3.62 \(\int x (a+b \sin (c+d x^3)) \, dx\)

Optimal. Leaf size=91 \[ \frac{i b e^{i c} x^2 \text{Gamma}\left (\frac{2}{3},-i d x^3\right )}{6 \left (-i d x^3\right )^{2/3}}-\frac{i b e^{-i c} x^2 \text{Gamma}\left (\frac{2}{3},i d x^3\right )}{6 \left (i d x^3\right )^{2/3}}+\frac{a x^2}{2} \]

[Out]

(a*x^2)/2 + ((I/6)*b*E^(I*c)*x^2*Gamma[2/3, (-I)*d*x^3])/((-I)*d*x^3)^(2/3) - ((I/6)*b*x^2*Gamma[2/3, I*d*x^3]
)/(E^(I*c)*(I*d*x^3)^(2/3))

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Rubi [A]  time = 0.0642006, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {14, 3389, 2218} \[ \frac{i b e^{i c} x^2 \text{Gamma}\left (\frac{2}{3},-i d x^3\right )}{6 \left (-i d x^3\right )^{2/3}}-\frac{i b e^{-i c} x^2 \text{Gamma}\left (\frac{2}{3},i d x^3\right )}{6 \left (i d x^3\right )^{2/3}}+\frac{a x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Sin[c + d*x^3]),x]

[Out]

(a*x^2)/2 + ((I/6)*b*E^(I*c)*x^2*Gamma[2/3, (-I)*d*x^3])/((-I)*d*x^3)^(2/3) - ((I/6)*b*x^2*Gamma[2/3, I*d*x^3]
)/(E^(I*c)*(I*d*x^3)^(2/3))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 3389

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 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 \left (a+b \sin \left (c+d x^3\right )\right ) \, dx &=\int \left (a x+b x \sin \left (c+d x^3\right )\right ) \, dx\\ &=\frac{a x^2}{2}+b \int x \sin \left (c+d x^3\right ) \, dx\\ &=\frac{a x^2}{2}+\frac{1}{2} (i b) \int e^{-i c-i d x^3} x \, dx-\frac{1}{2} (i b) \int e^{i c+i d x^3} x \, dx\\ &=\frac{a x^2}{2}+\frac{i b e^{i c} x^2 \Gamma \left (\frac{2}{3},-i d x^3\right )}{6 \left (-i d x^3\right )^{2/3}}-\frac{i b e^{-i c} x^2 \Gamma \left (\frac{2}{3},i d x^3\right )}{6 \left (i d x^3\right )^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.12647, size = 108, normalized size = 1.19 \[ \frac{x^2 \left (b \left (-i d x^3\right )^{2/3} (-\sin (c)-i \cos (c)) \text{Gamma}\left (\frac{2}{3},i d x^3\right )+i b \left (i d x^3\right )^{2/3} (\cos (c)+i \sin (c)) \text{Gamma}\left (\frac{2}{3},-i d x^3\right )+3 a \left (d^2 x^6\right )^{2/3}\right )}{6 \left (d^2 x^6\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*Sin[c + d*x^3]),x]

[Out]

(x^2*(3*a*(d^2*x^6)^(2/3) + b*((-I)*d*x^3)^(2/3)*Gamma[2/3, I*d*x^3]*((-I)*Cos[c] - Sin[c]) + I*b*(I*d*x^3)^(2
/3)*Gamma[2/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c])))/(6*(d^2*x^6)^(2/3))

