3.32 \(\int x^{3 a} \sin (x^{2 a}) \, dx\)

Optimal. Leaf size=115 \[ \frac {i x^{3 a+1} \left (-i x^{2 a}\right )^{-\frac {3 a+1}{2 a}} \operatorname {Gamma}\left (\frac {1}{2} \left (\frac {1}{a}+3\right ),-i x^{2 a}\right )}{4 a}-\frac {i x^{3 a+1} \left (i x^{2 a}\right )^{-\frac {3 a+1}{2 a}} \operatorname {Gamma}\left (\frac {1}{2} \left (\frac {1}{a}+3\right ),i x^{2 a}\right )}{4 a} \]

[Out]

1/4*I*x^(1+3*a)*GAMMA(3/2+1/2/a,-I*x^(2*a))/a/((-I*x^(2*a))^(1/2*(1+3*a)/a))-1/4*I*x^(1+3*a)*GAMMA(3/2+1/2/a,I
*x^(2*a))/a/((I*x^(2*a))^(1/2*(1+3*a)/a))

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Rubi [A]  time = 0.06, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3423, 2218} \[ \frac {i x^{3 a+1} \left (-i x^{2 a}\right )^{-\frac {3 a+1}{2 a}} \text {Gamma}\left (\frac {1}{2} \left (\frac {1}{a}+3\right ),-i x^{2 a}\right )}{4 a}-\frac {i x^{3 a+1} \left (i x^{2 a}\right )^{-\frac {3 a+1}{2 a}} \text {Gamma}\left (\frac {1}{2} \left (\frac {1}{a}+3\right ),i x^{2 a}\right )}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[x^(3*a)*Sin[x^(2*a)],x]

[Out]

((I/4)*x^(1 + 3*a)*Gamma[(3 + a^(-1))/2, (-I)*x^(2*a)])/(a*((-I)*x^(2*a))^((1 + 3*a)/(2*a))) - ((I/4)*x^(1 + 3
*a)*Gamma[(3 + a^(-1))/2, I*x^(2*a)])/(a*(I*x^(2*a))^((1 + 3*a)/(2*a)))

Rule 2218

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

Rule 3423

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, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.30, size = 142, normalized size = 1.23 \[ -\frac {x^{a+1} \left (x^{4 a}\right )^{-\frac {a+1}{2 a}} \left ((a+1) \left (-i x^{2 a}\right )^{\frac {a+1}{2 a}} \operatorname {Gamma}\left (\frac {a+1}{2 a},i x^{2 a}\right )+(a+1) \left (i x^{2 a}\right )^{\frac {a+1}{2 a}} \operatorname {Gamma}\left (\frac {a+1}{2 a},-i x^{2 a}\right )+4 a \left (x^{4 a}\right )^{\frac {a+1}{2 a}} \cos \left (x^{2 a}\right )\right )}{8 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3*a)*Sin[x^(2*a)],x]

[Out]

-1/8*(x^(1 + a)*(4*a*(x^(4*a))^((1 + a)/(2*a))*Cos[x^(2*a)] + (1 + a)*(I*x^(2*a))^((1 + a)/(2*a))*Gamma[(1 + a
)/(2*a), (-I)*x^(2*a)] + (1 + a)*((-I)*x^(2*a))^((1 + a)/(2*a))*Gamma[(1 + a)/(2*a), I*x^(2*a)]))/(a^2*(x^(4*a
))^((1 + a)/(2*a)))

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3 \, a} \sin \left (x^{2 \, a}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3*a)*sin(x^(2*a)),x, algorithm="fricas")

[Out]

integral(x^(3*a)*sin(x^(2*a)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3*a)*sin(x^(2*a)),x, algorithm="giac")

[Out]

integrate(x^(3*a)*sin(x^(2*a)), x)

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maple [C]  time = 0.16, size = 41, normalized size = 0.36 \[ \frac {x^{5 a +1} \hypergeom \left (\left [\frac {1}{4 a}+\frac {5}{4}\right ], \left [\frac {3}{2}, \frac {1}{4 a}+\frac {9}{4}\right ], -\frac {x^{4 a}}{4}\right )}{5 a +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3*a)*sin(x^(2*a)),x)

[Out]

1/(5*a+1)*x^(5*a+1)*hypergeom([5/4+1/4/a],[3/2,9/4+1/4/a],-1/4*x^(4*a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {x x^{a} \cos \left (x^{2 \, a}\right ) - \frac {{\left (a + 1\right )} x x^{a} \Gamma \left (\frac {1}{4 \, a} + \frac {1}{4}\right ) \,_1F_2\left (\begin {matrix} \frac {1}{4 \, a} + \frac {1}{4} \\ \frac {1}{2},\frac {1}{4 \, a} + \frac {5}{4} \end {matrix} ; -\frac {1}{4} \, x^{4 \, a} \right )}{4 \, a \Gamma \left (\frac {1}{4 \, a} + \frac {5}{4}\right )}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3*a)*sin(x^(2*a)),x, algorithm="maxima")

[Out]

-1/2*(x*x^a*cos(x^(2*a)) - (a + 1)*integrate(x^a*cos(x^(2*a)), x))/a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{3\,a}\,\sin \left (x^{2\,a}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3*a)*sin(x^(2*a)),x)

[Out]

int(x^(3*a)*sin(x^(2*a)), x)

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sympy [A]  time = 2.89, size = 54, normalized size = 0.47 \[ \frac {x x^{5 a} \Gamma \left (\frac {5}{4} + \frac {1}{4 a}\right ) {{}_{1}F_{2}\left (\begin {matrix} \frac {5}{4} + \frac {1}{4 a} \\ \frac {3}{2}, \frac {9}{4} + \frac {1}{4 a} \end {matrix}\middle | {- \frac {x^{4 a}}{4}} \right )}}{4 a \Gamma \left (\frac {9}{4} + \frac {1}{4 a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3*a)*sin(x**(2*a)),x)

[Out]

x*x**(5*a)*gamma(5/4 + 1/(4*a))*hyper((5/4 + 1/(4*a),), (3/2, 9/4 + 1/(4*a)), -x**(4*a)/4)/(4*a*gamma(9/4 + 1/
(4*a)))

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