3.1.17 \(\int x \cos ^3(a+b x^2) \, dx\) [17]

Optimal. Leaf size=33 \[ \frac {\sin \left (a+b x^2\right )}{2 b}-\frac {\sin ^3\left (a+b x^2\right )}{6 b} \]

[Out]

1/2*sin(b*x^2+a)/b-1/6*sin(b*x^2+a)^3/b

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Rubi [A]
time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3461, 2713} \begin {gather*} \frac {\sin \left (a+b x^2\right )}{2 b}-\frac {\sin ^3\left (a+b x^2\right )}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Cos[a + b*x^2]^3,x]

[Out]

Sin[a + b*x^2]/(2*b) - Sin[a + b*x^2]^3/(6*b)

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3461

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 \cos ^3\left (a+b x^2\right ) \, dx &=\frac {1}{2} \text {Subst}\left (\int \cos ^3(a+b x) \, dx,x,x^2\right )\\ &=-\frac {\text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin \left (a+b x^2\right )\right )}{2 b}\\ &=\frac {\sin \left (a+b x^2\right )}{2 b}-\frac {\sin ^3\left (a+b x^2\right )}{6 b}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 33, normalized size = 1.00 \begin {gather*} \frac {\sin \left (a+b x^2\right )}{2 b}-\frac {\sin ^3\left (a+b x^2\right )}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Cos[a + b*x^2]^3,x]

[Out]

Sin[a + b*x^2]/(2*b) - Sin[a + b*x^2]^3/(6*b)

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Maple [A]
time = 0.06, size = 26, normalized size = 0.79

method	result	size
derivativedivides	\(\frac {\left (2+\cos ^{2}\left (b \,x^{2}+a \right )\right ) \sin \left (b \,x^{2}+a \right )}{6 b}\)	\(26\)
default	\(\frac {\left (2+\cos ^{2}\left (b \,x^{2}+a \right )\right ) \sin \left (b \,x^{2}+a \right )}{6 b}\)	\(26\)
risch	\(\frac {3 \sin \left (b \,x^{2}+a \right )}{8 b}+\frac {\sin \left (3 b \,x^{2}+3 a \right )}{24 b}\)	\(31\)
norman	\(\frac {\frac {\tan \left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )}{b}+\frac {\tan ^{5}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )}{b}+\frac {2 \left (\tan ^{3}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )}{3 b}}{\left (1+\tan ^{2}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )^{3}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/b*(2+cos(b*x^2+a)^2)*sin(b*x^2+a)

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Maxima [A]
time = 0.29, size = 27, normalized size = 0.82 \begin {gather*} \frac {\sin \left (3 \, b x^{2} + 3 \, a\right ) + 9 \, \sin \left (b x^{2} + a\right )}{24 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(sin(3*b*x^2 + 3*a) + 9*sin(b*x^2 + a))/b

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Fricas [A]
time = 0.37, size = 25, normalized size = 0.76 \begin {gather*} \frac {{\left (\cos \left (b x^{2} + a\right )^{2} + 2\right )} \sin \left (b x^{2} + a\right )}{6 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(cos(b*x^2 + a)^2 + 2)*sin(b*x^2 + a)/b

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Sympy [A]
time = 0.18, size = 44, normalized size = 1.33 \begin {gather*} \begin {cases} \frac {\sin ^{3}{\left (a + b x^{2} \right )}}{3 b} + \frac {\sin {\left (a + b x^{2} \right )} \cos ^{2}{\left (a + b x^{2} \right )}}{2 b} & \text {for}\: b \neq 0 \\\frac {x^{2} \cos ^{3}{\left (a \right )}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((sin(a + b*x**2)**3/(3*b) + sin(a + b*x**2)*cos(a + b*x**2)**2/(2*b), Ne(b, 0)), (x**2*cos(a)**3/2,
True))

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Giac [A]
time = 0.45, size = 26, normalized size = 0.79 \begin {gather*} -\frac {\sin \left (b x^{2} + a\right )^{3} - 3 \, \sin \left (b x^{2} + a\right )}{6 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(sin(b*x^2 + a)^3 - 3*sin(b*x^2 + a))/b

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Mupad [B]
time = 0.29, size = 28, normalized size = 0.85 \begin {gather*} \frac {3\,\sin \left (b\,x^2+a\right )-{\sin \left (b\,x^2+a\right )}^3}{6\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cos(a + b*x^2)^3,x)

[Out]

(3*sin(a + b*x^2) - sin(a + b*x^2)^3)/(6*b)

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