Integrand size = 15, antiderivative size = 158 \[ \int x \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {12 b^2 n^2 x^2 \cos \left (a+b \log \left (c x^n\right )\right )}{16+40 b^2 n^2+9 b^4 n^4}+\frac {2 x^2 \cos ^3\left (a+b \log \left (c x^n\right )\right )}{4+9 b^2 n^2}+\frac {6 b^3 n^3 x^2 \sin \left (a+b \log \left (c x^n\right )\right )}{16+40 b^2 n^2+9 b^4 n^4}+\frac {3 b n x^2 \cos ^2\left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{4+9 b^2 n^2} \]
12*b^2*n^2*x^2*cos(a+b*ln(c*x^n))/(9*b^4*n^4+40*b^2*n^2+16)+2*x^2*cos(a+b* ln(c*x^n))^3/(9*b^2*n^2+4)+6*b^3*n^3*x^2*sin(a+b*ln(c*x^n))/(9*b^4*n^4+40* b^2*n^2+16)+3*b*n*x^2*cos(a+b*ln(c*x^n))^2*sin(a+b*ln(c*x^n))/(9*b^2*n^2+4 )
Time = 0.37 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.78 \[ \int x \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^2 \left (6 \left (4+9 b^2 n^2\right ) \cos \left (a+b \log \left (c x^n\right )\right )+2 \left (4+b^2 n^2\right ) \cos \left (3 \left (a+b \log \left (c x^n\right )\right )\right )+6 b n \left (4+5 b^2 n^2+\left (4+b^2 n^2\right ) \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )\right )}{4 \left (16+40 b^2 n^2+9 b^4 n^4\right )} \]
(x^2*(6*(4 + 9*b^2*n^2)*Cos[a + b*Log[c*x^n]] + 2*(4 + b^2*n^2)*Cos[3*(a + b*Log[c*x^n])] + 6*b*n*(4 + 5*b^2*n^2 + (4 + b^2*n^2)*Cos[2*(a + b*Log[c* x^n])])*Sin[a + b*Log[c*x^n]]))/(4*(16 + 40*b^2*n^2 + 9*b^4*n^4))
Time = 0.32 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4991, 4989}
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.
| \(\displaystyle \int x \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 4991 |
| \(\displaystyle \frac {6 b^2 n^2 \int x \cos \left (a+b \log \left (c x^n\right )\right )dx}{9 b^2 n^2+4}+\frac {2 x^2 \cos ^3\left (a+b \log \left (c x^n\right )\right )}{9 b^2 n^2+4}+\frac {3 b n x^2 \sin \left (a+b \log \left (c x^n\right )\right ) \cos ^2\left (a+b \log \left (c x^n\right )\right )}{9 b^2 n^2+4}\) |
| \(\Big \downarrow \) 4989 |
| \(\displaystyle \frac {2 x^2 \cos ^3\left (a+b \log \left (c x^n\right )\right )}{9 b^2 n^2+4}+\frac {3 b n x^2 \sin \left (a+b \log \left (c x^n\right )\right ) \cos ^2\left (a+b \log \left (c x^n\right )\right )}{9 b^2 n^2+4}+\frac {6 b^2 n^2 \left (\frac {b n x^2 \sin \left (a+b \log \left (c x^n\right )\right )}{b^2 n^2+4}+\frac {2 x^2 \cos \left (a+b \log \left (c x^n\right )\right )}{b^2 n^2+4}\right )}{9 b^2 n^2+4}\) |
(2*x^2*Cos[a + b*Log[c*x^n]]^3)/(4 + 9*b^2*n^2) + (3*b*n*x^2*Cos[a + b*Log [c*x^n]]^2*Sin[a + b*Log[c*x^n]])/(4 + 9*b^2*n^2) + (6*b^2*n^2*((2*x^2*Cos [a + b*Log[c*x^n]])/(4 + b^2*n^2) + (b*n*x^2*Sin[a + b*Log[c*x^n]])/(4 + b ^2*n^2)))/(4 + 9*b^2*n^2)
3.1.97.3.1 Defintions of rubi rules used
Int[Cos[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]*((e_.)*(x_))^(m_.), x_ Symbol] :> Simp[(m + 1)*(e*x)^(m + 1)*(Cos[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)*(Sin[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[Cos[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_)*((e_.)*(x_))^(m_. ), x_Symbol] :> Simp[(m + 1)*(e*x)^(m + 1)*(Cos[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)*Sin[d*(a + b*Log[c*x^n])]*(Cos[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*Cos[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 = 4.09 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.74
| method | result | size |
