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

Optimal. Leaf size=141 \[ \frac{3 a x^2 \sin (c+d x)}{d^2}-\frac{6 a \sin (c+d x)}{d^4}+\frac{6 a x \cos (c+d x)}{d^3}-\frac{a x^3 \cos (c+d x)}{d}+\frac{5 b x^4 \sin (c+d x)}{d^2}-\frac{60 b x^2 \sin (c+d x)}{d^4}+\frac{20 b x^3 \cos (c+d x)}{d^3}+\frac{120 b \sin (c+d x)}{d^6}-\frac{120 b x \cos (c+d x)}{d^5}-\frac{b x^5 \cos (c+d x)}{d} \]

[Out]

(-120*b*x*Cos[c + d*x])/d^5 + (6*a*x*Cos[c + d*x])/d^3 + (20*b*x^3*Cos[c + d*x])/d^3 - (a*x^3*Cos[c + d*x])/d
- (b*x^5*Cos[c + d*x])/d + (120*b*Sin[c + d*x])/d^6 - (6*a*Sin[c + d*x])/d^4 - (60*b*x^2*Sin[c + d*x])/d^4 + (
3*a*x^2*Sin[c + d*x])/d^2 + (5*b*x^4*Sin[c + d*x])/d^2

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Rubi [A]  time = 0.207666, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {3339, 3296, 2637} \[ \frac{3 a x^2 \sin (c+d x)}{d^2}-\frac{6 a \sin (c+d x)}{d^4}+\frac{6 a x \cos (c+d x)}{d^3}-\frac{a x^3 \cos (c+d x)}{d}+\frac{5 b x^4 \sin (c+d x)}{d^2}-\frac{60 b x^2 \sin (c+d x)}{d^4}+\frac{20 b x^3 \cos (c+d x)}{d^3}+\frac{120 b \sin (c+d x)}{d^6}-\frac{120 b x \cos (c+d x)}{d^5}-\frac{b x^5 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2)*Sin[c + d*x],x]

[Out]

(-120*b*x*Cos[c + d*x])/d^5 + (6*a*x*Cos[c + d*x])/d^3 + (20*b*x^3*Cos[c + d*x])/d^3 - (a*x^3*Cos[c + d*x])/d
- (b*x^5*Cos[c + d*x])/d + (120*b*Sin[c + d*x])/d^6 - (6*a*Sin[c + d*x])/d^4 - (60*b*x^2*Sin[c + d*x])/d^4 + (
3*a*x^2*Sin[c + d*x])/d^2 + (5*b*x^4*Sin[c + d*x])/d^2

Rule 3339

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*Sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[ExpandIntegran
d[Sin[c + d*x], (e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 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 2637

