Integrand size = 17, antiderivative size = 160 \[ \int x^2 \sin ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {2 b^3 n^3 x^3 \cos \left (a+b \log \left (c x^n\right )\right )}{3 \left (9+10 b^2 n^2+b^4 n^4\right )}+\frac {2 b^2 n^2 x^3 \sin \left (a+b \log \left (c x^n\right )\right )}{9+10 b^2 n^2+b^4 n^4}-\frac {b n x^3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^2\left (a+b \log \left (c x^n\right )\right )}{3 \left (1+b^2 n^2\right )}+\frac {x^3 \sin ^3\left (a+b \log \left (c x^n\right )\right )}{3 \left (1+b^2 n^2\right )} \]
-2/3*b^3*n^3*x^3*cos(a+b*ln(c*x^n))/(b^4*n^4+10*b^2*n^2+9)+2*b^2*n^2*x^3*s in(a+b*ln(c*x^n))/(b^4*n^4+10*b^2*n^2+9)-1/3*b*n*x^3*cos(a+b*ln(c*x^n))*si n(a+b*ln(c*x^n))^2/(b^2*n^2+1)+1/3*x^3*sin(a+b*ln(c*x^n))^3/(b^2*n^2+1)
Time = 0.41 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.76 \[ \int x^2 \sin ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^3 \left (-9 b n \left (1+b^2 n^2\right ) \cos \left (a+b \log \left (c x^n\right )\right )+b n \left (9+b^2 n^2\right ) \cos \left (3 \left (a+b \log \left (c x^n\right )\right )\right )-2 \left (-9-13 b^2 n^2+\left (9+b^2 n^2\right ) \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )\right )}{12 \left (9+10 b^2 n^2+b^4 n^4\right )} \]
(x^3*(-9*b*n*(1 + b^2*n^2)*Cos[a + b*Log[c*x^n]] + b*n*(9 + b^2*n^2)*Cos[3 *(a + b*Log[c*x^n])] - 2*(-9 - 13*b^2*n^2 + (9 + b^2*n^2)*Cos[2*(a + b*Log [c*x^n])])*Sin[a + b*Log[c*x^n]]))/(12*(9 + 10*b^2*n^2 + b^4*n^4))
Time = 0.32 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4990, 4988}
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.
\(\displaystyle \int x^2 \sin ^3\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 4990 |
\(\displaystyle \frac {2 b^2 n^2 \int x^2 \sin \left (a+b \log \left (c x^n\right )\right )dx}{3 \left (b^2 n^2+1\right )}+\frac {x^3 \sin ^3\left (a+b \log \left (c x^n\right )\right )}{3 \left (b^2 n^2+1\right )}-\frac {b n x^3 \sin ^2\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{3 \left (b^2 n^2+1\right )}\) |
\(\Big \downarrow \) 4988 |
\(\displaystyle \frac {x^3 \sin ^3\left (a+b \log \left (c x^n\right )\right )}{3 \left (b^2 n^2+1\right )}-\frac {b n x^3 \sin ^2\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{3 \left (b^2 n^2+1\right )}+\frac {2 b^2 n^2 \left (\frac {3 x^3 \sin \left (a+b \log \left (c x^n\right )\right )}{b^2 n^2+9}-\frac {b n x^3 \cos \left (a+b \log \left (c x^n\right )\right )}{b^2 n^2+9}\right )}{3 \left (b^2 n^2+1\right )}\) |
-1/3*(b*n*x^3*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]]^2)/(1 + b^2*n^2) + (x^3*Sin[a + b*Log[c*x^n]]^3)/(3*(1 + b^2*n^2)) + (2*b^2*n^2*(-((b*n*x^ 3*Cos[a + b*Log[c*x^n]])/(9 + b^2*n^2)) + (3*x^3*Sin[a + b*Log[c*x^n]])/(9 + b^2*n^2)))/(3*(1 + b^2*n^2))
