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

Optimal. Leaf size=106 \[ -\frac{b e^{i c} x \text{Gamma}\left (\frac{1}{3},-i d x^3\right )}{18 d \sqrt [3]{-i d x^3}}-\frac{b e^{-i c} x \text{Gamma}\left (\frac{1}{3},i d x^3\right )}{18 d \sqrt [3]{i d x^3}}+\frac{a x^4}{4}-\frac{b x \cos \left (c+d x^3\right )}{3 d} \]

[Out]

(a*x^4)/4 - (b*x*Cos[c + d*x^3])/(3*d) - (b*E^(I*c)*x*Gamma[1/3, (-I)*d*x^3])/(18*d*((-I)*d*x^3)^(1/3)) - (b*x
*Gamma[1/3, I*d*x^3])/(18*d*E^(I*c)*(I*d*x^3)^(1/3))

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Rubi [A]  time = 0.0723359, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {14, 3385, 3356, 2208} \[ -\frac{b e^{i c} x \text{Gamma}\left (\frac{1}{3},-i d x^3\right )}{18 d \sqrt [3]{-i d x^3}}-\frac{b e^{-i c} x \text{Gamma}\left (\frac{1}{3},i d x^3\right )}{18 d \sqrt [3]{i d x^3}}+\frac{a x^4}{4}-\frac{b x \cos \left (c+d x^3\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Sin[c + d*x^3]),x]

[Out]

(a*x^4)/4 - (b*x*Cos[c + d*x^3])/(3*d) - (b*E^(I*c)*x*Gamma[1/3, (-I)*d*x^3])/(18*d*((-I)*d*x^3)^(1/3)) - (b*x
*Gamma[1/3, I*d*x^3])/(18*d*E^(I*c)*(I*d*x^3)^(1/3))

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 3385

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> -Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Cos[c + d
*x^n])/(d*n), x] + Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e},
x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3356

Int[Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)], x_Symbol] :> Dist[1/2, Int[E^(-(c*I) - d*I*(e + f*x)^n), x],
x] + Dist[1/2, Int[E^(c*I + d*I*(e + f*x)^n), x], x] /; FreeQ[{c, d, e, f}, x] && IGtQ[n, 2]

Rule 2208

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> -Simp[(F^a*(c + d*x)*Gamma[1/n, -(b*(c + d*x)
^n*Log[F])])/(d*n*(-(b*(c + d*x)^n*Log[F]))^(1/n)), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rubi steps

\begin{align*} \int x^3 \left (a+b \sin \left (c+d x^3\right )\right ) \, dx &=\int \left (a x^3+b x^3 \sin \left (c+d x^3\right )\right ) \, dx\\ &=\frac{a x^4}{4}+b \int x^3 \sin \left (c+d x^3\right ) \, dx\\ &=\frac{a x^4}{4}-\frac{b x \cos \left (c+d x^3\right )}{3 d}+\frac{b \int \cos \left (c+d x^3\right ) \, dx}{3 d}\\ &=\frac{a x^4}{4}-\frac{b x \cos \left (c+d x^3\right )}{3 d}+\frac{b \int e^{-i c-i d x^3} \, dx}{6 d}+\frac{b \int e^{i c+i d x^3} \, dx}{6 d}\\ &=\frac{a x^4}{4}-\frac{b x \cos \left (c+d x^3\right )}{3 d}-\frac{b e^{i c} x \Gamma \left (\frac{1}{3},-i d x^3\right )}{18 d \sqrt [3]{-i d x^3}}-\frac{b e^{-i c} x \Gamma \left (\frac{1}{3},i d x^3\right )}{18 d \sqrt [3]{i d x^3}}\\ \end{align*}

Mathematica [A]  time = 0.191393, size = 124, normalized size = 1.17 \[ \frac{d x^7 \left (-2 b \sqrt [3]{-i d x^3} (\cos (c)-i \sin (c)) \text{Gamma}\left (\frac{1}{3},i d x^3\right )-2 b \sqrt [3]{i d x^3} (\cos (c)+i \sin (c)) \text{Gamma}\left (\frac{1}{3},-i d x^3\right )+3 \sqrt [3]{d^2 x^6} \left (3 a d x^3-4 b \cos \left (c+d x^3\right )\right )\right )}{36 \left (d^2 x^6\right )^{4/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*Sin[c + d*x^3]),x]

[Out]

(d*x^7*(3*(d^2*x^6)^(1/3)*(3*a*d*x^3 - 4*b*Cos[c + d*x^3]) - 2*b*((-I)*d*x^3)^(1/3)*Gamma[1/3, I*d*x^3]*(Cos[c
] - I*Sin[c]) - 2*b*(I*d*x^3)^(1/3)*Gamma[1/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c])))/(36*(d^2*x^6)^(4/3))

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Maple [F]  time = 0.06, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \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^3*(a+b*sin(d*x^3+c)),x)

