Integrand size = 15, antiderivative size = 70 \[ \int x^m \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {(1+m) x^{1+m} \cos \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2+b^2 n^2}+\frac {b n x^{1+m} \sin \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2+b^2 n^2} \]
(1+m)*x^(1+m)*cos(a+b*ln(c*x^n))/((1+m)^2+b^2*n^2)+b*n*x^(1+m)*sin(a+b*ln( c*x^n))/((1+m)^2+b^2*n^2)
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.76 \[ \int x^m \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{1+m} \left ((1+m) \cos \left (a+b \log \left (c x^n\right )\right )+b n \sin \left (a+b \log \left (c x^n\right )\right )\right )}{1+2 m+m^2+b^2 n^2} \]
(x^(1 + m)*((1 + m)*Cos[a + b*Log[c*x^n]] + b*n*Sin[a + b*Log[c*x^n]]))/(1 + 2*m + m^2 + b^2*n^2)
Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {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.
\(\displaystyle \int x^m \cos \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 4989 |
\(\displaystyle \frac {b n x^{m+1} \sin \left (a+b \log \left (c x^n\right )\right )}{b^2 n^2+(m+1)^2}+\frac {(m+1) x^{m+1} \cos \left (a+b \log \left (c x^n\right )\right )}{b^2 n^2+(m+1)^2}\) |
((1 + m)*x^(1 + m)*Cos[a + b*Log[c*x^n]])/((1 + m)^2 + b^2*n^2) + (b*n*x^( 1 + m)*Sin[a + b*Log[c*x^n]])/((1 + m)^2 + b^2*n^2)
3.2.26.3.1 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]
Time = 0.90 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \(\frac {x^{1+m} \left (\sin \left (a +b \ln \left (c \,x^{n}\right )\right ) b n +\cos \left (a +b \ln \left (c \,x^{n}\right )\right ) m +\cos \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}{b^{2} n^{2}+m^{2}+2 m +1}\) | \(63\) |
x^(1+m)/(b^2*n^2+m^2+2*m+1)*(sin(a+b*ln(c*x^n))*b*n+cos(a+b*ln(c*x^n))*m+c os(a+b*ln(c*x^n)))
Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.83 \[ \int x^m \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b n x x^{m} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + {\left (m + 1\right )} x x^{m} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{b^{2} n^{2} + m^{2} + 2 \, m + 1} \]
(b*n*x*x^m*sin(b*n*log(x) + b*log(c) + a) + (m + 1)*x*x^m*cos(b*n*log(x) + b*log(c) + a))/(b^2*n^2 + m^2 + 2*m + 1)
\[ \int x^m \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\begin {cases} \log {\left (x \right )} \cos {\left (a \right )} & \text {for}\: b = 0 \wedge m = -1 \\\int x^{m} \cos {\left (- a + \frac {i m \log {\left (c x^{n} \right )}}{n} + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = - \frac {i \left (m + 1\right )}{n} \\\int x^{m} \cos {\left (a + \frac {i m \log {\left (c x^{n} \right )}}{n} + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = \frac {i \left (m + 1\right )}{n} \\\frac {b n x x^{m} \sin {\left (a + b \log {\left (c x^{n} \right )} \right )}}{b^{2} n^{2} + m^{2} + 2 m + 1} + \frac {m x x^{m} \cos {\left (a + b \log {\left (c x^{n} \right )} \right )}}{b^{2} n^{2} + m^{2} + 2 m + 1} + \frac {x x^{m} \cos {\left (a + b \log {\left (c x^{n} \right )} \right )}}{b^{2} n^{2} + m^{2} + 2 m + 1} & \text {otherwise} \end {cases} \]
Piecewise((log(x)*cos(a), Eq(b, 0) & Eq(m, -1)), (Integral(x**m*cos(-a + I *m*log(c*x**n)/n + I*log(c*x**n)/n), x), Eq(b, -I*(m + 1)/n)), (Integral(x **m*cos(a + I*m*log(c*x**n)/n + I*log(c*x**n)/n), x), Eq(b, I*(m + 1)/n)), (b*n*x*x**m*sin(a + b*log(c*x**n))/(b**2*n**2 + m**2 + 2*m + 1) + m*x*x** m*cos(a + b*log(c*x**n))/(b**2*n**2 + m**2 + 2*m + 1) + x*x**m*cos(a + b*l og(c*x**n))/(b**2*n**2 + m**2 + 2*m + 1), True))
Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (70) = 140\).
