3.72 \(\int (c+d x)^3 \cos ^2(a+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=151 \[ \frac{2 d^2 (c+d x) \cos ^3(a+b x)}{9 b^3}+\frac{4 d^2 (c+d x) \cos (a+b x)}{3 b^3}+\frac{2 d (c+d x)^2 \sin (a+b x)}{3 b^2}+\frac{d (c+d x)^2 \sin (a+b x) \cos ^2(a+b x)}{3 b^2}+\frac{2 d^3 \sin ^3(a+b x)}{27 b^4}-\frac{14 d^3 \sin (a+b x)}{9 b^4}-\frac{(c+d x)^3 \cos ^3(a+b x)}{3 b} \]

[Out]

(4*d^2*(c + d*x)*Cos[a + b*x])/(3*b^3) + (2*d^2*(c + d*x)*Cos[a + b*x]^3)/(9*b^3) - ((c + d*x)^3*Cos[a + b*x]^
3)/(3*b) - (14*d^3*Sin[a + b*x])/(9*b^4) + (2*d*(c + d*x)^2*Sin[a + b*x])/(3*b^2) + (d*(c + d*x)^2*Cos[a + b*x
]^2*Sin[a + b*x])/(3*b^2) + (2*d^3*Sin[a + b*x]^3)/(27*b^4)

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Rubi [A]  time = 0.13308, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {4405, 3311, 3296, 2637, 2633} \[ \frac{2 d^2 (c+d x) \cos ^3(a+b x)}{9 b^3}+\frac{4 d^2 (c+d x) \cos (a+b x)}{3 b^3}+\frac{2 d (c+d x)^2 \sin (a+b x)}{3 b^2}+\frac{d (c+d x)^2 \sin (a+b x) \cos ^2(a+b x)}{3 b^2}+\frac{2 d^3 \sin ^3(a+b x)}{27 b^4}-\frac{14 d^3 \sin (a+b x)}{9 b^4}-\frac{(c+d x)^3 \cos ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Cos[a + b*x]^2*Sin[a + b*x],x]

[Out]

(4*d^2*(c + d*x)*Cos[a + b*x])/(3*b^3) + (2*d^2*(c + d*x)*Cos[a + b*x]^3)/(9*b^3) - ((c + d*x)^3*Cos[a + b*x]^
3)/(3*b) - (14*d^3*Sin[a + b*x])/(9*b^4) + (2*d*(c + d*x)^2*Sin[a + b*x])/(3*b^2) + (d*(c + d*x)^2*Cos[a + b*x
]^2*Sin[a + b*x])/(3*b^2) + (2*d^3*Sin[a + b*x]^3)/(27*b^4)

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 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

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 2637

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int (c+d x)^3 \cos ^2(a+b x) \sin (a+b x) \, dx &=-\frac{(c+d x)^3 \cos ^3(a+b x)}{3 b}+\frac{d \int (c+d x)^2 \cos ^3(a+b x) \, dx}{b}\\ &=\frac{2 d^2 (c+d x) \cos ^3(a+b x)}{9 b^3}-\frac{(c+d x)^3 \cos ^3(a+b x)}{3 b}+\frac{d (c+d x)^2 \cos ^2(a+b x) \sin (a+b x)}{3 b^2}+\frac{(2 d) \int (c+d x)^2 \cos (a+b x) \, dx}{3 b}-\frac{\left (2 d^3\right ) \int \cos ^3(a+b x) \, dx}{9 b^3}\\ &=\frac{2 d^2 (c+d x) \cos ^3(a+b x)}{9 b^3}-\frac{(c+d x)^3 \cos ^3(a+b x)}{3 b}+\frac{2 d (c+d x)^2 \sin (a+b x)}{3 b^2}+\frac{d (c+d x)^2 \cos ^2(a+b x) \sin (a+b x)}{3 b^2}-\frac{\left (4 d^2\right ) \int (c+d x) \sin (a+b x) \, dx}{3 b^2}+\frac{\left (2 d^3\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (a+b x)\right )}{9 b^4}\\ &=\frac{4 d^2 (c+d x) \cos (a+b x)}{3 b^3}+\frac{2 d^2 (c+d x) \cos ^3(a+b x)}{9 b^3}-\frac{(c+d x)^3 \cos ^3(a+b x)}{3 b}-\frac{2 d^3 \sin (a+b x)}{9 b^4}+\frac{2 d (c+d x)^2 \sin (a+b x)}{3 b^2}+\frac{d (c+d x)^2 \cos ^2(a+b x) \sin (a+b x)}{3 b^2}+\frac{2 d^3 \sin ^3(a+b x)}{27 b^4}-\frac{\left (4 d^3\right ) \int \cos (a+b x) \, dx}{3 b^3}\\ &=\frac{4 d^2 (c+d x) \cos (a+b x)}{3 b^3}+\frac{2 d^2 (c+d x) \cos ^3(a+b x)}{9 b^3}-\frac{(c+d x)^3 \cos ^3(a+b x)}{3 b}-\frac{14 d^3 \sin (a+b x)}{9 b^4}+\frac{2 d (c+d x)^2 \sin (a+b x)}{3 b^2}+\frac{d (c+d x)^2 \cos ^2(a+b x) \sin (a+b x)}{3 b^2}+\frac{2 d^3 \sin ^3(a+b x)}{27 b^4}\\ \end{align*}

