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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 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]

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*}

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Mathematica [A]  time = 0.19, size = 45, normalized size = 0.90 \[ \frac {2 b d (c+d x) \sin (a+b x)-\cos (a+b x) \left (b^2 (c+d x)^2-2 d^2\right )}{b^3} \]

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

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fricas [A]  time = 0.57, size = 63, normalized size = 1.26 \[ -\frac {{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right ) - 2 \, {\left (b d^{2} x + b c d\right )} \sin \left (b x + a\right )}{b^{3}} \]

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

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giac [A]  time = 1.05, size = 65, normalized size = 1.30 \[ -\frac {{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )}{b^{3}} + \frac {2 \, {\left (b d^{2} x + b c d\right )} \sin \left (b x + a\right )}{b^{3}} \]

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

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maple [B]  time = 0.02, size = 148, normalized size = 2.96 \[ \frac {\frac {d^{2} \left (-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )}{b^{2}}-\frac {2 a \,d^{2} \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{b^{2}}+\frac {2 c d \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{b}-\frac {a^{2} d^{2} \cos \left (b x +a \right )}{b^{2}}+\frac {2 a c d \cos \left (b x +a \right )}{b}-c^{2} \cos \left (b x +a \right )}{b} \]

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))

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maxima [B]  time = 0.66, size = 141, normalized size = 2.82 \[ -\frac {c^{2} \cos \left (b x + a\right ) - \frac {2 \, a c d \cos \left (b x + a\right )}{b} + \frac {a^{2} d^{2} \cos \left (b x + a\right )}{b^{2}} + \frac {2 \, {\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} c d}{b} - \frac {2 \, {\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} a d^{2}}{b^{2}} + \frac {{\left ({\left ({\left (b x + a\right )}^{2} - 2\right )} \cos \left (b x + a\right ) - 2 \, {\left (b x + a\right )} \sin \left (b x + a\right )\right )} d^{2}}{b^{2}}}{b} \]

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

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mupad [B]  time = 0.55, size = 84, normalized size = 1.68 \[ \frac {\cos \left (a+b\,x\right )\,\left (2\,d^2-b^2\,c^2\right )}{b^3}-\frac {d^2\,x^2\,\cos \left (a+b\,x\right )}{b}+\frac {2\,c\,d\,\sin \left (a+b\,x\right )}{b^2}+\frac {2\,d^2\,x\,\sin \left (a+b\,x\right )}{b^2}-\frac {2\,c\,d\,x\,\cos \left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)*(c + d*x)^2,x)

[Out]

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

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sympy [A]  time = 0.73, size = 112, normalized size = 2.24 \[ \begin {cases} - \frac {c^{2} \cos {\left (a + b x \right )}}{b} - \frac {2 c d x \cos {\left (a + b x \right )}}{b} - \frac {d^{2} x^{2} \cos {\left (a + b x \right )}}{b} + \frac {2 c d \sin {\left (a + b x \right )}}{b^{2}} + \frac {2 d^{2} x \sin {\left (a + b x \right )}}{b^{2}} + \frac {2 d^{2} \cos {\left (a + b x \right )}}{b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \sin {\relax (a )} & \text {otherwise} \end {cases} \]

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))

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