Integrand size = 33, antiderivative size = 117 \[ \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x^{1+m}}{2 (1+m)}-\frac {e^{-\frac {2 a \sqrt {-\frac {(1+m)^2}{n^2}} n}{1+m}} x^{1+m} \left (c x^n\right )^{\frac {1+m}{n}}}{8 (1+m)}-\frac {1}{4} e^{\frac {2 a \sqrt {-\frac {(1+m)^2}{n^2}} n}{1+m}} x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}} \log (x) \]
1/2*x^(1+m)/(1+m)-1/8*x^(1+m)*(c*x^n)^((1+m)/n)/exp(2*a*n*(-(1+m)^2/n^2)^( 1/2)/(1+m))/(1+m)-1/4*exp(2*a*n*(-(1+m)^2/n^2)^(1/2)/(1+m))*x^(1+m)*ln(x)/ ((c*x^n)^((1+m)/n))
\[ \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx \]
Time = 0.43 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.
\(\displaystyle \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 4996 |
\(\displaystyle \frac {x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (c x^n\right )^{\frac {m+1}{n}-1} \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 4992 |
\(\displaystyle -\frac {x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (-2 \left (c x^n\right )^{\frac {m+1}{n}-1}+e^{-\frac {2 a \sqrt {-\frac {(m+1)^2}{n^2}} n}{m+1}} \left (c x^n\right )^{\frac {2 (m+1)}{n}-1}+\frac {e^{\frac {2 a \sqrt {-\frac {(m+1)^2}{n^2}} n}{m+1}} x^{-n}}{c}\right )d\left (c x^n\right )}{4 n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (-\frac {n e^{-\frac {2 a n \sqrt {-\frac {(m+1)^2}{n^2}}}{m+1}} \left (c x^n\right )^{\frac {2 (m+1)}{n}}}{2 (m+1)}-e^{\frac {2 a n \sqrt {-\frac {(m+1)^2}{n^2}}}{m+1}} \log \left (c x^n\right )+\frac {2 n \left (c x^n\right )^{\frac {m+1}{n}}}{m+1}\right )}{4 n}\) |
(x^(1 + m)*((2*n*(c*x^n)^((1 + m)/n))/(1 + m) - (n*(c*x^n)^((2*(1 + m))/n) )/(2*E^((2*a*Sqrt[-((1 + m)^2/n^2)]*n)/(1 + m))*(1 + m)) - E^((2*a*Sqrt[-( (1 + m)^2/n^2)]*n)/(1 + m))*Log[c*x^n]))/(4*n*(c*x^n)^((1 + m)/n))
3.1.33.3.1 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])
\[\int x^{m} {\sin \left (a +\frac {\ln \left (c \,x^{n}\right ) \sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}}{2}\right )}^{2}d x\]
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.91 \[ \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=-\frac {{\left (2 \, {\left (m + 1\right )} e^{\left (-\frac {2 \, {\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )\right )}}{n}\right )} \log \left (x\right ) - 4 \, e^{\left (-\frac {{\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )}{n}\right )} + 1\right )} e^{\left (\frac {2 \, {\left ({\left (m + 1\right )} n \log \left (x\right ) - 2 i \, a n + {\left (m + 1\right )} \log \left (c\right )\right )}}{n} + \frac {2 i \, a n - {\left (m + 1\right )} \log \left (c\right )}{n}\right )}}{8 \, {\left (m + 1\right )}} \]
-1/8*(2*(m + 1)*e^(-2*((m + 1)*n*log(x) - 2*I*a*n + (m + 1)*log(c))/n)*log (x) - 4*e^(-((m + 1)*n*log(x) - 2*I*a*n + (m + 1)*log(c))/n) + 1)*e^(2*((m + 1)*n*log(x) - 2*I*a*n + (m + 1)*log(c))/n + (2*I*a*n - (m + 1)*log(c))/ n)/(m + 1)
\[ \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\int x^{m} \sin ^{2}{\left (a + \frac {\sqrt {- \frac {m^{2}}{n^{2}} - \frac {2 m}{n^{2}} - \frac {1}{n^{2}}} \log {\left (c x^{n} \right )}}{2} \right )}\, dx \]
Time = 0.26 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.48 \[ \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {4 \, {\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {m}{n} + \frac {1}{n}} x x^{m} - c^{\frac {2 \, m}{n} + \frac {2}{n}} x \cos \left (2 \, a\right ) e^{\left (m \log \left (x\right ) + \frac {m \log \left (x^{n}\right )}{n} + \frac {\log \left (x^{n}\right )}{n}\right )} - 2 \, {\left (\cos \left (2 \, a\right )^{3} + \cos \left (2 \, a\right ) \sin \left (2 \, a\right )^{2} + {\left (\cos \left (2 \, a\right )^{3} + \cos \left (2 \, a\right ) \sin \left (2 \, a\right )^{2}\right )} m\right )} \log \left (x\right )}{8 \, {\left ({\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {m}{n} + \frac {1}{n}} m + {\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{\frac {m}{n} + \frac {1}{n}}\right )}} \]
1/8*(4*(cos(2*a)^2 + sin(2*a)^2)*c^(m/n + 1/n)*x*x^m - c^(2*m/n + 2/n)*x*c os(2*a)*e^(m*log(x) + m*log(x^n)/n + log(x^n)/n) - 2*(cos(2*a)^3 + cos(2*a )*sin(2*a)^2 + (cos(2*a)^3 + cos(2*a)*sin(2*a)^2)*m)*log(x))/((cos(2*a)^2 + sin(2*a)^2)*c^(m/n + 1/n)*m + (cos(2*a)^2 + sin(2*a)^2)*c^(m/n + 1/n))
Result contains complex when optimal does not.