| parallelrisch | \(\frac {27 x^{2} \left (\frac {2 \left (b^{2} n^{2}+4\right ) \cos \left (3 b \ln \left (c \,x^{n}\right )+3 a \right )}{27}+\frac {b n \left (b^{2} n^{2}+4\right ) \sin \left (3 b \ln \left (c \,x^{n}\right )+3 a \right )}{9}+\left (\sin \left (a +b \ln \left (c \,x^{n}\right )\right ) b n +2 \cos \left (a +b \ln \left (c \,x^{n}\right )\right )\right ) \left (b^{2} n^{2}+\frac {4}{9}\right )\right )}{4 \left (9 b^{4} n^{4}+40 b^{2} n^{2}+16\right )}\) | \(117\) |
27/4*x^2*(2/27*(b^2*n^2+4)*cos(3*b*ln(c*x^n)+3*a)+1/9*b*n*(b^2*n^2+4)*sin( 3*b*ln(c*x^n)+3*a)+(sin(a+b*ln(c*x^n))*b*n+2*cos(a+b*ln(c*x^n)))*(b^2*n^2+ 4/9))/(9*b^4*n^4+40*b^2*n^2+16)
Time = 0.25 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.82 \[ \int x \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {12 \, b^{2} n^{2} x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + 2 \, {\left (b^{2} n^{2} + 4\right )} x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} + 3 \, {\left (2 \, b^{3} n^{3} x^{2} + {\left (b^{3} n^{3} + 4 \, b n\right )} x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2}\right )} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{9 \, b^{4} n^{4} + 40 \, b^{2} n^{2} + 16} \]
(12*b^2*n^2*x^2*cos(b*n*log(x) + b*log(c) + a) + 2*(b^2*n^2 + 4)*x^2*cos(b *n*log(x) + b*log(c) + a)^3 + 3*(2*b^3*n^3*x^2 + (b^3*n^3 + 4*b*n)*x^2*cos (b*n*log(x) + b*log(c) + a)^2)*sin(b*n*log(x) + b*log(c) + a))/(9*b^4*n^4 + 40*b^2*n^2 + 16)
\[ \int x \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\begin {cases} \int x \cos ^{3}{\left (a - \frac {2 i \log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = - \frac {2 i}{n} \\\int x \cos ^{3}{\left (a - \frac {2 i \log {\left (c x^{n} \right )}}{3 n} \right )}\, dx & \text {for}\: b = - \frac {2 i}{3 n} \\\int x \cos ^{3}{\left (a + \frac {2 i \log {\left (c x^{n} \right )}}{3 n} \right )}\, dx & \text {for}\: b = \frac {2 i}{3 n} \\\int x \cos ^{3}{\left (a + \frac {2 i \log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = \frac {2 i}{n} \\\frac {6 b^{3} n^{3} x^{2} \sin ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} + 40 b^{2} n^{2} + 16} + \frac {9 b^{3} n^{3} x^{2} \sin {\left (a + b \log {\left (c x^{n} \right )} \right )} \cos ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} + 40 b^{2} n^{2} + 16} + \frac {12 b^{2} n^{2} x^{2} \sin ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )} \cos {\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} + 40 b^{2} n^{2} + 16} + \frac {14 b^{2} n^{2} x^{2} \cos ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} + 40 b^{2} n^{2} + 16} + \frac {12 b n x^{2} \sin {\left (a + b \log {\left (c x^{n} \right )} \right )} \cos ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} + 40 b^{2} n^{2} + 16} + \frac {8 x^{2} \cos ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} + 40 b^{2} n^{2} + 16} & \text {otherwise} \end {cases} \]
Piecewise((Integral(x*cos(a - 2*I*log(c*x**n)/n)**3, x), Eq(b, -2*I/n)), ( Integral(x*cos(a - 2*I*log(c*x**n)/(3*n))**3, x), Eq(b, -2*I/(3*n))), (Int egral(x*cos(a + 2*I*log(c*x**n)/(3*n))**3, x), Eq(b, 2*I/(3*n))), (Integra l(x*cos(a + 2*I*log(c*x**n)/n)**3, x), Eq(b, 2*I/n)), (6*b**3*n**3*x**2*si n(a + b*log(c*x**n))**3/(9*b**4*n**4 + 40*b**2*n**2 + 16) + 9*b**3*n**3*x* *2*sin(a + b*log(c*x**n))*cos(a + b*log(c*x**n))**2/(9*b**4*n**4 + 40*b**2 *n**2 + 16) + 12*b**2*n**2*x**2*sin(a + b*log(c*x**n))**2*cos(a + b*log(c* x**n))/(9*b**4*n**4 + 40*b**2*n**2 + 16) + 14*b**2*n**2*x**2*cos(a + b*log (c*x**n))**3/(9*b**4*n**4 + 40*b**2*n**2 + 16) + 12*b*n*x**2*sin(a + b*log (c*x**n))*cos(a + b*log(c*x**n))**2/(9*b**4*n**4 + 40*b**2*n**2 + 16) + 8* x**2*cos(a + b*log(c*x**n))**3/(9*b**4*n**4 + 40*b**2*n**2 + 16), True))
Leaf count of result is larger than twice the leaf count of optimal. 1015 vs. \(2 (158) = 316\).
