Integrand size = 28, antiderivative size = 112 \[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=-\frac {e^{\frac {a (1+m)}{\sqrt {-(1+m)^2}}} x^{1+m} \left (c x^2\right )^{\frac {1+m}{2}}}{4 \sqrt {-(1+m)^2}}+\frac {e^{\frac {a \sqrt {-(1+m)^2}}{1+m}} (1+m) x^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)} \log (x)}{2 \sqrt {-(1+m)^2}} \] Output:
-1/4*exp(a*(1+m)/(-(1+m)^2)^(1/2))*x^(1+m)*(c*x^2)^(1/2+1/2*m)/(-(1+m)^2)^ (1/2)+1/2*exp(a*(-(1+m)^2)^(1/2)/(1+m))*(1+m)*x^(1+m)*(c*x^2)^(-1/2-1/2*m) *ln(x)/(-(1+m)^2)^(1/2)
\[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx \] Input:
Integrate[x^m*Sin[a + (Sqrt[-(1 + m)^2]*Log[c*x^2])/2],x]
Output:
Integrate[x^m*Sin[a + (Sqrt[-(1 + m)^2]*Log[c*x^2])/2], x]
Time = 0.37 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.89, 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.
\(\displaystyle \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(m+1)^2} \log \left (c x^2\right )\right ) \, dx\) |
\(\Big \downarrow \) 4996 |
\(\displaystyle \frac {1}{2} x^{m+1} \left (c x^2\right )^{\frac {1}{2} (-m-1)} \int \left (c x^2\right )^{\frac {m-1}{2}} \sin \left (a+\frac {1}{2} \sqrt {-(m+1)^2} \log \left (c x^2\right )\right )d\left (c x^2\right )\) |
\(\Big \downarrow \) 4992 |
\(\displaystyle \frac {(m+1) x^{m+1} \left (c x^2\right )^{\frac {1}{2} (-m-1)} \int \left (\frac {e^{\frac {a \sqrt {-(m+1)^2}}{m+1}}}{c x^2}-e^{\frac {a (m+1)}{\sqrt {-(m+1)^2}}} \left (c x^2\right )^m\right )d\left (c x^2\right )}{4 \sqrt {-(m+1)^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(m+1) x^{m+1} \left (c x^2\right )^{\frac {1}{2} (-m-1)} \left (e^{\frac {a \sqrt {-(m+1)^2}}{m+1}} \log \left (c x^2\right )-\frac {e^{\frac {a (m+1)}{\sqrt {-(m+1)^2}}} \left (c x^2\right )^{m+1}}{m+1}\right )}{4 \sqrt {-(m+1)^2}}\) |
Input:
Int[x^m*Sin[a + (Sqrt[-(1 + m)^2]*Log[c*x^2])/2],x]
Output:
((1 + m)*x^(1 + m)*(c*x^2)^((-1 - m)/2)*(-((E^((a*(1 + m))/Sqrt[-(1 + m)^2 ])*(c*x^2)^(1 + m))/(1 + m)) + E^((a*Sqrt[-(1 + m)^2])/(1 + m))*Log[c*x^2] ))/(4*Sqrt[-(1 + m)^2])
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])
\[\int x^{m} \sin \left (a +\frac {\sqrt {-\left (1+m \right )^{2}}\, \ln \left (c \,x^{2}\right )}{2}\right )d x\]
Input:
int(x^m*sin(a+1/2*(-(1+m)^2)^(1/2)*ln(c*x^2)),x)
Output:
int(x^m*sin(a+1/2*(-(1+m)^2)^(1/2)*ln(c*x^2)),x)
Result contains complex when optimal does not.
Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.46 \[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {{\left (i \, x^{2} x^{2 \, m} - 2 \, {\left (i \, m + i\right )} e^{\left (-{\left (m + 1\right )} \log \left (c\right ) + 2 i \, a\right )} \log \left (x\right )\right )} e^{\left (\frac {1}{2} \, {\left (m + 1\right )} \log \left (c\right ) - i \, a\right )}}{4 \, {\left (m + 1\right )}} \] Input:
integrate(x^m*sin(a+1/2*(-(1+m)^2)^(1/2)*log(c*x^2)),x, algorithm="fricas" )
Output:
1/4*(I*x^2*x^(2*m) - 2*(I*m + I)*e^(-(m + 1)*log(c) + 2*I*a)*log(x))*e^(1/ 2*(m + 1)*log(c) - I*a)/(m + 1)
\[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\int x^{m} \sin {\left (a + \frac {\sqrt {- m^{2} - 2 m - 1} \log {\left (c x^{2} \right )}}{2} \right )}\, dx \] Input:
integrate(x**m*sin(a+1/2*(-(1+m)**2)**(1/2)*ln(c*x**2)),x)
Output:
Integral(x**m*sin(a + sqrt(-m**2 - 2*m - 1)*log(c*x**2)/2), x)
Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.43 \[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {c^{m + 1} x^{2} x^{2 \, m} \sin \left (a\right ) + 2 \, {\left (m \sin \left (a\right ) + \sin \left (a\right )\right )} \log \left (x\right )}{4 \, {\left (c^{\frac {1}{2} \, m} m + c^{\frac {1}{2} \, m}\right )} \sqrt {c}} \] Input:
integrate(x^m*sin(a+1/2*(-(1+m)^2)^(1/2)*log(c*x^2)),x, algorithm="maxima" )
Output:
1/4*(c^(m + 1)*x^2*x^(2*m)*sin(a) + 2*(m*sin(a) + sin(a))*log(x))/((c^(1/2 *m)*m + c^(1/2*m))*sqrt(c))
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.69 \[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=-\frac {i \, m x x^{m} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - i \, a\right )} - i \, x x^{m} {\left | m + 1 \right |} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - i \, a\right )} - i \, m x x^{m} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + i \, a\right )} - i \, x x^{m} {\left | m + 1 \right |} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + i \, a\right )} + i \, x x^{m} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - i \, a\right )} - i \, x x^{m} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + i \, a\right )}}{2 \, {\left ({\left (m + 1\right )}^{2} - m^{2} - 2 \, m - 1\right )}} \] Input:
integrate(x^m*sin(a+1/2*(-(1+m)^2)^(1/2)*log(c*x^2)),x, algorithm="giac")
Output:
-1/2*(I*m*x*x^m*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log(x) - I*a) - I*x* x^m*abs(m + 1)*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log(x) - I*a) - I*m*x *x^m*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + I*a) - I*x*x^m*abs(m + 1)*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + I*a) + I*x*x^m*e^(1/2 *abs(m + 1)*log(c) + abs(m + 1)*log(x) - I*a) - I*x*x^m*e^(-1/2*abs(m + 1) *log(c) - abs(m + 1)*log(x) + I*a))/((m + 1)^2 - m^2 - 2*m - 1)
Time = 21.49 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.24 \[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {\frac {1}{c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}\,x\,x^m\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}\,1{}\mathrm {i}}{2\,m+2-\sqrt {-{\left (m+1\right )}^2}\,2{}\mathrm {i}}-\frac {c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}\,x\,x^m\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}\,1{}\mathrm {i}}{2\,m+2+\sqrt {-{\left (m+1\right )}^2}\,2{}\mathrm {i}} \] Input:
int(x^m*sin(a + (log(c*x^2)*(-(m + 1)^2)^(1/2))/2),x)
Output:
(1/c^(((- 2*m - m^2 - 1)^(1/2)*1i)/2)*x*x^m*exp(-a*1i)/(x^2)^(((- 2*m - m^ 2 - 1)^(1/2)*1i)/2)*1i)/(2*m - (-(m + 1)^2)^(1/2)*2i + 2) - (c^(((- 2*m - m^2 - 1)^(1/2)*1i)/2)*x*x^m*exp(a*1i)*(x^2)^(((- 2*m - m^2 - 1)^(1/2)*1i)/ 2)*1i)/(2*m + (-(m + 1)^2)^(1/2)*2i + 2)
Time = 0.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.49 \[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {x^{m} x \left (-\cos \left (\frac {\mathrm {log}\left (c \,x^{2}\right ) m}{2}+\frac {\mathrm {log}\left (c \,x^{2}\right )}{2}+a \right )+\sin \left (\frac {\mathrm {log}\left (c \,x^{2}\right ) m}{2}+\frac {\mathrm {log}\left (c \,x^{2}\right )}{2}+a \right )\right )}{2 m +2} \] Input:
int(x^m*sin(a+1/2*(-(1+m)^2)^(1/2)*log(c*x^2)),x)
Output:
(x**m*x*( - cos((log(c*x**2)*m + log(c*x**2) + 2*a)/2) + sin((log(c*x**2)* m + log(c*x**2) + 2*a)/2)))/(2*(m + 1))