Integrand size = 18, antiderivative size = 96 \[ \int x^3 \sqrt [3]{c \sin ^3(a+b x)} \, dx=-\frac {6 \sqrt [3]{c \sin ^3(a+b x)}}{b^4}+\frac {3 x^2 \sqrt [3]{c \sin ^3(a+b x)}}{b^2}+\frac {6 x \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b^3}-\frac {x^3 \cot (a+b x) \sqrt [3]{c \sin ^3(a+b x)}}{b} \] Output:
-6*(c*sin(b*x+a)^3)^(1/3)/b^4+3*x^2*(c*sin(b*x+a)^3)^(1/3)/b^2+6*x*cot(b*x +a)*(c*sin(b*x+a)^3)^(1/3)/b^3-x^3*cot(b*x+a)*(c*sin(b*x+a)^3)^(1/3)/b
Time = 0.61 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.49 \[ \int x^3 \sqrt [3]{c \sin ^3(a+b x)} \, dx=-\frac {\left (6-3 b^2 x^2+b x \left (-6+b^2 x^2\right ) \cot (a+b x)\right ) \sqrt [3]{c \sin ^3(a+b x)}}{b^4} \] Input:
Integrate[x^3*(c*Sin[a + b*x]^3)^(1/3),x]
Output:
-(((6 - 3*b^2*x^2 + b*x*(-6 + b^2*x^2)*Cot[a + b*x])*(c*Sin[a + b*x]^3)^(1 /3))/b^4)
Time = 0.63 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.86, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {7271, 3042, 3777, 3042, 3777, 25, 3042, 3777, 3042, 3117}
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.
\(\displaystyle \int x^3 \sqrt [3]{c \sin ^3(a+b x)} \, dx\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} \int x^3 \sin (a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} \int x^3 \sin (a+b x)dx\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} \left (\frac {3 \int x^2 \cos (a+b x)dx}{b}-\frac {x^3 \cos (a+b x)}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} \left (\frac {3 \int x^2 \sin \left (a+b x+\frac {\pi }{2}\right )dx}{b}-\frac {x^3 \cos (a+b x)}{b}\right )\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} \left (\frac {3 \left (\frac {2 \int -x \sin (a+b x)dx}{b}+\frac {x^2 \sin (a+b x)}{b}\right )}{b}-\frac {x^3 \cos (a+b x)}{b}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} \left (\frac {3 \left (\frac {x^2 \sin (a+b x)}{b}-\frac {2 \int x \sin (a+b x)dx}{b}\right )}{b}-\frac {x^3 \cos (a+b x)}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} \left (\frac {3 \left (\frac {x^2 \sin (a+b x)}{b}-\frac {2 \int x \sin (a+b x)dx}{b}\right )}{b}-\frac {x^3 \cos (a+b x)}{b}\right )\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} \left (\frac {3 \left (\frac {x^2 \sin (a+b x)}{b}-\frac {2 \left (\frac {\int \cos (a+b x)dx}{b}-\frac {x \cos (a+b x)}{b}\right )}{b}\right )}{b}-\frac {x^3 \cos (a+b x)}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} \left (\frac {3 \left (\frac {x^2 \sin (a+b x)}{b}-\frac {2 \left (\frac {\int \sin \left (a+b x+\frac {\pi }{2}\right )dx}{b}-\frac {x \cos (a+b x)}{b}\right )}{b}\right )}{b}-\frac {x^3 \cos (a+b x)}{b}\right )\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} \left (\frac {3 \left (\frac {x^2 \sin (a+b x)}{b}-\frac {2 \left (\frac {\sin (a+b x)}{b^2}-\frac {x \cos (a+b x)}{b}\right )}{b}\right )}{b}-\frac {x^3 \cos (a+b x)}{b}\right )\) |
