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

Optimal. Leaf size=72 \[ \frac{d \sin (a+b x) \cos ^3(a+b x)}{16 b^2}+\frac{3 d \sin (a+b x) \cos (a+b x)}{32 b^2}-\frac{(c+d x) \cos ^4(a+b x)}{4 b}+\frac{3 d x}{32 b} \]

[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)

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Rubi [A]  time = 0.0470773, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4405, 2635, 8} \[ \frac{d \sin (a+b x) \cos ^3(a+b x)}{16 b^2}+\frac{3 d \sin (a+b x) \cos (a+b x)}{32 b^2}-\frac{(c+d x) \cos ^4(a+b x)}{4 b}+\frac{3 d x}{32 b} \]

Antiderivative was successfully verified.

[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 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 2635

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] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (c+d x) \cos ^3(a+b x) \sin (a+b x) \, dx &=-\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]  time = 0.151772, size = 75, normalized size = 1.04 \[ \frac{d (\sin (2 (a+b x))-2 b x \cos (2 (a+b x)))}{16 b^2}+\frac{d (\sin (4 (a+b x))-4 b x \cos (4 (a+b x)))}{128 b^2}-\frac{c \cos ^4(a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*Cos[a + b*x]^4)/(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)

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Maple [A]  time = 0.02, size = 85, normalized size = 1.2 \begin{align*}{\frac{1}{b} \left ({\frac{d}{b} \left ( -{\frac{ \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{4}}+{\frac{\sin \left ( bx+a \right ) }{16} \left ( \left ( \cos \left ( bx+a \right ) \right ) ^{3}+{\frac{3\,\cos \left ( bx+a \right ) }{2}} \right ) }+{\frac{3\,bx}{32}}+{\frac{3\,a}{32}} \right ) }+{\frac{ad \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{4\,b}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{4}c}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.11205, size = 124, normalized size = 1.72 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 0.48835, size = 147, normalized size = 2.04 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 2.81213, size = 160, normalized size = 2.22 \begin{align*} \begin{cases} \frac{c \sin ^{4}{\left (a + b x \right )}}{4 b} + \frac{c \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{2 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((c*sin(a + b*x)**4/(4*b) + c*sin(a + b*x)**2*cos(a + b*x)**2/(2*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))

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Giac [A]  time = 1.11548, size = 101, normalized size = 1.4 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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