\(\int (c+d x) \cos ^3(a+b x) \sin (a+b x) \, dx\) [140]

Optimal result

Integrand size = 20, antiderivative size = 72 \[ \int (c+d x) \cos ^3(a+b x) \sin (a+b x) \, dx=\frac {3 d x}{32 b}-\frac {(c+d x) \cos ^4(a+b x)}{4 b}+\frac {3 d \cos (a+b x) \sin (a+b x)}{32 b^2}+\frac {d \cos ^3(a+b x) \sin (a+b x)}{16 b^2} \]

[Out]

3/32*d*x/b-1/4*(d*x+c)*cos(b*x+a)^4/b+3/32*d*cos(b*x+a)*sin(b*x+a)/b^2+1/16*d*cos(b*x+a)^3*sin(b*x+a)/b^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4490, 2715, 8} \[ \int (c+d x) \cos ^3(a+b x) \sin (a+b x) \, dx=\frac {d \sin (a+b x) \cos ^3(a+b x)}{16 b^2}+\frac {3 d \sin (a+b x) \cos (a+b x)}{32 b^2}-\frac {(c+d x) \cos ^4(a+b x)}{4 b}+\frac {3 d x}{32 b} \]

[In]

Int[(c + d*x)*Cos[a + b*x]^3*Sin[a + b*x],x]

[Out]

(3*d*x)/(32*b) - ((c + d*x)*Cos[a + b*x]^4)/(4*b) + (3*d*Cos[a + b*x]*Sin[a + b*x])/(32*b^2) + (d*Cos[a + b*x]
^3*Sin[a + b*x])/(16*b^2)

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 4490

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {(c+d x) \cos ^4(a+b x)}{4 b}+\frac {d \int \cos ^4(a+b x) \, dx}{4 b} \\ & = -\frac {(c+d x) \cos ^4(a+b x)}{4 b}+\frac {d \cos ^3(a+b x) \sin (a+b x)}{16 b^2}+\frac {(3 d) \int \cos ^2(a+b x) \, dx}{16 b} \\ & = -\frac {(c+d x) \cos ^4(a+b x)}{4 b}+\frac {3 d \cos (a+b x) \sin (a+b x)}{32 b^2}+\frac {d \cos ^3(a+b x) \sin (a+b x)}{16 b^2}+\frac {(3 d) \int 1 \, dx}{32 b} \\ & = \frac {3 d x}{32 b}-\frac {(c+d x) \cos ^4(a+b x)}{4 b}+\frac {3 d \cos (a+b x) \sin (a+b x)}{32 b^2}+\frac {d \cos ^3(a+b x) \sin (a+b x)}{16 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04 \[ \int (c+d x) \cos ^3(a+b x) \sin (a+b x) \, dx=-\frac {c \cos ^4(a+b x)}{4 b}+\frac {d (-2 b x \cos (2 (a+b x))+\sin (2 (a+b x)))}{16 b^2}+\frac {d (-4 b x \cos (4 (a+b x))+\sin (4 (a+b x)))}{128 b^2} \]

[In]

Integrate[(c + d*x)*Cos[a + b*x]^3*Sin[a + b*x],x]

[Out]

-1/4*(c*Cos[a + b*x]^4)/b + (d*(-2*b*x*Cos[2*(a + b*x)] + Sin[2*(a + b*x)]))/(16*b^2) + (d*(-4*b*x*Cos[4*(a +
b*x)] + Sin[4*(a + b*x)]))/(128*b^2)

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94

method	result	size
parallelrisch	\(\frac {-16 b \left (d x +c \right ) \cos \left (2 x b +2 a \right )-4 b \left (d x +c \right ) \cos \left (4 x b +4 a \right )+20 c b +8 d \sin \left (2 x b +2 a \right )+d \sin \left (4 x b +4 a \right )}{128 b^{2}}\)	\(68\)
risch	\(-\frac {\left (d x +c \right ) \cos \left (4 x b +4 a \right )}{32 b}+\frac {d \sin \left (4 x b +4 a \right )}{128 b^{2}}-\frac {\left (d x +c \right ) \cos \left (2 x b +2 a \right )}{8 b}+\frac {d \sin \left (2 x b +2 a \right )}{16 b^{2}}\)	\(70\)
derivativedivides	\(\frac {\frac {d a \cos \left (x b +a \right )^{4}}{4 b}-\frac {c \cos \left (x b +a \right )^{4}}{4}+\frac {d \left (-\frac {\left (x b +a \right ) \cos \left (x b +a \right )^{4}}{4}+\frac {\left (\cos \left (x b +a \right )^{3}+\frac {3 \cos \left (x b +a \right )}{2}\right ) \sin \left (x b +a \right )}{16}+\frac {3 x b}{32}+\frac {3 a}{32}\right )}{b}}{b}\)	\(85\)
default	\(\frac {\frac {d a \cos \left (x b +a \right )^{4}}{4 b}-\frac {c \cos \left (x b +a \right )^{4}}{4}+\frac {d \left (-\frac {\left (x b +a \right ) \cos \left (x b +a \right )^{4}}{4}+\frac {\left (\cos \left (x b +a \right )^{3}+\frac {3 \cos \left (x b +a \right )}{2}\right ) \sin \left (x b +a \right )}{16}+\frac {3 x b}{32}+\frac {3 a}{32}\right )}{b}}{b}\)	\(85\)
norman	\(\frac {\frac {5 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{16 b^{2}}-\frac {3 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}}{16 b^{2}}+\frac {3 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{5}}{16 b^{2}}-\frac {5 d \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{7}}{16 b^{2}}-\frac {5 d x}{32 b}+\frac {2 c \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{b}+\frac {2 c \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}}{b}+\frac {11 d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{8 b}-\frac {15 d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}}{16 b}+\frac {11 d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}}{8 b}-\frac {5 d x \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{8}}{32 b}}{\left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )^{4}}\)	\(197\)

