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

Optimal. Leaf size=50 \[ \frac {d \sin (a+b x) \cos (a+b x)}{4 b^2}+\frac {(c+d x) \sin ^2(a+b x)}{2 b}-\frac {d x}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4404, 2635, 8} \[ \frac {d \sin (a+b x) \cos (a+b x)}{4 b^2}+\frac {(c+d x) \sin ^2(a+b x)}{2 b}-\frac {d x}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

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

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) \cos (a+b x) \sin (a+b x) \, dx &=\frac {(c+d x) \sin ^2(a+b x)}{2 b}-\frac {d \int \sin ^2(a+b x) \, dx}{2 b}\\ &=\frac {d \cos (a+b x) \sin (a+b x)}{4 b^2}+\frac {(c+d x) \sin ^2(a+b x)}{2 b}-\frac {d \int 1 \, dx}{4 b}\\ &=-\frac {d x}{4 b}+\frac {d \cos (a+b x) \sin (a+b x)}{4 b^2}+\frac {(c+d x) \sin ^2(a+b x)}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 34, normalized size = 0.68 \[ \frac {d \sin (2 (a+b x))-2 b (c+d x) \cos (2 (a+b x))}{8 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.63, size = 42, normalized size = 0.84 \[ \frac {b d x - 2 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + d \cos \left (b x + a\right ) \sin \left (b x + a\right )}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.72, size = 38, normalized size = 0.76 \[ -\frac {{\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right )}{4 \, b^{2}} + \frac {d \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 74, normalized size = 1.48 \[ \frac {\frac {d \left (-\frac {\left (b x +a \right ) \left (\cos ^{2}\left (b x +a \right )\right )}{2}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{4}+\frac {b x}{4}+\frac {a}{4}\right )}{b}+\frac {d a \left (\cos ^{2}\left (b x +a \right )\right )}{2 b}-\frac {c \left (\cos ^{2}\left (b x +a \right )\right )}{2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 65, normalized size = 1.30 \[ -\frac {4 \, c \cos \left (b x + a\right )^{2} - \frac {4 \, a d \cos \left (b x + a\right )^{2}}{b} + \frac {{\left (2 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )} d}{b}}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.70, size = 47, normalized size = 0.94 \[ \frac {d\,\sin \left (2\,a+2\,b\,x\right )}{8\,b^2}-\frac {c\,\cos \left (2\,a+2\,b\,x\right )}{4\,b}-\frac {d\,x\,\cos \left (2\,a+2\,b\,x\right )}{4\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.47, size = 80, normalized size = 1.60 \[ \begin {cases} - \frac {c \cos ^{2}{\left (a + b x \right )}}{2 b} + \frac {d x \sin ^{2}{\left (a + b x \right )}}{4 b} - \frac {d x \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac {d \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{4 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin {\relax (a )} \cos {\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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