Integrand size = 28, antiderivative size = 112 \[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=-\frac {e^{\frac {a (1+m)}{\sqrt {-(1+m)^2}}} x^{1+m} \left (c x^2\right )^{\frac {1+m}{2}}}{4 \sqrt {-(1+m)^2}}+\frac {e^{\frac {a \sqrt {-(1+m)^2}}{1+m}} (1+m) x^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)} \log (x)}{2 \sqrt {-(1+m)^2}} \]
[Out]
Time = 0.21 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {4581, 4577} \[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {(m+1) e^{\frac {a \sqrt {-(m+1)^2}}{m+1}} x^{m+1} \log (x) \left (c x^2\right )^{\frac {1}{2} (-m-1)}}{2 \sqrt {-(m+1)^2}}-\frac {e^{\frac {a (m+1)}{\sqrt {-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac {m+1}{2}}}{4 \sqrt {-(m+1)^2}} \]
[In]
[Out]
Rule 4577
Rule 4581
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \left (x^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)}\right ) \text {Subst}\left (\int x^{-1+\frac {1+m}{2}} \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log (x)\right ) \, dx,x,c x^2\right ) \\ & = \frac {\left ((1+m) x^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)}\right ) \text {Subst}\left (\int \left (\frac {e^{\frac {a \sqrt {-(1+m)^2}}{1+m}}}{x}-e^{\frac {a (1+m)}{\sqrt {-(1+m)^2}}} x^m\right ) \, dx,x,c x^2\right )}{4 \sqrt {-(1+m)^2}} \\ & = -\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}} \\ \end{align*}
\[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx \]
[In]
[Out]
\[\int x^{m} \sin \left (a +\frac {\ln \left (c \,x^{2}\right ) \sqrt {-\left (1+m \right )^{2}}}{2}\right )d x\]
[In]
[Out]
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.46 \[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {{\left (i \, x^{2} x^{2 \, m} - 2 \, {\left (i \, m + i\right )} e^{\left (-{\left (m + 1\right )} \log \left (c\right ) + 2 i \, a\right )} \log \left (x\right )\right )} e^{\left (\frac {1}{2} \, {\left (m + 1\right )} \log \left (c\right ) - i \, a\right )}}{4 \, {\left (m + 1\right )}} \]
[In]
[Out]
\[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\int x^{m} \sin {\left (a + \frac {\sqrt {- m^{2} - 2 m - 1} \log {\left (c x^{2} \right )}}{2} \right )}\, dx \]
[In]
[Out]
none
Time = 0.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.43 \[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {c^{m + 1} x^{2} x^{2 \, m} \sin \left (a\right ) + 2 \, {\left (m \sin \left (a\right ) + \sin \left (a\right )\right )} \log \left (x\right )}{4 \, {\left (c^{\frac {1}{2} \, m} m + c^{\frac {1}{2} \, m}\right )} \sqrt {c}} \]
[In]
[Out]
Result contains complex when optimal does not.
Time = 0.58 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.69 \[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=-\frac {i \, m x x^{m} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - i \, a\right )} - i \, x x^{m} {\left | m + 1 \right |} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - i \, a\right )} - i \, m x x^{m} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + i \, a\right )} - i \, x x^{m} {\left | m + 1 \right |} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + i \, a\right )} + i \, x x^{m} e^{\left (\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) + {\left | m + 1 \right |} \log \left (x\right ) - i \, a\right )} - i \, x x^{m} e^{\left (-\frac {1}{2} \, {\left | m + 1 \right |} \log \left (c\right ) - {\left | m + 1 \right |} \log \left (x\right ) + i \, a\right )}}{2 \, {\left ({\left (m + 1\right )}^{2} - m^{2} - 2 \, m - 1\right )}} \]
[In]
[Out]
Time = 28.92 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.24 \[ \int x^m \sin \left (a+\frac {1}{2} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {\frac {1}{c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}\,x\,x^m\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}\,1{}\mathrm {i}}{2\,m+2-\sqrt {-{\left (m+1\right )}^2}\,2{}\mathrm {i}}-\frac {c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}\,x\,x^m\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}\,1{}\mathrm {i}}{2\,m+2+\sqrt {-{\left (m+1\right )}^2}\,2{}\mathrm {i}} \]
[In]
[Out]