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

Optimal. Leaf size=80 \[ \frac{a \sin (c+d x)}{d^2}-\frac{a x \cos (c+d x)}{d}+\frac{3 b x^2 \sin (c+d x)}{d^2}-\frac{6 b \sin (c+d x)}{d^4}+\frac{6 b x \cos (c+d x)}{d^3}-\frac{b x^3 \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.101841, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3339, 3296, 2637} \[ \frac{a \sin (c+d x)}{d^2}-\frac{a x \cos (c+d x)}{d}+\frac{3 b x^2 \sin (c+d x)}{d^2}-\frac{6 b \sin (c+d x)}{d^4}+\frac{6 b x \cos (c+d x)}{d^3}-\frac{b x^3 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^2)*Sin[c + d*x],x]

[Out]

(6*b*x*Cos[c + d*x])/d^3 - (a*x*Cos[c + d*x])/d - (b*x^3*Cos[c + d*x])/d - (6*b*Sin[c + d*x])/d^4 + (a*Sin[c +
d*x])/d^2 + (3*b*x^2*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 \left (a+b x^2\right ) \sin (c+d x) \, dx &=\int \left (a x \sin (c+d x)+b x^3 \sin (c+d x)\right ) \, dx\\ &=a \int x \sin (c+d x) \, dx+b \int x^3 \sin (c+d x) \, dx\\ &=-\frac{a x \cos (c+d x)}{d}-\frac{b x^3 \cos (c+d x)}{d}+\frac{a \int \cos (c+d x) \, dx}{d}+\frac{(3 b) \int x^2 \cos (c+d x) \, dx}{d}\\ &=-\frac{a x \cos (c+d x)}{d}-\frac{b x^3 \cos (c+d x)}{d}+\frac{a \sin (c+d x)}{d^2}+\frac{3 b x^2 \sin (c+d x)}{d^2}-\frac{(6 b) \int x \sin (c+d x) \, dx}{d^2}\\ &=\frac{6 b x \cos (c+d x)}{d^3}-\frac{a x \cos (c+d x)}{d}-\frac{b x^3 \cos (c+d x)}{d}+\frac{a \sin (c+d x)}{d^2}+\frac{3 b x^2 \sin (c+d x)}{d^2}-\frac{(6 b) \int \cos (c+d x) \, dx}{d^3}\\ &=\frac{6 b x \cos (c+d x)}{d^3}-\frac{a x \cos (c+d x)}{d}-\frac{b x^3 \cos (c+d x)}{d}-\frac{6 b \sin (c+d x)}{d^4}+\frac{a \sin (c+d x)}{d^2}+\frac{3 b x^2 \sin (c+d x)}{d^2}\\ \end{align*}

Mathematica [A]  time = 0.111948, size = 57, normalized size = 0.71 \[ \frac{\left (a d^2+3 b \left (d^2 x^2-2\right )\right ) \sin (c+d x)-d x \left (a d^2+b \left (d^2 x^2-6\right )\right ) \cos (c+d x)}{d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.006, size = 181, normalized size = 2.3 \begin{align*}{\frac{1}{{d}^{2}} \left ({\frac{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\,{\frac{cb \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}}}+a \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) +3\,{\frac{{c}^{2}b \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{2}}}+ac\cos \left ( dx+c \right ) +{\frac{{c}^{3}b\cos \left ( dx+c \right ) }{{d}^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.02718, size = 223, normalized size = 2.79 \begin{align*} \frac{a c \cos \left (d x + c\right ) + \frac{b c^{3} \cos \left (d x + c\right )}{d^{2}} -{\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a - \frac{3 \,{\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b c^{2}}{d^{2}} + \frac{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 )} b c}{d^{2}} - \frac{{\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}{d^{2}}}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.34065, size = 130, normalized size = 1.62 \begin{align*} -\frac{{\left (b d^{3} x^{3} +{\left (a d^{3} - 6 \, b d\right )} x\right )} \cos \left (d x + c\right ) -{\left (3 \, b d^{2} x^{2} + a d^{2} - 6 \, b\right )} \sin \left (d x + c\right )}{d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.43142, size = 99, normalized size = 1.24 \begin{align*} \begin{cases} - \frac{a x \cos{\left (c + d x \right )}}{d} + \frac{a \sin{\left (c + d x \right )}}{d^{2}} - \frac{b x^{3} \cos{\left (c + d x \right )}}{d} + \frac{3 b x^{2} \sin{\left (c + d x \right )}}{d^{2}} + \frac{6 b x \cos{\left (c + d x \right )}}{d^{3}} - \frac{6 b \sin{\left (c + d x \right )}}{d^{4}} & \text{for}\: d \neq 0 \\\left (\frac{a x^{2}}{2} + \frac{b x^{4}}{4}\right ) \sin{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.09518, size = 81, normalized size = 1.01 \begin{align*} -\frac{{\left (b d^{3} x^{3} + a d^{3} x - 6 \, b d x\right )} \cos \left (d x + c\right )}{d^{4}} + \frac{{\left (3 \, b d^{2} x^{2} + a d^{2} - 6 \, b\right )} \sin \left (d x + c\right )}{d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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