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

Optimal. Leaf size=135 \[ \frac{a^2 \sin (c+d x)}{d^2}-\frac{a^2 x \cos (c+d x)}{d}+\frac{4 a b x \sin (c+d x)}{d^2}+\frac{4 a b \cos (c+d x)}{d^3}-\frac{2 a b x^2 \cos (c+d x)}{d}+\frac{3 b^2 x^2 \sin (c+d x)}{d^2}-\frac{6 b^2 \sin (c+d x)}{d^4}+\frac{6 b^2 x \cos (c+d x)}{d^3}-\frac{b^2 x^3 \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.186066, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {6742, 3296, 2637, 2638} \[ \frac{a^2 \sin (c+d x)}{d^2}-\frac{a^2 x \cos (c+d x)}{d}+\frac{4 a b x \sin (c+d x)}{d^2}+\frac{4 a b \cos (c+d x)}{d^3}-\frac{2 a b x^2 \cos (c+d x)}{d}+\frac{3 b^2 x^2 \sin (c+d x)}{d^2}-\frac{6 b^2 \sin (c+d x)}{d^4}+\frac{6 b^2 x \cos (c+d x)}{d^3}-\frac{b^2 x^3 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x)^2*Sin[c + d*x],x]

[Out]

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

Rule 6742

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

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]

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

Mathematica [A]  time = 0.211745, size = 87, normalized size = 0.64 \[ \frac{\left (a^2 d^2+4 a b d^2 x+3 b^2 \left (d^2 x^2-2\right )\right ) \sin (c+d x)-d \left (a^2 d^2 x+2 a b \left (d^2 x^2-2\right )+b^2 x \left (d^2 x^2-6\right )\right ) \cos (c+d x)}{d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.008, size = 281, normalized size = 2.1 \begin{align*}{\frac{1}{{d}^{2}} \left ({\frac{{b}^{2} \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}}}+2\,{\frac{ab \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}}-3\,{\frac{{b}^{2}c \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}^{2} \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) -4\,{\frac{abc \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{{b}^{2}{c}^{2} \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{2}}}+{a}^{2}c\cos \left ( dx+c \right ) -2\,{\frac{ab{c}^{2}\cos \left ( dx+c \right ) }{d}}+{\frac{{b}^{2}{c}^{3}\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+a)^2*sin(d*x+c),x)

[Out]

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

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Maxima [A]  time = 1.03814, size = 350, normalized size = 2.59 \begin{align*} \frac{a^{2} c \cos \left (d x + c\right ) + \frac{b^{2} c^{3} \cos \left (d x + c\right )}{d^{2}} - \frac{2 \, a b c^{2} \cos \left (d x + c\right )}{d} -{\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a^{2} - \frac{3 \,{\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b^{2} c^{2}}{d^{2}} + \frac{4 \,{\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a b c}{d} + \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^{2} c}{d^{2}} - \frac{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 b}{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^{2}}{d^{2}}}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.65226, size = 200, normalized size = 1.48 \begin{align*} -\frac{{\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} x^{2} - 4 \, a b d +{\left (a^{2} d^{3} - 6 \, b^{2} d\right )} x\right )} \cos \left (d x + c\right ) -{\left (3 \, b^{2} d^{2} x^{2} + 4 \, a b d^{2} x + a^{2} d^{2} - 6 \, b^{2}\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+a)^2*sin(d*x+c),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.28016, size = 172, normalized size = 1.27 \begin{align*} \begin{cases} - \frac{a^{2} x \cos{\left (c + d x \right )}}{d} + \frac{a^{2} \sin{\left (c + d x \right )}}{d^{2}} - \frac{2 a b x^{2} \cos{\left (c + d x \right )}}{d} + \frac{4 a b x \sin{\left (c + d x \right )}}{d^{2}} + \frac{4 a b \cos{\left (c + d x \right )}}{d^{3}} - \frac{b^{2} x^{3} \cos{\left (c + d x \right )}}{d} + \frac{3 b^{2} x^{2} \sin{\left (c + d x \right )}}{d^{2}} + \frac{6 b^{2} x \cos{\left (c + d x \right )}}{d^{3}} - \frac{6 b^{2} \sin{\left (c + d x \right )}}{d^{4}} & \text{for}\: d \neq 0 \\\left (\frac{a^{2} x^{2}}{2} + \frac{2 a b x^{3}}{3} + \frac{b^{2} 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+a)**2*sin(d*x+c),x)

[Out]

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

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Giac [A]  time = 1.10855, size = 128, normalized size = 0.95 \begin{align*} -\frac{{\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} x^{2} + a^{2} d^{3} x - 6 \, b^{2} d x - 4 \, a b d\right )} \cos \left (d x + c\right )}{d^{4}} + \frac{{\left (3 \, b^{2} d^{2} x^{2} + 4 \, a b d^{2} x + a^{2} d^{2} - 6 \, b^{2}\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+a)^2*sin(d*x+c),x, algorithm="giac")

[Out]

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