Integrand size = 20, antiderivative size = 91 \[ \int x^3 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\frac {\left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{8 b^2}-\frac {x^2 \cot \left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 b}+\frac {1}{8} x^4 \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \] Output:
1/8*(c*sin(b*x^2+a)^3)^(2/3)/b^2-1/4*x^2*cot(b*x^2+a)*(c*sin(b*x^2+a)^3)^( 2/3)/b+1/8*x^4*csc(b*x^2+a)^2*(c*sin(b*x^2+a)^3)^(2/3)
Time = 0.49 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int x^3 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=-\frac {\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \left (\cos \left (2 \left (a+b x^2\right )\right )+2 b x^2 \left (-b x^2+\sin \left (2 \left (a+b x^2\right )\right )\right )\right )}{16 b^2} \] Input:
Integrate[x^3*(c*Sin[a + b*x^2]^3)^(2/3),x]
Output:
-1/16*(Csc[a + b*x^2]^2*(c*Sin[a + b*x^2]^3)^(2/3)*(Cos[2*(a + b*x^2)] + 2 *b*x^2*(-(b*x^2) + Sin[2*(a + b*x^2)])))/b^2
Time = 0.44 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {7271, 3860, 3042, 3791, 15}
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.
\(\displaystyle \int x^3 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \int x^3 \sin ^2\left (b x^2+a\right )dx\) |
\(\Big \downarrow \) 3860 |
\(\displaystyle \frac {1}{2} \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \int x^2 \sin ^2\left (b x^2+a\right )dx^2\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \int x^2 \sin \left (b x^2+a\right )^2dx^2\) |
\(\Big \downarrow \) 3791 |
\(\displaystyle \frac {1}{2} \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \left (\frac {\int x^2dx^2}{2}+\frac {\sin ^2\left (a+b x^2\right )}{4 b^2}-\frac {x^2 \sin \left (a+b x^2\right ) \cos \left (a+b x^2\right )}{2 b}\right )\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {1}{2} \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \left (\frac {\sin ^2\left (a+b x^2\right )}{4 b^2}-\frac {x^2 \sin \left (a+b x^2\right ) \cos \left (a+b x^2\right )}{2 b}+\frac {x^4}{4}\right )\) |
Input:
Int[x^3*(c*Sin[a + b*x^2]^3)^(2/3),x]
Output:
(Csc[a + b*x^2]^2*(c*Sin[a + b*x^2]^3)^(2/3)*(x^4/4 - (x^2*Cos[a + b*x^2]* Sin[a + b*x^2])/(2*b) + Sin[a + b*x^2]^2/(4*b^2)))/2
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Simp[b*(c + d*x)*Cos[e + f*x ]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^2*((n - 1)/n) Int[(c + d* x)*(b*Sin[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^ p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simplify[ (m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ (m + 1)/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])
Result contains complex when optimal does not.
Time = 1.31 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.20
method | result | size |
risch | \(-\frac {x^{4} \left (i c \,{\mathrm e}^{-3 i \left (b \,x^{2}+a \right )} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b \,x^{2}+a \right )}}{8 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}-\frac {i \left (2 b \,x^{2}+i\right ) \left (i c \,{\mathrm e}^{-3 i \left (b \,x^{2}+a \right )} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3}\right )^{\frac {2}{3}} {\mathrm e}^{4 i \left (b \,x^{2}+a \right )}}{32 b^{2} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}+\frac {i \left (i c \,{\mathrm e}^{-3 i \left (b \,x^{2}+a \right )} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3}\right )^{\frac {2}{3}} \left (2 b \,x^{2}-i\right )}{32 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2} b^{2}}\) | \(200\) |
Input:
int(x^3*(c*sin(b*x^2+a)^3)^(2/3),x,method=_RETURNVERBOSE)
Output:
-1/8*x^4/(exp(2*I*(b*x^2+a))-1)^2*(I*c*exp(-3*I*(b*x^2+a))*(exp(2*I*(b*x^2 +a))-1)^3)^(2/3)*exp(2*I*(b*x^2+a))-1/32*I/b^2*(2*b*x^2+I)/(exp(2*I*(b*x^2 +a))-1)^2*(I*c*exp(-3*I*(b*x^2+a))*(exp(2*I*(b*x^2+a))-1)^3)^(2/3)*exp(4*I *(b*x^2+a))+1/32*I*(I*c*exp(-3*I*(b*x^2+a))*(exp(2*I*(b*x^2+a))-1)^3)^(2/3 )/(exp(2*I*(b*x^2+a))-1)^2*(2*b*x^2-I)/b^2
Time = 0.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.05 \[ \int x^3 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=-\frac {{\left (2 \, b^{2} x^{4} - 4 \, b x^{2} \cos \left (b x^{2} + a\right ) \sin \left (b x^{2} + a\right ) - 2 \, \cos \left (b x^{2} + a\right )^{2} + 1\right )} \left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac {2}{3}}}{16 \, {\left (b^{2} \cos \left (b x^{2} + a\right )^{2} - b^{2}\right )}} \] Input:
integrate(x^3*(c*sin(b*x^2+a)^3)^(2/3),x, algorithm="fricas")
Output:
-1/16*(2*b^2*x^4 - 4*b*x^2*cos(b*x^2 + a)*sin(b*x^2 + a) - 2*cos(b*x^2 + a )^2 + 1)*(-(c*cos(b*x^2 + a)^2 - c)*sin(b*x^2 + a))^(2/3)/(b^2*cos(b*x^2 + a)^2 - b^2)
\[ \int x^3 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\int x^{3} \left (c \sin ^{3}{\left (a + b x^{2} \right )}\right )^{\frac {2}{3}}\, dx \] Input:
integrate(x**3*(c*sin(b*x**2+a)**3)**(2/3),x)
Output:
Integral(x**3*(c*sin(a + b*x**2)**3)**(2/3), x)
Time = 0.13 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.52 \[ \int x^3 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=-\frac {{\left (2 \, b^{2} x^{4} - 2 \, b x^{2} \sin \left (2 \, b x^{2} + 2 \, a\right ) - \cos \left (2 \, b x^{2} + 2 \, a\right )\right )} c^{\frac {2}{3}}}{32 \, b^{2}} \] Input:
integrate(x^3*(c*sin(b*x^2+a)^3)^(2/3),x, algorithm="maxima")
Output:
-1/32*(2*b^2*x^4 - 2*b*x^2*sin(2*b*x^2 + 2*a) - cos(2*b*x^2 + 2*a))*c^(2/3 )/b^2
\[ \int x^3 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\int { \left (c \sin \left (b x^{2} + a\right )^{3}\right )^{\frac {2}{3}} x^{3} \,d x } \] Input:
integrate(x^3*(c*sin(b*x^2+a)^3)^(2/3),x, algorithm="giac")
Output:
integrate((c*sin(b*x^2 + a)^3)^(2/3)*x^3, x)
Timed out. \[ \int x^3 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\int x^3\,{\left (c\,{\sin \left (b\,x^2+a\right )}^3\right )}^{2/3} \,d x \] Input:
int(x^3*(c*sin(a + b*x^2)^3)^(2/3),x)
Output:
int(x^3*(c*sin(a + b*x^2)^3)^(2/3), x)
\[ \int x^3 \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=c^{\frac {2}{3}} \left (\int \sin \left (b \,x^{2}+a \right )^{2} x^{3}d x \right ) \] Input:
int(x^3*(c*sin(b*x^2+a)^3)^(2/3),x)
Output:
c**(2/3)*int(sin(a + b*x**2)**2*x**3,x)