Integrand size = 28, antiderivative size = 74 \[ \int \frac {\sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {1}{2 x}+\frac {e^{2 a \sqrt {-\frac {1}{n^2}} n} \left (c x^n\right )^{-1/n}}{8 x}-\frac {e^{-2 a \sqrt {-\frac {1}{n^2}} n} \left (c x^n\right )^{\frac {1}{n}} \log (x)}{4 x} \] Output:
-1/2/x+1/8*exp(2*a*(-1/n^2)^(1/2)*n)/x/((c*x^n)^(1/n))-1/4*(c*x^n)^(1/n)*l n(x)/exp(2*a*(-1/n^2)^(1/2)*n)/x
\[ \int \frac {\sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )}{x^2} \, dx=\int \frac {\sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )}{x^2} \, dx \] Input:
Integrate[Sin[a + (Sqrt[-n^(-2)]*Log[c*x^n])/2]^2/x^2,x]
Output:
Integrate[Sin[a + (Sqrt[-n^(-2)]*Log[c*x^n])/2]^2/x^2, x]
Time = 0.34 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.19, 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.
\(\displaystyle \int \frac {\sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 4996 |
\(\displaystyle \frac {\left (c x^n\right )^{\frac {1}{n}} \int \left (c x^n\right )^{-1-\frac {1}{n}} \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )d\left (c x^n\right )}{n x}\) |
\(\Big \downarrow \) 4992 |
\(\displaystyle -\frac {\left (c x^n\right )^{\frac {1}{n}} \int \left (\frac {e^{-2 a \sqrt {-\frac {1}{n^2}} n} x^{-n}}{c}-2 \left (c x^n\right )^{-\frac {n+1}{n}}+e^{2 a \sqrt {-\frac {1}{n^2}} n} \left (c x^n\right )^{-\frac {n+2}{n}}\right )d\left (c x^n\right )}{4 n x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (c x^n\right )^{\frac {1}{n}} \left (\frac {1}{2} n e^{2 a \sqrt {-\frac {1}{n^2}} n} \left (c x^n\right )^{-2/n}-e^{-2 a \sqrt {-\frac {1}{n^2}} n} \log \left (c x^n\right )-2 n \left (c x^n\right )^{-1/n}\right )}{4 n x}\) |
Input:
Int[Sin[a + (Sqrt[-n^(-2)]*Log[c*x^n])/2]^2/x^2,x]
Output:
((c*x^n)^n^(-1)*((E^(2*a*Sqrt[-n^(-2)]*n)*n)/(2*(c*x^n)^(2/n)) - (2*n)/(c* x^n)^n^(-1) - Log[c*x^n]/E^(2*a*Sqrt[-n^(-2)]*n)))/(4*n*x)
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])
Time = 6.98 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(\frac {\left (n -\ln \left (c \,x^{n}\right )\right ) \cos \left (\sqrt {-\frac {1}{n^{2}}}\, \ln \left (c \,x^{n}\right )+2 a \right )+n \left (\sqrt {-\frac {1}{n^{2}}}\, \ln \left (c \,x^{n}\right ) \sin \left (\sqrt {-\frac {1}{n^{2}}}\, \ln \left (c \,x^{n}\right )+2 a \right )-2\right )}{4 x n}\) | \(77\) |
Input:
int(sin(a+1/2*(-1/n^2)^(1/2)*ln(c*x^n))^2/x^2,x,method=_RETURNVERBOSE)
Output:
1/4*((n-ln(c*x^n))*cos((-1/n^2)^(1/2)*ln(c*x^n)+2*a)+n*((-1/n^2)^(1/2)*ln( c*x^n)*sin((-1/n^2)^(1/2)*ln(c*x^n)+2*a)-2))/x/n
Result contains complex when optimal does not.
Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.84 \[ \int \frac {\sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {{\left (2 \, x^{2} \log \left (x\right ) + 4 \, x e^{\left (\frac {2 i \, a n - \log \left (c\right )}{n}\right )} - e^{\left (\frac {2 \, {\left (2 i \, a n - \log \left (c\right )\right )}}{n}\right )}\right )} e^{\left (-\frac {2 i \, a n - \log \left (c\right )}{n}\right )}}{8 \, x^{2}} \] Input:
integrate(sin(a+1/2*(-1/n^2)^(1/2)*log(c*x^n))^2/x^2,x, algorithm="fricas" )
Output:
-1/8*(2*x^2*log(x) + 4*x*e^((2*I*a*n - log(c))/n) - e^(2*(2*I*a*n - log(c) )/n))*e^(-(2*I*a*n - log(c))/n)/x^2
Time = 13.99 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.42 \[ \int \frac {\sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )}{x^2} \, dx=\frac {\sqrt {- \frac {1}{n^{2}}} \log {\left (c x^{n} \right )} \sin {\left (2 a + \sqrt {- \frac {1}{n^{2}}} \log {\left (c x^{n} \right )} \right )}}{4 x} + \frac {\cos {\left (2 a + \sqrt {- \frac {1}{n^{2}}} \log {\left (c x^{n} \right )} \right )}}{4 x} - \frac {1}{2 x} - \frac {\log {\left (c x^{n} \right )} \cos {\left (2 a + \sqrt {- \frac {1}{n^{2}}} \log {\left (c x^{n} \right )} \right )}}{4 n x} \] Input:
integrate(sin(a+1/2*(-1/n**2)**(1/2)*ln(c*x**n))**2/x**2,x)
Output:
sqrt(-1/n**2)*log(c*x**n)*sin(2*a + sqrt(-1/n**2)*log(c*x**n))/(4*x) + cos (2*a + sqrt(-1/n**2)*log(c*x**n))/(4*x) - 1/(2*x) - log(c*x**n)*cos(2*a + sqrt(-1/n**2)*log(c*x**n))/(4*n*x)
Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.65 \[ \int \frac {\sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {2 \, c^{\frac {2}{n}} x^{3} \cos \left (2 \, a\right ) \log \left (x\right ) + 4 \, c^{\left (\frac {1}{n}\right )} x^{2} - x \cos \left (2 \, a\right )}{8 \, c^{\left (\frac {1}{n}\right )} x^{3}} \] Input:
integrate(sin(a+1/2*(-1/n^2)^(1/2)*log(c*x^n))^2/x^2,x, algorithm="maxima" )
Output:
-1/8*(2*c^(2/n)*x^3*cos(2*a)*log(x) + 4*c^(1/n)*x^2 - x*cos(2*a))/(c^(1/n) *x^3)
\[ \int \frac {\sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )}{x^2} \, dx=\int { \frac {\sin \left (\frac {1}{2} \, \sqrt {-\frac {1}{n^{2}}} \log \left (c x^{n}\right ) + a\right )^{2}}{x^{2}} \,d x } \] Input:
integrate(sin(a+1/2*(-1/n^2)^(1/2)*log(c*x^n))^2/x^2,x, algorithm="giac")
Output:
integrate(sin(1/2*sqrt(-1/n^2)*log(c*x^n) + a)^2/x^2, x)
Timed out. \[ \int \frac {\sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )}{x^2} \, dx=\int \frac {{\sin \left (a+\frac {\ln \left (c\,x^n\right )\,\sqrt {-\frac {1}{n^2}}}{2}\right )}^2}{x^2} \,d x \] Input:
int(sin(a + (log(c*x^n)*(-1/n^2)^(1/2))/2)^2/x^2,x)
Output:
int(sin(a + (log(c*x^n)*(-1/n^2)^(1/2))/2)^2/x^2, x)
Time = 0.17 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.15 \[ \int \frac {\sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )}{x^2} \, dx=\frac {2 \cos \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +2 a n}{2 n}\right ) \mathrm {log}\left (x^{n} c \right ) \sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +2 a n}{2 n}\right ) i -2 \cos \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +2 a n}{2 n}\right ) \sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +2 a n}{2 n}\right ) i n +2 \,\mathrm {log}\left (x^{n} c \right ) {\sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +2 a n}{2 n}\right )}^{2}-\mathrm {log}\left (x^{n} c \right )-4 {\sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +2 a n}{2 n}\right )}^{2} n}{4 n x} \] Input:
int(sin(a+1/2*(-1/n^2)^(1/2)*log(c*x^n))^2/x^2,x)
Output:
(2*cos((log(x**n*c)*i + 2*a*n)/(2*n))*log(x**n*c)*sin((log(x**n*c)*i + 2*a *n)/(2*n))*i - 2*cos((log(x**n*c)*i + 2*a*n)/(2*n))*sin((log(x**n*c)*i + 2 *a*n)/(2*n))*i*n + 2*log(x**n*c)*sin((log(x**n*c)*i + 2*a*n)/(2*n))**2 - l og(x**n*c) - 4*sin((log(x**n*c)*i + 2*a*n)/(2*n))**2*n)/(4*n*x)