Mathematica [A]  time = 0.960798, size = 127, normalized size = 0.84 \[ \frac{-27 b (c+d x) \cos (a+b x) \left (b^2 (c+d x)^2-6 d^2\right )-3 b (c+d x) \cos (3 (a+b x)) \left (3 b^2 (c+d x)^2-2 d^2\right )+2 d \sin (a+b x) \left (\cos (2 (a+b x)) \left (9 b^2 (c+d x)^2-2 d^2\right )+45 b^2 (c+d x)^2-82 d^2\right )}{108 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^3*Cos[a + b*x]^2*Sin[a + b*x],x]

[Out]

(-27*b*(c + d*x)*(-6*d^2 + b^2*(c + d*x)^2)*Cos[a + b*x] - 3*b*(c + d*x)*(-2*d^2 + 3*b^2*(c + d*x)^2)*Cos[3*(a
+ b*x)] + 2*d*(-82*d^2 + 45*b^2*(c + d*x)^2 + (-2*d^2 + 9*b^2*(c + d*x)^2)*Cos[2*(a + b*x)])*Sin[a + b*x])/(1
08*b^4)

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Maple [B]  time = 0.018, size = 447, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*cos(b*x+a)^2*sin(b*x+a),x)

[Out]

1/b*(1/b^3*d^3*(-1/3*(b*x+a)^3*cos(b*x+a)^3+1/3*(b*x+a)^2*(2+cos(b*x+a)^2)*sin(b*x+a)-4/3*sin(b*x+a)+4/3*(b*x+
a)*cos(b*x+a)+2/9*(b*x+a)*cos(b*x+a)^3-2/27*(2+cos(b*x+a)^2)*sin(b*x+a))-3/b^3*a*d^3*(-1/3*(b*x+a)^2*cos(b*x+a
)^3+2/9*(b*x+a)*(2+cos(b*x+a)^2)*sin(b*x+a)+2/27*cos(b*x+a)^3+4/9*cos(b*x+a))+3/b^2*c*d^2*(-1/3*(b*x+a)^2*cos(
b*x+a)^3+2/9*(b*x+a)*(2+cos(b*x+a)^2)*sin(b*x+a)+2/27*cos(b*x+a)^3+4/9*cos(b*x+a))+3/b^3*a^2*d^3*(-1/3*(b*x+a)
*cos(b*x+a)^3+1/9*(2+cos(b*x+a)^2)*sin(b*x+a))-6/b^2*a*c*d^2*(-1/3*(b*x+a)*cos(b*x+a)^3+1/9*(2+cos(b*x+a)^2)*s
in(b*x+a))+3/b*c^2*d*(-1/3*(b*x+a)*cos(b*x+a)^3+1/9*(2+cos(b*x+a)^2)*sin(b*x+a))+1/3/b^3*a^3*d^3*cos(b*x+a)^3-
1/b^2*a^2*c*d^2*cos(b*x+a)^3+1/b*a*c^2*d*cos(b*x+a)^3-1/3*c^3*cos(b*x+a)^3)

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Maxima [B]  time = 1.17515, size = 682, normalized size = 4.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*cos(b*x+a)^2*sin(b*x+a),x, algorithm="maxima")

[Out]