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Maple [F]  time = 0.05, size = 0, normalized size = 0. \begin{align*} \int x \left ( a+b\sin \left ( d{x}^{3}+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.15415, size = 375, normalized size = 4.12 \begin{align*} \frac{1}{2} \, a x^{2} + \frac{\left (x^{3}{\left | d \right |}\right )^{\frac{1}{3}}{\left ({\left ({\left (-i \, \Gamma \left (\frac{2}{3}, i \, d x^{3}\right ) + i \, \Gamma \left (\frac{2}{3}, -i \, d x^{3}\right )\right )} \cos \left (\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right ) +{\left (-i \, \Gamma \left (\frac{2}{3}, i \, d x^{3}\right ) + i \, \Gamma \left (\frac{2}{3}, -i \, d x^{3}\right )\right )} \cos \left (-\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right ) -{\left (\Gamma \left (\frac{2}{3}, i \, d x^{3}\right ) + \Gamma \left (\frac{2}{3}, -i \, d x^{3}\right )\right )} \sin \left (\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right ) +{\left (\Gamma \left (\frac{2}{3}, i \, d x^{3}\right ) + \Gamma \left (\frac{2}{3}, -i \, d x^{3}\right )\right )} \sin \left (-\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right )\right )} \cos \left (c\right ) -{\left ({\left (\Gamma \left (\frac{2}{3}, i \, d x^{3}\right ) + \Gamma \left (\frac{2}{3}, -i \, d x^{3}\right )\right )} \cos \left (\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right ) +{\left (\Gamma \left (\frac{2}{3}, i \, d x^{3}\right ) + \Gamma \left (\frac{2}{3}, -i \, d x^{3}\right )\right )} \cos \left (-\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right ) -{\left (i \, \Gamma \left (\frac{2}{3}, i \, d x^{3}\right ) - i \, \Gamma \left (\frac{2}{3}, -i \, d x^{3}\right )\right )} \sin \left (\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right ) -{\left (-i \, \Gamma \left (\frac{2}{3}, i \, d x^{3}\right ) + i \, \Gamma \left (\frac{2}{3}, -i \, d x^{3}\right )\right )} \sin \left (-\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right )\right )} \sin \left (c\right )\right )} b}{12 \, x{\left | d \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2 + 1/12*(x^3*abs(d))^(1/3)*(((-I*gamma(2/3, I*d*x^3) + I*gamma(2/3, -I*d*x^3))*cos(1/3*pi + 2/3*arcta
n2(0, d)) + (-I*gamma(2/3, I*d*x^3) + I*gamma(2/3, -I*d*x^3))*cos(-1/3*pi + 2/3*arctan2(0, d)) - (gamma(2/3, I
*d*x^3) + gamma(2/3, -I*d*x^3))*sin(1/3*pi + 2/3*arctan2(0, d)) + (gamma(2/3, I*d*x^3) + gamma(2/3, -I*d*x^3))
*sin(-1/3*pi + 2/3*arctan2(0, d)))*cos(c) - ((gamma(2/3, I*d*x^3) + gamma(2/3, -I*d*x^3))*cos(1/3*pi + 2/3*arc
tan2(0, d)) + (gamma(2/3, I*d*x^3) + gamma(2/3, -I*d*x^3))*cos(-1/3*pi + 2/3*arctan2(0, d)) - (I*gamma(2/3, I*
d*x^3) - I*gamma(2/3, -I*d*x^3))*sin(1/3*pi + 2/3*arctan2(0, d)) - (-I*gamma(2/3, I*d*x^3) + I*gamma(2/3, -I*d
*x^3))*sin(-1/3*pi + 2/3*arctan2(0, d)))*sin(c))*b/(x*abs(d))

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Fricas [A]  time = 1.73832, size = 149, normalized size = 1.64 \begin{align*} \frac{3 \, a d x^{2} - b \left (i \, d\right )^{\frac{1}{3}} e^{\left (-i \, c\right )} \Gamma \left (\frac{2}{3}, i \, d x^{3}\right ) - b \left (-i \, d\right )^{\frac{1}{3}} e^{\left (i \, c\right )} \Gamma \left (\frac{2}{3}, -i \, d x^{3}\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*a*d*x^2 - b*(I*d)^(1/3)*e^(-I*c)*gamma(2/3, I*d*x^3) - b*(-I*d)^(1/3)*e^(I*c)*gamma(2/3, -I*d*x^3))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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