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

Optimal. Leaf size=175 \[ -\frac{40 d^2 (c+d x) \sin (a+b x)}{9 b^3}-\frac{2 d^2 (c+d x) \sin (a+b x) \cos ^2(a+b x)}{9 b^3}+\frac{d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}+\frac{2 d (c+d x)^2 \cos (a+b x)}{b^2}-\frac{2 d^3 \cos ^3(a+b x)}{27 b^4}-\frac{40 d^3 \cos (a+b x)}{9 b^4}+\frac{2 (c+d x)^3 \sin (a+b x)}{3 b}+\frac{(c+d x)^3 \sin (a+b x) \cos ^2(a+b x)}{3 b} \]

[Out]

(-40*d^3*Cos[a + b*x])/(9*b^4) + (2*d*(c + d*x)^2*Cos[a + b*x])/b^2 - (2*d^3*Cos[a + b*x]^3)/(27*b^4) + (d*(c
+ d*x)^2*Cos[a + b*x]^3)/(3*b^2) - (40*d^2*(c + d*x)*Sin[a + b*x])/(9*b^3) + (2*(c + d*x)^3*Sin[a + b*x])/(3*b
) - (2*d^2*(c + d*x)*Cos[a + b*x]^2*Sin[a + b*x])/(9*b^3) + ((c + d*x)^3*Cos[a + b*x]^2*Sin[a + b*x])/(3*b)

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Rubi [A]  time = 0.156228, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3311, 3296, 2638, 3310} \[ -\frac{40 d^2 (c+d x) \sin (a+b x)}{9 b^3}-\frac{2 d^2 (c+d x) \sin (a+b x) \cos ^2(a+b x)}{9 b^3}+\frac{d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}+\frac{2 d (c+d x)^2 \cos (a+b x)}{b^2}-\frac{2 d^3 \cos ^3(a+b x)}{27 b^4}-\frac{40 d^3 \cos (a+b x)}{9 b^4}+\frac{2 (c+d x)^3 \sin (a+b x)}{3 b}+\frac{(c+d x)^3 \sin (a+b x) \cos ^2(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Cos[a + b*x]^3,x]

[Out]

(-40*d^3*Cos[a + b*x])/(9*b^4) + (2*d*(c + d*x)^2*Cos[a + b*x])/b^2 - (2*d^3*Cos[a + b*x]^3)/(27*b^4) + (d*(c
+ d*x)^2*Cos[a + b*x]^3)/(3*b^2) - (40*d^2*(c + d*x)*Sin[a + b*x])/(9*b^3) + (2*(c + d*x)^3*Sin[a + b*x])/(3*b
) - (2*d^2*(c + d*x)*Cos[a + b*x]^2*Sin[a + b*x])/(9*b^3) + ((c + d*x)^3*Cos[a + b*x]^2*Sin[a + b*x])/(3*b)

