Integrand size = 28, antiderivative size = 101 \[ \int x^m \cos \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {e^{\frac {a (1+m)}{\sqrt {-\frac {(1+m)^2}{n^2}} n}} x^{1+m} \left (c x^n\right )^{\frac {1+m}{n}}}{4 (1+m)}+\frac {1}{2} e^{\frac {a \sqrt {-\frac {(1+m)^2}{n^2}} n}{1+m}} x^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}} \log (x) \] Output:
exp(a*(1+m)/(-(1+m)^2/n^2)^(1/2)/n)*x^(1+m)*(c*x^n)^((1+m)/n)/(4+4*m)+1/2* exp(a*(-(1+m)^2/n^2)^(1/2)*n/(1+m))*x^(1+m)*ln(x)/((c*x^n)^((1+m)/n))
\[ \int x^m \cos \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\int x^m \cos \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx \] Input:
Integrate[x^m*Cos[a + Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n]],x]
Output:
Integrate[x^m*Cos[a + Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n]], x]
Time = 0.41 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {4997, 4993, 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.
\(\displaystyle \int x^m \cos \left (a+\sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 4997 |
\(\displaystyle \frac {x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (c x^n\right )^{\frac {m+1}{n}-1} \cos \left (a+\sqrt {-\frac {(m+1)^2}{n^2}} \log \left (c x^n\right )\right )d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 4993 |
\(\displaystyle \frac {x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (e^{\frac {a (m+1)}{\sqrt {-\frac {(m+1)^2}{n^2}} n}} \left (c x^n\right )^{\frac {2 (m+1)}{n}-1}+\frac {e^{\frac {a \sqrt {-\frac {(m+1)^2}{n^2}} n}{m+1}} x^{-n}}{c}\right )d\left (c x^n\right )}{2 n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (\frac {n e^{\frac {a (m+1)}{n \sqrt {-\frac {(m+1)^2}{n^2}}}} \left (c x^n\right )^{\frac {2 (m+1)}{n}}}{2 (m+1)}+e^{\frac {a n \sqrt {-\frac {(m+1)^2}{n^2}}}{m+1}} \log \left (c x^n\right )\right )}{2 n}\) |
Input:
Int[x^m*Cos[a + Sqrt[-((1 + m)^2/n^2)]*Log[c*x^n]],x]
Output:
(x^(1 + m)*((E^((a*(1 + m))/(Sqrt[-((1 + m)^2/n^2)]*n))*n*(c*x^n)^((2*(1 + m))/n))/(2*(1 + m)) + E^((a*Sqrt[-((1 + m)^2/n^2)]*n)/(1 + m))*Log[c*x^n] ))/(2*n*(c*x^n)^((1 + m)/n))
Int[Cos[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[1/2^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[Cos[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_ .), x_Symbol] :> Simp[(e*x)^(m + 1)/(e*n*(c*x^n)^((m + 1)/n)) Subst[Int[x ^((m + 1)/n - 1)*Cos[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])
\[\int x^{m} \cos \left (a +\sqrt {-\frac {\left (1+m \right )^{2}}{n^{2}}}\, \ln \left (c \,x^{n}\right )\right )d x\]
Input:
int(x^m*cos(a+(-(1+m)^2/n^2)^(1/2)*ln(c*x^n)),x)
Output:
int(x^m*cos(a+(-(1+m)^2/n^2)^(1/2)*ln(c*x^n)),x)
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.59 \[ \int x^m \cos \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {{\left (x^{2} x^{2 \, m} + 2 \, {\left (m + 1\right )} e^{\left (\frac {2 \, {\left (i \, a n - {\left (m + 1\right )} \log \left (c\right )\right )}}{n}\right )} \log \left (x\right )\right )} e^{\left (-\frac {i \, a n - {\left (m + 1\right )} \log \left (c\right )}{n}\right )}}{4 \, {\left (m + 1\right )}} \] Input:
integrate(x^m*cos(a+(-(1+m)^2/n^2)^(1/2)*log(c*x^n)),x, algorithm="fricas" )
Output:
1/4*(x^2*x^(2*m) + 2*(m + 1)*e^(2*(I*a*n - (m + 1)*log(c))/n)*log(x))*e^(- (I*a*n - (m + 1)*log(c))/n)/(m + 1)
\[ \int x^m \cos \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\int x^{m} \cos {\left (a + \sqrt {- \frac {m^{2}}{n^{2}} - \frac {2 m}{n^{2}} - \frac {1}{n^{2}}} \log {\left (c x^{n} \right )} \right )}\, dx \] Input:
integrate(x**m*cos(a+(-(1+m)**2/n**2)**(1/2)*ln(c*x**n)),x)
Output:
Integral(x**m*cos(a + sqrt(-m**2/n**2 - 2*m/n**2 - 1/n**2)*log(c*x**n)), x )
Time = 0.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.81 \[ \int x^m \cos \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {c^{\frac {2 \, m}{n} + \frac {2}{n}} x \cos \left (a\right ) e^{\left (m \log \left (x\right ) + \frac {m \log \left (x^{n}\right )}{n} + \frac {\log \left (x^{n}\right )}{n}\right )} + 2 \, {\left (m \cos \left (a\right ) + \cos \left (a\right )\right )} \log \left (x\right )}{4 \, {\left (c^{\frac {m}{n} + \frac {1}{n}} m + c^{\frac {m}{n} + \frac {1}{n}}\right )}} \] Input:
integrate(x^m*cos(a+(-(1+m)^2/n^2)^(1/2)*log(c*x^n)),x, algorithm="maxima" )
Output:
1/4*(c^(2*m/n + 2/n)*x*cos(a)*e^(m*log(x) + m*log(x^n)/n + log(x^n)/n) + 2 *(m*cos(a) + cos(a))*log(x))/(c^(m/n + 1/n)*m + c^(m/n + 1/n))
Result contains complex when optimal does not.
