Integrand size = 17, antiderivative size = 202 \[ \int x^2 \sin ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {8 b^4 n^4 x^3}{81+180 b^2 n^2+64 b^4 n^4}-\frac {24 b^3 n^3 x^3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{81+180 b^2 n^2+64 b^4 n^4}+\frac {36 b^2 n^2 x^3 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{81+180 b^2 n^2+64 b^4 n^4}-\frac {4 b n x^3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{9+16 b^2 n^2}+\frac {3 x^3 \sin ^4\left (a+b \log \left (c x^n\right )\right )}{9+16 b^2 n^2} \]
8*b^4*n^4*x^3/(64*b^4*n^4+180*b^2*n^2+81)-24*b^3*n^3*x^3*cos(a+b*ln(c*x^n) )*sin(a+b*ln(c*x^n))/(64*b^4*n^4+180*b^2*n^2+81)+36*b^2*n^2*x^3*sin(a+b*ln (c*x^n))^2/(64*b^4*n^4+180*b^2*n^2+81)-4*b*n*x^3*cos(a+b*ln(c*x^n))*sin(a+ b*ln(c*x^n))^3/(16*b^2*n^2+9)+3*x^3*sin(a+b*ln(c*x^n))^4/(16*b^2*n^2+9)
Time = 0.38 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.85 \[ \int x^2 \sin ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^3 \left (81+180 b^2 n^2+64 b^4 n^4-12 \left (9+16 b^2 n^2\right ) \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+3 \left (9+4 b^2 n^2\right ) \cos \left (4 \left (a+b \log \left (c x^n\right )\right )\right )-72 b n \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-128 b^3 n^3 \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+36 b n \sin \left (4 \left (a+b \log \left (c x^n\right )\right )\right )+16 b^3 n^3 \sin \left (4 \left (a+b \log \left (c x^n\right )\right )\right )\right )}{8 \left (81+180 b^2 n^2+64 b^4 n^4\right )} \]
(x^3*(81 + 180*b^2*n^2 + 64*b^4*n^4 - 12*(9 + 16*b^2*n^2)*Cos[2*(a + b*Log [c*x^n])] + 3*(9 + 4*b^2*n^2)*Cos[4*(a + b*Log[c*x^n])] - 72*b*n*Sin[2*(a + b*Log[c*x^n])] - 128*b^3*n^3*Sin[2*(a + b*Log[c*x^n])] + 36*b*n*Sin[4*(a + b*Log[c*x^n])] + 16*b^3*n^3*Sin[4*(a + b*Log[c*x^n])]))/(8*(81 + 180*b^ 2*n^2 + 64*b^4*n^4))
Time = 0.36 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4990, 4990, 15}
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.
\(\displaystyle \int x^2 \sin ^4\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 4990 |
\(\displaystyle \frac {12 b^2 n^2 \int x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )dx}{16 b^2 n^2+9}+\frac {3 x^3 \sin ^4\left (a+b \log \left (c x^n\right )\right )}{16 b^2 n^2+9}-\frac {4 b n x^3 \sin ^3\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{16 b^2 n^2+9}\) |
\(\Big \downarrow \) 4990 |
\(\displaystyle \frac {12 b^2 n^2 \left (\frac {2 b^2 n^2 \int x^2dx}{4 b^2 n^2+9}+\frac {3 x^3 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+9}-\frac {2 b n x^3 \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+9}\right )}{16 b^2 n^2+9}+\frac {3 x^3 \sin ^4\left (a+b \log \left (c x^n\right )\right )}{16 b^2 n^2+9}-\frac {4 b n x^3 \sin ^3\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{16 b^2 n^2+9}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {3 x^3 \sin ^4\left (a+b \log \left (c x^n\right )\right )}{16 b^2 n^2+9}-\frac {4 b n x^3 \sin ^3\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{16 b^2 n^2+9}+\frac {12 b^2 n^2 \left (\frac {3 x^3 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+9}-\frac {2 b n x^3 \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+9}+\frac {2 b^2 n^2 x^3}{3 \left (4 b^2 n^2+9\right )}\right )}{16 b^2 n^2+9}\) |
