Integrand size = 18, antiderivative size = 65 \[ \int x \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=-\frac {\cot \left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3}}{4 b}+\frac {1}{4} x^2 \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \] Output:
-1/4*cot(b*x^2+a)*(c*sin(b*x^2+a)^3)^(2/3)/b+1/4*x^2*csc(b*x^2+a)^2*(c*sin (b*x^2+a)^3)^(2/3)
Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85 \[ \int x \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\frac {\csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \left (2 \left (a+b x^2\right )-\sin \left (2 \left (a+b x^2\right )\right )\right )}{8 b} \] Input:
Integrate[x*(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)*(2*(a + b*x^2) - Sin[2*(a + b *x^2)]))/(8*b)
Time = 0.39 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {7266, 3042, 3686, 3042, 3115, 24}
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.
\(\displaystyle \int x \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx\) |
\(\Big \downarrow \) 7266 |
\(\displaystyle \frac {1}{2} \int \left (c \sin ^3\left (b x^2+a\right )\right )^{2/3}dx^2\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \left (c \sin \left (b x^2+a\right )^3\right )^{2/3}dx^2\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \frac {1}{2} \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \int \sin ^2\left (b x^2+a\right )dx^2\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \int \sin \left (b x^2+a\right )^2dx^2\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{2} \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \left (\frac {\int 1dx^2}{2}-\frac {\sin \left (a+b x^2\right ) \cos \left (a+b x^2\right )}{2 b}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{2} \csc ^2\left (a+b x^2\right ) \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \left (\frac {x^2}{2}-\frac {\sin \left (a+b x^2\right ) \cos \left (a+b x^2\right )}{2 b}\right )\) |
Input:
Int[x*(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^2/2 - (Cos[a + b*x^2]*Sin[ a + b*x^2])/(2*b)))/2
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^ n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Si n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function OfQ[x^(m + 1), u, x]
Result contains complex when optimal does not.
Time = 1.79 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.80
method | result | size |
risch | \(-\frac {x^{2} \left (i c \,{\mathrm e}^{-3 i \left (b \,x^{2}+a \right )} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b \,x^{2}+a \right )}}{4 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}-\frac {i \left (i c \,{\mathrm e}^{-3 i \left (b \,x^{2}+a \right )} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3}\right )^{\frac {2}{3}} {\mathrm e}^{4 i \left (b \,x^{2}+a \right )}}{16 b \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}+\frac {i \left (i c \,{\mathrm e}^{-3 i \left (b \,x^{2}+a \right )} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3}\right )^{\frac {2}{3}}}{16 b \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{2}}\) | \(182\) |
Input:
int(x*(c*sin(b*x^2+a)^3)^(2/3),x,method=_RETURNVERBOSE)
Output:
-1/4*x^2/(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))-1/16*I/b/(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/ 16*I/b/(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)
Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.11 \[ \int x \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=-\frac {{\left (b x^{2} - \cos \left (b x^{2} + a\right ) \sin \left (b x^{2} + a\right )\right )} \left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac {2}{3}}}{4 \, {\left (b \cos \left (b x^{2} + a\right )^{2} - b\right )}} \] Input:
integrate(x*(c*sin(b*x^2+a)^3)^(2/3),x, algorithm="fricas")
Output:
-1/4*(b*x^2 - cos(b*x^2 + a)*sin(b*x^2 + a))*(-(c*cos(b*x^2 + a)^2 - c)*si n(b*x^2 + a))^(2/3)/(b*cos(b*x^2 + a)^2 - b)
\[ \int x \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\int x \left (c \sin ^{3}{\left (a + b x^{2} \right )}\right )^{\frac {2}{3}}\, dx \] Input:
integrate(x*(c*sin(b*x**2+a)**3)**(2/3),x)
Output:
Integral(x*(c*sin(a + b*x**2)**3)**(2/3), x)
Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.43 \[ \int x \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=-\frac {{\left (2 \, b x^{2} - \sin \left (2 \, b x^{2} + 2 \, a\right )\right )} c^{\frac {2}{3}}}{16 \, b} \] Input:
integrate(x*(c*sin(b*x^2+a)^3)^(2/3),x, algorithm="maxima")
Output:
-1/16*(2*b*x^2 - sin(2*b*x^2 + 2*a))*c^(2/3)/b
\[ \int x \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\int { \left (c \sin \left (b x^{2} + a\right )^{3}\right )^{\frac {2}{3}} x \,d x } \] Input:
integrate(x*(c*sin(b*x^2+a)^3)^(2/3),x, algorithm="giac")
Output:
integrate((c*sin(b*x^2 + a)^3)^(2/3)*x, x)
Timed out. \[ \int x \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\int x\,{\left (c\,{\sin \left (b\,x^2+a\right )}^3\right )}^{2/3} \,d x \] Input:
int(x*(c*sin(a + b*x^2)^3)^(2/3),x)
Output:
int(x*(c*sin(a + b*x^2)^3)^(2/3), x)
Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.49 \[ \int x \left (c \sin ^3\left (a+b x^2\right )\right )^{2/3} \, dx=\frac {c^{\frac {2}{3}} \left (-\cos \left (b \,x^{2}+a \right ) \sin \left (b \,x^{2}+a \right )+b \,x^{2}\right )}{4 b} \] Input:
int(x*(c*sin(b*x^2+a)^3)^(2/3),x)
Output:
(c**(2/3)*( - cos(a + b*x**2)*sin(a + b*x**2) + b*x**2))/(4*b)