Integrand size = 28, antiderivative size = 133 \[ \int x^m \sin \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=-\frac {e^{\frac {a (1+m)}{\sqrt {-\frac {(1+m)^2}{n^2}} n}} x^{1+m} \left (c x^n\right )^{\frac {1+m}{n}}}{4 \sqrt {-\frac {(1+m)^2}{n^2}} n}+\frac {e^{\frac {a \sqrt {-\frac {(1+m)^2}{n^2}} n}{1+m}} (1+m) x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}} \log (x)}{2 \sqrt {-\frac {(1+m)^2}{n^2}} n} \]
-1/4*exp(a*(1+m)/n/(-(1+m)^2/n^2)^(1/2))*x^(1+m)*(c*x^n)^((1+m)/n)/n/(-(1+ m)^2/n^2)^(1/2)+1/2*exp(a*n*(-(1+m)^2/n^2)^(1/2)/(1+m))*(1+m)*x^(1+m)*ln(x )/n/((c*x^n)^((1+m)/n))/(-(1+m)^2/n^2)^(1/2)
\[ \int x^m \sin \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\int x^m \sin \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx \]
Time = 0.43 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {4996, 4992, 2009}
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.
\(\displaystyle \int x^m \sin \left (a+\sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 4996 |
\(\displaystyle \frac {x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (c x^n\right )^{\frac {m+1}{n}-1} \sin \left (a+\sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 4992 |
\(\displaystyle \frac {(m+1) x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (\frac {e^{\frac {a \sqrt {-\frac {(m+1)^2}{n^2}} n}{m+1}} x^{-n}}{c}-e^{\frac {a (m+1)}{\sqrt {-\frac {(m+1)^2}{n^2}} n}} \left (c x^n\right )^{\frac {2 (m+1)}{n}-1}\right )d\left (c x^n\right )}{2 n^2 \sqrt {-\frac {(m+1)^2}{n^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(m+1) x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (e^{\frac {a n \sqrt {-\frac {(m+1)^2}{n^2}}}{m+1}} \log \left (c x^n\right )-\frac {n e^{\frac {a (m+1)}{n \sqrt {-\frac {(m+1)^2}{n^2}}}} \left (c x^n\right )^{\frac {2 (m+1)}{n}}}{2 (m+1)}\right )}{2 n^2 \sqrt {-\frac {(m+1)^2}{n^2}}}\) |
((1 + m)*x^(1 + m)*(-1/2*(E^((a*(1 + m))/(Sqrt[-((1 + m)^2/n^2)]*n))*n*(c* x^n)^((2*(1 + m))/n))/(1 + m) + E^((a*Sqrt[-((1 + m)^2/n^2)]*n)/(1 + m))*L og[c*x^n]))/(2*Sqrt[-((1 + m)^2/n^2)]*n^2*(c*x^n)^((1 + m)/n))
3.1.26.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Simp[(m + 1)^p/(2^p*b^p*d^p*p^p) Int[ExpandIntegrand[(e*x)^m*(E^(a*b*d ^2*(p/(m + 1)))/x^((m + 1)/p) - x^((m + 1)/p)/E^(a*b*d^2*(p/(m + 1))))^p, x ], x], x] /; FreeQ[{a, b, d, e, m}, x] && IGtQ[p, 0] && EqQ[b^2*d^2*p^2 + ( m + 1)^2, 0]
Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_ .), x_Symbol] :> Simp[(e*x)^(m + 1)/(e*n*(c*x^n)^((m + 1)/n)) Subst[Int[x ^((m + 1)/n - 1)*Sin[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])
\[\int x^{m} \sin \left (a +\ln \left (c \,x^{n}\right ) \sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}\right )d x\]
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.47 \[ \int x^m \sin \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {{\left (i \, x^{2} x^{2 \, m} - 2 \, {\left (i \, m + i\right )} e^{\left (\frac {2 \, {\left (i \, a n - {\left (m + 1\right )} \log \left (c\right )\right )}}{n}\right )} \log \left (x\right )\right )} e^{\left (-\frac {i \, a n - {\left (m + 1\right )} \log \left (c\right )}{n}\right )}}{4 \, {\left (m + 1\right )}} \]
1/4*(I*x^2*x^(2*m) - 2*(I*m + I)*e^(2*(I*a*n - (m + 1)*log(c))/n)*log(x))* e^(-(I*a*n - (m + 1)*log(c))/n)/(m + 1)
\[ \int x^m \sin \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\int x^{m} \sin {\left (a + \sqrt {- \frac {m^{2}}{n^{2}} - \frac {2 m}{n^{2}} - \frac {1}{n^{2}}} \log {\left (c x^{n} \right )} \right )}\, dx \]
Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.62 \[ \int x^m \sin \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {c^{\frac {2 \, m}{n} + \frac {2}{n}} x e^{\left (m \log \left (x\right ) + \frac {m \log \left (x^{n}\right )}{n} + \frac {\log \left (x^{n}\right )}{n}\right )} \sin \left (a\right ) + 2 \, {\left (m \sin \left (a\right ) + \sin \left (a\right )\right )} \log \left (x\right )}{4 \, {\left (c^{\frac {m}{n} + \frac {1}{n}} m + c^{\frac {m}{n} + \frac {1}{n}}\right )}} \]
1/4*(c^(2*m/n + 2/n)*x*e^(m*log(x) + m*log(x^n)/n + log(x^n)/n)*sin(a) + 2 *(m*sin(a) + sin(a))*log(x))/(c^(m/n + 1/n)*m + c^(m/n + 1/n))
Result contains complex when optimal does not.
Time = 1.06 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.05 \[ \int x^m \sin \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {-i \, m n^{2} x x^{m} e^{\left (i \, a - \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + i \, m n^{2} x x^{m} e^{\left (-i \, a + \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} - i \, n^{2} x x^{m} e^{\left (i \, a - \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} - i \, n x x^{m} {\left | m n + n \right |} e^{\left (i \, a - \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + i \, n^{2} x x^{m} e^{\left (-i \, a + \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} - i \, n x x^{m} {\left | m n + n \right |} e^{\left (-i \, a + \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )}}{2 \, {\left (m^{2} n^{2} + 2 \, m n^{2} - {\left (m n + n\right )}^{2} + n^{2}\right )}} \]
1/2*(-I*m*n^2*x*x^m*e^(I*a - (n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c)) /n^2) + I*m*n^2*x*x^m*e^(-I*a + (n*abs(m*n + n)*log(x) + abs(m*n + n)*log( c))/n^2) - I*n^2*x*x^m*e^(I*a - (n*abs(m*n + n)*log(x) + abs(m*n + n)*log( c))/n^2) - I*n*x*x^m*abs(m*n + n)*e^(I*a - (n*abs(m*n + n)*log(x) + abs(m* n + n)*log(c))/n^2) + I*n^2*x*x^m*e^(-I*a + (n*abs(m*n + n)*log(x) + abs(m *n + n)*log(c))/n^2) - I*n*x*x^m*abs(m*n + n)*e^(-I*a + (n*abs(m*n + n)*lo g(x) + abs(m*n + n)*log(c))/n^2))/(m^2*n^2 + 2*m*n^2 - (m*n + n)^2 + n^2)
Time = 28.79 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02 \[ \int x^m \sin \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x\,x^m\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}}\,1{}\mathrm {i}}{2\,m+2-n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,2{}\mathrm {i}}-\frac {x\,x^m\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,m+2+n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,2{}\mathrm {i}} \]