[Out]

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

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Maxima [B]  time = 1.13706, size = 396, normalized size = 3.74 \begin{align*} \frac{1}{4} \, a x^{4} - \frac{{\left (12 \, \left (x^{3}{\left | d \right |}\right )^{\frac{1}{3}} x \cos \left (d x^{3} + c\right ) +{\left ({\left ({\left (\Gamma \left (\frac{1}{3}, i \, d x^{3}\right ) + \Gamma \left (\frac{1}{3}, -i \, d x^{3}\right )\right )} \cos \left (\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right ) +{\left (\Gamma \left (\frac{1}{3}, i \, d x^{3}\right ) + \Gamma \left (\frac{1}{3}, -i \, d x^{3}\right )\right )} \cos \left (-\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right ) +{\left (-i \, \Gamma \left (\frac{1}{3}, i \, d x^{3}\right ) + i \, \Gamma \left (\frac{1}{3}, -i \, d x^{3}\right )\right )} \sin \left (\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right ) +{\left (i \, \Gamma \left (\frac{1}{3}, i \, d x^{3}\right ) - i \, \Gamma \left (\frac{1}{3}, -i \, d x^{3}\right )\right )} \sin \left (-\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right )\right )} \cos \left (c\right ) +{\left ({\left (-i \, \Gamma \left (\frac{1}{3}, i \, d x^{3}\right ) + i \, \Gamma \left (\frac{1}{3}, -i \, d x^{3}\right )\right )} \cos \left (\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right ) +{\left (-i \, \Gamma \left (\frac{1}{3}, i \, d x^{3}\right ) + i \, \Gamma \left (\frac{1}{3}, -i \, d x^{3}\right )\right )} \cos \left (-\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right ) -{\left (\Gamma \left (\frac{1}{3}, i \, d x^{3}\right ) + \Gamma \left (\frac{1}{3}, -i \, d x^{3}\right )\right )} \sin \left (\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right ) +{\left (\Gamma \left (\frac{1}{3}, i \, d x^{3}\right ) + \Gamma \left (\frac{1}{3}, -i \, d x^{3}\right )\right )} \sin \left (-\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right )\right )} \sin \left (c\right )\right )} x\right )} b}{36 \, \left (x^{3}{\left | d \right |}\right )^{\frac{1}{3}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*x^4 - 1/36*(12*(x^3*abs(d))^(1/3)*x*cos(d*x^3 + c) + (((gamma(1/3, I*d*x^3) + gamma(1/3, -I*d*x^3))*cos(
1/6*pi + 1/3*arctan2(0, d)) + (gamma(1/3, I*d*x^3) + gamma(1/3, -I*d*x^3))*cos(-1/6*pi + 1/3*arctan2(0, d)) +
(-I*gamma(1/3, I*d*x^3) + I*gamma(1/3, -I*d*x^3))*sin(1/6*pi + 1/3*arctan2(0, d)) + (I*gamma(1/3, I*d*x^3) - I
*gamma(1/3, -I*d*x^3))*sin(-1/6*pi + 1/3*arctan2(0, d)))*cos(c) + ((-I*gamma(1/3, I*d*x^3) + I*gamma(1/3, -I*d
*x^3))*cos(1/6*pi + 1/3*arctan2(0, d)) + (-I*gamma(1/3, I*d*x^3) + I*gamma(1/3, -I*d*x^3))*cos(-1/6*pi + 1/3*a
rctan2(0, d)) - (gamma(1/3, I*d*x^3) + gamma(1/3, -I*d*x^3))*sin(1/6*pi + 1/3*arctan2(0, d)) + (gamma(1/3, I*d
*x^3) + gamma(1/3, -I*d*x^3))*sin(-1/6*pi + 1/3*arctan2(0, d)))*sin(c))*x)*b/((x^3*abs(d))^(1/3)*d)

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Fricas [A]  time = 1.8875, size = 201, normalized size = 1.9 \begin{align*} \frac{9 \, a d^{2} x^{4} - 12 \, b d x \cos \left (d x^{3} + c\right ) + 2 i \, b \left (i \, d\right )^{\frac{2}{3}} e^{\left (-i \, c\right )} \Gamma \left (\frac{1}{3}, i \, d x^{3}\right ) - 2 i \, b \left (-i \, d\right )^{\frac{2}{3}} e^{\left (i \, c\right )} \Gamma \left (\frac{1}{3}, -i \, d x^{3}\right )}{36 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*(9*a*d^2*x^4 - 12*b*d*x*cos(d*x^3 + c) + 2*I*b*(I*d)^(2/3)*e^(-I*c)*gamma(1/3, I*d*x^3) - 2*I*b*(-I*d)^(2
/3)*e^(I*c)*gamma(1/3, -I*d*x^3))/d^2

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

[Out]

Integral(x**3*(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^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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