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

Optimal. Leaf size=111 \[ \frac{2 a x \sin (c+d x)}{d^2}+\frac{2 a \cos (c+d x)}{d^3}-\frac{a x^2 \cos (c+d x)}{d}+\frac{4 b x^3 \sin (c+d x)}{d^2}+\frac{12 b x^2 \cos (c+d x)}{d^3}-\frac{24 b x \sin (c+d x)}{d^4}-\frac{24 b \cos (c+d x)}{d^5}-\frac{b x^4 \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.16329, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {3339, 3296, 2638} \[ \frac{2 a x \sin (c+d x)}{d^2}+\frac{2 a \cos (c+d x)}{d^3}-\frac{a x^2 \cos (c+d x)}{d}+\frac{4 b x^3 \sin (c+d x)}{d^2}+\frac{12 b x^2 \cos (c+d x)}{d^3}-\frac{24 b x \sin (c+d x)}{d^4}-\frac{24 b \cos (c+d x)}{d^5}-\frac{b x^4 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^2)*Sin[c + d*x],x]

[Out]

(-24*b*Cos[c + d*x])/d^5 + (2*a*Cos[c + d*x])/d^3 + (12*b*x^2*Cos[c + d*x])/d^3 - (a*x^2*Cos[c + d*x])/d - (b*
x^4*Cos[c + d*x])/d - (24*b*x*Sin[c + d*x])/d^4 + (2*a*x*Sin[c + d*x])/d^2 + (4*b*x^3*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 2638

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

Rubi steps

\begin{align*} \int x^2 \left (a+b x^2\right ) \sin (c+d x) \, dx &=\int \left (a x^2 \sin (c+d x)+b x^4 \sin (c+d x)\right ) \, dx\\ &=a \int x^2 \sin (c+d x) \, dx+b \int x^4 \sin (c+d x) \, dx\\ &=-\frac{a x^2 \cos (c+d x)}{d}-\frac{b x^4 \cos (c+d x)}{d}+\frac{(2 a) \int x \cos (c+d x) \, dx}{d}+\frac{(4 b) \int x^3 \cos (c+d x) \, dx}{d}\\ &=-\frac{a x^2 \cos (c+d x)}{d}-\frac{b x^4 \cos (c+d x)}{d}+\frac{2 a x \sin (c+d x)}{d^2}+\frac{4 b x^3 \sin (c+d x)}{d^2}-\frac{(2 a) \int \sin (c+d x) \, dx}{d^2}-\frac{(12 b) \int x^2 \sin (c+d x) \, dx}{d^2}\\ &=\frac{2 a \cos (c+d x)}{d^3}+\frac{12 b x^2 \cos (c+d x)}{d^3}-\frac{a x^2 \cos (c+d x)}{d}-\frac{b x^4 \cos (c+d x)}{d}+\frac{2 a x \sin (c+d x)}{d^2}+\frac{4 b x^3 \sin (c+d x)}{d^2}-\frac{(24 b) \int x \cos (c+d x) \, dx}{d^3}\\ &=\frac{2 a \cos (c+d x)}{d^3}+\frac{12 b x^2 \cos (c+d x)}{d^3}-\frac{a x^2 \cos (c+d x)}{d}-\frac{b x^4 \cos (c+d x)}{d}-\frac{24 b x \sin (c+d x)}{d^4}+\frac{2 a x \sin (c+d x)}{d^2}+\frac{4 b x^3 \sin (c+d x)}{d^2}+\frac{(24 b) \int \sin (c+d x) \, dx}{d^4}\\ &=-\frac{24 b \cos (c+d x)}{d^5}+\frac{2 a \cos (c+d x)}{d^3}+\frac{12 b x^2 \cos (c+d x)}{d^3}-\frac{a x^2 \cos (c+d x)}{d}-\frac{b x^4 \cos (c+d x)}{d}-\frac{24 b x \sin (c+d x)}{d^4}+\frac{2 a x \sin (c+d x)}{d^2}+\frac{4 b x^3 \sin (c+d x)}{d^2}\\ \end{align*}

Mathematica [A]  time = 0.143472, size = 75, normalized size = 0.68 \[ \frac{2 d x \left (a d^2+2 b \left (d^2 x^2-6\right )\right ) \sin (c+d x)-\left (a d^2 \left (d^2 x^2-2\right )+b \left (d^4 x^4-12 d^2 x^2+24\right )\right ) \cos (c+d x)}{d^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a*d^2*(-2 + d^2*x^2) + b*(24 - 12*d^2*x^2 + d^4*x^4))*Cos[c + d*x]) + 2*d*x*(a*d^2 + 2*b*(-6 + d^2*x^2))*S
in[c + d*x])/d^5