Time = 3.37 (sec) , antiderivative size = 498, normalized size of antiderivative = 4.26 \[ \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=-\frac {m^{2} n^{2} x x^{m} e^{\left (2 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 )} + m^{2} n^{2} x x^{m} e^{\left (-2 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 \, m^{2} n^{2} x x^{m} + 2 \, m n^{2} x x^{m} e^{\left (2 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 )} + m n x x^{m} {\left | m n + n \right |} e^{\left (2 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 \, m n^{2} x x^{m} e^{\left (-2 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 )} - m n x x^{m} {\left | m n + n \right |} e^{\left (-2 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 )} - 4 \, m n^{2} x x^{m} + n^{2} x x^{m} e^{\left (2 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 )} + n x x^{m} {\left | m n + n \right |} e^{\left (2 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 )} + n^{2} x x^{m} e^{\left (-2 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 )} - n x x^{m} {\left | m n + n \right |} e^{\left (-2 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 n + n\right )}^{2} x x^{m} - 2 \, n^{2} x x^{m}}{4 \, {\left (m^{3} n^{2} + 3 \, m^{2} n^{2} - {\left (m n + n\right )}^{2} m + 3 \, m n^{2} - {\left (m n + n\right )}^{2} + n^{2}\right )}} \]
-1/4*(m^2*n^2*x*x^m*e^(2*I*a - (n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c ))/n^2) + m^2*n^2*x*x^m*e^(-2*I*a + (n*abs(m*n + n)*log(x) + abs(m*n + n)* log(c))/n^2) - 2*m^2*n^2*x*x^m + 2*m*n^2*x*x^m*e^(2*I*a - (n*abs(m*n + n)* log(x) + abs(m*n + n)*log(c))/n^2) + m*n*x*x^m*abs(m*n + n)*e^(2*I*a - (n* abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) + 2*m*n^2*x*x^m*e^(-2*I*a + (n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - m*n*x*x^m*abs(m*n + n)*e^(-2*I*a + (n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^2) - 4*m*n ^2*x*x^m + n^2*x*x^m*e^(2*I*a - (n*abs(m*n + n)*log(x) + abs(m*n + n)*log( c))/n^2) + n*x*x^m*abs(m*n + n)*e^(2*I*a - (n*abs(m*n + n)*log(x) + abs(m* n + n)*log(c))/n^2) + n^2*x*x^m*e^(-2*I*a + (n*abs(m*n + n)*log(x) + abs(m *n + n)*log(c))/n^2) - n*x*x^m*abs(m*n + n)*e^(-2*I*a + (n*abs(m*n + n)*lo g(x) + abs(m*n + n)*log(c))/n^2) + 2*(m*n + n)^2*x*x^m - 2*n^2*x*x^m)/(m^3 *n^2 + 3*m^2*n^2 - (m*n + n)^2*m + 3*m*n^2 - (m*n + n)^2 + n^2)
Time = 28.00 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.24 \[ \int x^m \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x\,x^m}{2\,m+2}-\frac {x\,x^m\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}}}{4\,m+4-n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,4{}\mathrm {i}}-\frac {x\,x^m\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}}{4\,m+4+n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,4{}\mathrm {i}} \]