[In]

int((d*x+c)*cos(b*x+a)^3*sin(b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/128*(-16*b*(d*x+c)*cos(2*b*x+2*a)-4*b*(d*x+c)*cos(4*b*x+4*a)+20*c*b+8*d*sin(2*b*x+2*a)+d*sin(4*b*x+4*a))/b^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81 \[ \int (c+d x) \cos ^3(a+b x) \sin (a+b x) \, dx=-\frac {8 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{4} - 3 \, b d x - {\left (2 \, d \cos \left (b x + a\right )^{3} + 3 \, d \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{32 \, b^{2}} \]

[In]

integrate((d*x+c)*cos(b*x+a)^3*sin(b*x+a),x, algorithm="fricas")

[Out]

-1/32*(8*(b*d*x + b*c)*cos(b*x + a)^4 - 3*b*d*x - (2*d*cos(b*x + a)^3 + 3*d*cos(b*x + a))*sin(b*x + a))/b^2

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.92 \[ \int (c+d x) \cos ^3(a+b x) \sin (a+b x) \, dx=\begin {cases} - \frac {c \cos ^{4}{\left (a + b x \right )}}{4 b} + \frac {3 d x \sin ^{4}{\left (a + b x \right )}}{32 b} + \frac {3 d x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{16 b} - \frac {5 d x \cos ^{4}{\left (a + b x \right )}}{32 b} + \frac {3 d \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{32 b^{2}} + \frac {5 d \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{32 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin {\left (a \right )} \cos ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate((d*x+c)*cos(b*x+a)**3*sin(b*x+a),x)

[Out]

Piecewise((-c*cos(a + b*x)**4/(4*b) + 3*d*x*sin(a + b*x)**4/(32*b) + 3*d*x*sin(a + b*x)**2*cos(a + b*x)**2/(16
*b) - 5*d*x*cos(a + b*x)**4/(32*b) + 3*d*sin(a + b*x)**3*cos(a + b*x)/(32*b**2) + 5*d*sin(a + b*x)*cos(a + b*x
)**3/(32*b**2), Ne(b, 0)), ((c*x + d*x**2/2)*sin(a)*cos(a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.28 \[ \int (c+d x) \cos ^3(a+b x) \sin (a+b x) \, dx=-\frac {32 \, c \cos \left (b x + a\right )^{4} - \frac {32 \, a d \cos \left (b x + a\right )^{4}}{b} + \frac {{\left (4 \, {\left (b x + a\right )} \cos \left (4 \, b x + 4 \, a\right ) + 16 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (4 \, b x + 4 \, a\right ) - 8 \, \sin \left (2 \, b x + 2 \, a\right )\right )} d}{b}}{128 \, b} \]

[In]

integrate((d*x+c)*cos(b*x+a)^3*sin(b*x+a),x, algorithm="maxima")

[Out]

-1/128*(32*c*cos(b*x + a)^4 - 32*a*d*cos(b*x + a)^4/b + (4*(b*x + a)*cos(4*b*x + 4*a) + 16*(b*x + a)*cos(2*b*x
+ 2*a) - sin(4*b*x + 4*a) - 8*sin(2*b*x + 2*a))*d/b)/b

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04 \[ \int (c+d x) \cos ^3(a+b x) \sin (a+b x) \, dx=-\frac {{\left (b d x + b c\right )} \cos \left (4 \, b x + 4 \, a\right )}{32 \, b^{2}} - \frac {{\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right )}{8 \, b^{2}} + \frac {d \sin \left (4 \, b x + 4 \, a\right )}{128 \, b^{2}} + \frac {d \sin \left (2 \, b x + 2 \, a\right )}{16 \, b^{2}} \]

[In]

integrate((d*x+c)*cos(b*x+a)^3*sin(b*x+a),x, algorithm="giac")

[Out]

-1/32*(b*d*x + b*c)*cos(4*b*x + 4*a)/b^2 - 1/8*(b*d*x + b*c)*cos(2*b*x + 2*a)/b^2 + 1/128*d*sin(4*b*x + 4*a)/b
^2 + 1/16*d*sin(2*b*x + 2*a)/b^2

Mupad [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.31 \[ \int (c+d x) \cos ^3(a+b x) \sin (a+b x) \, dx=\frac {4\,d\,\sin \left (2\,a+2\,b\,x\right )+\frac {d\,\sin \left (4\,a+4\,b\,x\right )}{2}+4\,b\,c\,{\sin \left (2\,a+2\,b\,x\right )}^2+16\,b\,c\,{\sin \left (a+b\,x\right )}^2+8\,b\,d\,x\,\left (2\,{\sin \left (a+b\,x\right )}^2-1\right )+2\,b\,d\,x\,\left (2\,{\sin \left (2\,a+2\,b\,x\right )}^2-1\right )}{64\,b^2} \]

[In]

int(cos(a + b*x)^3*sin(a + b*x)*(c + d*x),x)

[Out]

(4*d*sin(2*a + 2*b*x) + (d*sin(4*a + 4*b*x))/2 + 4*b*c*sin(2*a + 2*b*x)^2 + 16*b*c*sin(a + b*x)^2 + 8*b*d*x*(2
*sin(a + b*x)^2 - 1) + 2*b*d*x*(2*sin(2*a + 2*b*x)^2 - 1))/(64*b^2)