3.4.34 \(\int x^m (c \sin ^3(a+b x))^{2/3} \, dx\) [334]

Optimal. Leaf size=169 \[ \frac {x^{1+m} \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 (1+m)}+\frac {i 2^{-3-m} e^{2 i a} x^m (-i b x)^{-m} \csc ^2(a+b x) \Gamma (1+m,-2 i b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{b}-\frac {i 2^{-3-m} e^{-2 i a} x^m (i b x)^{-m} \csc ^2(a+b x) \Gamma (1+m,2 i b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{b} \]

[Out]

1/2*x^(1+m)*csc(b*x+a)^2*(c*sin(b*x+a)^3)^(2/3)/(1+m)+I*2^(-3-m)*exp(2*I*a)*x^m*csc(b*x+a)^2*GAMMA(1+m,-2*I*b*
x)*(c*sin(b*x+a)^3)^(2/3)/b/((-I*b*x)^m)-I*2^(-3-m)*x^m*csc(b*x+a)^2*GAMMA(1+m,2*I*b*x)*(c*sin(b*x+a)^3)^(2/3)
/b/exp(2*I*a)/((I*b*x)^m)

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Rubi [A]
time = 0.22, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6852, 3393, 3388, 2212} \begin {gather*} \frac {i e^{2 i a} 2^{-m-3} x^m (-i b x)^{-m} \csc ^2(a+b x) \text {Gamma}(m+1,-2 i b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{b}-\frac {i e^{-2 i a} 2^{-m-3} x^m (i b x)^{-m} \csc ^2(a+b x) \text {Gamma}(m+1,2 i b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{b}+\frac {x^{m+1} \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^m*(c*Sin[a + b*x]^3)^(2/3),x]

[Out]

(x^(1 + m)*Csc[a + b*x]^2*(c*Sin[a + b*x]^3)^(2/3))/(2*(1 + m)) + (I*2^(-3 - m)*E^((2*I)*a)*x^m*Csc[a + b*x]^2
*Gamma[1 + m, (-2*I)*b*x]*(c*Sin[a + b*x]^3)^(2/3))/(b*((-I)*b*x)^m) - (I*2^(-3 - m)*x^m*Csc[a + b*x]^2*Gamma[
1 + m, (2*I)*b*x]*(c*Sin[a + b*x]^3)^(2/3))/(b*E^((2*I)*a)*(I*b*x)^m)

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3388

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 6852

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[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]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int x^m \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 x^m \sin ^2(a+b x) \, dx\\ &=\left (\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int \left (\frac {x^m}{2}-\frac {1}{2} x^m \cos (2 a+2 b x)\right ) \, dx\\ &=\frac {x^{1+m} \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 (1+m)}-\frac {1}{2} \left (\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int x^m \cos (2 a+2 b x) \, dx\\ &=\frac {x^{1+m} \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 (1+m)}-\frac {1}{4} \left (\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int e^{-i (2 a+2 b x)} x^m \, dx-\frac {1}{4} \left (\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int e^{i (2 a+2 b x)} x^m \, dx\\ &=\frac {x^{1+m} \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{2 (1+m)}+\frac {i 2^{-3-m} e^{2 i a} x^m (-i b x)^{-m} \csc ^2(a+b x) \Gamma (1+m,-2 i b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{b}-\frac {i 2^{-3-m} e^{-2 i a} x^m (i b x)^{-m} \csc ^2(a+b x) \Gamma (1+m,2 i b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{b}\\ \end {align*}

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Mathematica [A]
time = 0.36, size = 142, normalized size = 0.84 \begin {gather*} \frac {2^{-3-m} x^m \left (b^2 x^2\right )^{-m} \csc ^2(a+b x) \left (2^{2+m} b x \left (b^2 x^2\right )^m-i (1+m) (-i b x)^m \Gamma (1+m,2 i b x) (\cos (a)-i \sin (a))^2+i (1+m) (i b x)^m \Gamma (1+m,-2 i b x) (\cos (a)+i \sin (a))^2\right ) \left (c \sin ^3(a+b x)\right )^{2/3}}{b (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^m*(c*Sin[a + b*x]^3)^(2/3),x]

[Out]

(2^(-3 - m)*x^m*Csc[a + b*x]^2*(2^(2 + m)*b*x*(b^2*x^2)^m - I*(1 + m)*((-I)*b*x)^m*Gamma[1 + m, (2*I)*b*x]*(Co
s[a] - I*Sin[a])^2 + I*(1 + m)*(I*b*x)^m*Gamma[1 + m, (-2*I)*b*x]*(Cos[a] + I*Sin[a])^2)*(c*Sin[a + b*x]^3)^(2
/3))/(b*(1 + m)*(b^2*x^2)^m)

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int x^{m} \left (c \left (\sin ^{3}\left (b x +a \right )\right )\right )^{\frac {2}{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((m + 1)*integrate(x^m*cos(2*b*x + 2*a), x) - e^(m*log(x) + log(x)))*c^(2/3)/(m + 1)

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Fricas [A]
time = 0.10, size = 112, normalized size = 0.66 \begin {gather*} -\frac {{\left (4 \, b x x^{m} - {\left (i \, m + i\right )} e^{\left (-m \log \left (2 i \, b\right ) - 2 i \, a\right )} \Gamma \left (m + 1, 2 i \, b x\right ) - {\left (-i \, m - i\right )} e^{\left (-m \log \left (-2 i \, b\right ) + 2 i \, a\right )} \Gamma \left (m + 1, -2 i \, b x\right )\right )} \left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{\frac {2}{3}}}{8 \, {\left ({\left (b m + b\right )} \cos \left (b x + a\right )^{2} - b m - b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*b*x*x^m - (I*m + I)*e^(-m*log(2*I*b) - 2*I*a)*gamma(m + 1, 2*I*b*x) - (-I*m - I)*e^(-m*log(-2*I*b) + 2
*I*a)*gamma(m + 1, -2*I*b*x))*(-(c*cos(b*x + a)^2 - c)*sin(b*x + a))^(2/3)/((b*m + b)*cos(b*x + a)^2 - b*m - b
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^m\,{\left (c\,{\sin \left (a+b\,x\right )}^3\right )}^{2/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(c*sin(a + b*x)^3)^(2/3),x)

[Out]

int(x^m*(c*sin(a + b*x)^3)^(2/3), x)

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