Integrand size = 17, antiderivative size = 210 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {3 b^4 n^4}{4 \left (1+5 b^2 n^2+4 b^4 n^4\right ) x^2}-\frac {3 b^3 n^3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right ) x^2}-\frac {3 b^2 n^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right ) x^2}-\frac {b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{\left (1+4 b^2 n^2\right ) x^2}-\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+4 b^2 n^2\right ) x^2} \]
-3/4*b^4*n^4/(4*b^4*n^4+5*b^2*n^2+1)/x^2-3/2*b^3*n^3*cos(a+b*ln(c*x^n))*si n(a+b*ln(c*x^n))/(4*b^4*n^4+5*b^2*n^2+1)/x^2-3/2*b^2*n^2*sin(a+b*ln(c*x^n) )^2/(4*b^4*n^4+5*b^2*n^2+1)/x^2-b*n*cos(a+b*ln(c*x^n))*sin(a+b*ln(c*x^n))^ 3/(4*b^2*n^2+1)/x^2-1/2*sin(a+b*ln(c*x^n))^4/(4*b^2*n^2+1)/x^2
Time = 0.38 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.80 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {3+15 b^2 n^2+12 b^4 n^4-4 \left (1+4 b^2 n^2\right ) \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+\left (1+b^2 n^2\right ) \cos \left (4 \left (a+b \log \left (c x^n\right )\right )\right )+4 b n \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+16 b^3 n^3 \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-2 b n \sin \left (4 \left (a+b \log \left (c x^n\right )\right )\right )-2 b^3 n^3 \sin \left (4 \left (a+b \log \left (c x^n\right )\right )\right )}{16 \left (1+5 b^2 n^2+4 b^4 n^4\right ) x^2} \]
-1/16*(3 + 15*b^2*n^2 + 12*b^4*n^4 - 4*(1 + 4*b^2*n^2)*Cos[2*(a + b*Log[c* x^n])] + (1 + b^2*n^2)*Cos[4*(a + b*Log[c*x^n])] + 4*b*n*Sin[2*(a + b*Log[ c*x^n])] + 16*b^3*n^3*Sin[2*(a + b*Log[c*x^n])] - 2*b*n*Sin[4*(a + b*Log[c *x^n])] - 2*b^3*n^3*Sin[4*(a + b*Log[c*x^n])])/((1 + 5*b^2*n^2 + 4*b^4*n^4 )*x^2)
Time = 0.34 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.92, 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.
| \(\displaystyle \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx\) |
| \(\Big \downarrow \) 4990 |
| \(\displaystyle \frac {3 b^2 n^2 \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^3}dx}{4 b^2 n^2+1}-\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 x^2 \left (4 b^2 n^2+1\right )}-\frac {b n \sin ^3\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{x^2 \left (4 b^2 n^2+1\right )}\) |
| \(\Big \downarrow \) 4990 |
| \(\displaystyle \frac {3 b^2 n^2 \left (\frac {b^2 n^2 \int \frac {1}{x^3}dx}{2 \left (b^2 n^2+1\right )}-\frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 x^2 \left (b^2 n^2+1\right )}-\frac {b n \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{2 x^2 \left (b^2 n^2+1\right )}\right )}{4 b^2 n^2+1}-\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 x^2 \left (4 b^2 n^2+1\right )}-\frac {b n \sin ^3\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{x^2 \left (4 b^2 n^2+1\right )}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle -\frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 x^2 \left (4 b^2 n^2+1\right )}-\frac {b n \sin ^3\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{x^2 \left (4 b^2 n^2+1\right )}+\frac {3 b^2 n^2 \left (-\frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 x^2 \left (b^2 n^2+1\right )}-\frac {b n \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{2 x^2 \left (b^2 n^2+1\right )}-\frac {b^2 n^2}{4 x^2 \left (b^2 n^2+1\right )}\right )}{4 b^2 n^2+1}\) |
-((b*n*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]]^3)/((1 + 4*b^2*n^2)*x^2 )) - Sin[a + b*Log[c*x^n]]^4/(2*(1 + 4*b^2*n^2)*x^2) + (3*b^2*n^2*(-1/4*(b ^2*n^2)/((1 + b^2*n^2)*x^2) - (b*n*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x ^n]])/(2*(1 + b^2*n^2)*x^2) - Sin[a + b*Log[c*x^n]]^2/(2*(1 + b^2*n^2)*x^2 )))/(1 + 4*b^2*n^2)
3.1.24.3.1 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]
Time = 18.99 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.95
| method | result | size |
| parallelrisch | \(\frac {-12 b^{4} n^{4}-16 b^{3} n^{3} \sin \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )+2 b^{3} n^{3} \sin \left (4 b \ln \left (c \,x^{n}\right )+4 a \right )-b^{2} n^{2} \cos \left (4 b \ln \left (c \,x^{n}\right )+4 a \right )+16 b^{2} n^{2} \cos \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )-15 b^{2} n^{2}-4 b n \sin \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )+2 b n \sin \left (4 b \ln \left (c \,x^{n}\right )+4 a \right )-\cos \left (4 b \ln \left (c \,x^{n}\right )+4 a \right )+4 \cos \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )-3}{16 x^{2} \left (4 b^{4} n^{4}+5 b^{2} n^{2}+1\right )}\) | \(200\) |
1/16*(-12*b^4*n^4-16*b^3*n^3*sin(2*b*ln(c*x^n)+2*a)+2*b^3*n^3*sin(4*b*ln(c *x^n)+4*a)-b^2*n^2*cos(4*b*ln(c*x^n)+4*a)+16*b^2*n^2*cos(2*b*ln(c*x^n)+2*a )-15*b^2*n^2-4*b*n*sin(2*b*ln(c*x^n)+2*a)+2*b*n*sin(4*b*ln(c*x^n)+4*a)-cos (4*b*ln(c*x^n)+4*a)+4*cos(2*b*ln(c*x^n)+2*a)-3)/x^2/(4*b^4*n^4+5*b^2*n^2+1 )
Time = 0.26 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {3 \, b^{4} n^{4} + 2 \, {\left (b^{2} n^{2} + 1\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{4} + 8 \, b^{2} n^{2} - 2 \, {\left (5 \, b^{2} n^{2} + 2\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - 2 \, {\left (2 \, {\left (b^{3} n^{3} + b n\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - {\left (5 \, b^{3} n^{3} + 2 \, b n\right )} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + 2}{4 \, {\left (4 \, b^{4} n^{4} + 5 \, b^{2} n^{2} + 1\right )} x^{2}} \]
-1/4*(3*b^4*n^4 + 2*(b^2*n^2 + 1)*cos(b*n*log(x) + b*log(c) + a)^4 + 8*b^2 *n^2 - 2*(5*b^2*n^2 + 2)*cos(b*n*log(x) + b*log(c) + a)^2 - 2*(2*(b^3*n^3 + b*n)*cos(b*n*log(x) + b*log(c) + a)^3 - (5*b^3*n^3 + 2*b*n)*cos(b*n*log( x) + b*log(c) + a))*sin(b*n*log(x) + b*log(c) + a) + 2)/((4*b^4*n^4 + 5*b^ 2*n^2 + 1)*x^2)
Result contains complex when optimal does not.