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 2638

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

Rule 3310

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

Rubi steps

\begin{align*} \int (c+d x)^3 \cos ^3(a+b x) \, dx &=\frac{d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}+\frac{(c+d x)^3 \cos ^2(a+b x) \sin (a+b x)}{3 b}+\frac{2}{3} \int (c+d x)^3 \cos (a+b x) \, dx-\frac{\left (2 d^2\right ) \int (c+d x) \cos ^3(a+b x) \, dx}{3 b^2}\\ &=-\frac{2 d^3 \cos ^3(a+b x)}{27 b^4}+\frac{d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}+\frac{2 (c+d x)^3 \sin (a+b x)}{3 b}-\frac{2 d^2 (c+d x) \cos ^2(a+b x) \sin (a+b x)}{9 b^3}+\frac{(c+d x)^3 \cos ^2(a+b x) \sin (a+b x)}{3 b}-\frac{(2 d) \int (c+d x)^2 \sin (a+b x) \, dx}{b}-\frac{\left (4 d^2\right ) \int (c+d x) \cos (a+b x) \, dx}{9 b^2}\\ &=\frac{2 d (c+d x)^2 \cos (a+b x)}{b^2}-\frac{2 d^3 \cos ^3(a+b x)}{27 b^4}+\frac{d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}-\frac{4 d^2 (c+d x) \sin (a+b x)}{9 b^3}+\frac{2 (c+d x)^3 \sin (a+b x)}{3 b}-\frac{2 d^2 (c+d x) \cos ^2(a+b x) \sin (a+b x)}{9 b^3}+\frac{(c+d x)^3 \cos ^2(a+b x) \sin (a+b x)}{3 b}-\frac{\left (4 d^2\right ) \int (c+d x) \cos (a+b x) \, dx}{b^2}+\frac{\left (4 d^3\right ) \int \sin (a+b x) \, dx}{9 b^3}\\ &=-\frac{4 d^3 \cos (a+b x)}{9 b^4}+\frac{2 d (c+d x)^2 \cos (a+b x)}{b^2}-\frac{2 d^3 \cos ^3(a+b x)}{27 b^4}+\frac{d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}-\frac{40 d^2 (c+d x) \sin (a+b x)}{9 b^3}+\frac{2 (c+d x)^3 \sin (a+b x)}{3 b}-\frac{2 d^2 (c+d x) \cos ^2(a+b x) \sin (a+b x)}{9 b^3}+\frac{(c+d x)^3 \cos ^2(a+b x) \sin (a+b x)}{3 b}+\frac{\left (4 d^3\right ) \int \sin (a+b x) \, dx}{b^3}\\ &=-\frac{40 d^3 \cos (a+b x)}{9 b^4}+\frac{2 d (c+d x)^2 \cos (a+b x)}{b^2}-\frac{2 d^3 \cos ^3(a+b x)}{27 b^4}+\frac{d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}-\frac{40 d^2 (c+d x) \sin (a+b x)}{9 b^3}+\frac{2 (c+d x)^3 \sin (a+b x)}{3 b}-\frac{2 d^2 (c+d x) \cos ^2(a+b x) \sin (a+b x)}{9 b^3}+\frac{(c+d x)^3 \cos ^2(a+b x) \sin (a+b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.968956, size = 121, normalized size = 0.69 \[ \frac{243 d \cos (a+b x) \left (b^2 (c+d x)^2-2 d^2\right )+d \cos (3 (a+b x)) \left (9 b^2 (c+d x)^2-2 d^2\right )+6 b (c+d x) \sin (a+b x) \left (\cos (2 (a+b x)) \left (3 b^2 (c+d x)^2-2 d^2\right )+15 b^2 (c+d x)^2-82 d^2\right )}{108 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(243*d*(-2*d^2 + b^2*(c + d*x)^2)*Cos[a + b*x] + d*(-2*d^2 + 9*b^2*(c + d*x)^2)*Cos[3*(a + b*x)] + 6*b*(c + d*
x)*(-82*d^2 + 15*b^2*(c + d*x)^2 + (-2*d^2 + 3*b^2*(c + d*x)^2)*Cos[2*(a + b*x)])*Sin[a + b*x])/(108*b^4)

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Maple [B]  time = 0.029, size = 560, normalized size = 3.2 \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)^3,x)

[Out]

1/b*(1/b^3*d^3*(1/3*(b*x+a)^3*(2+cos(b*x+a)^2)*sin(b*x+a)+2*(b*x+a)^2*cos(b*x+a)-40/9*cos(b*x+a)-4*(b*x+a)*sin
(b*x+a)+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)-3/b^3*a*d^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^2*c*d^2*(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)*co
s(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^2*d^3*(1/3*(b*x+a)*(2+cos(b*x+a)^2
)*sin(b*x+a)+1/9*cos(b*x+a)^3+2/3*cos(b*x+a))-6/b^2*a*c*d^2*(1/3*(b*x+a)*(2+cos(b*x+a)^2)*sin(b*x+a)+1/9*cos(b
*x+a)^3+2/3*cos(b*x+a))+3/b*c^2*d*(1/3*(b*x+a)*(2+cos(b*x+a)^2)*sin(b*x+a)+1/9*cos(b*x+a)^3+2/3*cos(b*x+a))-1/
3/b^3*a^3*d^3*(2+cos(b*x+a)^2)*sin(b*x+a)+1/b^2*a^2*c*d^2*(2+cos(b*x+a)^2)*sin(b*x+a)-1/b*a*c^2*d*(2+cos(b*x+a
)^2)*sin(b*x+a)+1/3*c^3*(2+cos(b*x+a)^2)*sin(b*x+a))

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Maxima [B]  time = 1.11295, size = 722, normalized size = 4.13 \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)^3,x, algorithm="maxima")

[Out]

