Integrand size = 20, antiderivative size = 188 \[ \int x^3 \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3} \, dx=\frac {1}{8} x^4 \csc ^2\left (a+b x^n\right ) \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3}+\frac {4^{-1-\frac {2}{n}} e^{2 i a} x^4 \left (-i b x^n\right )^{-4/n} \csc ^2\left (a+b x^n\right ) \Gamma \left (\frac {4}{n},-2 i b x^n\right ) \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3}}{n}+\frac {4^{-1-\frac {2}{n}} e^{-2 i a} x^4 \left (i b x^n\right )^{-4/n} \csc ^2\left (a+b x^n\right ) \Gamma \left (\frac {4}{n},2 i b x^n\right ) \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3}}{n} \] Output:
1/8*x^4*csc(a+b*x^n)^2*(c*sin(a+b*x^n)^3)^(2/3)+4^(-1-2/n)*exp(2*I*a)*x^4* csc(a+b*x^n)^2*GAMMA(4/n,-2*I*b*x^n)*(c*sin(a+b*x^n)^3)^(2/3)/n/((-I*b*x^n )^(4/n))+4^(-1-2/n)*x^4*csc(a+b*x^n)^2*GAMMA(4/n,2*I*b*x^n)*(c*sin(a+b*x^n )^3)^(2/3)/exp(2*I*a)/n/((I*b*x^n)^(4/n))
Time = 0.80 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.86 \[ \int x^3 \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3} \, dx=\frac {2^{-3-\frac {4}{n}} e^{-2 i a} x^4 \left (b^2 x^{2 n}\right )^{-4/n} \csc ^2\left (a+b x^n\right ) \left (16^{\frac {1}{n}} e^{2 i a} n \left (b^2 x^{2 n}\right )^{4/n}+2 e^{4 i a} \left (i b x^n\right )^{4/n} \Gamma \left (\frac {4}{n},-2 i b x^n\right )+2 \left (-i b x^n\right )^{4/n} \Gamma \left (\frac {4}{n},2 i b x^n\right )\right ) \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3}}{n} \] Input:
Integrate[x^3*(c*Sin[a + b*x^n]^3)^(2/3),x]
Output:
(2^(-3 - 4/n)*x^4*Csc[a + b*x^n]^2*(16^n^(-1)*E^((2*I)*a)*n*(b^2*x^(2*n))^ (4/n) + 2*E^((4*I)*a)*(I*b*x^n)^(4/n)*Gamma[4/n, (-2*I)*b*x^n] + 2*((-I)*b *x^n)^(4/n)*Gamma[4/n, (2*I)*b*x^n])*(c*Sin[a + b*x^n]^3)^(2/3))/(E^((2*I) *a)*n*(b^2*x^(2*n))^(4/n))
Time = 0.52 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.73, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {7271, 3906, 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.
\(\displaystyle \int x^3 \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3} \, dx\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \csc ^2\left (a+b x^n\right ) \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3} \int x^3 \sin ^2\left (b x^n+a\right )dx\) |
\(\Big \downarrow \) 3906 |
\(\displaystyle \csc ^2\left (a+b x^n\right ) \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3} \int \left (\frac {x^3}{2}-\frac {1}{2} x^3 \cos \left (2 b x^n+2 a\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \csc ^2\left (a+b x^n\right ) \left (\frac {e^{2 i a} 4^{-\frac {2}{n}-1} x^4 \left (-i b x^n\right )^{-4/n} \Gamma \left (\frac {4}{n},-2 i b x^n\right )}{n}+\frac {e^{-2 i a} 4^{-\frac {2}{n}-1} x^4 \left (i b x^n\right )^{-4/n} \Gamma \left (\frac {4}{n},2 i b x^n\right )}{n}+\frac {x^4}{8}\right ) \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3}\) |
Input:
Int[x^3*(c*Sin[a + b*x^n]^3)^(2/3),x]
Output:
Csc[a + b*x^n]^2*(x^4/8 + (4^(-1 - 2/n)*E^((2*I)*a)*x^4*Gamma[4/n, (-2*I)* b*x^n])/(n*((-I)*b*x^n)^(4/n)) + (4^(-1 - 2/n)*x^4*Gamma[4/n, (2*I)*b*x^n] )/(E^((2*I)*a)*n*(I*b*x^n)^(4/n)))*(c*Sin[a + b*x^n]^3)^(2/3)
Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x _Symbol] :> Int[ExpandTrigReduce[(e*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 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])
\[\int x^{3} {\left (c \sin \left (a +b \,x^{n}\right )^{3}\right )}^{\frac {2}{3}}d x\]
Input:
int(x^3*(c*sin(a+b*x^n)^3)^(2/3),x)
Output:
int(x^3*(c*sin(a+b*x^n)^3)^(2/3),x)
\[ \int x^3 \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3} \, dx=\int { \left (c \sin \left (b x^{n} + a\right )^{3}\right )^{\frac {2}{3}} x^{3} \,d x } \] Input:
integrate(x^3*(c*sin(a+b*x^n)^3)^(2/3),x, algorithm="fricas")
Output:
integral((-(c*cos(b*x^n + a)^2 - c)*sin(b*x^n + a))^(2/3)*x^3, x)
Timed out. \[ \int x^3 \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3} \, dx=\text {Timed out} \] Input:
integrate(x**3*(c*sin(a+b*x**n)**3)**(2/3),x)
Output:
Timed out
\[ \int x^3 \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3} \, dx=\int { \left (c \sin \left (b x^{n} + a\right )^{3}\right )^{\frac {2}{3}} x^{3} \,d x } \] Input:
integrate(x^3*(c*sin(a+b*x^n)^3)^(2/3),x, algorithm="maxima")
Output:
-1/16*(x^4 - 4*integrate(x^3*cos(2*b*x^n + 2*a), x))*c^(2/3)
\[ \int x^3 \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3} \, dx=\int { \left (c \sin \left (b x^{n} + a\right )^{3}\right )^{\frac {2}{3}} x^{3} \,d x } \] Input:
integrate(x^3*(c*sin(a+b*x^n)^3)^(2/3),x, algorithm="giac")
Output:
integrate((c*sin(b*x^n + a)^3)^(2/3)*x^3, x)
Timed out. \[ \int x^3 \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3} \, dx=\int x^3\,{\left (c\,{\sin \left (a+b\,x^n\right )}^3\right )}^{2/3} \,d x \] Input:
int(x^3*(c*sin(a + b*x^n)^3)^(2/3),x)
Output:
int(x^3*(c*sin(a + b*x^n)^3)^(2/3), x)
\[ \int x^3 \left (c \sin ^3\left (a+b x^n\right )\right )^{2/3} \, dx=c^{\frac {2}{3}} \left (\int \sin \left (x^{n} b +a \right )^{2} x^{3}d x \right ) \] Input:
int(x^3*(c*sin(a+b*x^n)^3)^(2/3),x)
Output:
c**(2/3)*int(sin(x**n*b + a)**2*x**3,x)