\(\int (c+d x) \cos (a+b x) \sin (a+b x) \, dx\) [5]

Optimal result

Integrand size = 18, antiderivative size = 50 \[ \int (c+d x) \cos (a+b x) \sin (a+b x) \, dx=-\frac {d x}{4 b}+\frac {d \cos (a+b x) \sin (a+b x)}{4 b^2}+\frac {(c+d x) \sin ^2(a+b x)}{2 b} \]

[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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4489, 2715, 8} \[ \int (c+d x) \cos (a+b x) \sin (a+b x) \, dx=\frac {d \sin (a+b x) \cos (a+b x)}{4 b^2}+\frac {(c+d x) \sin ^2(a+b x)}{2 b}-\frac {d x}{4 b} \]

[In]

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

[Out]

-1/4*(d*x)/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 2715

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] && Integ
erQ[2*n]

Rule 4489

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*} \text {integral}& = \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] (verified)

Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.68 \[ \int (c+d x) \cos (a+b x) \sin (a+b x) \, dx=\frac {-2 b (c+d x) \cos (2 (a+b x))+d \sin (2 (a+b x))}{8 b^2} \]

[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)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.72

method	result	size
risch	\(-\frac {\left (d x +c \right ) \cos \left (2 x b +2 a \right )}{4 b}+\frac {d \sin \left (2 x b +2 a \right )}{8 b^{2}}\)	\(36\)
parallelrisch	\(\frac {-2 b \left (d x +c \right ) \cos \left (2 x b +2 a \right )+2 c b +d \sin \left (2 x b +2 a \right )}{8 b^{2}}\)	\(39\)
derivativedivides	\(\frac {\frac {d a \cos \left (x b +a \right )^{2}}{2 b}-\frac {c \cos \left (x b +a \right )^{2}}{2}+\frac {d \left (-\frac {\left (x b +a \right ) \cos \left (x b +a \right )^{2}}{2}+\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{4}+\frac {x b}{4}+\frac {a}{4}\right )}{b}}{b}\)	\(74\)
default	\(\frac {\frac {d a \cos \left (x b +a \right )^{2}}{2 b}-\frac {c \cos \left (x b +a \right )^{2}}{2}+\frac {d \left (-\frac {\left (x b +a \right ) \cos \left (x b +a \right )^{2}}{2}+\frac {\cos \left (x b +a \right ) \sin \left (x b +a \right )}{4}+\frac {x b}{4}+\frac {a}{4}\right )}{b}}{b}\)	\(74\)
norman	\(\frac {\frac {d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{2 b^{2}}-\frac {d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}}{2 b^{2}}-\frac {d x}{4 b}+\frac {2 c \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{b}+\frac {3 d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{2 b}-\frac {d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}}{4 b}}{\left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )^{2}}\)	\(110\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.84 \[ \int (c+d x) \cos (a+b x) \sin (a+b x) \, dx=\frac {b d x - 2 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + d \cos \left (b x + a\right ) \sin \left (b x + a\right )}{4 \, b^{2}} \]

[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

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.60 \[ \int (c+d x) \cos (a+b x) \sin (a+b x) \, dx=\begin {cases} \frac {c \sin ^{2}{\left (a + b x \right )}}{2 b} + \frac {d x \sin ^{2}{\left (a + b x \right )}}{4 b} - \frac {d x \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac {d \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{4 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin {\left (a \right )} \cos {\left (a \right )} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.30 \[ \int (c+d x) \cos (a+b x) \sin (a+b x) \, dx=-\frac {4 \, c \cos \left (b x + a\right )^{2} - \frac {4 \, a d \cos \left (b x + a\right )^{2}}{b} + \frac {{\left (2 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )} d}{b}}{8 \, b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76 \[ \int (c+d x) \cos (a+b x) \sin (a+b x) \, dx=-\frac {{\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right )}{4 \, b^{2}} + \frac {d \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{2}} \]

[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

Mupad [B] (verification not implemented)

Time = 22.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94 \[ \int (c+d x) \cos (a+b x) \sin (a+b x) \, dx=\frac {d\,\sin \left (2\,a+2\,b\,x\right )}{8\,b^2}-\frac {c\,\cos \left (2\,a+2\,b\,x\right )}{4\,b}-\frac {d\,x\,\cos \left (2\,a+2\,b\,x\right )}{4\,b} \]

[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)