3.1.13.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)], x_ Symbol] :> Simp[(m + 1)*(e*x)^(m + 1)*(Sin[d*(a + b*Log[c*x^n])]/(b^2*d^2*e *n^2 + e*(m + 1)^2)), x] - Simp[b*d*n*(e*x)^(m + 1)*(Cos[d*(a + b*Log[c*x^n ])]/(b^2*d^2*e*n^2 + e*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, e, m, n}, x] & & NeQ[b^2*d^2*n^2 + (m + 1)^2, 0]
Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_ ), x_Symbol] :> Simp[(m + 1)*(e*x)^(m + 1)*(Sin[d*(a + b*Log[c*x^n])]^p/(b^ 2*d^2*e*n^2*p^2 + e*(m + 1)^2)), x] + (-Simp[b*d*n*p*(e*x)^(m + 1)*Cos[d*(a + b*Log[c*x^n])]*(Sin[d*(a + b*Log[c*x^n])]^(p - 1)/(b^2*d^2*e*n^2*p^2 + e *(m + 1)^2)), x] + Simp[b^2*d^2*n^2*p*((p - 1)/(b^2*d^2*n^2*p^2 + (m + 1)^2 )) Int[(e*x)^m*Sin[d*(a + b*Log[c*x^n])]^(p - 2), x], x]) /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 1] && NeQ[b^2*d^2*n^2*p^2 + (m + 1)^2, 0]
\[\int x^{2} {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}^{3}d x\]
Time = 0.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.86 \[ \int x^2 \sin ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {{\left (b^{3} n^{3} + 9 \, b n\right )} x^{3} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - 3 \, {\left (b^{3} n^{3} + 3 \, b n\right )} x^{3} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - {\left ({\left (b^{2} n^{2} + 9\right )} x^{3} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - {\left (7 \, b^{2} n^{2} + 9\right )} x^{3}\right )} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{3 \, {\left (b^{4} n^{4} + 10 \, b^{2} n^{2} + 9\right )}} \]
1/3*((b^3*n^3 + 9*b*n)*x^3*cos(b*n*log(x) + b*log(c) + a)^3 - 3*(b^3*n^3 + 3*b*n)*x^3*cos(b*n*log(x) + b*log(c) + a) - ((b^2*n^2 + 9)*x^3*cos(b*n*lo g(x) + b*log(c) + a)^2 - (7*b^2*n^2 + 9)*x^3)*sin(b*n*log(x) + b*log(c) + a))/(b^4*n^4 + 10*b^2*n^2 + 9)
Timed out. \[ \int x^2 \sin ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1008 vs. \(2 (154) = 308\).
Time = 0.25 (sec) , antiderivative size = 1008, normalized size of antiderivative = 6.30 \[ \int x^2 \sin ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]
1/24*(((b^3*cos(6*b*log(c))*cos(3*b*log(c)) + b^3*sin(6*b*log(c))*sin(3*b* log(c)) + b^3*cos(3*b*log(c)))*n^3 - (b^2*cos(3*b*log(c))*sin(6*b*log(c)) - b^2*cos(6*b*log(c))*sin(3*b*log(c)) + b^2*sin(3*b*log(c)))*n^2 + 9*(b*co s(6*b*log(c))*cos(3*b*log(c)) + b*sin(6*b*log(c))*sin(3*b*log(c)) + b*cos( 3*b*log(c)))*n - 9*cos(3*b*log(c))*sin(6*b*log(c)) + 9*cos(6*b*log(c))*sin (3*b*log(c)) - 9*sin(3*b*log(c)))*x^3*cos(3*b*log(x^n) + 3*a) - 9*((b^3*co s(4*b*log(c))*cos(3*b*log(c)) + b^3*cos(3*b*log(c))*cos(2*b*log(c)) + b^3* sin(4*b*log(c))*sin(3*b*log(c)) + b^3*sin(3*b*log(c))*sin(2*b*log(c)))*n^3 - 3*(b^2*cos(3*b*log(c))*sin(4*b*log(c)) - b^2*cos(4*b*log(c))*sin(3*b*lo g(c)) + b^2*cos(2*b*log(c))*sin(3*b*log(c)) - b^2*cos(3*b*log(c))*sin(2*b* log(c)))*n^2 + (b*cos(4*b*log(c))*cos(3*b*log(c)) + b*cos(3*b*log(c))*cos( 2*b*log(c)) + b*sin(4*b*log(c))*sin(3*b*log(c)) + b*sin(3*b*log(c))*sin(2* b*log(c)))*n - 3*cos(3*b*log(c))*sin(4*b*log(c)) + 3*cos(4*b*log(c))*sin(3 *b*log(c)) - 3*cos(2*b*log(c))*sin(3*b*log(c)) + 3*cos(3*b*log(c))*sin(2*b *log(c)))*x^3*cos(b*log(x^n) + a) - ((b^3*cos(3*b*log(c))*sin(6*b*log(c)) - b^3*cos(6*b*log(c))*sin(3*b*log(c)) + b^3*sin(3*b*log(c)))*n^3 + (b^2*co s(6*b*log(c))*cos(3*b*log(c)) + b^2*sin(6*b*log(c))*sin(3*b*log(c)) + b^2* cos(3*b*log(c)))*n^2 + 9*(b*cos(3*b*log(c))*sin(6*b*log(c)) - b*cos(6*b*lo g(c))*sin(3*b*log(c)) + b*sin(3*b*log(c)))*n + 9*cos(6*b*log(c))*cos(3*b*l og(c)) + 9*sin(6*b*log(c))*sin(3*b*log(c)) + 9*cos(3*b*log(c)))*x^3*sin...
Leaf count of result is larger than twice the leaf count of optimal. 18085 vs. \(2 (154) = 308\).
Time = 1.40 (sec) , antiderivative size = 18085, normalized size of antiderivative = 113.03 \[ \int x^2 \sin ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]
1/24*(b^3*n^3*x^3*e^(3/2*pi*b*n*sgn(x) - 3/2*pi*b*n + 3/2*pi*b*sgn(c) - 3/ 2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs (x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a)^2*tan(1/2*a)^2 - 9*b^3*n^3*x^3*e^(1 /2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(3/2*b*n*lo g(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c )))^2*tan(3/2*a)^2*tan(1/2*a)^2 - 9*b^3*n^3*x^3*e^(-1/2*pi*b*n*sgn(x) + 1/ 2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log (abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a)^2*ta n(1/2*a)^2 + b^3*n^3*x^3*e^(-3/2*pi*b*n*sgn(x) + 3/2*pi*b*n - 3/2*pi*b*sgn (c) + 3/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n *log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a)^2*tan(1/2*a)^2 + b^3*n^3*x^ 3*e^(3/2*pi*b*n*sgn(x) - 3/2*pi*b*n + 3/2*pi*b*sgn(c) - 3/2*pi*b)*tan(3/2* b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log (abs(c)))^2*tan(3/2*a)^2 + 9*b^3*n^3*x^3*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)) )^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a)^2 + 9*b^3*n^ 3*x^3*e^(-1/2*pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan (3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2* b*log(abs(c)))^2*tan(3/2*a)^2 + b^3*n^3*x^3*e^(-3/2*pi*b*n*sgn(x) + 3/2*pi *b*n - 3/2*pi*b*sgn(c) + 3/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(...
Time = 28.06 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.76 \[ \int x^2 \sin ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {x^3\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}}\,3{}\mathrm {i}}{-24+b\,n\,8{}\mathrm {i}}-\frac {3\,x^3\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}}{8\,b\,n-24{}\mathrm {i}}+\frac {x^3\,{\mathrm {e}}^{-a\,3{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,3{}\mathrm {i}}}\,1{}\mathrm {i}}{-24+b\,n\,24{}\mathrm {i}}+\frac {x^3\,{\mathrm {e}}^{a\,3{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,3{}\mathrm {i}}}{24\,b\,n-24{}\mathrm {i}} \]