3.338 \(\int (c \sin ^3(a+b x))^{2/3} \, dx\)

Optimal. Leaf size=55 \[ \frac {1}{2} x \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}-\frac {\cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b} \]

[Out]

-1/2*cot(b*x+a)*(c*sin(b*x+a)^3)^(2/3)/b+1/2*x*csc(b*x+a)^2*(c*sin(b*x+a)^3)^(2/3)

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Rubi [A]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3207, 2635, 8} \[ \frac {1}{2} x \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}-\frac {\cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x]^3)^(2/3),x]

[Out]

-(Cot[a + b*x]*(c*Sin[a + b*x]^3)^(2/3))/(2*b) + (x*Csc[a + b*x]^2*(c*Sin[a + b*x]^3)^(2/3))/2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

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] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[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
]])

Rubi steps

\begin {align*} \int \left (c \sin ^3(a+b x)\right )^{2/3} \, dx &=\left (\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int \sin ^2(a+b x) \, dx\\ &=-\frac {\cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b}+\frac {1}{2} \left (\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int 1 \, dx\\ &=-\frac {\cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 b}+\frac {1}{2} x \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 47, normalized size = 0.85 \[ \frac {(2 (a+b x)-\sin (2 (a+b x))) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x]^3)^(2/3),x]

[Out]

(Csc[a + b*x]^2*(c*Sin[a + b*x]^3)^(2/3)*(2*(a + b*x) - Sin[2*(a + b*x)]))/(4*b)

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fricas [A]  time = 0.66, size = 60, normalized size = 1.09 \[ -\frac {{\left (b x - \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{\frac {2}{3}}}{2 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a)^3)^(2/3),x, algorithm="fricas")

[Out]

-1/2*(b*x - cos(b*x + a)*sin(b*x + a))*(-(c*cos(b*x + a)^2 - c)*sin(b*x + a))^(2/3)/(b*cos(b*x + a)^2 - b)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \sin \left (b x + a\right )^{3}\right )^{\frac {2}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a)^3)^(2/3),x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a)^3)^(2/3), x)

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maple [C]  time = 0.27, size = 158, normalized size = 2.87 \[ -\frac {x \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b x +a \right )}}{2 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{4 i \left (b x +a \right )}}{8 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}}}{8 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(b*x+a)^3)^(2/3),x)

[Out]

-1/2*x/(exp(2*I*(b*x+a))-1)^2*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(2/3)*exp(2*I*(b*x+a))-1/8*I/b/(e
xp(2*I*(b*x+a))-1)^2*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(2/3)*exp(4*I*(b*x+a))+1/8*I*(I*c*(exp(2*I
*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(2/3)/(exp(2*I*(b*x+a))-1)^2/b

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maxima [B]  time = 0.52, size = 116, normalized size = 2.11 \[ \frac {c^{\frac {2}{3}} \arctan \left (\frac {\sin \left (b x + a\right )}{\cos \left (b x + a\right ) + 1}\right ) - \frac {\frac {c^{\frac {2}{3}} \sin \left (b x + a\right )}{\cos \left (b x + a\right ) + 1} - \frac {c^{\frac {2}{3}} \sin \left (b x + a\right )^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}}}{\frac {2 \, \sin \left (b x + a\right )^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {\sin \left (b x + a\right )^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 1}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a)^3)^(2/3),x, algorithm="maxima")

[Out]

(c^(2/3)*arctan(sin(b*x + a)/(cos(b*x + a) + 1)) - (c^(2/3)*sin(b*x + a)/(cos(b*x + a) + 1) - c^(2/3)*sin(b*x
+ a)^3/(cos(b*x + a) + 1)^3)/(2*sin(b*x + a)^2/(cos(b*x + a) + 1)^2 + sin(b*x + a)^4/(cos(b*x + a) + 1)^4 + 1)
)/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (c\,{\sin \left (a+b\,x\right )}^3\right )}^{2/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a + b*x)^3)^(2/3),x)

[Out]

int((c*sin(a + b*x)^3)^(2/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \sin ^{3}{\left (a + b x \right )}\right )^{\frac {2}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a)**3)**(2/3),x)

[Out]

Integral((c*sin(a + b*x)**3)**(2/3), x)

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