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

Optimal. Leaf size=51 \[ -\frac{d \sin ^3(a+b x)}{9 b^2}+\frac{d \sin (a+b x)}{3 b^2}-\frac{(c+d x) \cos ^3(a+b x)}{3 b} \]

[Out]

-((c + d*x)*Cos[a + b*x]^3)/(3*b) + (d*Sin[a + b*x])/(3*b^2) - (d*Sin[a + b*x]^3)/(9*b^2)

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Rubi [A]  time = 0.0342699, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {4405, 2633} \[ -\frac{d \sin ^3(a+b x)}{9 b^2}+\frac{d \sin (a+b x)}{3 b^2}-\frac{(c+d x) \cos ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c + d*x)*Cos[a + b*x]^3)/(3*b) + (d*Sin[a + b*x])/(3*b^2) - (d*Sin[a + b*x]^3)/(9*b^2)

Rule 4405

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int (c+d x) \cos ^2(a+b x) \sin (a+b x) \, dx &=-\frac{(c+d x) \cos ^3(a+b x)}{3 b}+\frac{d \int \cos ^3(a+b x) \, dx}{3 b}\\ &=-\frac{(c+d x) \cos ^3(a+b x)}{3 b}-\frac{d \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (a+b x)\right )}{3 b^2}\\ &=-\frac{(c+d x) \cos ^3(a+b x)}{3 b}+\frac{d \sin (a+b x)}{3 b^2}-\frac{d \sin ^3(a+b x)}{9 b^2}\\ \end{align*}

Mathematica [A]  time = 0.153405, size = 71, normalized size = 1.39 \[ \frac{d (\sin (a+b x)-b x \cos (a+b x))}{4 b^2}+\frac{d (\sin (3 (a+b x))-3 b x \cos (3 (a+b x)))}{36 b^2}-\frac{c \cos ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*Cos[a + b*x]^3)/(3*b) + (d*(-(b*x*Cos[a + b*x]) + Sin[a + b*x]))/(4*b^2) + (d*(-3*b*x*Cos[3*(a + b*x)] + S
in[3*(a + b*x)]))/(36*b^2)

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Maple [A]  time = 0.018, size = 71, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ({\frac{d}{b} \left ( -{\frac{ \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{3}}+{\frac{ \left ( 2+ \left ( \cos \left ( bx+a \right ) \right ) ^{2} \right ) \sin \left ( bx+a \right ) }{9}} \right ) }+{\frac{ad \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{3\,b}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{3}c}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(d/b*(-1/3*(b*x+a)*cos(b*x+a)^3+1/9*(2+cos(b*x+a)^2)*sin(b*x+a))+1/3/b*d*a*cos(b*x+a)^3-1/3*cos(b*x+a)^3*c
)

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Maxima [A]  time = 1.11705, size = 116, normalized size = 2.27 \begin{align*} -\frac{12 \, c \cos \left (b x + a\right )^{3} - \frac{12 \, a d \cos \left (b x + a\right )^{3}}{b} + \frac{{\left (3 \,{\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) + 9 \,{\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (3 \, b x + 3 \, a\right ) - 9 \, \sin \left (b x + a\right )\right )} d}{b}}{36 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/36*(12*c*cos(b*x + a)^3 - 12*a*d*cos(b*x + a)^3/b + (3*(b*x + a)*cos(3*b*x + 3*a) + 9*(b*x + a)*cos(b*x + a
) - sin(3*b*x + 3*a) - 9*sin(b*x + a))*d/b)/b

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Fricas [A]  time = 0.474018, size = 112, normalized size = 2.2 \begin{align*} -\frac{3 \,{\left (b d x + b c\right )} \cos \left (b x + a\right )^{3} -{\left (d \cos \left (b x + a\right )^{2} + 2 \, d\right )} \sin \left (b x + a\right )}{9 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*(3*(b*d*x + b*c)*cos(b*x + a)^3 - (d*cos(b*x + a)^2 + 2*d)*sin(b*x + a))/b^2

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Sympy [A]  time = 1.13508, size = 85, normalized size = 1.67 \begin{align*} \begin{cases} - \frac{c \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac{d x \cos ^{3}{\left (a + b x \right )}}{3 b} + \frac{2 d \sin ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac{d \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b^{2}} & \text{for}\: b \neq 0 \\\left (c x + \frac{d x^{2}}{2}\right ) \sin{\left (a \right )} \cos ^{2}{\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)**2*sin(b*x+a),x)

[Out]

Piecewise((-c*cos(a + b*x)**3/(3*b) - d*x*cos(a + b*x)**3/(3*b) + 2*d*sin(a + b*x)**3/(9*b**2) + d*sin(a + b*x
)*cos(a + b*x)**2/(3*b**2), Ne(b, 0)), ((c*x + d*x**2/2)*sin(a)*cos(a)**2, True))

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Giac [A]  time = 1.20394, size = 93, normalized size = 1.82 \begin{align*} -\frac{{\left (b d x + b c\right )} \cos \left (3 \, b x + 3 \, a\right )}{12 \, b^{2}} - \frac{{\left (b d x + b c\right )} \cos \left (b x + a\right )}{4 \, b^{2}} + \frac{d \sin \left (3 \, b x + 3 \, a\right )}{36 \, b^{2}} + \frac{d \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)^2*sin(b*x+a),x, algorithm="giac")

[Out]

-1/12*(b*d*x + b*c)*cos(3*b*x + 3*a)/b^2 - 1/4*(b*d*x + b*c)*cos(b*x + a)/b^2 + 1/36*d*sin(3*b*x + 3*a)/b^2 +
1/4*d*sin(b*x + a)/b^2