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

Optimal. Leaf size=49 \[ -\frac {2 d^2 \sin (a+b x)}{b^3}+\frac {2 d (c+d x) \cos (a+b x)}{b^2}+\frac {(c+d x)^2 \sin (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 49, 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, 2637} \[ \frac {2 d (c+d x) \cos (a+b x)}{b^2}-\frac {2 d^2 \sin (a+b x)}{b^3}+\frac {(c+d x)^2 \sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Cos[a + b*x],x]

[Out]

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 \cos (a+b x) \, dx &=\frac {(c+d x)^2 \sin (a+b x)}{b}-\frac {(2 d) \int (c+d x) \sin (a+b x) \, dx}{b}\\ &=\frac {2 d (c+d x) \cos (a+b x)}{b^2}+\frac {(c+d x)^2 \sin (a+b x)}{b}-\frac {\left (2 d^2\right ) \int \cos (a+b x) \, dx}{b^2}\\ &=\frac {2 d (c+d x) \cos (a+b x)}{b^2}-\frac {2 d^2 \sin (a+b x)}{b^3}+\frac {(c+d x)^2 \sin (a+b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 44, normalized size = 0.90 \[ \frac {\sin (a+b x) \left (b^2 (c+d x)^2-2 d^2\right )+2 b d (c+d x) \cos (a+b x)}{b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*Cos[a + b*x],x]

[Out]

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

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fricas [A]  time = 0.57, size = 62, normalized size = 1.27 \[ \frac {2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \sin \left (b x + a\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.40, size = 64, normalized size = 1.31 \[ \frac {2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )}{b^{3}} + \frac {{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \sin \left (b x + a\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 143, normalized size = 2.92 \[ \frac {\frac {d^{2} \left (\left (b x +a \right )^{2} \sin \left (b x +a \right )-2 \sin \left (b x +a \right )+2 \left (b x +a \right ) \cos \left (b x +a \right )\right )}{b^{2}}-\frac {2 a \,d^{2} \left (\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )\right )}{b^{2}}+\frac {2 c d \left (\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )\right )}{b}+\frac {a^{2} d^{2} \sin \left (b x +a \right )}{b^{2}}-\frac {2 a c d \sin \left (b x +a \right )}{b}+c^{2} \sin \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.71, size = 136, normalized size = 2.78 \[ \frac {c^{2} \sin \left (b x + a\right ) - \frac {2 \, a c d \sin \left (b x + a\right )}{b} + \frac {a^{2} d^{2} \sin \left (b x + a\right )}{b^{2}} + \frac {2 \, {\left ({\left (b x + a\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right )\right )} c d}{b} - \frac {2 \, {\left ({\left (b x + a\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right )\right )} a d^{2}}{b^{2}} + \frac {{\left (2 \, {\left (b x + a\right )} \cos \left (b x + a\right ) + {\left ({\left (b x + a\right )}^{2} - 2\right )} \sin \left (b x + a\right )\right )} d^{2}}{b^{2}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.12, size = 84, normalized size = 1.71 \[ \frac {d^2\,x^2\,\sin \left (a+b\,x\right )}{b}-\frac {\sin \left (a+b\,x\right )\,\left (2\,d^2-b^2\,c^2\right )}{b^3}+\frac {2\,c\,d\,\cos \left (a+b\,x\right )}{b^2}+\frac {2\,d^2\,x\,\cos \left (a+b\,x\right )}{b^2}+\frac {2\,c\,d\,x\,\sin \left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*(c + d*x)^2,x)

[Out]

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

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sympy [A]  time = 0.54, size = 112, normalized size = 2.29 \[ \begin {cases} \frac {c^{2} \sin {\left (a + b x \right )}}{b} + \frac {2 c d x \sin {\left (a + b x \right )}}{b} + \frac {d^{2} x^{2} \sin {\left (a + b x \right )}}{b} + \frac {2 c d \cos {\left (a + b x \right )}}{b^{2}} + \frac {2 d^{2} x \cos {\left (a + b x \right )}}{b^{2}} - \frac {2 d^{2} \sin {\left (a + b x \right )}}{b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \cos {\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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