Input:
Int[x^3*(c*Sin[a + b*x]^3)^(1/3),x]
Output:
Csc[a + b*x]*(c*Sin[a + b*x]^3)^(1/3)*(-((x^3*Cos[a + b*x])/b) + (3*((x^2* Sin[a + b*x])/b - (2*(-((x*Cos[a + b*x])/b) + Sin[a + b*x]/b^2))/b))/b)
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 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 = 1.21 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.57
method | result | size |
risch | \(-\frac {i \left (b^{3} x^{3}+3 i x^{2} b^{2}-6 b x -6 i\right ) \left (i c \,{\mathrm e}^{-3 i \left (b x +a \right )} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3}\right )^{\frac {1}{3}} {\mathrm e}^{2 i \left (b x +a \right )}}{2 b^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {i \left (i c \,{\mathrm e}^{-3 i \left (b x +a \right )} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3}\right )^{\frac {1}{3}} \left (b^{3} x^{3}-3 i x^{2} b^{2}-6 b x +6 i\right )}{2 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right ) b^{4}}\) | \(151\) |
Input:
int(x^3*(c*sin(b*x+a)^3)^(1/3),x,method=_RETURNVERBOSE)
Output:
-1/2*I/b^4*(b^3*x^3+3*I*x^2*b^2-6*b*x-6*I)/(exp(2*I*(b*x+a))-1)*(I*c*exp(- 3*I*(b*x+a))*(exp(2*I*(b*x+a))-1)^3)^(1/3)*exp(2*I*(b*x+a))-1/2*I*(I*c*exp (-3*I*(b*x+a))*(exp(2*I*(b*x+a))-1)^3)^(1/3)/(exp(2*I*(b*x+a))-1)*(b^3*x^3 -3*I*x^2*b^2-6*b*x+6*I)/b^4
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.77 \[ \int x^3 \sqrt [3]{c \sin ^3(a+b x)} \, dx=-\frac {{\left ({\left (b^{3} x^{3} - 6 \, b x\right )} \cos \left (b x + a\right ) - 3 \, {\left (b^{2} x^{2} - 2\right )} \sin \left (b x + a\right )\right )} \left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{\frac {1}{3}}}{b^{4} \sin \left (b x + a\right )} \] Input:
integrate(x^3*(c*sin(b*x+a)^3)^(1/3),x, algorithm="fricas")
Output:
-((b^3*x^3 - 6*b*x)*cos(b*x + a) - 3*(b^2*x^2 - 2)*sin(b*x + a))*(-(c*cos( b*x + a)^2 - c)*sin(b*x + a))^(1/3)/(b^4*sin(b*x + a))
Time = 1.66 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.34 \[ \int x^3 \sqrt [3]{c \sin ^3(a+b x)} \, dx=\begin {cases} \frac {x^{4} \sqrt [3]{c \sin ^{3}{\left (a \right )}}}{4} & \text {for}\: b = 0 \\0 & \text {for}\: a = - b x \vee a = - b x + \pi \\- \frac {x^{3} \sqrt [3]{c \sin ^{3}{\left (a + b x \right )}} \cos {\left (a + b x \right )}}{b \sin {\left (a + b x \right )}} + \frac {3 x^{2} \sqrt [3]{c \sin ^{3}{\left (a + b x \right )}}}{b^{2}} + \frac {6 x \sqrt [3]{c \sin ^{3}{\left (a + b x \right )}} \cos {\left (a + b x \right )}}{b^{3} \sin {\left (a + b x \right )}} - \frac {6 \sqrt [3]{c \sin ^{3}{\left (a + b x \right )}}}{b^{4}} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*(c*sin(b*x+a)**3)**(1/3),x)
Output:
