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

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {14, 3460, 3377, 2717} \begin {gather*} \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} \end {gather*}

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 2717

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

Rule 3377

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 3460

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]))

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 \text {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 \text {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*}

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Mathematica [A]
time = 0.00, size = 44, normalized size = 1.00 \begin {gather*} \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 {gather*}

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 = 0.03, size = 40, normalized size = 0.91

method	result	size
risch	\(\frac {a \,x^{4}}{4}-\frac {b \,x^{2} \cos \left (d \,x^{2}+c \right )}{2 d}+\frac {b \sin \left (d \,x^{2}+c \right )}{2 d^{2}}\)	\(39\)
default	\(\frac {a \,x^{4}}{4}+b \left (-\frac {x^{2} \cos \left (d \,x^{2}+c \right )}{2 d}+\frac {\sin \left (d \,x^{2}+c \right )}{2 d^{2}}\right )\)	\(40\)
norman	\(\frac {\frac {b \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{d^{2}}+\frac {a \,x^{4}}{4}+\frac {a \,x^{4} \left (\tan ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{4}-\frac {b \,x^{2}}{2 d}+\frac {b \,x^{2} \left (\tan ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{2 d}}{1+\tan ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.28, size = 37, normalized size = 0.84 \begin {gather*} \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 {gather*}

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 = 0.37, size = 40, normalized size = 0.91 \begin {gather*} \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 {gather*}

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 = 0.18, size = 49, normalized size = 1.11 \begin {gather*} \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 {gather*}

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 = 6.21, size = 75, normalized size = 1.70 \begin {gather*} \frac {{\left (d x^{2} + c\right )}^{2} a - 2 \, {\left (d x^{2} + c\right )} b \cos \left (d x^{2} + c\right ) + 2 \, b \sin \left (d x^{2} + c\right )}{4 \, d^{2}} - \frac {{\left (d x^{2} + c\right )} a c - b c \cos \left (d x^{2} + c\right )}{2 \, d^{2}} \end {gather*}

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

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Mupad [B]
time = 0.09, size = 38, normalized size = 0.86 \begin {gather*} \frac {a\,x^4}{4}+\frac {\frac {b\,\sin \left (d\,x^2+c\right )}{2}-\frac {b\,d\,x^2\,\cos \left (d\,x^2+c\right )}{2}}{d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*sin(c + d*x^2)),x)

[Out]

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

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