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

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

[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)))

________________________________________________________________________________________

Rubi [A]  time = 0.0560074, antiderivative size = 115, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.293061, size = 142, normalized size = 1.23 \[ -\frac{x^{a+1} \left (x^{4 a}\right )^{-\frac{a+1}{2 a}} \left ((a+1) \left (-i x^{2 a}\right )^{\frac{a+1}{2 a}} \text{Gamma}\left (\frac{a+1}{2 a},i x^{2 a}\right )+(a+1) \left (i x^{2 a}\right )^{\frac{a+1}{2 a}} \text{Gamma}\left (\frac{a+1}{2 a},-i x^{2 a}\right )+4 a \left (x^{4 a}\right )^{\frac{a+1}{2 a}} \cos \left (x^{2 a}\right )\right )}{8 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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)]))/(8*a^2*(x^(4*a))
^((1 + a)/(2*a)))

________________________________________________________________________________________

Maple [C]  time = 0.089, size = 41, normalized size = 0.4 \begin{align*}{\frac{{x}^{5\,a+1}}{5\,a+1}{\mbox{$_1$F$_2$}({\frac{5}{4}}+{\frac{1}{4\,a}};\,{\frac{3}{2}},{\frac{9}{4}}+{\frac{1}{4\,a}};\,-{\frac{{x}^{4\,a}}{4}})}} \end{align*}

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))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{x x^{a} \cos \left (x^{2 \, a}\right ) -{\left (a + 1\right )} \int x^{a} \cos \left (x^{2 \, a}\right )\,{d x}}{2 \, a} \end{align*}

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{3 \, a} \sin \left (x^{2 \, a}\right ), x\right ) \end{align*}

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)

________________________________________________________________________________________

Sympy [A]  time = 5.48852, size = 54, normalized size = 0.47 \begin{align*} \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 )} \end{align*}

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)))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3 \, a} \sin \left (x^{2 \, a}\right )\,{d x} \end{align*}

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)