3.3 \(\int x (a+b x) \sin (c+d x) \, dx\)

Optimal. Leaf size=65 \[ \frac{a \sin (c+d x)}{d^2}-\frac{a x \cos (c+d x)}{d}+\frac{2 b x \sin (c+d x)}{d^2}+\frac{2 b \cos (c+d x)}{d^3}-\frac{b x^2 \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.10529, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {6742, 3296, 2637, 2638} \[ \frac{a \sin (c+d x)}{d^2}-\frac{a x \cos (c+d x)}{d}+\frac{2 b x \sin (c+d x)}{d^2}+\frac{2 b \cos (c+d x)}{d^3}-\frac{b x^2 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x)*Sin[c + d*x],x]

[Out]

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

Mathematica [A]  time = 0.100959, size = 45, normalized size = 0.69 \[ \frac{d (a+2 b x) \sin (c+d x)-\left (a d^2 x+b \left (d^2 x^2-2\right )\right ) \cos (c+d x)}{d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x)*Sin[c + d*x],x]

[Out]

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

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Maple [A]  time = 0.004, size = 121, normalized size = 1.9 \begin{align*}{\frac{1}{{d}^{2}} \left ({\frac{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}}+a \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) -2\,{\frac{cb \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{d}}+ac\cos \left ( dx+c \right ) -{\frac{{c}^{2}b\cos \left ( dx+c \right ) }{d}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01656, size = 158, normalized size = 2.43 \begin{align*} \frac{a c \cos \left (d x + c\right ) - \frac{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 + \frac{2 \,{\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b c}{d} - \frac{{\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}{d}}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.61627, size = 108, normalized size = 1.66 \begin{align*} -\frac{{\left (b d^{2} x^{2} + a d^{2} x - 2 \, b\right )} \cos \left (d x + c\right ) -{\left (2 \, b d x + a d\right )} \sin \left (d x + c\right )}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.603914, size = 82, normalized size = 1.26 \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^{2} \cos{\left (c + d x \right )}}{d} + \frac{2 b x \sin{\left (c + d x \right )}}{d^{2}} + \frac{2 b \cos{\left (c + d x \right )}}{d^{3}} & \text{for}\: d \neq 0 \\\left (\frac{a x^{2}}{2} + \frac{b x^{3}}{3}\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)*sin(d*x+c),x)

[Out]

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

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Giac [A]  time = 1.10656, size = 66, normalized size = 1.02 \begin{align*} -\frac{{\left (b d^{2} x^{2} + a d^{2} x - 2 \, b\right )} \cos \left (d x + c\right )}{d^{3}} + \frac{{\left (2 \, b d x + a d\right )} \sin \left (d x + c\right )}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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