Time = 0.24 (sec) , antiderivative size = 313, normalized size of antiderivative = 4.47 \[ \int x^m \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {{\left ({\left (\cos \left (2 \, b \log \left (c\right )\right ) \cos \left (b \log \left (c\right )\right ) + \sin \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) + \cos \left (b \log \left (c\right )\right )\right )} m + {\left (b \cos \left (b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) - b \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) + b \sin \left (b \log \left (c\right )\right )\right )} n + \cos \left (2 \, b \log \left (c\right )\right ) \cos \left (b \log \left (c\right )\right ) + \sin \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) + \cos \left (b \log \left (c\right )\right )\right )} x x^{m} \cos \left (b \log \left (x^{n}\right ) + a\right ) - {\left ({\left (\cos \left (b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) - \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) + \sin \left (b \log \left (c\right )\right )\right )} m - {\left (b \cos \left (2 \, b \log \left (c\right )\right ) \cos \left (b \log \left (c\right )\right ) + b \sin \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) + b \cos \left (b \log \left (c\right )\right )\right )} n + \cos \left (b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) - \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (b \log \left (c\right )\right ) + \sin \left (b \log \left (c\right )\right )\right )} x x^{m} \sin \left (b \log \left (x^{n}\right ) + a\right )}{2 \, {\left ({\left (\cos \left (b \log \left (c\right )\right )^{2} + \sin \left (b \log \left (c\right )\right )^{2}\right )} m^{2} + {\left (b^{2} \cos \left (b \log \left (c\right )\right )^{2} + b^{2} \sin \left (b \log \left (c\right )\right )^{2}\right )} n^{2} + 2 \, {\left (\cos \left (b \log \left (c\right )\right )^{2} + \sin \left (b \log \left (c\right )\right )^{2}\right )} m + \cos \left (b \log \left (c\right )\right )^{2} + \sin \left (b \log \left (c\right )\right )^{2}\right )}} \]
1/2*(((cos(2*b*log(c))*cos(b*log(c)) + sin(2*b*log(c))*sin(b*log(c)) + cos (b*log(c)))*m + (b*cos(b*log(c))*sin(2*b*log(c)) - b*cos(2*b*log(c))*sin(b *log(c)) + b*sin(b*log(c)))*n + cos(2*b*log(c))*cos(b*log(c)) + sin(2*b*lo g(c))*sin(b*log(c)) + cos(b*log(c)))*x*x^m*cos(b*log(x^n) + a) - ((cos(b*l og(c))*sin(2*b*log(c)) - cos(2*b*log(c))*sin(b*log(c)) + sin(b*log(c)))*m - (b*cos(2*b*log(c))*cos(b*log(c)) + b*sin(2*b*log(c))*sin(b*log(c)) + b*c os(b*log(c)))*n + cos(b*log(c))*sin(2*b*log(c)) - cos(2*b*log(c))*sin(b*lo g(c)) + sin(b*log(c)))*x*x^m*sin(b*log(x^n) + a))/((cos(b*log(c))^2 + sin( b*log(c))^2)*m^2 + (b^2*cos(b*log(c))^2 + b^2*sin(b*log(c))^2)*n^2 + 2*(co s(b*log(c))^2 + sin(b*log(c))^2)*m + cos(b*log(c))^2 + sin(b*log(c))^2)
Leaf count of result is larger than twice the leaf count of optimal. 5162 vs. \(2 (70) = 140\).
Time = 0.48 (sec) , antiderivative size = 5162, normalized size of antiderivative = 73.74 \[ \int x^m \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]
1/2*(2*b*n*x*abs(x)^m*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(1/4*pi*m*sg n(x) - 1/4*pi*m)^2*tan(1/2*a) + 2*b*n*x*abs(x)^m*e^(-1/2*pi*b*n*sgn(x) + 1 /2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*lo g(abs(c)))^2*tan(1/4*pi*m*sgn(x) - 1/4*pi*m)^2*tan(1/2*a) - 2*b*n*x*abs(x) ^m*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(1/2 *b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(1/4*pi*m*sgn(x) - 1/4*pi*m)*ta n(1/2*a)^2 + 2*b*n*x*abs(x)^m*e^(-1/2*pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi* b*sgn(c) + 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(1/ 4*pi*m*sgn(x) - 1/4*pi*m)*tan(1/2*a)^2 + 2*b*n*x*abs(x)^m*e^(1/2*pi*b*n*sg n(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))*tan(1/4*pi*m*sgn(x) - 1/4*pi*m)^2*tan(1/2*a)^2 + 2*b*n* x*abs(x)^m*e^(-1/2*pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b )*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))*tan(1/4*pi*m*sgn(x) - 1/4*p i*m)^2*tan(1/2*a)^2 - m*x*abs(x)^m*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2 *pi*b*sgn(c) - 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*ta n(1/4*pi*m*sgn(x) - 1/4*pi*m)^2*tan(1/2*a)^2 - m*x*abs(x)^m*e^(-1/2*pi*b*n *sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(1/4*pi*m*sgn(x) - 1/4*pi*m)^2*tan(1/2*a)^2 - x *abs(x)^m*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*...
Time = 27.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int x^m \cos \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x\,x^m\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}}{2\,m+2+b\,n\,2{}\mathrm {i}}+\frac {x\,x^m\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}}\,1{}\mathrm {i}}{m\,2{}\mathrm {i}+2\,b\,n+2{}\mathrm {i}} \]