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

Rubi steps

\begin{align*} \int x^3 \left (a+b x^2\right ) \sin (c+d x) \, dx &=\int \left (a x^3 \sin (c+d x)+b x^5 \sin (c+d x)\right ) \, dx\\ &=a \int x^3 \sin (c+d x) \, dx+b \int x^5 \sin (c+d x) \, dx\\ &=-\frac{a x^3 \cos (c+d x)}{d}-\frac{b x^5 \cos (c+d x)}{d}+\frac{(3 a) \int x^2 \cos (c+d x) \, dx}{d}+\frac{(5 b) \int x^4 \cos (c+d x) \, dx}{d}\\ &=-\frac{a x^3 \cos (c+d x)}{d}-\frac{b x^5 \cos (c+d x)}{d}+\frac{3 a x^2 \sin (c+d x)}{d^2}+\frac{5 b x^4 \sin (c+d x)}{d^2}-\frac{(6 a) \int x \sin (c+d x) \, dx}{d^2}-\frac{(20 b) \int x^3 \sin (c+d x) \, dx}{d^2}\\ &=\frac{6 a x \cos (c+d x)}{d^3}+\frac{20 b x^3 \cos (c+d x)}{d^3}-\frac{a x^3 \cos (c+d x)}{d}-\frac{b x^5 \cos (c+d x)}{d}+\frac{3 a x^2 \sin (c+d x)}{d^2}+\frac{5 b x^4 \sin (c+d x)}{d^2}-\frac{(6 a) \int \cos (c+d x) \, dx}{d^3}-\frac{(60 b) \int x^2 \cos (c+d x) \, dx}{d^3}\\ &=\frac{6 a x \cos (c+d x)}{d^3}+\frac{20 b x^3 \cos (c+d x)}{d^3}-\frac{a x^3 \cos (c+d x)}{d}-\frac{b x^5 \cos (c+d x)}{d}-\frac{6 a \sin (c+d x)}{d^4}-\frac{60 b x^2 \sin (c+d x)}{d^4}+\frac{3 a x^2 \sin (c+d x)}{d^2}+\frac{5 b x^4 \sin (c+d x)}{d^2}+\frac{(120 b) \int x \sin (c+d x) \, dx}{d^4}\\ &=-\frac{120 b x \cos (c+d x)}{d^5}+\frac{6 a x \cos (c+d x)}{d^3}+\frac{20 b x^3 \cos (c+d x)}{d^3}-\frac{a x^3 \cos (c+d x)}{d}-\frac{b x^5 \cos (c+d x)}{d}-\frac{6 a \sin (c+d x)}{d^4}-\frac{60 b x^2 \sin (c+d x)}{d^4}+\frac{3 a x^2 \sin (c+d x)}{d^2}+\frac{5 b x^4 \sin (c+d x)}{d^2}+\frac{(120 b) \int \cos (c+d x) \, dx}{d^5}\\ &=-\frac{120 b x \cos (c+d x)}{d^5}+\frac{6 a x \cos (c+d x)}{d^3}+\frac{20 b x^3 \cos (c+d x)}{d^3}-\frac{a x^3 \cos (c+d x)}{d}-\frac{b x^5 \cos (c+d x)}{d}+\frac{120 b \sin (c+d x)}{d^6}-\frac{6 a \sin (c+d x)}{d^4}-\frac{60 b x^2 \sin (c+d x)}{d^4}+\frac{3 a x^2 \sin (c+d x)}{d^2}+\frac{5 b x^4 \sin (c+d x)}{d^2}\\ \end{align*}

Mathematica [A]  time = 0.167586, size = 92, normalized size = 0.65 \[ \frac{\left (3 a d^2 \left (d^2 x^2-2\right )+5 b \left (d^4 x^4-12 d^2 x^2+24\right )\right ) \sin (c+d x)-d x \left (a d^2 \left (d^2 x^2-6\right )+b \left (d^4 x^4-20 d^2 x^2+120\right )\right ) \cos (c+d x)}{d^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x^2)*Sin[c + d*x],x]

[Out]

(-(d*x*(a*d^2*(-6 + d^2*x^2) + b*(120 - 20*d^2*x^2 + d^4*x^4))*Cos[c + d*x]) + (3*a*d^2*(-2 + d^2*x^2) + 5*b*(
24 - 12*d^2*x^2 + d^4*x^4))*Sin[c + d*x])/d^6

