3.4.18 \(\int x^m \sqrt [3]{c \sin ^3(a+b x^2)} \, dx\) [318]

Optimal. Leaf size=153 \[ \frac {1}{4} i e^{i a} x^{1+m} \left (-i b x^2\right )^{\frac {1}{2} (-1-m)} \csc \left (a+b x^2\right ) \Gamma \left (\frac {1+m}{2},-i b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\frac {1}{4} i e^{-i a} x^{1+m} \left (i b x^2\right )^{\frac {1}{2} (-1-m)} \csc \left (a+b x^2\right ) \Gamma \left (\frac {1+m}{2},i b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \]

[Out]

1/4*I*exp(I*a)*x^(1+m)*(-I*b*x^2)^(-1/2-1/2*m)*csc(b*x^2+a)*GAMMA(1/2+1/2*m,-I*b*x^2)*(c*sin(b*x^2+a)^3)^(1/3)
-1/4*I*x^(1+m)*(I*b*x^2)^(-1/2-1/2*m)*csc(b*x^2+a)*GAMMA(1/2+1/2*m,I*b*x^2)*(c*sin(b*x^2+a)^3)^(1/3)/exp(I*a)

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Rubi [A]
time = 0.20, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6852, 3470, 2250} \begin {gather*} \frac {1}{4} i e^{i a} x^{m+1} \left (-i b x^2\right )^{\frac {1}{2} (-m-1)} \csc \left (a+b x^2\right ) \text {Gamma}\left (\frac {m+1}{2},-i b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\frac {1}{4} i e^{-i a} x^{m+1} \left (i b x^2\right )^{\frac {1}{2} (-m-1)} \csc \left (a+b x^2\right ) \text {Gamma}\left (\frac {m+1}{2},i b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^m*(c*Sin[a + b*x^2]^3)^(1/3),x]

[Out]

(I/4)*E^(I*a)*x^(1 + m)*((-I)*b*x^2)^((-1 - m)/2)*Csc[a + b*x^2]*Gamma[(1 + m)/2, (-I)*b*x^2]*(c*Sin[a + b*x^2
]^3)^(1/3) - ((I/4)*x^(1 + m)*(I*b*x^2)^((-1 - m)/2)*Csc[a + b*x^2]*Gamma[(1 + m)/2, I*b*x^2]*(c*Sin[a + b*x^2
]^3)^(1/3))/E^(I*a)

Rule 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 3470

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[I/2, Int[(e*x)^m*E^((-c)*I - d*I*x^n),
x], x] - Dist[I/2, Int[(e*x)^m*E^(c*I + d*I*x^n), x], x] /; FreeQ[{c, d, e, m}, x] && IGtQ[n, 0]

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

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Mathematica [A]
time = 0.20, size = 138, normalized size = 0.90 \begin {gather*} \frac {1}{4} i x^{1+m} \left (b^2 x^4\right )^{\frac {1}{2} (-1-m)} \csc \left (a+b x^2\right ) \left (-\left (-i b x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},i b x^2\right ) (\cos (a)-i \sin (a))+\left (i b x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},-i b x^2\right ) (\cos (a)+i \sin (a))\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^m*(c*Sin[a + b*x^2]^3)^(1/3),x]

[Out]

(I/4)*x^(1 + m)*(b^2*x^4)^((-1 - m)/2)*Csc[a + b*x^2]*(-(((-I)*b*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, I*b*x^2]*(C
os[a] - I*Sin[a])) + (I*b*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, (-I)*b*x^2]*(Cos[a] + I*Sin[a]))*(c*Sin[a + b*x^2]
^3)^(1/3)

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int x^{m} \left (c \left (\sin ^{3}\left (b \,x^{2}+a \right )\right )\right )^{\frac {1}{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(c*sin(b*x^2+a)^3)^(1/3),x)

[Out]

int(x^m*(c*sin(b*x^2+a)^3)^(1/3),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^2+a)^3)^(1/3),x, algorithm="maxima")

[Out]

integrate((c*sin(b*x^2 + a)^3)^(1/3)*x^m, x)

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Fricas [A]
time = 0.10, size = 98, normalized size = 0.64 \begin {gather*} -\frac {{\left (e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (i \, b\right ) - i \, a\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, i \, b x^{2}\right ) + e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-i \, b\right ) + i \, a\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -i \, b x^{2}\right )\right )} \left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac {1}{3}}}{4 \, b \sin \left (b x^{2} + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(c*sin(b*x^2+a)^3)^(1/3),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(c*sin(b*x**2+a)**3)**(1/3),x)

[Out]

Integral(x**m*(c*sin(a + b*x**2)**3)**(1/3), 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^2+a)^3)^(1/3),x, algorithm="giac")

[Out]

integrate((c*sin(b*x^2 + a)^3)^(1/3)*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 (b\,x^2+a\right )}^3\right )}^{1/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(c*sin(a + b*x^2)^3)^(1/3),x)

[Out]

int(x^m*(c*sin(a + b*x^2)^3)^(1/3), x)

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