-1/108*(36*c^3*cos(b*x + a)^3 - 108*a*c^2*d*cos(b*x + a)^3/b + 108*a^2*c*d^2*cos(b*x + a)^3/b^2 - 36*a^3*d^3*c
os(b*x + a)^3/b^3 + 9*(3*(b*x + a)*cos(3*b*x + 3*a) + 9*(b*x + a)*cos(b*x + a) - sin(3*b*x + 3*a) - 9*sin(b*x
+ a))*c^2*d/b - 18*(3*(b*x + a)*cos(3*b*x + 3*a) + 9*(b*x + a)*cos(b*x + a) - sin(3*b*x + 3*a) - 9*sin(b*x + a
))*a*c*d^2/b^2 + 9*(3*(b*x + a)*cos(3*b*x + 3*a) + 9*(b*x + a)*cos(b*x + a) - sin(3*b*x + 3*a) - 9*sin(b*x + a
))*a^2*d^3/b^3 + 3*((9*(b*x + a)^2 - 2)*cos(3*b*x + 3*a) + 27*((b*x + a)^2 - 2)*cos(b*x + a) - 6*(b*x + a)*sin
(3*b*x + 3*a) - 54*(b*x + a)*sin(b*x + a))*c*d^2/b^2 - 3*((9*(b*x + a)^2 - 2)*cos(3*b*x + 3*a) + 27*((b*x + a)
^2 - 2)*cos(b*x + a) - 6*(b*x + a)*sin(3*b*x + 3*a) - 54*(b*x + a)*sin(b*x + a))*a*d^3/b^3 + (3*(3*(b*x + a)^3
- 2*b*x - 2*a)*cos(3*b*x + 3*a) + 27*((b*x + a)^3 - 6*b*x - 6*a)*cos(b*x + a) - (9*(b*x + a)^2 - 2)*sin(3*b*x
+ 3*a) - 81*((b*x + a)^2 - 2)*sin(b*x + a))*d^3/b^3)/b

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Fricas [A]  time = 0.496975, size = 404, normalized size = 2.68 \begin{align*} -\frac{3 \,{\left (3 \, b^{3} d^{3} x^{3} + 9 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{3} - 2 \, b c d^{2} +{\left (9 \, b^{3} c^{2} d - 2 \, b d^{3}\right )} x\right )} \cos \left (b x + a\right )^{3} - 36 \,{\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right ) -{\left (18 \, b^{2} d^{3} x^{2} + 36 \, b^{2} c d^{2} x + 18 \, b^{2} c^{2} d - 40 \, d^{3} +{\left (9 \, b^{2} d^{3} x^{2} + 18 \, b^{2} c d^{2} x + 9 \, b^{2} c^{2} d - 2 \, d^{3}\right )} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{27 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*cos(b*x+a)^2*sin(b*x+a),x, algorithm="fricas")

[Out]

-1/27*(3*(3*b^3*d^3*x^3 + 9*b^3*c*d^2*x^2 + 3*b^3*c^3 - 2*b*c*d^2 + (9*b^3*c^2*d - 2*b*d^3)*x)*cos(b*x + a)^3
- 36*(b*d^3*x + b*c*d^2)*cos(b*x + a) - (18*b^2*d^3*x^2 + 36*b^2*c*d^2*x + 18*b^2*c^2*d - 40*d^3 + (9*b^2*d^3*
x^2 + 18*b^2*c*d^2*x + 9*b^2*c^2*d - 2*d^3)*cos(b*x + a)^2)*sin(b*x + a))/b^4