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Maple [B]  time = 0.007, size = 302, normalized size = 2.7 \begin{align*}{\frac{1}{{d}^{3}} \left ({\frac{b \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}}}-4\,{\frac{cb \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}}}+a \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 ) +6\,{\frac{{c}^{2}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}}}-2\,ac \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) -4\,{\frac{{c}^{3}b \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{2}}}-a{c}^{2}\cos \left ( dx+c \right ) -{\frac{b{c}^{4}\cos \left ( dx+c \right ) }{{d}^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^3*(1/d^2*b*(-(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))-4/d^2*b*c*(-(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))+a*(-(d*
x+c)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x+c))+6/d^2*b*c^2*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)
*sin(d*x+c))-2*a*c*(sin(d*x+c)-(d*x+c)*cos(d*x+c))-4/d^2*b*c^3*(sin(d*x+c)-(d*x+c)*cos(d*x+c))-a*c^2*cos(d*x+c
)-1/d^2*b*c^4*cos(d*x+c))

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Maxima [B]  time = 1.04283, size = 348, normalized size = 3.14 \begin{align*} -\frac{a c^{2} \cos \left (d x + c\right ) + \frac{b c^{4} \cos \left (d x + c\right )}{d^{2}} - 2 \,{\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a c - \frac{4 \,{\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b c^{3}}{d^{2}} +{\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 + \frac{6 \,{\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^{2}}{d^{2}} - \frac{4 \,{\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}{d^{2}} + \frac{{\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}{d^{2}}}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.35728, size = 171, normalized size = 1.54 \begin{align*} -\frac{{\left (b d^{4} x^{4} - 2 \, a d^{2} +{\left (a d^{4} - 12 \, b d^{2}\right )} x^{2} + 24 \, b\right )} \cos \left (d x + c\right ) - 2 \,{\left (2 \, b d^{3} x^{3} +{\left (a d^{3} - 12 \, b d\right )} x\right )} \sin \left (d x + c\right )}{d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((b*d^4*x^4 - 2*a*d^2 + (a*d^4 - 12*b*d^2)*x^2 + 24*b)*cos(d*x + c) - 2*(2*b*d^3*x^3 + (a*d^3 - 12*b*d)*x)*si
n(d*x + c))/d^5

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Sympy [A]  time = 2.9895, size = 134, normalized size = 1.21 \begin{align*} \begin{cases} - \frac{a x^{2} \cos{\left (c + d x \right )}}{d} + \frac{2 a x \sin{\left (c + d x \right )}}{d^{2}} + \frac{2 a \cos{\left (c + d x \right )}}{d^{3}} - \frac{b x^{4} \cos{\left (c + d x \right )}}{d} + \frac{4 b x^{3} \sin{\left (c + d x \right )}}{d^{2}} + \frac{12 b x^{2} \cos{\left (c + d x \right )}}{d^{3}} - \frac{24 b x \sin{\left (c + d x \right )}}{d^{4}} - \frac{24 b \cos{\left (c + d x \right )}}{d^{5}} & \text{for}\: d \neq 0 \\\left (\frac{a x^{3}}{3} + \frac{b x^{5}}{5}\right ) \sin{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a*x**2*cos(c + d*x)/d + 2*a*x*sin(c + d*x)/d**2 + 2*a*cos(c + d*x)/d**3 - b*x**4*cos(c + d*x)/d +
4*b*x**3*sin(c + d*x)/d**2 + 12*b*x**2*cos(c + d*x)/d**3 - 24*b*x*sin(c + d*x)/d**4 - 24*b*cos(c + d*x)/d**5,
Ne(d, 0)), ((a*x**3/3 + b*x**5/5)*sin(c), True))

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Giac [A]  time = 1.16785, size = 107, normalized size = 0.96 \begin{align*} -\frac{{\left (b d^{4} x^{4} + a d^{4} x^{2} - 12 \, b d^{2} x^{2} - 2 \, a d^{2} + 24 \, b\right )} \cos \left (d x + c\right )}{d^{5}} + \frac{2 \,{\left (2 \, b d^{3} x^{3} + a d^{3} x - 12 \, b d x\right )} \sin \left (d x + c\right )}{d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*d^4*x^4 + a*d^4*x^2 - 12*b*d^2*x^2 - 2*a*d^2 + 24*b)*cos(d*x + c)/d^5 + 2*(2*b*d^3*x^3 + a*d^3*x - 12*b*d*
x)*sin(d*x + c)/d^5