Piecewise((x**4*(c*sin(a)**3)**(1/3)/4, Eq(b, 0)), (0, Eq(a, -b*x) | Eq(a, -b*x + pi)), (-x**3*(c*sin(a + b*x)**3)**(1/3)*cos(a + b*x)/(b*sin(a + b* x)) + 3*x**2*(c*sin(a + b*x)**3)**(1/3)/b**2 + 6*x*(c*sin(a + b*x)**3)**(1 /3)*cos(a + b*x)/(b**3*sin(a + b*x)) - 6*(c*sin(a + b*x)**3)**(1/3)/b**4, True))
Time = 0.17 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.52 \[ \int x^3 \sqrt [3]{c \sin ^3(a+b x)} \, dx=\frac {3 \, {\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} a^{2} c^{\frac {1}{3}} - 3 \, {\left ({\left ({\left (b x + a\right )}^{2} - 2\right )} \cos \left (b x + a\right ) - 2 \, {\left (b x + a\right )} \sin \left (b x + a\right )\right )} a c^{\frac {1}{3}} + \frac {4 \, a^{3} c^{\frac {1}{3}}}{\frac {\sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1} + {\left ({\left ({\left (b x + a\right )}^{3} - 6 \, b x - 6 \, a\right )} \cos \left (b x + a\right ) - 3 \, {\left ({\left (b x + a\right )}^{2} - 2\right )} \sin \left (b x + a\right )\right )} c^{\frac {1}{3}}}{2 \, b^{4}} \] Input:
integrate(x^3*(c*sin(b*x+a)^3)^(1/3),x, algorithm="maxima")
Output:
1/2*(3*((b*x + a)*cos(b*x + a) - sin(b*x + a))*a^2*c^(1/3) - 3*(((b*x + a) ^2 - 2)*cos(b*x + a) - 2*(b*x + a)*sin(b*x + a))*a*c^(1/3) + 4*a^3*c^(1/3) /(sin(b*x + a)^2/(cos(b*x + a) + 1)^2 + 1) + (((b*x + a)^3 - 6*b*x - 6*a)* cos(b*x + a) - 3*((b*x + a)^2 - 2)*sin(b*x + a))*c^(1/3))/b^4
\[ \int x^3 \sqrt [3]{c \sin ^3(a+b x)} \, dx=\int { \left (c \sin \left (b x + a\right )^{3}\right )^{\frac {1}{3}} x^{3} \,d x } \] Input:
integrate(x^3*(c*sin(b*x+a)^3)^(1/3),x, algorithm="giac")
Output:
integrate((c*sin(b*x + a)^3)^(1/3)*x^3, x)
Time = 40.91 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.14 \[ \int x^3 \sqrt [3]{c \sin ^3(a+b x)} \, dx=\frac {2^{1/3}\,{\left (c\,\left (3\,\sin \left (a+b\,x\right )-\sin \left (3\,a+3\,b\,x\right )\right )\right )}^{1/3}\,\left (3\,b^2\,x^2-12\,{\sin \left (a+b\,x\right )}^2+6\,b\,x\,\sin \left (2\,a+2\,b\,x\right )+3\,b^2\,x^2\,\left (2\,{\sin \left (a+b\,x\right )}^2-1\right )-b^3\,x^3\,\sin \left (2\,a+2\,b\,x\right )\right )}{4\,b^4\,{\sin \left (a+b\,x\right )}^2} \] Input:
int(x^3*(c*sin(a + b*x)^3)^(1/3),x)
Output:
(2^(1/3)*(c*(3*sin(a + b*x) - sin(3*a + 3*b*x)))^(1/3)*(3*b^2*x^2 - 12*sin (a + b*x)^2 + 6*b*x*sin(2*a + 2*b*x) + 3*b^2*x^2*(2*sin(a + b*x)^2 - 1) - b^3*x^3*sin(2*a + 2*b*x)))/(4*b^4*sin(a + b*x)^2)
Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.56 \[ \int x^3 \sqrt [3]{c \sin ^3(a+b x)} \, dx=\frac {c^{\frac {1}{3}} \left (-\cos \left (b x +a \right ) b^{3} x^{3}+6 \cos \left (b x +a \right ) b x +3 \sin \left (b x +a \right ) b^{2} x^{2}-6 \sin \left (b x +a \right )\right )}{b^{4}} \] Input:
int(x^3*(c*sin(b*x+a)^3)^(1/3),x)
Output:
(c**(1/3)*( - cos(a + b*x)*b**3*x**3 + 6*cos(a + b*x)*b*x + 3*sin(a + b*x) *b**2*x**2 - 6*sin(a + b*x)))/b**4