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Maple [B]  time = 0.007, size = 449, normalized size = 3.2 \begin{align*}{\frac{1}{{d}^{4}} \left ({\frac{b \left ( - \left ( dx+c \right ) ^{5}\cos \left ( dx+c \right ) +5\, \left ( dx+c \right ) ^{4}\sin \left ( dx+c \right ) +20\, \left ( dx+c \right ) ^{3}\cos \left ( dx+c \right ) -60\, \left ( dx+c \right ) ^{2}\sin \left ( dx+c \right ) +120\,\sin \left ( dx+c \right ) -120\, \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{2}}}-5\,{\frac{cb \left ( - \left ( dx+c \right ) ^{4}\cos \left ( dx+c \right ) +4\, \left ( dx+c \right ) ^{3}\sin \left ( dx+c \right ) +12\, \left ( dx+c \right ) ^{2}\cos \left ( dx+c \right ) -24\,\cos \left ( dx+c \right ) -24\, \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) }{{d}^{2}}}+a \left ( - \left ( dx+c \right ) ^{3}\cos \left ( dx+c \right ) +3\, \left ( dx+c \right ) ^{2}\sin \left ( dx+c \right ) -6\,\sin \left ( dx+c \right ) +6\, \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) +10\,{\frac{{c}^{2}b \left ( - \left ( dx+c \right ) ^{3}\cos \left ( dx+c \right ) +3\, \left ( dx+c \right ) ^{2}\sin \left ( dx+c \right ) -6\,\sin \left ( dx+c \right ) +6\, \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{2}}}-3\,ac \left ( - \left ( dx+c \right ) ^{2}\cos \left ( dx+c \right ) +2\,\cos \left ( dx+c \right ) +2\, \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) -10\,{\frac{{c}^{3}b \left ( - \left ( dx+c \right ) ^{2}\cos \left ( dx+c \right ) +2\,\cos \left ( dx+c \right ) +2\, \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) }{{d}^{2}}}+3\,a{c}^{2} \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) +5\,{\frac{b{c}^{4} \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{2}}}+a{c}^{3}\cos \left ( dx+c \right ) +{\frac{b{c}^{5}\cos \left ( dx+c \right ) }{{d}^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^4*(1/d^2*b*(-(d*x+c)^5*cos(d*x+c)+5*(d*x+c)^4*sin(d*x+c)+20*(d*x+c)^3*cos(d*x+c)-60*(d*x+c)^2*sin(d*x+c)+1
20*sin(d*x+c)-120*(d*x+c)*cos(d*x+c))-5/d^2*b*c*(-(d*x+c)^4*cos(d*x+c)+4*(d*x+c)^3*sin(d*x+c)+12*(d*x+c)^2*cos
(d*x+c)-24*cos(d*x+c)-24*(d*x+c)*sin(d*x+c))+a*(-(d*x+c)^3*cos(d*x+c)+3*(d*x+c)^2*sin(d*x+c)-6*sin(d*x+c)+6*(d
*x+c)*cos(d*x+c))+10/d^2*b*c^2*(-(d*x+c)^3*cos(d*x+c)+3*(d*x+c)^2*sin(d*x+c)-6*sin(d*x+c)+6*(d*x+c)*cos(d*x+c)
)-3*a*c*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x+c))-10/d^2*b*c^3*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*
x+c)+2*(d*x+c)*sin(d*x+c))+3*a*c^2*(sin(d*x+c)-(d*x+c)*cos(d*x+c))+5/d^2*b*c^4*(sin(d*x+c)-(d*x+c)*cos(d*x+c))
+a*c^3*cos(d*x+c)+1/d^2*b*c^5*cos(d*x+c))

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Maxima [B]  time = 1.07076, size = 502, normalized size = 3.56 \begin{align*} \frac{a c^{3} \cos \left (d x + c\right ) + \frac{b c^{5} \cos \left (d x + c\right )}{d^{2}} - 3 \,{\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a c^{2} - \frac{5 \,{\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b c^{4}}{d^{2}} + 3 \,{\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \,{\left (d x + c\right )} \sin \left (d x + c\right )\right )} a c + \frac{10 \,{\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \,{\left (d x + c\right )} \sin \left (d x + c\right )\right )} b c^{3}}{d^{2}} -{\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \,{\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} a - \frac{10 \,{\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \,{\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} b c^{2}}{d^{2}} + \frac{5 \,{\left ({\left ({\left (d x + c\right )}^{4} - 12 \,{\left (d x + c\right )}^{2} + 24\right )} \cos \left (d x + c\right ) - 4 \,{\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \sin \left (d x + c\right )\right )} b c}{d^{2}} - \frac{{\left ({\left ({\left (d x + c\right )}^{5} - 20 \,{\left (d x + c\right )}^{3} + 120 \, d x + 120 \, c\right )} \cos \left (d x + c\right ) - 5 \,{\left ({\left (d x + c\right )}^{4} - 12 \,{\left (d x + c\right )}^{2} + 24\right )} \sin \left (d x + c\right )\right )} b}{d^{2}}}{d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*c^3*cos(d*x + c) + b*c^5*cos(d*x + c)/d^2 - 3*((d*x + c)*cos(d*x + c) - sin(d*x + c))*a*c^2 - 5*((d*x + c)*
cos(d*x + c) - sin(d*x + c))*b*c^4/d^2 + 3*(((d*x + c)^2 - 2)*cos(d*x + c) - 2*(d*x + c)*sin(d*x + c))*a*c + 1
0*(((d*x + c)^2 - 2)*cos(d*x + c) - 2*(d*x + c)*sin(d*x + c))*b*c^3/d^2 - (((d*x + c)^3 - 6*d*x - 6*c)*cos(d*x
+ c) - 3*((d*x + c)^2 - 2)*sin(d*x + c))*a - 10*(((d*x + c)^3 - 6*d*x - 6*c)*cos(d*x + c) - 3*((d*x + c)^2 -
2)*sin(d*x + c))*b*c^2/d^2 + 5*(((d*x + c)^4 - 12*(d*x + c)^2 + 24)*cos(d*x + c) - 4*((d*x + c)^3 - 6*d*x - 6*
c)*sin(d*x + c))*b*c/d^2 - (((d*x + c)^5 - 20*(d*x + c)^3 + 120*d*x + 120*c)*cos(d*x + c) - 5*((d*x + c)^4 - 1
2*(d*x + c)^2 + 24)*sin(d*x + c))*b/d^2)/d^4

