Integrand size = 20, antiderivative size = 153 \[ \int x^m \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=\frac {1}{4} i e^{i a} x^{1+m} \left (-i b x^2\right )^{\frac {1}{2} (-1-m)} \csc \left (a+b x^2\right ) \Gamma \left (\frac {1+m}{2},-i b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\frac {1}{4} i e^{-i a} x^{1+m} \left (i b x^2\right )^{\frac {1}{2} (-1-m)} \csc \left (a+b x^2\right ) \Gamma \left (\frac {1+m}{2},i b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \]
1/4*I*exp(I*a)*x^(1+m)*(-I*b*x^2)^(-1/2-1/2*m)*csc(b*x^2+a)*GAMMA(1/2+1/2* m,-I*b*x^2)*(c*sin(b*x^2+a)^3)^(1/3)-1/4*I*x^(1+m)*(I*b*x^2)^(-1/2-1/2*m)* csc(b*x^2+a)*GAMMA(1/2+1/2*m,I*b*x^2)*(c*sin(b*x^2+a)^3)^(1/3)/exp(I*a)
Time = 0.33 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.90 \[ \int x^m \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=\frac {1}{4} i x^{1+m} \left (b^2 x^4\right )^{\frac {1}{2} (-1-m)} \csc \left (a+b x^2\right ) \left (-\left (-i b x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},i b x^2\right ) (\cos (a)-i \sin (a))+\left (i b x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},-i b x^2\right ) (\cos (a)+i \sin (a))\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \]
(I/4)*x^(1 + m)*(b^2*x^4)^((-1 - m)/2)*Csc[a + b*x^2]*(-(((-I)*b*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, I*b*x^2]*(Cos[a] - I*Sin[a])) + (I*b*x^2)^((1 + m )/2)*Gamma[(1 + m)/2, (-I)*b*x^2]*(Cos[a] + I*Sin[a]))*(c*Sin[a + b*x^2]^3 )^(1/3)
Time = 0.49 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {7271, 3870, 2648}
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.
\(\displaystyle \int x^m \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \int x^m \sin \left (b x^2+a\right )dx\) |
\(\Big \downarrow \) 3870 |
\(\displaystyle \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \left (\frac {1}{2} i \int e^{-i b x^2-i a} x^mdx-\frac {1}{2} i \int e^{i b x^2+i a} x^mdx\right )\) |
\(\Big \downarrow \) 2648 |
\(\displaystyle \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \left (\frac {1}{4} i e^{i a} x^{m+1} \left (-i b x^2\right )^{\frac {1}{2} (-m-1)} \Gamma \left (\frac {m+1}{2},-i b x^2\right )-\frac {1}{4} i e^{-i a} x^{m+1} \left (i b x^2\right )^{\frac {1}{2} (-m-1)} \Gamma \left (\frac {m+1}{2},i b x^2\right )\right )\) |
Csc[a + b*x^2]*((I/4)*E^(I*a)*x^(1 + m)*((-I)*b*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, (-I)*b*x^2] - ((I/4)*x^(1 + m)*(I*b*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, I*b*x^2])/E^(I*a))*(c*Sin[a + b*x^2]^3)^(1/3)
3.4.18.3.1 Defintions of rubi rules used
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_ .), x_Symbol] :> Simp[(-F^a)*((e + f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[ F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; FreeQ[{F , a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]
Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[I/2 Int[(e*x)^m*E^((-c)*I - d*I*x^n), x], x] - Simp[I/2 Int[(e*x)^m*E^(c*I + d*I*x^n), x], x] /; FreeQ[{c, d, e, m}, x] && IGtQ[n, 0]
Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*v^m)^ FracPart[p]/v^(m*FracPart[p])) Int[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] && !IntegerQ[p] && !FreeQ[v, x] && !(EqQ[a, 1] && EqQ[m, 1]) && !(Eq Q[v, x] && EqQ[m, 1])
\[\int x^{m} {\left (c \left (\sin ^{3}\left (b \,x^{2}+a \right )\right )\right )}^{\frac {1}{3}}d x\]
Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.64 \[ \int x^m \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=-\frac {{\left (e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (i \, b\right ) - i \, a\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, i \, b x^{2}\right ) + e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-i \, b\right ) + i \, a\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -i \, b x^{2}\right )\right )} \left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac {1}{3}}}{4 \, b \sin \left (b x^{2} + a\right )} \]
-1/4*(e^(-1/2*(m - 1)*log(I*b) - I*a)*gamma(1/2*m + 1/2, I*b*x^2) + e^(-1/ 2*(m - 1)*log(-I*b) + I*a)*gamma(1/2*m + 1/2, -I*b*x^2))*(-(c*cos(b*x^2 + a)^2 - c)*sin(b*x^2 + a))^(1/3)/(b*sin(b*x^2 + a))
\[ \int x^m \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=\int x^{m} \sqrt [3]{c \sin ^{3}{\left (a + b x^{2} \right )}}\, dx \]
\[ \int x^m \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=\int { \left (c \sin \left (b x^{2} + a\right )^{3}\right )^{\frac {1}{3}} x^{m} \,d x } \]
\[ \int x^m \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=\int { \left (c \sin \left (b x^{2} + a\right )^{3}\right )^{\frac {1}{3}} x^{m} \,d x } \]
Timed out. \[ \int x^m \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \, dx=\int x^m\,{\left (c\,{\sin \left (b\,x^2+a\right )}^3\right )}^{1/3} \,d x \]