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]

-(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)

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Rubi [A]  time = 0.0259538, antiderivative size = 50, normalized size of antiderivative = 1., 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 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]

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 8

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

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*}

Mathematica [A]  time = 0.100934, 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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Maple [A]  time = 0.014, size = 74, normalized size = 1.5 \begin{align*}{\frac{1}{b} \left ({\frac{d}{b} \left ( -{\frac{ \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{2}}+{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{4}}+{\frac{bx}{4}}+{\frac{a}{4}} \right ) }+{\frac{ad \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{2\,b}}-{\frac{c \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{2}} \right ) } \end{align*}

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

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Maxima [A]  time = 1.1213, size = 88, normalized size = 1.76 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 0.464452, size = 108, normalized size = 2.16 \begin{align*} \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}} \end{align*}

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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Sympy [A]  time = 0.599258, size = 80, normalized size = 1.6 \begin{align*} \begin{cases} \frac{c \sin ^{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{\left (a \right )} \cos{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

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*sin(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)*co
s(a + b*x)/(4*b**2), Ne(b, 0)), ((c*x + d*x**2/2)*sin(a)*cos(a), True))

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Giac [A]  time = 1.14172, size = 51, normalized size = 1.02 \begin{align*} -\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}} \end{align*}

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