Integrand size = 24, antiderivative size = 68 \[ \int \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x}{2}-\frac {1}{8} e^{-2 a \sqrt {-\frac {1}{n^2}} n} x \left (c x^n\right )^{\frac {1}{n}}-\frac {1}{4} e^{2 a \sqrt {-\frac {1}{n^2}} n} x \left (c x^n\right )^{-1/n} \log (x) \] Output:
1/2*x-1/8*x*(c*x^n)^(1/n)/exp(2*a*(-1/n^2)^(1/2)*n)-1/4*exp(2*a*(-1/n^2)^( 1/2)*n)*x*ln(x)/((c*x^n)^(1/n))
\[ \int \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right ) \, dx=\int \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right ) \, dx \] Input:
Integrate[Sin[a + (Sqrt[-n^(-2)]*Log[c*x^n])/2]^2,x]
Output:
Integrate[Sin[a + (Sqrt[-n^(-2)]*Log[c*x^n])/2]^2, x]
Time = 0.31 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.26, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4986, 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.
\(\displaystyle \int \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 4986 |
\(\displaystyle \frac {x \left (c x^n\right )^{-1/n} \int \left (c x^n\right )^{\frac {1}{n}-1} \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 4992 |
\(\displaystyle -\frac {x \left (c x^n\right )^{-1/n} \int \left (-2 \left (c x^n\right )^{\frac {1}{n}-1}+e^{-2 a \sqrt {-\frac {1}{n^2}} n} \left (c x^n\right )^{\frac {2}{n}-1}+\frac {e^{2 a \sqrt {-\frac {1}{n^2}} n} x^{-n}}{c}\right )d\left (c x^n\right )}{4 n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \left (c x^n\right )^{-1/n} \left (-\frac {1}{2} n e^{-2 a \sqrt {-\frac {1}{n^2}} n} \left (c x^n\right )^{2/n}-e^{2 a \sqrt {-\frac {1}{n^2}} n} \log \left (c x^n\right )+2 n \left (c x^n\right )^{\frac {1}{n}}\right )}{4 n}\) |
Input:
Int[Sin[a + (Sqrt[-n^(-2)]*Log[c*x^n])/2]^2,x]
Output:
(x*(2*n*(c*x^n)^n^(-1) - (n*(c*x^n)^(2/n))/(2*E^(2*a*Sqrt[-n^(-2)]*n)) - E ^(2*a*Sqrt[-n^(-2)]*n)*Log[c*x^n]))/(4*n*(c*x^n)^n^(-1))
Int[Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Si mp[x/(n*(c*x^n)^(1/n)) Subst[Int[x^(1/n - 1)*Sin[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, b, c, d, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])
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]
\[\int {\sin \left (a +\frac {\sqrt {-\frac {1}{n^{2}}}\, \ln \left (c \,x^{n}\right )}{2}\right )}^{2}d x\]
Input:
int(sin(a+1/2*(-1/n^2)^(1/2)*ln(c*x^n))^2,x)
Output:
int(sin(a+1/2*(-1/n^2)^(1/2)*ln(c*x^n))^2,x)
Result contains complex when optimal does not.
Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84 \[ \int \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right ) \, dx=-\frac {1}{8} \, {\left (x^{2} - 4 \, x e^{\left (\frac {2 i \, a n - \log \left (c\right )}{n}\right )} + 2 \, e^{\left (\frac {2 \, {\left (2 i \, a n - \log \left (c\right )\right )}}{n}\right )} \log \left (x\right )\right )} e^{\left (-\frac {2 i \, a n - \log \left (c\right )}{n}\right )} \] Input:
integrate(sin(a+1/2*(-1/n^2)^(1/2)*log(c*x^n))^2,x, algorithm="fricas")
Output:
-1/8*(x^2 - 4*x*e^((2*I*a*n - log(c))/n) + 2*e^(2*(2*I*a*n - log(c))/n)*lo g(x))*e^(-(2*I*a*n - log(c))/n)
\[ \int \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right ) \, dx=\int \sin ^{2}{\left (a + \frac {\sqrt {- \frac {1}{n^{2}}} \log {\left (c x^{n} \right )}}{2} \right )}\, dx \] Input:
integrate(sin(a+1/2*(-1/n**2)**(1/2)*ln(c*x**n))**2,x)
Output:
Integral(sin(a + sqrt(-1/n**2)*log(c*x**n)/2)**2, x)
Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.60 \[ \int \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right ) \, dx=-\frac {c^{\frac {2}{n}} x^{2} \cos \left (2 \, a\right ) - 4 \, c^{\left (\frac {1}{n}\right )} x + 2 \, \cos \left (2 \, a\right ) \log \left (x\right )}{8 \, c^{\left (\frac {1}{n}\right )}} \] Input:
integrate(sin(a+1/2*(-1/n^2)^(1/2)*log(c*x^n))^2,x, algorithm="maxima")
Output:
-1/8*(c^(2/n)*x^2*cos(2*a) - 4*c^(1/n)*x + 2*cos(2*a)*log(x))/c^(1/n)
Time = 0.40 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.01 \[ \int \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right ) \, dx=+\infty \] Input:
integrate(sin(a+1/2*(-1/n^2)^(1/2)*log(c*x^n))^2,x, algorithm="giac")
Output:
+Infinity
Time = 20.00 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.26 \[ \int \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x}{2}-\frac {x\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{\sqrt {-\frac {1}{n^2}}\,1{}\mathrm {i}}}\,1{}\mathrm {i}}{4\,n\,\sqrt {-\frac {1}{n^2}}+4{}\mathrm {i}}+\frac {x\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{\sqrt {-\frac {1}{n^2}}\,1{}\mathrm {i}}\,1{}\mathrm {i}}{4\,n\,\sqrt {-\frac {1}{n^2}}-4{}\mathrm {i}} \] Input:
int(sin(a + (log(c*x^n)*(-1/n^2)^(1/2))/2)^2,x)
Output:
x/2 - (x*exp(-a*2i)/(c*x^n)^((-1/n^2)^(1/2)*1i)*1i)/(4*n*(-1/n^2)^(1/2) + 4i) + (x*exp(a*2i)*(c*x^n)^((-1/n^2)^(1/2)*1i)*1i)/(4*n*(-1/n^2)^(1/2) - 4 i)
Time = 0.17 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.31 \[ \int \sin ^2\left (a+\frac {1}{2} \sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right ) \, dx=\frac {x \left (-2 \cos \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +2 a n}{2 n}\right ) \mathrm {log}\left (x^{n} c \right ) \sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +2 a n}{2 n}\right ) i -2 \cos \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +2 a n}{2 n}\right ) \sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +2 a n}{2 n}\right ) i n +2 \,\mathrm {log}\left (x^{n} c \right ) {\sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +2 a n}{2 n}\right )}^{2}-\mathrm {log}\left (x^{n} c \right )+4 {\sin \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +2 a n}{2 n}\right )}^{2} n \right )}{4 n} \] Input:
int(sin(a+1/2*(-1/n^2)^(1/2)*log(c*x^n))^2,x)
Output:
(x*( - 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 - log(x**n*c) + 4*sin((log(x**n*c)*i + 2*a*n)/(2*n))**2*n))/(4*n)