Integrand size = 30, antiderivative size = 218 \[ \int x^m \sin ^3\left (a+\frac {1}{6} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {9 e^{\frac {a \sqrt {-(1+m)^2}}{1+m}} x^{1+m} \left (c x^2\right )^{\frac {1}{6} (-1-m)}}{16 \sqrt {-(1+m)^2}}-\frac {9 e^{\frac {a (1+m)}{\sqrt {-(1+m)^2}}} x^{1+m} \left (c x^2\right )^{\frac {1+m}{6}}}{32 \sqrt {-(1+m)^2}}+\frac {e^{\frac {3 a (1+m)}{\sqrt {-(1+m)^2}}} x^{1+m} \left (c x^2\right )^{\frac {1+m}{2}}}{16 \sqrt {-(1+m)^2}}-\frac {e^{-\frac {3 a (1+m)}{\sqrt {-(1+m)^2}}} (1+m) x^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)} \log (x)}{8 \sqrt {-(1+m)^2}} \]
9/16*exp(a*(-(1+m)^2)^(1/2)/(1+m))*x^(1+m)*(c*x^2)^(-1/6-1/6*m)/(-(1+m)^2) ^(1/2)-9/32*exp(a*(1+m)/(-(1+m)^2)^(1/2))*x^(1+m)*(c*x^2)^(1/6+1/6*m)/(-(1 +m)^2)^(1/2)+1/16*exp(3*a*(1+m)/(-(1+m)^2)^(1/2))*x^(1+m)*(c*x^2)^(1/2+1/2 *m)/(-(1+m)^2)^(1/2)-1/8*(1+m)*x^(1+m)*(c*x^2)^(-1/2-1/2*m)*ln(x)/exp(3*a* (1+m)/(-(1+m)^2)^(1/2))/(-(1+m)^2)^(1/2)
\[ \int x^m \sin ^3\left (a+\frac {1}{6} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\int x^m \sin ^3\left (a+\frac {1}{6} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx \]
Time = 0.46 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.
\(\displaystyle \int x^m \sin ^3\left (a+\frac {1}{6} \sqrt {-(m+1)^2} \log \left (c x^2\right )\right ) \, dx\) |
\(\Big \downarrow \) 4996 |
\(\displaystyle \frac {1}{2} x^{m+1} \left (c x^2\right )^{\frac {1}{2} (-m-1)} \int \left (c x^2\right )^{\frac {m-1}{2}} \sin ^3\left (a+\frac {1}{6} \sqrt {-(m+1)^2} \log \left (c x^2\right )\right )d\left (c x^2\right )\) |
\(\Big \downarrow \) 4992 |
\(\displaystyle \frac {\sqrt {-(m+1)^2} x^{m+1} \left (c x^2\right )^{\frac {1}{2} (-m-1)} \int \left (-3 e^{\frac {a \sqrt {-(m+1)^2}}{m+1}} \left (c x^2\right )^{\frac {m-2}{3}}-e^{\frac {3 a (m+1)}{\sqrt {-(m+1)^2}}} \left (c x^2\right )^m+3 e^{\frac {a (m+1)}{\sqrt {-(m+1)^2}}} \left (c x^2\right )^{\frac {1}{3} (2 m-1)}+\frac {e^{-\frac {3 a (m+1)}{\sqrt {-(m+1)^2}}}}{c x^2}\right )d\left (c x^2\right )}{16 (m+1)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {-(m+1)^2} x^{m+1} \left (c x^2\right )^{\frac {1}{2} (-m-1)} \left (-\frac {9 e^{\frac {a \sqrt {-(m+1)^2}}{m+1}} \left (c x^2\right )^{\frac {m+1}{3}}}{m+1}+\frac {9 e^{\frac {a (m+1)}{\sqrt {-(m+1)^2}}} \left (c x^2\right )^{\frac {2 (m+1)}{3}}}{2 (m+1)}-\frac {e^{\frac {3 a (m+1)}{\sqrt {-(m+1)^2}}} \left (c x^2\right )^{m+1}}{m+1}+e^{-\frac {3 a (m+1)}{\sqrt {-(m+1)^2}}} \log \left (c x^2\right )\right )}{16 (m+1)}\) |
(Sqrt[-(1 + m)^2]*x^(1 + m)*(c*x^2)^((-1 - m)/2)*((-9*E^((a*Sqrt[-(1 + m)^ 2])/(1 + m))*(c*x^2)^((1 + m)/3))/(1 + m) + (9*E^((a*(1 + m))/Sqrt[-(1 + m )^2])*(c*x^2)^((2*(1 + m))/3))/(2*(1 + m)) - (E^((3*a*(1 + m))/Sqrt[-(1 + m)^2])*(c*x^2)^(1 + m))/(1 + m) + Log[c*x^2]/E^((3*a*(1 + m))/Sqrt[-(1 + m )^2])))/(16*(1 + m))
3.1.51.3.1 Defintions of rubi rules used
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])
\[\int x^{m} {\sin \left (a +\frac {\ln \left (c \,x^{2}\right ) \sqrt {-\left (1+m \right )^{2}}}{6}\right )}^{3}d x\]