Time = 0.96 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.64 \[ \int x^m \cos \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {m n^{2} x x^{m} e^{\left (i \, a - \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + m n^{2} x x^{m} e^{\left (-i \, a + \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + n^{2} x x^{m} e^{\left (i \, a - \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + n x x^{m} {\left | m n + n \right |} e^{\left (i \, a - \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} + n^{2} x x^{m} e^{\left (-i \, a + \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )} - n x x^{m} {\left | m n + n \right |} e^{\left (-i \, a + \frac {n {\left | m n + n \right |} \log \left (x\right ) + {\left | m n + n \right |} \log \left (c\right )}{n^{2}}\right )}}{2 \, {\left (m^{2} n^{2} + 2 \, m n^{2} - {\left (m n + n\right )}^{2} + n^{2}\right )}} \] Input:
integrate(x^m*cos(a+(-(1+m)^2/n^2)^(1/2)*log(c*x^n)),x, algorithm="giac")
Output:
1/2*(m*n^2*x*x^m*e^(I*a - (n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n^ 2) + m*n^2*x*x^m*e^(-I*a + (n*abs(m*n + n)*log(x) + abs(m*n + n)*log(c))/n ^2) + n^2*x*x^m*e^(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^(I*a - (n*abs(m*n + n)*log(x) + abs(m*n + n)*lo g(c))/n^2) + n^2*x*x^m*e^(-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^(-I*a + (n*abs(m*n + n)*log(x) + abs(m* n + n)*log(c))/n^2))/(m^2*n^2 + 2*m*n^2 - (m*n + n)^2 + n^2)
Time = 21.95 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.30 \[ \int x^m \cos \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x\,x^m\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}}}{2\,m+2-n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,2{}\mathrm {i}}+\frac {x\,x^m\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{\sqrt {-\frac {2\,m}{n^2}-\frac {1}{n^2}-\frac {m^2}{n^2}}\,1{}\mathrm {i}}}{2\,m+2+n\,\sqrt {-\frac {{\left (m+1\right )}^2}{n^2}}\,2{}\mathrm {i}} \] Input:
int(x^m*cos(a + log(c*x^n)*(-(m + 1)^2/n^2)^(1/2)),x)
Output:
(x*x^m*exp(-a*1i)/(c*x^n)^((- (2*m)/n^2 - 1/n^2 - m^2/n^2)^(1/2)*1i))/(2*m - n*(-(m + 1)^2/n^2)^(1/2)*2i + 2) + (x*x^m*exp(a*1i)*(c*x^n)^((- (2*m)/n ^2 - 1/n^2 - m^2/n^2)^(1/2)*1i))/(2*m + n*(-(m + 1)^2/n^2)^(1/2)*2i + 2)
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.58 \[ \int x^m \cos \left (a+\sqrt {-\frac {(1+m)^2}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x^{m} x \left (\cos \left (\frac {\mathrm {log}\left (x^{n} c \right ) m +\mathrm {log}\left (x^{n} c \right )+a n}{n}\right )+\sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) m +\mathrm {log}\left (x^{n} c \right )+a n}{n}\right )\right )}{2 m +2} \] Input:
int(x^m*cos(a+(-(1+m)^2/n^2)^(1/2)*log(c*x^n)),x)
Output:
(x**m*x*(cos((log(x**n*c)*m + log(x**n*c) + a*n)/n) + sin((log(x**n*c)*m + log(x**n*c) + a*n)/n)))/(2*(m + 1))