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

Optimal. Leaf size=27 \[ \frac{d \cos (a+b x)}{b^2}+\frac{(c+d x) \sin (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.0164785, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3296, 2638} \[ \frac{d \cos (a+b x)}{b^2}+\frac{(c+d x) \sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Cos[a + b*x],x]

[Out]

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

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]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0561993, size = 26, normalized size = 0.96 \[ \frac{b (c+d x) \sin (a+b x)+d \cos (a+b x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Cos[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.029, size = 51, normalized size = 1.9 \begin{align*}{\frac{1}{b} \left ({\frac{d \left ( \cos \left ( bx+a \right ) + \left ( bx+a \right ) \sin \left ( bx+a \right ) \right ) }{b}}-{\frac{da\sin \left ( bx+a \right ) }{b}}+c\sin \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.959584, size = 68, normalized size = 2.52 \begin{align*} \frac{c \sin \left (b x + a\right ) - \frac{a d \sin \left (b x + a\right )}{b} + \frac{{\left ({\left (b x + a\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right )\right )} d}{b}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.05773, size = 69, normalized size = 2.56 \begin{align*} \frac{d \cos \left (b x + a\right ) +{\left (b d x + b c\right )} \sin \left (b x + a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.233663, size = 46, normalized size = 1.7 \begin{align*} \begin{cases} \frac{c \sin{\left (a + b x \right )}}{b} + \frac{d x \sin{\left (a + b x \right )}}{b} + \frac{d \cos{\left (a + b x \right )}}{b^{2}} & \text{for}\: b \neq 0 \\\left (c x + \frac{d x^{2}}{2}\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),x)

[Out]

Piecewise((c*sin(a + b*x)/b + d*x*sin(a + b*x)/b + d*cos(a + b*x)/b**2, Ne(b, 0)), ((c*x + d*x**2/2)*cos(a), T
rue))

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Giac [A]  time = 1.09986, size = 41, normalized size = 1.52 \begin{align*} \frac{d \cos \left (b x + a\right )}{b^{2}} + \frac{{\left (b d x + b c\right )} \sin \left (b x + a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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