{k}-\frac {x^2 \cos (k x)}{k}\right )+\frac {2 \left (\cos (k x)-\frac {1}{3} \cos ^3(k x)\right )}{9 k^3}+\frac {2 x \sin ^3(k x)}{9 k^2}-\frac {x^2 \sin ^2(k x) \cos (k x)}{3 k}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2}{3} \left (\frac {2 \left (\frac {x \sin (k x)}{k}-\frac {\int \sin (k x)dx}{k}\right )}{k}-\frac {x^2 \cos (k x)}{k}\right )+\frac {2 \left (\cos (k x)-\frac {1}{3} \cos ^3(k x)\right )}{9 k^3}+\frac {2 x \sin ^3(k x)}{9 k^2}-\frac {x^2 \sin ^2(k x) \cos (k x)}{3 k}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} \left (\frac {2 \left (\frac {x \sin (k x)}{k}-\frac {\int \sin (k x)dx}{k}\right )}{k}-\frac {x^2 \cos (k x)}{k}\right )+\frac {2 \left (\cos (k x)-\frac {1}{3} \cos ^3(k x)\right )}{9 k^3}+\frac {2 x \sin ^3(k x)}{9 k^2}-\frac {x^2 \sin ^2(k x) \cos (k x)}{3 k}\) |
\(\Big \downarrow \) 3118 |
\(\displaystyle \frac {2 \left (\cos (k x)-\frac {1}{3} \cos ^3(k x)\right )}{9 k^3}+\frac {2}{3} \left (\frac {2 \left (\frac {\cos (k x)}{k^2}+\frac {x \sin (k x)}{k}\right )}{k}-\frac {x^2 \cos (k x)}{k}\right )+\frac {2 x \sin ^3(k x)}{9 k^2}-\frac {x^2 \sin ^2(k x) \cos (k x)}{3 k}\) |
(2*(Cos[k*x] - Cos[k*x]^3/3))/(9*k^3) - (x^2*Cos[k*x]*Sin[k*x]^2)/(3*k) + (2*x*Sin[k*x]^3)/(9*k^2) + (2*(-((x^2*Cos[k*x])/k) + (2*(Cos[k*x]/k^2 + (x *Sin[k*x])/k))/k))/3
3.2.44.3.1 Defintions of rubi rules used
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo l] :> Simp[d*m*(c + d*x)^(m - 1)*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Sim p[b*(c + d*x)^m*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^ 2*((n - 1)/n) Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[d^2 *m*((m - 1)/(f^2*n^2)) Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]
Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.72
method | result | size |
risch | \(-\frac {3 \left (x^{2} k^{2}-2\right ) \cos \left (k x \right )}{4 k^{3}}+\frac {3 x \sin \left (k x \right )}{2 k^{2}}+\frac {\left (9 x^{2} k^{2}-2\right ) \cos \left (3 k x \right )}{108 k^{3}}-\frac {x \sin \left (3 k x \right )}{18 k^{2}}\) | \(61\) |
derivativedivides | \(\frac {-\frac {k^{2} x^{2} \left (2+\sin ^{2}\left (k x \right )\right ) \cos \left (k x \right )}{3}+\frac {4 \cos \left (k x \right )}{3}+\frac {4 k x \sin \left (k x \right )}{3}+\frac {2 k x \left (\sin ^{3}\left (k x \right )\right )}{9}+\frac {2 \left (2+\sin ^{2}\left (k x \right )\right ) \cos \left (k x \right )}{27}}{k^{3}}\) | \(64\) |
default | \(\frac {-\frac {k^{2} x^{2} \left (2+\sin ^{2}\left (k x \right )\right ) \cos \left (k x \right )}{3}+\frac {4 \cos \left (k x \right )}{3}+\frac {4 k x \sin \left (k x \right )}{3}+\frac {2 k x \left (\sin ^{3}\left (k x \right )\right )}{9}+\frac {2 \left (2+\sin ^{2}\left (k x \right )\right ) \cos \left (k x \right )}{27}}{k^{3}}\) | \(64\) |