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.45 \[ \int x^m \sin ^3\left (a+\frac {1}{6} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=-\frac {{\left (4 \, {\left (-i \, m - i\right )} e^{\left (-{\left (m + 1\right )} \log \left (c\right ) - 2 \, {\left (m + 1\right )} \log \left (x\right ) + 6 i \, a\right )} \log \left (x\right ) - 9 i \, e^{\left (-\frac {1}{3} \, {\left (m + 1\right )} \log \left (c\right ) - \frac {2}{3} \, {\left (m + 1\right )} \log \left (x\right ) + 2 i \, a\right )} + 18 i \, e^{\left (-\frac {2}{3} \, {\left (m + 1\right )} \log \left (c\right ) - \frac {4}{3} \, {\left (m + 1\right )} \log \left (x\right ) + 4 i \, a\right )} + 2 i\right )} e^{\left (\frac {1}{2} \, {\left (m + 1\right )} \log \left (c\right ) + 2 \, {\left (m + 1\right )} \log \left (x\right ) - 3 i \, a\right )}}{32 \, {\left (m + 1\right )}} \]
-1/32*(4*(-I*m - I)*e^(-(m + 1)*log(c) - 2*(m + 1)*log(x) + 6*I*a)*log(x) - 9*I*e^(-1/3*(m + 1)*log(c) - 2/3*(m + 1)*log(x) + 2*I*a) + 18*I*e^(-2/3* (m + 1)*log(c) - 4/3*(m + 1)*log(x) + 4*I*a) + 2*I)*e^(1/2*(m + 1)*log(c) + 2*(m + 1)*log(x) - 3*I*a)/(m + 1)
\[ \int x^m \sin ^3\left (a+\frac {1}{6} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\int x^{m} \sin ^{3}{\left (a + \frac {\sqrt {- m^{2} - 2 m - 1} \log {\left (c x^{2} \right )}}{6} \right )}\, dx \]
Time = 0.22 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.94 \[ \int x^m \sin ^3\left (a+\frac {1}{6} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\frac {9 \, {\left (\cos \left (2 \, a\right ) \sin \left (3 \, a\right ) - \cos \left (3 \, a\right ) \sin \left (2 \, a\right )\right )} c^{\frac {5}{6} \, m + \frac {5}{6}} x^{\frac {5}{3}} x^{\frac {4}{3} \, m} + 18 \, {\left (\cos \left (3 \, a\right ) \sin \left (4 \, a\right ) - \cos \left (4 \, a\right ) \sin \left (3 \, a\right )\right )} c^{\frac {1}{2} \, m + \frac {1}{2}} x x^{\frac {2}{3} \, m} - 2 \, {\left (c^{\frac {7}{6} \, m + 1} x^{2} x^{2 \, m} \sin \left (3 \, a\right ) + 2 \, {\left ({\left (\cos \left (3 \, a\right )^{2} \sin \left (3 \, a\right ) + \sin \left (3 \, a\right )^{3}\right )} c^{\frac {1}{6} \, m} m + {\left (\cos \left (3 \, a\right )^{2} \sin \left (3 \, a\right ) + \sin \left (3 \, a\right )^{3}\right )} c^{\frac {1}{6} \, m}\right )} \log \left (x\right )\right )} c^{\frac {1}{6}} x^{\frac {1}{3}}}{32 \, {\left ({\left (\cos \left (3 \, a\right )^{2} + \sin \left (3 \, a\right )^{2}\right )} c^{\frac {2}{3} \, m} m + {\left (\cos \left (3 \, a\right )^{2} + \sin \left (3 \, a\right )^{2}\right )} c^{\frac {2}{3} \, m}\right )} c^{\frac {2}{3}} x^{\frac {1}{3}}} \]
1/32*(9*(cos(2*a)*sin(3*a) - cos(3*a)*sin(2*a))*c^(5/6*m + 5/6)*x^(5/3)*x^ (4/3*m) + 18*(cos(3*a)*sin(4*a) - cos(4*a)*sin(3*a))*c^(1/2*m + 1/2)*x*x^( 2/3*m) - 2*(c^(7/6*m + 1)*x^2*x^(2*m)*sin(3*a) + 2*((cos(3*a)^2*sin(3*a) + sin(3*a)^3)*c^(1/6*m)*m + (cos(3*a)^2*sin(3*a) + sin(3*a)^3)*c^(1/6*m))*l og(x))*c^(1/6)*x^(1/3))/(((cos(3*a)^2 + sin(3*a)^2)*c^(2/3*m)*m + (cos(3*a )^2 + sin(3*a)^2)*c^(2/3*m))*c^(2/3)*x^(1/3))
Result contains complex when optimal does not.
