3.1 \(\int x^3 \cos (a+b x^2) \, dx\)

Optimal. Leaf size=34 \[ \frac{\cos \left (a+b x^2\right )}{2 b^2}+\frac{x^2 \sin \left (a+b x^2\right )}{2 b} \]

[Out]

Cos[a + b*x^2]/(2*b^2) + (x^2*Sin[a + b*x^2])/(2*b)

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Rubi [A]  time = 0.0317191, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3380, 3296, 2638} \[ \frac{\cos \left (a+b x^2\right )}{2 b^2}+\frac{x^2 \sin \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Cos[a + b*x^2],x]

[Out]

Cos[a + b*x^2]/(2*b^2) + (x^2*Sin[a + b*x^2])/(2*b)

Rule 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[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 2638

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

Rubi steps

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

Mathematica [A]  time = 0.0492611, size = 29, normalized size = 0.85 \[ \frac{b x^2 \sin \left (a+b x^2\right )+\cos \left (a+b x^2\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Cos[a + b*x^2],x]

[Out]

(Cos[a + b*x^2] + b*x^2*Sin[a + b*x^2])/(2*b^2)

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Maple [A]  time = 0.023, size = 31, normalized size = 0.9 \begin{align*}{\frac{\cos \left ( b{x}^{2}+a \right ) }{2\,{b}^{2}}}+{\frac{{x}^{2}\sin \left ( b{x}^{2}+a \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cos(b*x^2+a),x)

[Out]

1/2*cos(b*x^2+a)/b^2+1/2*x^2*sin(b*x^2+a)/b

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Maxima [A]  time = 1.33306, size = 36, normalized size = 1.06 \begin{align*} \frac{b x^{2} \sin \left (b x^{2} + a\right ) + \cos \left (b x^{2} + a\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2*sin(b*x^2 + a) + cos(b*x^2 + a))/b^2

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Fricas [A]  time = 1.56881, size = 66, normalized size = 1.94 \begin{align*} \frac{b x^{2} \sin \left (b x^{2} + a\right ) + \cos \left (b x^{2} + a\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2*sin(b*x^2 + a) + cos(b*x^2 + a))/b^2

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Sympy [A]  time = 1.08675, size = 36, normalized size = 1.06 \begin{align*} \begin{cases} \frac{x^{2} \sin{\left (a + b x^{2} \right )}}{2 b} + \frac{\cos{\left (a + b x^{2} \right )}}{2 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{4} \cos{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*cos(b*x**2+a),x)

[Out]

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

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Giac [A]  time = 1.12968, size = 36, normalized size = 1.06 \begin{align*} \frac{b x^{2} \sin \left (b x^{2} + a\right ) + \cos \left (b x^{2} + a\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2*sin(b*x^2 + a) + cos(b*x^2 + a))/b^2