norman | \(\frac {-\frac {2 x^{2}}{3 k}+\frac {80}{27 k^{3}}+\frac {8 x \tan \left (\frac {k x}{2}\right )}{3 k^{2}}+\frac {64 x \left (\tan ^{3}\left (\frac {k x}{2}\right )\right )}{9 k^{2}}+\frac {8 x \left (\tan ^{5}\left (\frac {k x}{2}\right )\right )}{3 k^{2}}-\frac {2 x^{2} \left (\tan ^{2}\left (\frac {k x}{2}\right )\right )}{k}+\frac {2 x^{2} \left (\tan ^{4}\left (\frac {k x}{2}\right )\right )}{k}+\frac {2 x^{2} \left (\tan ^{6}\left (\frac {k x}{2}\right )\right )}{3 k}+\frac {8 \left (\tan ^{4}\left (\frac {k x}{2}\right )\right )}{3 k^{3}}+\frac {56 \left (\tan ^{2}\left (\frac {k x}{2}\right )\right )}{9 k^{3}}}{\left (1+\tan ^{2}\left (\frac {k x}{2}\right )\right )^{3}}\) | \(133\) |
-3/4*(k^2*x^2-2)/k^3*cos(k*x)+3/2*x*sin(k*x)/k^2+1/108*(9*k^2*x^2-2)/k^3*c os(3*k*x)-1/18*x/k^2*sin(3*k*x)
Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int x^2 \sin ^3(k x) \, dx=\frac {{\left (9 \, k^{2} x^{2} - 2\right )} \cos \left (k x\right )^{3} - 3 \, {\left (9 \, k^{2} x^{2} - 14\right )} \cos \left (k x\right ) - 6 \, {\left (k x \cos \left (k x\right )^{2} - 7 \, k x\right )} \sin \left (k x\right )}{27 \, k^{3}} \]
1/27*((9*k^2*x^2 - 2)*cos(k*x)^3 - 3*(9*k^2*x^2 - 14)*cos(k*x) - 6*(k*x*co s(k*x)^2 - 7*k*x)*sin(k*x))/k^3
Time = 0.34 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.18 \[ \int x^2 \sin ^3(k x) \, dx=\begin {cases} - \frac {x^{2} \sin ^{2}{\left (k x \right )} \cos {\left (k x \right )}}{k} - \frac {2 x^{2} \cos ^{3}{\left (k x \right )}}{3 k} + \frac {14 x \sin ^{3}{\left (k x \right )}}{9 k^{2}} + \frac {4 x \sin {\left (k x \right )} \cos ^{2}{\left (k x \right )}}{3 k^{2}} + \frac {14 \sin ^{2}{\left (k x \right )} \cos {\left (k x \right )}}{9 k^{3}} + \frac {40 \cos ^{3}{\left (k x \right )}}{27 k^{3}} & \text {for}\: k \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((-x**2*sin(k*x)**2*cos(k*x)/k - 2*x**2*cos(k*x)**3/(3*k) + 14*x* sin(k*x)**3/(9*k**2) + 4*x*sin(k*x)*cos(k*x)**2/(3*k**2) + 14*sin(k*x)**2* cos(k*x)/(9*k**3) + 40*cos(k*x)**3/(27*k**3), Ne(k, 0)), (0, True))
Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.65 \[ \int x^2 \sin ^3(k x) \, dx=-\frac {6 \, k x \sin \left (3 \, k x\right ) - 162 \, k x \sin \left (k x\right ) - {\left (9 \, k^{2} x^{2} - 2\right )} \cos \left (3 \, k x\right ) + 81 \, {\left (k^{2} x^{2} - 2\right )} \cos \left (k x\right )}{108 \, k^{3}} \]
-1/108*(6*k*x*sin(3*k*x) - 162*k*x*sin(k*x) - (9*k^2*x^2 - 2)*cos(3*k*x) + 81*(k^2*x^2 - 2)*cos(k*x))/k^3
Time = 0.31 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int x^2 \sin ^3(k x) \, dx=-\frac {x \sin \left (3 \, k x\right )}{18 \, k^{2}} + \frac {3 \, x \sin \left (k x\right )}{2 \, k^{2}} + \frac {{\left (9 \, k^{2} x^{2} - 2\right )} \cos \left (3 \, k x\right )}{108 \, k^{3}} - \frac {3 \, {\left (k^{2} x^{2} - 2\right )} \cos \left (k x\right )}{4 \, k^{3}} \]
-1/18*x*sin(3*k*x)/k^2 + 3/2*x*sin(k*x)/k^2 + 1/108*(9*k^2*x^2 - 2)*cos(3* k*x)/k^3 - 3/4*(k^2*x^2 - 2)*cos(k*x)/k^3
Time = 0.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.79 \[ \int x^2 \sin ^3(k x) \, dx=\frac {\frac {14\,\cos \left (k\,x\right )}{9}-\frac {2\,{\cos \left (k\,x\right )}^3}{27}+k\,\left (\frac {14\,x\,\sin \left (k\,x\right )}{9}-\frac {2\,x\,{\cos \left (k\,x\right )}^2\,\sin \left (k\,x\right )}{9}\right )+k^2\,\left (\frac {x^2\,{\cos \left (k\,x\right )}^3}{3}-x^2\,\cos \left (k\,x\right )\right )}{k^3} \]