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

Optimal. Leaf size=51 \[ \frac {\cos ^2\left (a+b x^2\right )}{8 b^2}+\frac {x^2 \sin \left (a+b x^2\right ) \cos \left (a+b x^2\right )}{4 b}+\frac {x^4}{8} \]

[Out]

1/8*x^4+1/8*cos(b*x^2+a)^2/b^2+1/4*x^2*cos(b*x^2+a)*sin(b*x^2+a)/b

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Rubi [A]  time = 0.05, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3380, 3310, 30} \[ \frac {\cos ^2\left (a+b x^2\right )}{8 b^2}+\frac {x^2 \sin \left (a+b x^2\right ) \cos \left (a+b x^2\right )}{4 b}+\frac {x^4}{8} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Cos[a + b*x^2]^2,x]

[Out]

x^4/8 + Cos[a + b*x^2]^2/(8*b^2) + (x^2*Cos[a + b*x^2]*Sin[a + b*x^2])/(4*b)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3310

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

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]))

Rubi steps

\begin {align*} \int x^3 \cos ^2\left (a+b x^2\right ) \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x \cos ^2(a+b x) \, dx,x,x^2\right )\\ &=\frac {\cos ^2\left (a+b x^2\right )}{8 b^2}+\frac {x^2 \cos \left (a+b x^2\right ) \sin \left (a+b x^2\right )}{4 b}+\frac {1}{4} \operatorname {Subst}\left (\int x \, dx,x,x^2\right )\\ &=\frac {x^4}{8}+\frac {\cos ^2\left (a+b x^2\right )}{8 b^2}+\frac {x^2 \cos \left (a+b x^2\right ) \sin \left (a+b x^2\right )}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 40, normalized size = 0.78 \[ \frac {2 b x^2 \left (\sin \left (2 \left (a+b x^2\right )\right )+b x^2\right )+\cos \left (2 \left (a+b x^2\right )\right )}{16 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Cos[a + b*x^2]^2,x]

[Out]

(Cos[2*(a + b*x^2)] + 2*b*x^2*(b*x^2 + Sin[2*(a + b*x^2)]))/(16*b^2)

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fricas [A]  time = 0.94, size = 45, normalized size = 0.88 \[ \frac {b^{2} x^{4} + 2 \, b x^{2} \cos \left (b x^{2} + a\right ) \sin \left (b x^{2} + a\right ) + \cos \left (b x^{2} + a\right )^{2}}{8 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b^2*x^4 + 2*b*x^2*cos(b*x^2 + a)*sin(b*x^2 + a) + cos(b*x^2 + a)^2)/b^2

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giac [A]  time = 0.43, size = 55, normalized size = 1.08 \[ \frac {2 \, b x^{2} \sin \left (2 \, b x^{2} + 2 \, a\right ) + 2 \, {\left (b x^{2} + a\right )}^{2} - 4 \, {\left (b x^{2} + a\right )} a + \cos \left (2 \, b x^{2} + 2 \, a\right )}{16 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*b*x^2*sin(2*b*x^2 + 2*a) + 2*(b*x^2 + a)^2 - 4*(b*x^2 + a)*a + cos(2*b*x^2 + 2*a))/b^2

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maple [A]  time = 0.04, size = 42, normalized size = 0.82 \[ \frac {x^{4}}{8}+\frac {x^{2} \sin \left (2 b \,x^{2}+2 a \right )}{8 b}+\frac {\cos \left (2 b \,x^{2}+2 a \right )}{16 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x^4+1/8/b*x^2*sin(2*b*x^2+2*a)+1/16/b^2*cos(2*b*x^2+2*a)

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maxima [A]  time = 0.90, size = 42, normalized size = 0.82 \[ \frac {2 \, b^{2} x^{4} + 2 \, b x^{2} \sin \left (2 \, b x^{2} + 2 \, a\right ) + \cos \left (2 \, b x^{2} + 2 \, a\right )}{16 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*b^2*x^4 + 2*b*x^2*sin(2*b*x^2 + 2*a) + cos(2*b*x^2 + 2*a))/b^2

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mupad [B]  time = 0.15, size = 41, normalized size = 0.80 \[ \frac {\cos \left (2\,b\,x^2+2\,a\right )}{16\,b^2}+\frac {x^4}{8}+\frac {x^2\,\sin \left (2\,b\,x^2+2\,a\right )}{8\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cos(a + b*x^2)^2,x)

[Out]

cos(2*a + 2*b*x^2)/(16*b^2) + x^4/8 + (x^2*sin(2*a + 2*b*x^2))/(8*b)

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sympy [A]  time = 1.55, size = 78, normalized size = 1.53 \[ \begin {cases} \frac {x^{4} \sin ^{2}{\left (a + b x^{2} \right )}}{8} + \frac {x^{4} \cos ^{2}{\left (a + b x^{2} \right )}}{8} + \frac {x^{2} \sin {\left (a + b x^{2} \right )} \cos {\left (a + b x^{2} \right )}}{4 b} + \frac {\cos ^{2}{\left (a + b x^{2} \right )}}{8 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{4} \cos ^{2}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**4*sin(a + b*x**2)**2/8 + x**4*cos(a + b*x**2)**2/8 + x**2*sin(a + b*x**2)*cos(a + b*x**2)/(4*b)
+ cos(a + b*x**2)**2/(8*b**2), Ne(b, 0)), (x**4*cos(a)**2/4, True))

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