Integrand size = 17, antiderivative size = 201 \[ \int x^m \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {6 b^2 (1+m) n^2 x^{1+m} \cos \left (a+b \log \left (c x^n\right )\right )}{\left ((1+m)^2+b^2 n^2\right ) \left ((1+m)^2+9 b^2 n^2\right )}+\frac {(1+m) x^{1+m} \cos ^3\left (a+b \log \left (c x^n\right )\right )}{(1+m)^2+9 b^2 n^2}+\frac {6 b^3 n^3 x^{1+m} \sin \left (a+b \log \left (c x^n\right )\right )}{\left ((1+m)^2+b^2 n^2\right ) \left ((1+m)^2+9 b^2 n^2\right )}+\frac {3 b n x^{1+m} \cos ^2\left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2+9 b^2 n^2} \]
6*b^2*(1+m)*n^2*x^(1+m)*cos(a+b*ln(c*x^n))/((1+m)^2+b^2*n^2)/((1+m)^2+9*b^ 2*n^2)+(1+m)*x^(1+m)*cos(a+b*ln(c*x^n))^3/((1+m)^2+9*b^2*n^2)+6*b^3*n^3*x^ (1+m)*sin(a+b*ln(c*x^n))/((1+m)^2+b^2*n^2)/((1+m)^2+9*b^2*n^2)+3*b*n*x^(1+ m)*cos(a+b*ln(c*x^n))^2*sin(a+b*ln(c*x^n))/((1+m)^2+9*b^2*n^2)
Time = 1.59 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.45 \[ \int x^m \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{4} x^{1+m} \left (-\frac {3 \sin (b n \log (x)) \left (-b n \cos \left (a-b n \log (x)+b \log \left (c x^n\right )\right )+(1+m) \sin \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )}{1+2 m+m^2+b^2 n^2}+\frac {3 \cos (b n \log (x)) \left ((1+m) \cos \left (a-b n \log (x)+b \log \left (c x^n\right )\right )+b n \sin \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )}{1+2 m+m^2+b^2 n^2}-\frac {\sin (3 b n \log (x)) \left (-3 b n \cos \left (3 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )+(1+m) \sin \left (3 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{1+2 m+m^2+9 b^2 n^2}+\frac {\cos (3 b n \log (x)) \left ((1+m) \cos \left (3 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )+3 b n \sin \left (3 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{1+2 m+m^2+9 b^2 n^2}\right ) \]
(x^(1 + m)*((-3*Sin[b*n*Log[x]]*(-(b*n*Cos[a - b*n*Log[x] + b*Log[c*x^n]]) + (1 + m)*Sin[a - b*n*Log[x] + b*Log[c*x^n]]))/(1 + 2*m + m^2 + b^2*n^2) + (3*Cos[b*n*Log[x]]*((1 + m)*Cos[a - b*n*Log[x] + b*Log[c*x^n]] + b*n*Sin [a - b*n*Log[x] + b*Log[c*x^n]]))/(1 + 2*m + m^2 + b^2*n^2) - (Sin[3*b*n*L og[x]]*(-3*b*n*Cos[3*(a - b*n*Log[x] + b*Log[c*x^n])] + (1 + m)*Sin[3*(a - b*n*Log[x] + b*Log[c*x^n])]))/(1 + 2*m + m^2 + 9*b^2*n^2) + (Cos[3*b*n*Lo g[x]]*((1 + m)*Cos[3*(a - b*n*Log[x] + b*Log[c*x^n])] + 3*b*n*Sin[3*(a - b *n*Log[x] + b*Log[c*x^n])]))/(1 + 2*m + m^2 + 9*b^2*n^2)))/4
Time = 0.36 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.91, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, 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.