Time = 2.86 (sec) , antiderivative size = 1297, normalized size of antiderivative = 5.95 \[ \int x^m \sin ^3\left (a+\frac {1}{6} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=\text {Too large to display} \]
1/8*(I*(m + 1)^2*m*x*x^m*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log(x) - 3* I*a) - 9*I*m^3*x*x^m*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log(x) - 3*I*a) - I*(m + 1)^2*x*x^m*abs(m + 1)*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log( x) - 3*I*a) + 9*I*m^2*x*x^m*abs(m + 1)*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log(x) - 3*I*a) - 27*I*(m + 1)^2*m*x*x^m*e^(1/6*abs(m + 1)*log(c) + 1/3 *abs(m + 1)*log(x) - I*a) + 27*I*m^3*x*x^m*e^(1/6*abs(m + 1)*log(c) + 1/3* abs(m + 1)*log(x) - I*a) + 9*I*(m + 1)^2*x*x^m*abs(m + 1)*e^(1/6*abs(m + 1 )*log(c) + 1/3*abs(m + 1)*log(x) - I*a) - 9*I*m^2*x*x^m*abs(m + 1)*e^(1/6* abs(m + 1)*log(c) + 1/3*abs(m + 1)*log(x) - I*a) + 27*I*(m + 1)^2*m*x*x^m* e^(-1/6*abs(m + 1)*log(c) - 1/3*abs(m + 1)*log(x) + I*a) - 27*I*m^3*x*x^m* e^(-1/6*abs(m + 1)*log(c) - 1/3*abs(m + 1)*log(x) + I*a) + 9*I*(m + 1)^2*x *x^m*abs(m + 1)*e^(-1/6*abs(m + 1)*log(c) - 1/3*abs(m + 1)*log(x) + I*a) - 9*I*m^2*x*x^m*abs(m + 1)*e^(-1/6*abs(m + 1)*log(c) - 1/3*abs(m + 1)*log(x ) + I*a) - I*(m + 1)^2*m*x*x^m*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log( x) + 3*I*a) + 9*I*m^3*x*x^m*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 3*I*a) - I*(m + 1)^2*x*x^m*abs(m + 1)*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 3*I*a) + 9*I*m^2*x*x^m*abs(m + 1)*e^(-1/2*abs(m + 1)*log(c) - abs(m + 1)*log(x) + 3*I*a) + I*(m + 1)^2*x*x^m*e^(1/2*abs(m + 1)*log(c) + abs(m + 1)*log(x) - 3*I*a) - 27*I*m^2*x*x^m*e^(1/2*abs(m + 1)*log(c) + a bs(m + 1)*log(x) - 3*I*a) + 18*I*m*x*x^m*abs(m + 1)*e^(1/2*abs(m + 1)*l...
Time = 29.61 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.33 \[ \int x^m \sin ^3\left (a+\frac {1}{6} \sqrt {-(1+m)^2} \log \left (c x^2\right )\right ) \, dx=-\frac {\frac {1}{c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}\,x\,x^m\,{\mathrm {e}}^{-a\,3{}\mathrm {i}}\,\frac {1}{{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}}\,1{}\mathrm {i}}{8\,m+8-\sqrt {-{\left (m+1\right )}^2}\,8{}\mathrm {i}}+\frac {c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}\,x\,x^m\,{\mathrm {e}}^{a\,3{}\mathrm {i}}\,{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{2}}\,1{}\mathrm {i}}{8\,m+8+\sqrt {-{\left (m+1\right )}^2}\,8{}\mathrm {i}}-\frac {\frac {1}{c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{6}}}\,x\,x^m\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{6}}}\,\left (27\,m+27+\sqrt {-{\left (m+1\right )}^2}\,9{}\mathrm {i}\right )\,1{}\mathrm {i}}{64\,{\left (m\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^2}+\frac {c^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{6}}\,x\,x^m\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (x^2\right )}^{\frac {\sqrt {-m^2-2\,m-1}\,1{}\mathrm {i}}{6}}\,\left (27\,m+27-\sqrt {-{\left (m+1\right )}^2}\,9{}\mathrm {i}\right )\,1{}\mathrm {i}}{64\,{\left (m\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^2} \]
(c^(((- 2*m - m^2 - 1)^(1/2)*1i)/2)*x*x^m*exp(a*3i)*(x^2)^(((- 2*m - m^2 - 1)^(1/2)*1i)/2)*1i)/(8*m + (-(m + 1)^2)^(1/2)*8i + 8) - (1/c^(((- 2*m - m ^2 - 1)^(1/2)*1i)/2)*x*x^m*exp(-a*3i)/(x^2)^(((- 2*m - m^2 - 1)^(1/2)*1i)/ 2)*1i)/(8*m - (-(m + 1)^2)^(1/2)*8i + 8) - (1/c^(((- 2*m - m^2 - 1)^(1/2)* 1i)/6)*x*x^m*exp(-a*1i)/(x^2)^(((- 2*m - m^2 - 1)^(1/2)*1i)/6)*(27*m + (-( m + 1)^2)^(1/2)*9i + 27)*1i)/(64*(m*1i + 1i)^2) + (c^(((- 2*m - m^2 - 1)^( 1/2)*1i)/6)*x*x^m*exp(a*1i)*(x^2)^(((- 2*m - m^2 - 1)^(1/2)*1i)/6)*(27*m - (-(m + 1)^2)^(1/2)*9i + 27)*1i)/(64*(m*1i + 1i)^2)