3.1.62 \(\int x (a+b \sin (c+d x^3)) \, dx\) [62]

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

[Out]

1/2*a*x^2+1/6*I*b*exp(I*c)*x^2*GAMMA(2/3,-I*d*x^3)/(-I*d*x^3)^(2/3)-1/6*I*b*x^2*GAMMA(2/3,I*d*x^3)/exp(I*c)/(I
*d*x^3)^(2/3)

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 91, normalized size of antiderivative = 1.00, 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, 3470, 2250} \begin {gather*} \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} \end {gather*}

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 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]

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 = 0.09, size = 108, normalized size = 1.19 \begin {gather*} \frac {x^2 \left (3 a \left (d^2 x^6\right )^{2/3}+b \left (-i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},i d x^3\right ) (-i \cos (c)-\sin (c))+i b \left (i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},-i d x^3\right ) (\cos (c)+i \sin (c))\right )}{6 \left (d^2 x^6\right )^{2/3}} \end {gather*}

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))

________________________________________________________________________________________

Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int x \left (a +b \sin \left (d \,x^{3}+c \right )\right )\, dx\]

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)

________________________________________________________________________________________

Maxima [A]
time = 0.33, size = 93, normalized size = 1.02 \begin {gather*} \frac {1}{2} \, a x^{2} - \frac {\left (d x^{3}\right )^{\frac {1}{3}} {\left ({\left ({\left (\sqrt {3} + i\right )} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) - {\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} b}{12 \, d x} \end {gather*}

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*(d*x^3)^(1/3)*(((sqrt(3) + I)*gamma(2/3, I*d*x^3) + (sqrt(3) - I)*gamma(2/3, -I*d*x^3))*cos(c
) - ((I*sqrt(3) - 1)*gamma(2/3, I*d*x^3) + (-I*sqrt(3) - 1)*gamma(2/3, -I*d*x^3))*sin(c))*b/(d*x)

________________________________________________________________________________________

Fricas [A]
time = 0.13, size = 53, normalized size = 0.58 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*sin(c + d*x^3)),x)

[Out]

int(x*(a + b*sin(c + d*x^3)), x)

________________________________________________________________________________________