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Fricas [A]  time = 1.33239, size = 209, normalized size = 1.48 \begin{align*} -\frac{{\left (b d^{5} x^{5} +{\left (a d^{5} - 20 \, b d^{3}\right )} x^{3} - 6 \,{\left (a d^{3} - 20 \, b d\right )} x\right )} \cos \left (d x + c\right ) -{\left (5 \, b d^{4} x^{4} - 6 \, a d^{2} + 3 \,{\left (a d^{4} - 20 \, b d^{2}\right )} x^{2} + 120 \, b\right )} \sin \left (d x + c\right )}{d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((b*d^5*x^5 + (a*d^5 - 20*b*d^3)*x^3 - 6*(a*d^3 - 20*b*d)*x)*cos(d*x + c) - (5*b*d^4*x^4 - 6*a*d^2 + 3*(a*d^4
- 20*b*d^2)*x^2 + 120*b)*sin(d*x + c))/d^6

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Sympy [A]  time = 5.0927, size = 168, normalized size = 1.19 \begin{align*} \begin{cases} - \frac{a x^{3} \cos{\left (c + d x \right )}}{d} + \frac{3 a x^{2} \sin{\left (c + d x \right )}}{d^{2}} + \frac{6 a x \cos{\left (c + d x \right )}}{d^{3}} - \frac{6 a \sin{\left (c + d x \right )}}{d^{4}} - \frac{b x^{5} \cos{\left (c + d x \right )}}{d} + \frac{5 b x^{4} \sin{\left (c + d x \right )}}{d^{2}} + \frac{20 b x^{3} \cos{\left (c + d x \right )}}{d^{3}} - \frac{60 b x^{2} \sin{\left (c + d x \right )}}{d^{4}} - \frac{120 b x \cos{\left (c + d x \right )}}{d^{5}} + \frac{120 b \sin{\left (c + d x \right )}}{d^{6}} & \text{for}\: d \neq 0 \\\left (\frac{a x^{4}}{4} + \frac{b x^{6}}{6}\right ) \sin{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b*x**2+a)*sin(d*x+c),x)

[Out]

Piecewise((-a*x**3*cos(c + d*x)/d + 3*a*x**2*sin(c + d*x)/d**2 + 6*a*x*cos(c + d*x)/d**3 - 6*a*sin(c + d*x)/d*
*4 - b*x**5*cos(c + d*x)/d + 5*b*x**4*sin(c + d*x)/d**2 + 20*b*x**3*cos(c + d*x)/d**3 - 60*b*x**2*sin(c + d*x)
/d**4 - 120*b*x*cos(c + d*x)/d**5 + 120*b*sin(c + d*x)/d**6, Ne(d, 0)), ((a*x**4/4 + b*x**6/6)*sin(c), True))

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Giac [A]  time = 1.14615, size = 131, normalized size = 0.93 \begin{align*} -\frac{{\left (b d^{5} x^{5} + a d^{5} x^{3} - 20 \, b d^{3} x^{3} - 6 \, a d^{3} x + 120 \, b d x\right )} \cos \left (d x + c\right )}{d^{6}} + \frac{{\left (5 \, b d^{4} x^{4} + 3 \, a d^{4} x^{2} - 60 \, b d^{2} x^{2} - 6 \, a d^{2} + 120 \, b\right )} \sin \left (d x + c\right )}{d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*d^5*x^5 + a*d^5*x^3 - 20*b*d^3*x^3 - 6*a*d^3*x + 120*b*d*x)*cos(d*x + c)/d^6 + (5*b*d^4*x^4 + 3*a*d^4*x^2
- 60*b*d^2*x^2 - 6*a*d^2 + 120*b)*sin(d*x + c)/d^6