Time = 98.44 (sec) , antiderivative size = 1066, normalized size of antiderivative = 5.08 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\text {Too large to display} \]
Piecewise((I*sin(4*a - 4*I*log(c*x**n)/n)/(24*x**2) + cos(2*a - 2*I*log(c* x**n)/n)/(8*x**2) + cos(4*a - 4*I*log(c*x**n)/n)/(48*x**2) - 3/(16*x**2) - I*log(c*x**n)*sin(2*a - 2*I*log(c*x**n)/n)/(4*n*x**2) - log(c*x**n)*cos(2 *a - 2*I*log(c*x**n)/n)/(4*n*x**2), Eq(b, -I/n)), (I*sin(2*a - I*log(c*x** n)/n)/(6*x**2) + I*sin(4*a - 2*I*log(c*x**n)/n)/(32*x**2) + cos(2*a - I*lo g(c*x**n)/n)/(3*x**2) - 3/(16*x**2) + I*log(c*x**n)*sin(4*a - 2*I*log(c*x* *n)/n)/(16*n*x**2) + log(c*x**n)*cos(4*a - 2*I*log(c*x**n)/n)/(16*n*x**2), Eq(b, -I/(2*n))), (-I*sin(2*a + I*log(c*x**n)/n)/(6*x**2) + cos(2*a + I*l og(c*x**n)/n)/(3*x**2) - cos(4*a + 2*I*log(c*x**n)/n)/(32*x**2) - 3/(16*x* *2) - I*log(c*x**n)*sin(4*a + 2*I*log(c*x**n)/n)/(16*n*x**2) + log(c*x**n) *cos(4*a + 2*I*log(c*x**n)/n)/(16*n*x**2), Eq(b, I/(2*n))), (-I*sin(4*a + 4*I*log(c*x**n)/n)/(24*x**2) + cos(2*a + 2*I*log(c*x**n)/n)/(8*x**2) + cos (4*a + 4*I*log(c*x**n)/n)/(48*x**2) - 3/(16*x**2) + I*log(c*x**n)*sin(2*a + 2*I*log(c*x**n)/n)/(4*n*x**2) - log(c*x**n)*cos(2*a + 2*I*log(c*x**n)/n) /(4*n*x**2), Eq(b, I/n)), (-3*b**4*n**4*sin(a + b*log(c*x**n))**4/(16*b**4 *n**4*x**2 + 20*b**2*n**2*x**2 + 4*x**2) - 6*b**4*n**4*sin(a + b*log(c*x** n))**2*cos(a + b*log(c*x**n))**2/(16*b**4*n**4*x**2 + 20*b**2*n**2*x**2 + 4*x**2) - 3*b**4*n**4*cos(a + b*log(c*x**n))**4/(16*b**4*n**4*x**2 + 20*b* *2*n**2*x**2 + 4*x**2) - 10*b**3*n**3*sin(a + b*log(c*x**n))**3*cos(a + b* log(c*x**n))/(16*b**4*n**4*x**2 + 20*b**2*n**2*x**2 + 4*x**2) - 6*b**3*...
Leaf count of result is larger than twice the leaf count of optimal. 1082 vs. \(2 (202) = 404\).
Time = 0.26 (sec) , antiderivative size = 1082, normalized size of antiderivative = 5.15 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\text {Too large to display} \]
-1/32*(24*(b^4*cos(4*b*log(c))^2 + b^4*sin(4*b*log(c))^2)*n^4 + 30*(b^2*co s(4*b*log(c))^2 + b^2*sin(4*b*log(c))^2)*n^2 + 6*cos(4*b*log(c))^2 - (2*(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 - (b^2*cos(8*b*log(c))*cos(4*b*log(c)) + b^2*sin (8*b*log(c))*sin(4*b*log(c)) + b^2*cos(4*b*log(c)))*n^2 + 2*(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 - cos(8*b*log(c))*cos(4*b*log(c)) - sin(8*b*log(c))*sin(4*b*log(c)) - cos(4*b*log(c)))*cos(4*b*log(x^n) + 4*a) + 4*(4*(b^3*cos(4*b*log(c))*sin (6*b*log(c)) - b^3*cos(6*b*log(c))*sin(4*b*log(c)) + b^3*cos(2*b*log(c))*s in(4*b*log(c)) - b^3*cos(4*b*log(c))*sin(2*b*log(c)))*n^3 - 4*(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*log(c))*sin(2*b*log(c)))*n^2 + (b *cos(4*b*log(c))*sin(6*b*log(c)) - b*cos(6*b*log(c))*sin(4*b*log(c)) + b*c os(2*b*log(c))*sin(4*b*log(c)) - b*cos(4*b*log(c))*sin(2*b*log(c)))*n - co s(6*b*log(c))*cos(4*b*log(c)) - cos(4*b*log(c))*cos(2*b*log(c)) - sin(6*b* log(c))*sin(4*b*log(c)) - sin(4*b*log(c))*sin(2*b*log(c)))*cos(2*b*log(x^n ) + 2*a) + 6*sin(4*b*log(c))^2 - (2*(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*cos(4*b*log(c)))*n^3 + (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*s in(4*b*log(c)))*n^2 + 2*(b*cos(8*b*log(c))*cos(4*b*log(c)) + b*sin(8*b*...
\[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {\sin \left (b \log \left (c x^{n}\right ) + a\right )^{4}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int \frac {{\sin \left (a+b\,\ln \left (c\,x^n\right )\right )}^4}{x^3} \,d x \]