(-4*b*n*x^3*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]]^3)/(9 + 16*b^2*n^2 ) + (3*x^3*Sin[a + b*Log[c*x^n]]^4)/(9 + 16*b^2*n^2) + (12*b^2*n^2*((2*b^2 *n^2*x^3)/(3*(9 + 4*b^2*n^2)) - (2*b*n*x^3*Cos[a + b*Log[c*x^n]]*Sin[a + b *Log[c*x^n]])/(9 + 4*b^2*n^2) + (3*x^3*Sin[a + b*Log[c*x^n]]^2)/(9 + 4*b^2 *n^2)))/(9 + 16*b^2*n^2)
3.1.19.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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]
\[\int x^{2} {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}^{4}d x\]
Time = 0.26 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.88 \[ \int x^2 \sin ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {3 \, {\left (4 \, b^{2} n^{2} + 9\right )} x^{3} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{4} - 6 \, {\left (10 \, b^{2} n^{2} + 9\right )} x^{3} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + {\left (8 \, b^{4} n^{4} + 48 \, b^{2} n^{2} + 27\right )} x^{3} + 4 \, {\left ({\left (4 \, b^{3} n^{3} + 9 \, b n\right )} x^{3} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - {\left (10 \, b^{3} n^{3} + 9 \, b n\right )} x^{3} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{64 \, b^{4} n^{4} + 180 \, b^{2} n^{2} + 81} \]
(3*(4*b^2*n^2 + 9)*x^3*cos(b*n*log(x) + b*log(c) + a)^4 - 6*(10*b^2*n^2 + 9)*x^3*cos(b*n*log(x) + b*log(c) + a)^2 + (8*b^4*n^4 + 48*b^2*n^2 + 27)*x^ 3 + 4*((4*b^3*n^3 + 9*b*n)*x^3*cos(b*n*log(x) + b*log(c) + a)^3 - (10*b^3* n^3 + 9*b*n)*x^3*cos(b*n*log(x) + b*log(c) + a))*sin(b*n*log(x) + b*log(c) + a))/(64*b^4*n^4 + 180*b^2*n^2 + 81)
Timed out. \[ \int x^2 \sin ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1107 vs. \(2 (202) = 404\).
Time = 0.25 (sec) , antiderivative size = 1107, normalized size of antiderivative = 5.48 \[ \int x^2 \sin ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]
1/16*((16*(b^3*cos(4*b*log(c))*sin(8*b*log(c)) - b^3*cos(8*b*log(c))*sin(4 *b*log(c)) + b^3*sin(4*b*log(c)))*n^3 + 12*(b^2*cos(8*b*log(c))*cos(4*b*lo g(c)) + b^2*sin(8*b*log(c))*sin(4*b*log(c)) + b^2*cos(4*b*log(c)))*n^2 + 3 6*(b*cos(4*b*log(c))*sin(8*b*log(c)) - b*cos(8*b*log(c))*sin(4*b*log(c)) + b*sin(4*b*log(c)))*n + 27*cos(8*b*log(c))*cos(4*b*log(c)) + 27*sin(8*b*lo g(c))*sin(4*b*log(c)) + 27*cos(4*b*log(c)))*x^3*cos(4*b*log(x^n) + 4*a) - 4*(32*(b^3*cos(4*b*log(c))*sin(6*b*log(c)) - b^3*cos(6*b*log(c))*sin(4*b*l og(c)) + b^3*cos(2*b*log(c))*sin(4*b*log(c)) - b^3*cos(4*b*log(c))*sin(2*b *log(c)))*n^3 + 48*(b^2*cos(6*b*log(c))*cos(4*b*log(c)) + b^2*cos(4*b*log( c))*cos(2*b*log(c)) + b^2*sin(6*b*log(c))*sin(4*b*log(c)) + b^2*sin(4*b*lo g(c))*sin(2*b*log(c)))*n^2 + 18*(b*cos(4*b*log(c))*sin(6*b*log(c)) - b*cos (6*b*log(c))*sin(4*b*log(c)) + b*cos(2*b*log(c))*sin(4*b*log(c)) - b*cos(4 *b*log(c))*sin(2*b*log(c)))*n + 27*cos(6*b*log(c))*cos(4*b*log(c)) + 27*co s(4*b*log(c))*cos(2*b*log(c)) + 27*sin(6*b*log(c))*sin(4*b*log(c)) + 27*si n(4*b*log(c))*sin(2*b*log(c)))*x^3*cos(2*b*log(x^n) + 2*a) + (16*(b^3*cos( 8*b*log(c))*cos(4*b*log(c)) + b^3*sin(8*b*log(c))*sin(4*b*log(c)) + b^3*co s(4*b*log(c)))*n^3 - 12*(b^2*cos(4*b*log(c))*sin(8*b*log(c)) - b^2*cos(8*b *log(c))*sin(4*b*log(c)) + b^2*sin(4*b*log(c)))*n^2 + 36*(b*cos(8*b*log(c) )*cos(4*b*log(c)) + b*sin(8*b*log(c))*sin(4*b*log(c)) + b*cos(4*b*log(c))) *n - 27*cos(4*b*log(c))*sin(8*b*log(c)) + 27*cos(8*b*log(c))*sin(4*b*lo...
Leaf count of result is larger than twice the leaf count of optimal. 17035 vs. \(2 (202) = 404\).
Time = 1.30 (sec) , antiderivative size = 17035, normalized size of antiderivative = 84.33 \[ \int x^2 \sin ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]
1/8*x^3 + 1/16*(256*b^3*n^3*x^3*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*l og(abs(c)))^2*tan(2*a)^2*tan(a) + 256*b^3*n^3*x^3*e^(-pi*b*n*sgn(x) + pi*b *n - pi*b*sgn(c) + pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b* n*log(abs(x)) + b*log(abs(c)))^2*tan(2*a)^2*tan(a) - 32*b^3*n^3*x^3*e^(2*p i*b*n*sgn(x) - 2*pi*b*n + 2*pi*b*sgn(c) - 2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(2*a)*tan(a)^ 2 - 32*b^3*n^3*x^3*e^(-2*pi*b*n*sgn(x) + 2*pi*b*n - 2*pi*b*sgn(c) + 2*pi*b )*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(a bs(c)))^2*tan(2*a)*tan(a)^2 + 256*b^3*n^3*x^3*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log (abs(x)) + b*log(abs(c)))*tan(2*a)^2*tan(a)^2 + 256*b^3*n^3*x^3*e^(-pi*b*n *sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs (c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))*tan(2*a)^2*tan(a)^2 - 32*b^3* n^3*x^3*e^(2*pi*b*n*sgn(x) - 2*pi*b*n + 2*pi*b*sgn(c) - 2*pi*b)*tan(2*b*n* log(abs(x)) + 2*b*log(abs(c)))*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan( 2*a)^2*tan(a)^2 - 32*b^3*n^3*x^3*e^(-2*pi*b*n*sgn(x) + 2*pi*b*n - 2*pi*b*s gn(c) + 2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))*tan(b*n*log(abs(x )) + b*log(abs(c)))^2*tan(2*a)^2*tan(a)^2 + 12*b^2*n^2*x^3*e^(2*pi*b*n*sgn (x) - 2*pi*b*n + 2*pi*b*sgn(c) - 2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*lo...
Time = 27.24 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.63 \[ \int x^2 \sin ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^3}{8}-\frac {x^3\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}}\,1{}\mathrm {i}}{8\,b\,n+12{}\mathrm {i}}-\frac {x^3\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}}{12+b\,n\,8{}\mathrm {i}}+\frac {x^3\,{\mathrm {e}}^{-a\,4{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,4{}\mathrm {i}}}\,1{}\mathrm {i}}{64\,b\,n+48{}\mathrm {i}}+\frac {x^3\,{\mathrm {e}}^{a\,4{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,4{}\mathrm {i}}}{48+b\,n\,64{}\mathrm {i}} \]