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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, 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, 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 2637

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

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 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]))

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 = 0.00, size = 44, normalized size = 1.00 \[ \frac {a x^4}{4}+\frac {b \sin \left (c+d x^2\right )}{2 d^2}-\frac {b x^2 \cos \left (c+d x^2\right )}{2 d} \]

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)

________________________________________________________________________________________

fricas [A]  time = 0.66, size = 40, normalized size = 0.91 \[ \frac {a d^{2} x^{4} - 2 \, b d x^{2} \cos \left (d x^{2} + c\right ) + 2 \, b \sin \left (d x^{2} + c\right )}{4 \, d^{2}} \]

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

________________________________________________________________________________________

giac [A]  time = 0.42, size = 61, normalized size = 1.39 \[ \frac {\frac {{\left ({\left (d x^{2} + c\right )}^{2} - 2 \, {\left (d x^{2} + c\right )} c\right )} a}{d} - \frac {2 \, {\left (d x^{2} \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right )} b}{d}}{4 \, d} \]

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

________________________________________________________________________________________

maple [A]  time = 0.02, size = 40, normalized size = 0.91 \[ \frac {a \,x^{4}}{4}+b \left (-\frac {x^{2} \cos \left (d \,x^{2}+c \right )}{2 d}+\frac {\sin \left (d \,x^{2}+c \right )}{2 d^{2}}\right ) \]

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))

________________________________________________________________________________________

maxima [A]  time = 0.89, size = 37, normalized size = 0.84 \[ \frac {1}{4} \, a x^{4} - \frac {{\left (d x^{2} \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right )} b}{2 \, d^{2}} \]

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

________________________________________________________________________________________

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

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

________________________________________________________________________________________

sympy [A]  time = 1.05, size = 49, normalized size = 1.11 \[ \begin {cases} \frac {a x^{4}}{4} - \frac {b x^{2} \cos {\left (c + d x^{2} \right )}}{2 d} + \frac {b \sin {\left (c + d x^{2} \right )}}{2 d^{2}} & \text {for}\: d \neq 0 \\\frac {x^{4} \left (a + b \sin {\relax (c )}\right )}{4} & \text {otherwise} \end {cases} \]

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))

________________________________________________________________________________________