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Sympy [A]  time = 5.38453, size = 391, normalized size = 2.59 \begin{align*} \begin{cases} - \frac{c^{3} \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac{c^{2} d x \cos ^{3}{\left (a + b x \right )}}{b} - \frac{c d^{2} x^{2} \cos ^{3}{\left (a + b x \right )}}{b} - \frac{d^{3} x^{3} \cos ^{3}{\left (a + b x \right )}}{3 b} + \frac{2 c^{2} d \sin ^{3}{\left (a + b x \right )}}{3 b^{2}} + \frac{c^{2} d \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b^{2}} + \frac{4 c d^{2} x \sin ^{3}{\left (a + b x \right )}}{3 b^{2}} + \frac{2 c d^{2} x \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b^{2}} + \frac{2 d^{3} x^{2} \sin ^{3}{\left (a + b x \right )}}{3 b^{2}} + \frac{d^{3} x^{2} \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b^{2}} + \frac{4 c d^{2} \sin ^{2}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{3 b^{3}} + \frac{14 c d^{2} \cos ^{3}{\left (a + b x \right )}}{9 b^{3}} + \frac{4 d^{3} x \sin ^{2}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{3 b^{3}} + \frac{14 d^{3} x \cos ^{3}{\left (a + b x \right )}}{9 b^{3}} - \frac{40 d^{3} \sin ^{3}{\left (a + b x \right )}}{27 b^{4}} - \frac{14 d^{3} \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{9 b^{4}} & \text{for}\: b \neq 0 \\\left (c^{3} x + \frac{3 c^{2} d x^{2}}{2} + c d^{2} x^{3} + \frac{d^{3} x^{4}}{4}\right ) \sin{\left (a \right )} \cos ^{2}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*cos(b*x+a)**2*sin(b*x+a),x)

[Out]

Piecewise((-c**3*cos(a + b*x)**3/(3*b) - c**2*d*x*cos(a + b*x)**3/b - c*d**2*x**2*cos(a + b*x)**3/b - d**3*x**
3*cos(a + b*x)**3/(3*b) + 2*c**2*d*sin(a + b*x)**3/(3*b**2) + c**2*d*sin(a + b*x)*cos(a + b*x)**2/b**2 + 4*c*d
**2*x*sin(a + b*x)**3/(3*b**2) + 2*c*d**2*x*sin(a + b*x)*cos(a + b*x)**2/b**2 + 2*d**3*x**2*sin(a + b*x)**3/(3
*b**2) + d**3*x**2*sin(a + b*x)*cos(a + b*x)**2/b**2 + 4*c*d**2*sin(a + b*x)**2*cos(a + b*x)/(3*b**3) + 14*c*d
**2*cos(a + b*x)**3/(9*b**3) + 4*d**3*x*sin(a + b*x)**2*cos(a + b*x)/(3*b**3) + 14*d**3*x*cos(a + b*x)**3/(9*b
**3) - 40*d**3*sin(a + b*x)**3/(27*b**4) - 14*d**3*sin(a + b*x)*cos(a + b*x)**2/(9*b**4), Ne(b, 0)), ((c**3*x
+ 3*c**2*d*x**2/2 + c*d**2*x**3 + d**3*x**4/4)*sin(a)*cos(a)**2, True))

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Giac [A]  time = 1.20255, size = 312, normalized size = 2.07 \begin{align*} -\frac{{\left (3 \, b^{3} d^{3} x^{3} + 9 \, b^{3} c d^{2} x^{2} + 9 \, b^{3} c^{2} d x + 3 \, b^{3} c^{3} - 2 \, b d^{3} x - 2 \, b c d^{2}\right )} \cos \left (3 \, b x + 3 \, a\right )}{36 \, b^{4}} - \frac{{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3} - 6 \, b d^{3} x - 6 \, b c d^{2}\right )} \cos \left (b x + a\right )}{4 \, b^{4}} + \frac{{\left (9 \, b^{2} d^{3} x^{2} + 18 \, b^{2} c d^{2} x + 9 \, b^{2} c^{2} d - 2 \, d^{3}\right )} \sin \left (3 \, b x + 3 \, a\right )}{108 \, b^{4}} + \frac{3 \,{\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d - 2 \, d^{3}\right )} \sin \left (b x + a\right )}{4 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*cos(b*x+a)^2*sin(b*x+a),x, algorithm="giac")

[Out]

-1/36*(3*b^3*d^3*x^3 + 9*b^3*c*d^2*x^2 + 9*b^3*c^2*d*x + 3*b^3*c^3 - 2*b*d^3*x - 2*b*c*d^2)*cos(3*b*x + 3*a)/b
^4 - 1/4*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3 - 6*b*d^3*x - 6*b*c*d^2)*cos(b*x + a)/b^4 +
1/108*(9*b^2*d^3*x^2 + 18*b^2*c*d^2*x + 9*b^2*c^2*d - 2*d^3)*sin(3*b*x + 3*a)/b^4 + 3/4*(b^2*d^3*x^2 + 2*b^2*c
*d^2*x + b^2*c^2*d - 2*d^3)*sin(b*x + a)/b^4