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

Optimal. Leaf size=96 \[ \frac {2 a \cos (c+d x)}{d^3}+\frac {2 a x \sin (c+d x)}{d^2}-\frac {a x^2 \cos (c+d x)}{d}-\frac {6 b \sin (c+d x)}{d^4}+\frac {6 b x \cos (c+d x)}{d^3}+\frac {3 b x^2 \sin (c+d x)}{d^2}-\frac {b x^3 \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6742, 3296, 2638, 2637} \[ \frac {2 a x \sin (c+d x)}{d^2}+\frac {2 a \cos (c+d x)}{d^3}-\frac {a x^2 \cos (c+d x)}{d}+\frac {3 b x^2 \sin (c+d x)}{d^2}-\frac {6 b \sin (c+d x)}{d^4}+\frac {6 b x \cos (c+d x)}{d^3}-\frac {b x^3 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x)*Sin[c + d*x],x]

[Out]

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

Rule 2637

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 65, normalized size = 0.68 \[ \frac {\left (2 a d^2 x+3 b \left (d^2 x^2-2\right )\right ) \sin (c+d x)-d \left (a \left (d^2 x^2-2\right )+b x \left (d^2 x^2-6\right )\right ) \cos (c+d x)}{d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 67, normalized size = 0.70 \[ -\frac {{\left (b d^{3} x^{3} + a d^{3} x^{2} - 6 \, b d x - 2 \, a d\right )} \cos \left (d x + c\right ) - {\left (3 \, b d^{2} x^{2} + 2 \, a d^{2} x - 6 \, b\right )} \sin \left (d x + c\right )}{d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.42, size = 68, normalized size = 0.71 \[ -\frac {{\left (b d^{3} x^{3} + a d^{3} x^{2} - 6 \, b d x - 2 \, a d\right )} \cos \left (d x + c\right )}{d^{4}} + \frac {{\left (3 \, b d^{2} x^{2} + 2 \, a d^{2} x - 6 \, b\right )} \sin \left (d x + c\right )}{d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 225, normalized size = 2.34 \[ \frac {\frac {b \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \left (d x +c \right ) \cos \left (d x +c \right )\right )}{d}+a \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )-\frac {3 b c \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d}-2 a c \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )+\frac {3 b \,c^{2} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d}-a \,c^{2} \cos \left (d x +c \right )+\frac {b \,c^{3} \cos \left (d x +c \right )}{d}}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.47, size = 201, normalized size = 2.09 \[ -\frac {a c^{2} \cos \left (d x + c\right ) - \frac {b c^{3} \cos \left (d x + c\right )}{d} - 2 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a c + \frac {3 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b c^{2}}{d} + {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} a - \frac {3 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} b c}{d} + \frac {{\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} b}{d}}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.62, size = 92, normalized size = 0.96 \[ \frac {3\,b\,x^2\,\sin \left (c+d\,x\right )+2\,a\,x\,\sin \left (c+d\,x\right )}{d^2}+\frac {2\,a\,\cos \left (c+d\,x\right )+6\,b\,x\,\cos \left (c+d\,x\right )}{d^3}-\frac {a\,x^2\,\cos \left (c+d\,x\right )+b\,x^3\,\cos \left (c+d\,x\right )}{d}-\frac {6\,b\,\sin \left (c+d\,x\right )}{d^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sin(c + d*x)*(a + b*x),x)

[Out]

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

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sympy [A]  time = 1.15, size = 117, normalized size = 1.22 \[ \begin {cases} - \frac {a x^{2} \cos {\left (c + d x \right )}}{d} + \frac {2 a x \sin {\left (c + d x \right )}}{d^{2}} + \frac {2 a \cos {\left (c + d x \right )}}{d^{3}} - \frac {b x^{3} \cos {\left (c + d x \right )}}{d} + \frac {3 b x^{2} \sin {\left (c + d x \right )}}{d^{2}} + \frac {6 b x \cos {\left (c + d x \right )}}{d^{3}} - \frac {6 b \sin {\left (c + d x \right )}}{d^{4}} & \text {for}\: d \neq 0 \\\left (\frac {a x^{3}}{3} + \frac {b x^{4}}{4}\right ) \sin {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x+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**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**3/3 + b*x**4/4
)*sin(c), True))

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