3.2 \(\int x^3 (a+b \sin (c+d x^2)) \, dx\)

Optimal. Leaf size=44 \[ \frac{a x^4}{4}+\frac{b \sin \left (c+d x^2\right )}{2 d^2}-\frac{b x^2 \cos \left (c+d x^2\right )}{2 d} \]

[Out]

(a*x^4)/4 - (b*x^2*Cos[c + d*x^2])/(2*d) + (b*Sin[c + d*x^2])/(2*d^2)

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Rubi [A]  time = 0.0425759, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {14, 3379, 3296, 2637} \[ \frac{a x^4}{4}+\frac{b \sin \left (c+d x^2\right )}{2 d^2}-\frac{b x^2 \cos \left (c+d x^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Sin[c + d*x^2]),x]

[Out]

(a*x^4)/4 - (b*x^2*Cos[c + d*x^2])/(2*d) + (b*Sin[c + d*x^2])/(2*d^2)

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 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x^3 \left (a+b \sin \left (c+d x^2\right )\right ) \, dx &=\int \left (a x^3+b x^3 \sin \left (c+d x^2\right )\right ) \, dx\\ &=\frac{a x^4}{4}+b \int x^3 \sin \left (c+d x^2\right ) \, dx\\ &=\frac{a x^4}{4}+\frac{1}{2} b \operatorname{Subst}\left (\int x \sin (c+d x) \, dx,x,x^2\right )\\ &=\frac{a x^4}{4}-\frac{b x^2 \cos \left (c+d x^2\right )}{2 d}+\frac{b \operatorname{Subst}\left (\int \cos (c+d x) \, dx,x,x^2\right )}{2 d}\\ &=\frac{a x^4}{4}-\frac{b x^2 \cos \left (c+d x^2\right )}{2 d}+\frac{b \sin \left (c+d x^2\right )}{2 d^2}\\ \end{align*}

Mathematica [A]  time = 0.0042806, size = 44, normalized size = 1. \[ \frac{a x^4}{4}+\frac{b \sin \left (c+d x^2\right )}{2 d^2}-\frac{b x^2 \cos \left (c+d x^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*Sin[c + d*x^2]),x]

[Out]

(a*x^4)/4 - (b*x^2*Cos[c + d*x^2])/(2*d) + (b*Sin[c + d*x^2])/(2*d^2)

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Maple [A]  time = 0.007, size = 40, normalized size = 0.9 \begin{align*}{\frac{a{x}^{4}}{4}}+b \left ( -{\frac{{x}^{2}\cos \left ( d{x}^{2}+c \right ) }{2\,d}}+{\frac{\sin \left ( d{x}^{2}+c \right ) }{2\,{d}^{2}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*sin(d*x^2+c)),x)

[Out]

1/4*a*x^4+b*(-1/2/d*x^2*cos(d*x^2+c)+1/2/d^2*sin(d*x^2+c))

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Maxima [A]  time = 0.976055, size = 50, normalized size = 1.14 \begin{align*} \frac{1}{4} \, a x^{4} - \frac{{\left (d x^{2} \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right )} b}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*sin(d*x^2+c)),x, algorithm="maxima")

[Out]

1/4*a*x^4 - 1/2*(d*x^2*cos(d*x^2 + c) - sin(d*x^2 + c))*b/d^2

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Fricas [A]  time = 1.92282, size = 93, normalized size = 2.11 \begin{align*} \frac{a d^{2} x^{4} - 2 \, b d x^{2} \cos \left (d x^{2} + c\right ) + 2 \, b \sin \left (d x^{2} + c\right )}{4 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*sin(d*x^2+c)),x, algorithm="fricas")

[Out]

1/4*(a*d^2*x^4 - 2*b*d*x^2*cos(d*x^2 + c) + 2*b*sin(d*x^2 + c))/d^2

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Sympy [A]  time = 0.986867, size = 49, normalized size = 1.11 \begin{align*} \begin{cases} \frac{a x^{4}}{4} - \frac{b x^{2} \cos{\left (c + d x^{2} \right )}}{2 d} + \frac{b \sin{\left (c + d x^{2} \right )}}{2 d^{2}} & \text{for}\: d \neq 0 \\\frac{x^{4} \left (a + b \sin{\left (c \right )}\right )}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*sin(d*x**2+c)),x)

[Out]

Piecewise((a*x**4/4 - b*x**2*cos(c + d*x**2)/(2*d) + b*sin(c + d*x**2)/(2*d**2), Ne(d, 0)), (x**4*(a + b*sin(c
))/4, True))

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Giac [A]  time = 1.17592, size = 82, normalized size = 1.86 \begin{align*} \frac{\frac{{\left ({\left (d x^{2} + c\right )}^{2} - 2 \,{\left (d x^{2} + c\right )} c\right )} a}{d} - \frac{2 \,{\left (d x^{2} \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right )} b}{d}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*sin(d*x^2+c)),x, algorithm="giac")

[Out]

1/4*(((d*x^2 + c)^2 - 2*(d*x^2 + c)*c)*a/d - 2*(d*x^2*cos(d*x^2 + c) - sin(d*x^2 + c))*b/d)/d