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

Optimal. Leaf size=50 \[ \frac{2 d (c+d x) \sin (a+b x)}{b^2}+\frac{2 d^2 \cos (a+b x)}{b^3}-\frac{(c+d x)^2 \cos (a+b x)}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0389583, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3296, 2638} \[ \frac{2 d (c+d x) \sin (a+b x)}{b^2}+\frac{2 d^2 \cos (a+b x)}{b^3}-\frac{(c+d x)^2 \cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Sin[a + b*x],x]

[Out]

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

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

Mathematica [A]  time = 0.170771, size = 45, normalized size = 0.9 \[ \frac{2 b d (c+d x) \sin (a+b x)-\cos (a+b x) \left (b^2 (c+d x)^2-2 d^2\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*Sin[a + b*x],x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.006, size = 148, normalized size = 3. \begin{align*}{\frac{1}{b} \left ({\frac{{d}^{2} \left ( - \left ( bx+a \right ) ^{2}\cos \left ( bx+a \right ) +2\,\cos \left ( bx+a \right ) +2\, \left ( bx+a \right ) \sin \left ( bx+a \right ) \right ) }{{b}^{2}}}-2\,{\frac{a{d}^{2} \left ( \sin \left ( bx+a \right ) - \left ( bx+a \right ) \cos \left ( bx+a \right ) \right ) }{{b}^{2}}}+2\,{\frac{cd \left ( \sin \left ( bx+a \right ) - \left ( bx+a \right ) \cos \left ( bx+a \right ) \right ) }{b}}-{\frac{{a}^{2}{d}^{2}\cos \left ( bx+a \right ) }{{b}^{2}}}+2\,{\frac{acd\cos \left ( bx+a \right ) }{b}}-{c}^{2}\cos \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.03117, size = 190, normalized size = 3.8 \begin{align*} -\frac{c^{2} \cos \left (b x + a\right ) - \frac{2 \, a c d \cos \left (b x + a\right )}{b} + \frac{a^{2} d^{2} \cos \left (b x + a\right )}{b^{2}} + \frac{2 \,{\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} c d}{b} - \frac{2 \,{\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} a d^{2}}{b^{2}} + \frac{{\left ({\left ({\left (b x + a\right )}^{2} - 2\right )} \cos \left (b x + a\right ) - 2 \,{\left (b x + a\right )} \sin \left (b x + a\right )\right )} d^{2}}{b^{2}}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.6819, size = 138, normalized size = 2.76 \begin{align*} -\frac{{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right ) - 2 \,{\left (b d^{2} x + b c d\right )} \sin \left (b x + a\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.636426, size = 112, normalized size = 2.24 \begin{align*} \begin{cases} - \frac{c^{2} \cos{\left (a + b x \right )}}{b} - \frac{2 c d x \cos{\left (a + b x \right )}}{b} - \frac{d^{2} x^{2} \cos{\left (a + b x \right )}}{b} + \frac{2 c d \sin{\left (a + b x \right )}}{b^{2}} + \frac{2 d^{2} x \sin{\left (a + b x \right )}}{b^{2}} + \frac{2 d^{2} \cos{\left (a + b x \right )}}{b^{3}} & \text{for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac{d^{2} x^{3}}{3}\right ) \sin{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.15278, size = 88, normalized size = 1.76 \begin{align*} -\frac{{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )}{b^{3}} + \frac{2 \,{\left (b d^{2} x + b c d\right )} \sin \left (b x + a\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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