\(\displaystyle \int x^m \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 4991 |
\(\displaystyle \frac {6 b^2 n^2 \int x^m \cos \left (a+b \log \left (c x^n\right )\right )dx}{9 b^2 n^2+(m+1)^2}+\frac {(m+1) x^{m+1} \cos ^3\left (a+b \log \left (c x^n\right )\right )}{9 b^2 n^2+(m+1)^2}+\frac {3 b n x^{m+1} \sin \left (a+b \log \left (c x^n\right )\right ) \cos ^2\left (a+b \log \left (c x^n\right )\right )}{9 b^2 n^2+(m+1)^2}\) |
\(\Big \downarrow \) 4989 |
\(\displaystyle \frac {(m+1) x^{m+1} \cos ^3\left (a+b \log \left (c x^n\right )\right )}{9 b^2 n^2+(m+1)^2}+\frac {3 b n x^{m+1} \sin \left (a+b \log \left (c x^n\right )\right ) \cos ^2\left (a+b \log \left (c x^n\right )\right )}{9 b^2 n^2+(m+1)^2}+\frac {6 b^2 n^2 \left (\frac {b n x^{m+1} \sin \left (a+b \log \left (c x^n\right )\right )}{b^2 n^2+(m+1)^2}+\frac {(m+1) x^{m+1} \cos \left (a+b \log \left (c x^n\right )\right )}{b^2 n^2+(m+1)^2}\right )}{9 b^2 n^2+(m+1)^2}\) |
((1 + m)*x^(1 + m)*Cos[a + b*Log[c*x^n]]^3)/((1 + m)^2 + 9*b^2*n^2) + (3*b *n*x^(1 + m)*Cos[a + b*Log[c*x^n]]^2*Sin[a + b*Log[c*x^n]])/((1 + m)^2 + 9 *b^2*n^2) + (6*b^2*n^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 )))/((1 + m)^2 + 9*b^2*n^2)
3.2.24.3.1 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 = 28.45 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.80
method | result | size |
parallelrisch | \(\frac {3 \left (\frac {\left (1+m \right ) \left (b^{2} n^{2}+m^{2}+2 m +1\right ) \cos \left (3 b \ln \left (c \,x^{n}\right )+3 a \right )}{27}+\frac {b n \left (b^{2} n^{2}+m^{2}+2 m +1\right ) \sin \left (3 b \ln \left (c \,x^{n}\right )+3 a \right )}{9}+\left (\left (1+m \right ) \cos \left (a +b \ln \left (c \,x^{n}\right )\right )+\sin \left (a +b \ln \left (c \,x^{n}\right )\right ) b n \right ) \left (b^{2} n^{2}+\frac {1}{9} m^{2}+\frac {2}{9} m +\frac {1}{9}\right )\right ) x^{1+m}}{4 \left (b^{2} n^{2}+m^{2}+2 m +1\right ) \left (b^{2} n^{2}+\frac {1}{9} m^{2}+\frac {2}{9} m +\frac {1}{9}\right )}\) | \(160\) |
3/4*(1/27*(1+m)*(b^2*n^2+m^2+2*m+1)*cos(3*b*ln(c*x^n)+3*a)+1/9*b*n*(b^2*n^ 2+m^2+2*m+1)*sin(3*b*ln(c*x^n)+3*a)+((1+m)*cos(a+b*ln(c*x^n))+sin(a+b*ln(c *x^n))*b*n)*(b^2*n^2+1/9*m^2+2/9*m+1/9))*x^(1+m)/(b^2*n^2+m^2+2*m+1)/(b^2* n^2+1/9*m^2+2/9*m+1/9)
Time = 0.27 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.95 \[ \int x^m \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {3 \, {\left (2 \, b^{3} n^{3} x + {\left (b^{3} n^{3} + {\left (b m^{2} + 2 \, b m + b\right )} n\right )} x \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2}\right )} x^{m} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + {\left (6 \, {\left (b^{2} m + b^{2}\right )} n^{2} x \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + {\left (m^{3} + {\left (b^{2} m + b^{2}\right )} n^{2} + 3 \, m^{2} + 3 \, m + 1\right )} x \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3}\right )} x^{m}}{9 \, b^{4} n^{4} + m^{4} + 4 \, m^{3} + 10 \, {\left (b^{2} m^{2} + 2 \, b^{2} m + b^{2}\right )} n^{2} + 6 \, m^{2} + 4 \, m + 1} \]
(3*(2*b^3*n^3*x + (b^3*n^3 + (b*m^2 + 2*b*m + b)*n)*x*cos(b*n*log(x) + b*l og(c) + a)^2)*x^m*sin(b*n*log(x) + b*log(c) + a) + (6*(b^2*m + b^2)*n^2*x* cos(b*n*log(x) + b*log(c) + a) + (m^3 + (b^2*m + b^2)*n^2 + 3*m^2 + 3*m + 1)*x*cos(b*n*log(x) + b*log(c) + a)^3)*x^m)/(9*b^4*n^4 + m^4 + 4*m^3 + 10* (b^2*m^2 + 2*b^2*m + b^2)*n^2 + 6*m^2 + 4*m + 1)
Timed out. \[ \int x^m \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 2352 vs. \(2 (201) = 402\).
