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

Optimal. Leaf size=89 \[ -\frac {d^2 \sin ^2(a+b x)}{4 b^3}+\frac {d (c+d x) \sin (a+b x) \cos (a+b x)}{2 b^2}+\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}-\frac {c d x}{2 b}-\frac {d^2 x^2}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4404, 3310} \[ \frac {d (c+d x) \sin (a+b x) \cos (a+b x)}{2 b^2}-\frac {d^2 \sin ^2(a+b x)}{4 b^3}+\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}-\frac {c d x}{2 b}-\frac {d^2 x^2}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 4404

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[((c +
d*x)^m*Sin[a + b*x]^(n + 1))/(b*(n + 1)), x] - Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n +
1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 50, normalized size = 0.56 \[ \frac {\cos (2 (a+b x)) \left (d^2-2 b^2 (c+d x)^2\right )+2 b d (c+d x) \sin (2 (a+b x))}{8 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d^2 - 2*b^2*(c + d*x)^2)*Cos[2*(a + b*x)] + 2*b*d*(c + d*x)*Sin[2*(a + b*x)])/(8*b^3)

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fricas [A]  time = 0.50, size = 92, normalized size = 1.03 \[ \frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x - {\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )^{2} + 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )}{4 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 73, normalized size = 0.82 \[ -\frac {{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right )}{8 \, b^{3}} + \frac {{\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )}{4 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 - d^2)*cos(2*b*x + 2*a)/b^3 + 1/4*(b*d^2*x + b*c*d)*sin(2*b*x +
2*a)/b^3

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maple [B]  time = 0.01, size = 215, normalized size = 2.42 \[ \frac {\frac {d^{2} \left (-\frac {\left (b x +a \right )^{2} \left (\cos ^{2}\left (b x +a \right )\right )}{2}+\left (b x +a \right ) \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{4}-\frac {\left (\sin ^{2}\left (b x +a \right )\right )}{4}\right )}{b^{2}}-\frac {2 a \,d^{2} \left (-\frac {\left (b x +a \right ) \left (\cos ^{2}\left (b x +a \right )\right )}{2}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{4}+\frac {b x}{4}+\frac {a}{4}\right )}{b^{2}}+\frac {2 c d \left (-\frac {\left (b x +a \right ) \left (\cos ^{2}\left (b x +a \right )\right )}{2}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{4}+\frac {b x}{4}+\frac {a}{4}\right )}{b}-\frac {a^{2} d^{2} \left (\cos ^{2}\left (b x +a \right )\right )}{2 b^{2}}+\frac {a c d \left (\cos ^{2}\left (b x +a \right )\right )}{b}-\frac {c^{2} \left (\cos ^{2}\left (b x +a \right )\right )}{2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.36, size = 171, normalized size = 1.92 \[ -\frac {4 \, c^{2} \cos \left (b x + a\right )^{2} - \frac {8 \, a c d \cos \left (b x + a\right )^{2}}{b} + \frac {4 \, a^{2} d^{2} \cos \left (b x + a\right )^{2}}{b^{2}} + \frac {2 \, {\left (2 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )} c d}{b} - \frac {2 \, {\left (2 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )} a d^{2}}{b^{2}} + \frac {{\left ({\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{2}}{b^{2}}}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*c^2*cos(b*x + a)^2 - 8*a*c*d*cos(b*x + a)^2/b + 4*a^2*d^2*cos(b*x + a)^2/b^2 + 2*(2*(b*x + a)*cos(2*b*
x + 2*a) - sin(2*b*x + 2*a))*c*d/b - 2*(2*(b*x + a)*cos(2*b*x + 2*a) - sin(2*b*x + 2*a))*a*d^2/b^2 + ((2*(b*x
+ a)^2 - 1)*cos(2*b*x + 2*a) - 2*(b*x + a)*sin(2*b*x + 2*a))*d^2/b^2)/b

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mupad [B]  time = 0.16, size = 100, normalized size = 1.12 \[ \frac {\cos \left (2\,a+2\,b\,x\right )\,\left (\frac {d^2}{4}-\frac {b^2\,c^2}{2}\right )}{2\,b^3}+\frac {d^2\,x\,\sin \left (2\,a+2\,b\,x\right )}{4\,b^2}-\frac {d^2\,x^2\,\cos \left (2\,a+2\,b\,x\right )}{4\,b}+\frac {c\,d\,\sin \left (2\,a+2\,b\,x\right )}{4\,b^2}-\frac {c\,d\,x\,\cos \left (2\,a+2\,b\,x\right )}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.05, size = 175, normalized size = 1.97 \[ \begin {cases} - \frac {c^{2} \cos ^{2}{\left (a + b x \right )}}{2 b} + \frac {c d x \sin ^{2}{\left (a + b x \right )}}{2 b} - \frac {c d x \cos ^{2}{\left (a + b x \right )}}{2 b} + \frac {d^{2} x^{2} \sin ^{2}{\left (a + b x \right )}}{4 b} - \frac {d^{2} x^{2} \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac {c d \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b^{2}} + \frac {d^{2} x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b^{2}} + \frac {d^{2} \cos ^{2}{\left (a + b x \right )}}{4 b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \sin {\relax (a )} \cos {\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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