Time = 0.26 (sec) , antiderivative size = 1015, normalized size of antiderivative = 6.42 \[ \int x \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]
1/8*((3*(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 + 2*(b^2*cos(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 + 12*( b*cos(3*b*log(c))*sin(6*b*log(c)) - b*cos(6*b*log(c))*sin(3*b*log(c)) + b* sin(3*b*log(c)))*n + 8*cos(6*b*log(c))*cos(3*b*log(c)) + 8*sin(6*b*log(c)) *sin(3*b*log(c)) + 8*cos(3*b*log(c)))*x^2*cos(3*b*log(x^n) + 3*a) + 3*(9*( b^3*cos(3*b*log(c))*sin(4*b*log(c)) - b^3*cos(4*b*log(c))*sin(3*b*log(c)) + b^3*cos(2*b*log(c))*sin(3*b*log(c)) - b^3*cos(3*b*log(c))*sin(2*b*log(c) ))*n^3 + 18*(b^2*cos(4*b*log(c))*cos(3*b*log(c)) + b^2*cos(3*b*log(c))*cos (2*b*log(c)) + b^2*sin(4*b*log(c))*sin(3*b*log(c)) + b^2*sin(3*b*log(c))*s in(2*b*log(c)))*n^2 + 4*(b*cos(3*b*log(c))*sin(4*b*log(c)) - b*cos(4*b*log (c))*sin(3*b*log(c)) + b*cos(2*b*log(c))*sin(3*b*log(c)) - b*cos(3*b*log(c ))*sin(2*b*log(c)))*n + 8*cos(4*b*log(c))*cos(3*b*log(c)) + 8*cos(3*b*log( c))*cos(2*b*log(c)) + 8*sin(4*b*log(c))*sin(3*b*log(c)) + 8*sin(3*b*log(c) )*sin(2*b*log(c)))*x^2*cos(b*log(x^n) + a) + (3*(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 - 2*(b^2*cos(3*b*log(c))*sin(6*b*log(c)) - b^2*cos(6*b*log(c))*sin(3*b*l og(c)) + b^2*sin(3*b*log(c)))*n^2 + 12*(b*cos(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 - 8*cos(3*b*log (c))*sin(6*b*log(c)) + 8*cos(6*b*log(c))*sin(3*b*log(c)) - 8*sin(3*b*lo...
Leaf count of result is larger than twice the leaf count of optimal. 18069 vs. \(2 (158) = 316\).
Time = 1.23 (sec) , antiderivative size = 18069, normalized size of antiderivative = 114.36 \[ \int x \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]
-1/4*(27*b^3*n^3*x^2*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*tan(1/2*a) + 27*b^3*n^3*x^2*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(ab s(c)))^2*tan(3/2*a)^2*tan(1/2*a) + 3*b^3*n^3*x^2*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)*tan( 1/2*a)^2 + 3*b^3*n^3*x^2*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)*tan(1/2*a)^2 + 27*b^3*n^3*x ^2*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*lo g(abs(c)))*tan(3/2*a)^2*tan(1/2*a)^2 + 27*b^3*n^3*x^2*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)))*tan(3/2*a)^ 2*tan(1/2*a)^2 + 3*b^3*n^3*x^2*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)))*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 + 3*b^3*n ^3*x^2*e^(-3/2*pi*b*n*sgn(x) + 3/2*pi*b*n - 3/2*pi*b*sgn(c) + 3/2*pi*b)...
Time = 26.99 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.77 \[ \int x \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^2\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}}\,3{}\mathrm {i}}{8\,b\,n+16{}\mathrm {i}}+\frac {3\,x^2\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}}{16+b\,n\,8{}\mathrm {i}}+\frac {x^2\,{\mathrm {e}}^{-a\,3{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,3{}\mathrm {i}}}\,1{}\mathrm {i}}{24\,b\,n+16{}\mathrm {i}}+\frac {x^2\,{\mathrm {e}}^{a\,3{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,3{}\mathrm {i}}}{16+b\,n\,24{}\mathrm {i}} \]