Time = 0.36 (sec) , antiderivative size = 2352, normalized size of antiderivative = 11.70 \[ \int x^m \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]
1/8*(((cos(6*b*log(c))*cos(3*b*log(c)) + sin(6*b*log(c))*sin(3*b*log(c)) + cos(3*b*log(c)))*m^3 + 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 + 3*(cos(6*b*log(c)) *cos(3*b*log(c)) + sin(6*b*log(c))*sin(3*b*log(c)) + cos(3*b*log(c)))*m^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)) + (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)))*m)*n^2 + 3*(cos(6*b*log (c))*cos(3*b*log(c)) + sin(6*b*log(c))*sin(3*b*log(c)) + cos(3*b*log(c)))* m + 3*((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)))*m^2 + b*cos(3*b*log(c))*sin(6*b*log(c)) - b*cos(6 *b*log(c))*sin(3*b*log(c)) + 2*(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)))*m + b*sin(3*b*log(c)))*n + cos(6*b*log(c))*cos(3*b*log(c)) + sin(6*b*log(c))*sin(3*b*log(c)) + cos( 3*b*log(c)))*x*x^m*cos(3*b*log(x^n) + 3*a) + 3*((cos(4*b*log(c))*cos(3*b*l og(c)) + cos(3*b*log(c))*cos(2*b*log(c)) + sin(4*b*log(c))*sin(3*b*log(c)) + sin(3*b*log(c))*sin(2*b*log(c)))*m^3 + 9*(b^3*cos(3*b*log(c))*sin(4*b*l og(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 + 3*(cos(4*b*log(c))*c os(3*b*log(c)) + cos(3*b*log(c))*cos(2*b*log(c)) + sin(4*b*log(c))*sin(3*b *log(c)) + sin(3*b*log(c))*sin(2*b*log(c)))*m^2 + 9*(b^2*cos(4*b*log(c)...
Leaf count of result is larger than twice the leaf count of optimal. 159584 vs. \(2 (201) = 402\).
Time = 4.73 (sec) , antiderivative size = 159584, normalized size of antiderivative = 793.95 \[ \int x^m \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]
1/8*(54*b^3*n^3*x*abs(x)^m*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sg n(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(1/4*pi*m*sgn(x) - 1/4*pi*m)^2*tan (3/2*a)^2*tan(1/2*a) + 54*b^3*n^3*x*abs(x)^m*e^(-1/2*pi*b*n*sgn(x) + 1/2*p i*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(ab s(c)))^2*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(3/2*a)^2*tan(1/2*a) + 6*b^3*n^3*x*abs(x)^m*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*t an(1/4*pi*m*sgn(x) - 1/4*pi*m)^2*tan(3/2*a)*tan(1/2*a)^2 + 6*b^3*n^3*x*abs (x)^m*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(1/4*pi*m*sgn(x) - 1/4*pi*m)^2*tan(3/2*a)*tan(1/2*a)^2 - 6*b^3*n^3*x*abs(x)^m*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*l og(abs(x)) + 1/2*b*log(abs(c)))^2*tan(1/4*pi*m*sgn(x) - 1/4*pi*m)*tan(3/2* a)^2*tan(1/2*a)^2 - 54*b^3*n^3*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(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(1/4*pi*m*sgn(x) - 1/4*pi*m)*tan(3/2*a)^2*tan(1/2*a)^2 + 54*b^3*n^3*x*abs(x)^m*e^(-1/2*pi*...
Time = 28.56 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.70 \[ \int x^m \cos ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {3\,x\,x^m\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}}{8\,m+8+b\,n\,8{}\mathrm {i}}+\frac {x\,x^m\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}}\,3{}\mathrm {i}}{m\,8{}\mathrm {i}+8\,b\,n+8{}\mathrm {i}}+\frac {x\,x^m\,{\mathrm {e}}^{a\,3{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,3{}\mathrm {i}}}{8\,m+8+b\,n\,24{}\mathrm {i}}+\frac {x\,x^m\,{\mathrm {e}}^{-a\,3{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,3{}\mathrm {i}}}\,1{}\mathrm {i}}{m\,8{}\mathrm {i}+24\,b\,n+8{}\mathrm {i}} \]