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

Optimal. Leaf size=28 \[ \frac {d \sin (a+b x)}{b^2}-\frac {(c+d x) \cos (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 28, normalized size of antiderivative = 1.00, 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, 2637} \[ \frac {d \sin (a+b x)}{b^2}-\frac {(c+d x) \cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sin[a + b*x],x]

[Out]

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 27, normalized size = 0.96 \[ \frac {d \sin (a+b x)-b (c+d x) \cos (a+b x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Sin[a + b*x],x]

[Out]

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

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fricas [A]  time = 0.69, size = 30, normalized size = 1.07 \[ -\frac {{\left (b d x + b c\right )} \cos \left (b x + a\right ) - d \sin \left (b x + a\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.88, size = 31, normalized size = 1.11 \[ -\frac {{\left (b d x + b c\right )} \cos \left (b x + a\right )}{b^{2}} + \frac {d \sin \left (b x + a\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 52, normalized size = 1.86 \[ \frac {\frac {d \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{b}+\frac {d a \cos \left (b x +a \right )}{b}-c \cos \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 53, normalized size = 1.89 \[ -\frac {c \cos \left (b x + a\right ) - \frac {a d \cos \left (b x + a\right )}{b} + \frac {{\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} d}{b}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.53, size = 35, normalized size = 1.25 \[ \frac {d\,\sin \left (a+b\,x\right )}{b^2}-\frac {c\,\cos \left (a+b\,x\right )+d\,x\,\cos \left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)*(c + d*x),x)

[Out]

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

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sympy [A]  time = 0.25, size = 46, normalized size = 1.64 \[ \begin {cases} - \frac {c \cos {\left (a + b x \right )}}{b} - \frac {d x \cos {\left (a + b x \right )}}{b} + \frac {d \sin {\left (a + b x \right )}}{b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin {\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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