-1/108*(36*(sin(b*x + a)^3 - 3*sin(b*x + a))*c^3 - 108*(sin(b*x + a)^3 - 3*sin(b*x + a))*a*c^2*d/b + 108*(sin(
b*x + a)^3 - 3*sin(b*x + a))*a^2*c*d^2/b^2 - 36*(sin(b*x + a)^3 - 3*sin(b*x + a))*a^3*d^3/b^3 - 9*(3*(b*x + a)
*sin(3*b*x + 3*a) + 27*(b*x + a)*sin(b*x + a) + cos(3*b*x + 3*a) + 27*cos(b*x + a))*c^2*d/b + 18*(3*(b*x + a)*
sin(3*b*x + 3*a) + 27*(b*x + a)*sin(b*x + a) + cos(3*b*x + 3*a) + 27*cos(b*x + a))*a*c*d^2/b^2 - 9*(3*(b*x + a
)*sin(3*b*x + 3*a) + 27*(b*x + a)*sin(b*x + a) + cos(3*b*x + 3*a) + 27*cos(b*x + a))*a^2*d^3/b^3 - 3*(6*(b*x +
a)*cos(3*b*x + 3*a) + 162*(b*x + 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))*c*d^2/b^2 + 3*(6*(b*x + a)*cos(3*b*x + 3*a) + 162*(b*x + 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))*a*d^3/b^3 - ((9*(b*x + a)^2 - 2)*cos(3*b*x + 3*a) + 243
*((b*x + a)^2 - 2)*cos(b*x + a) + 3*(3*(b*x + a)^3 - 2*b*x - 2*a)*sin(3*b*x + 3*a) + 81*((b*x + a)^3 - 6*b*x -
6*a)*sin(b*x + a))*d^3/b^3)/b

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Fricas [A]  time = 1.50399, size = 494, normalized size = 2.82 \begin{align*} \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 )} \cos \left (b x + a\right )^{3} + 6 \,{\left (9 \, b^{2} d^{3} x^{2} + 18 \, b^{2} c d^{2} x + 9 \, b^{2} c^{2} d - 20 \, d^{3}\right )} \cos \left (b x + a\right ) + 3 \,{\left (6 \, b^{3} d^{3} x^{3} + 18 \, b^{3} c d^{2} x^{2} + 6 \, b^{3} c^{3} - 40 \, b c d^{2} +{\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 )^{2} + 2 \,{\left (9 \, b^{3} c^{2} d - 20 \, b d^{3}\right )} x\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)^3,x, algorithm="fricas")

[Out]

1/27*((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)^3 + 6*(9*b^2*d^3*x^2 + 18*b^2*c*d^2*
x + 9*b^2*c^2*d - 20*d^3)*cos(b*x + a) + 3*(6*b^3*d^3*x^3 + 18*b^3*c*d^2*x^2 + 6*b^3*c^3 - 40*b*c*d^2 + (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)^2 + 2*(9*b^3*c^2*
d - 20*b*d^3)*x)*sin(b*x + a))/b^4

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Sympy [A]  time = 5.38698, size = 495, normalized size = 2.83 \begin{align*} \begin{cases} \frac{2 c^{3} \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac{c^{3} \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} + \frac{2 c^{2} d x \sin ^{3}{\left (a + b x \right )}}{b} + \frac{3 c^{2} d x \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} + \frac{2 c d^{2} x^{2} \sin ^{3}{\left (a + b x \right )}}{b} + \frac{3 c d^{2} x^{2} \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} + \frac{2 d^{3} x^{3} \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac{d^{3} x^{3} \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} + \frac{2 c^{2} d \sin ^{2}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{b^{2}} + \frac{7 c^{2} d \cos ^{3}{\left (a + b x \right )}}{3 b^{2}} + \frac{4 c d^{2} x \sin ^{2}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{b^{2}} + \frac{14 c d^{2} x \cos ^{3}{\left (a + b x \right )}}{3 b^{2}} + \frac{2 d^{3} x^{2} \sin ^{2}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{b^{2}} + \frac{7 d^{3} x^{2} \cos ^{3}{\left (a + b x \right )}}{3 b^{2}} - \frac{40 c d^{2} \sin ^{3}{\left (a + b x \right )}}{9 b^{3}} - \frac{14 c d^{2} \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b^{3}} - \frac{40 d^{3} x \sin ^{3}{\left (a + b x \right )}}{9 b^{3}} - \frac{14 d^{3} x \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b^{3}} - \frac{40 d^{3} \sin ^{2}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{9 b^{4}} - \frac{122 d^{3} \cos ^{3}{\left (a + b x \right )}}{27 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 ) \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)**3*cos(b*x+a)**3,x)

[Out]

Piecewise((2*c**3*sin(a + b*x)**3/(3*b) + c**3*sin(a + b*x)*cos(a + b*x)**2/b + 2*c**2*d*x*sin(a + b*x)**3/b +
3*c**2*d*x*sin(a + b*x)*cos(a + b*x)**2/b + 2*c*d**2*x**2*sin(a + b*x)**3/b + 3*c*d**2*x**2*sin(a + b*x)*cos(
a + b*x)**2/b + 2*d**3*x**3*sin(a + b*x)**3/(3*b) + d**3*x**3*sin(a + b*x)*cos(a + b*x)**2/b + 2*c**2*d*sin(a
+ b*x)**2*cos(a + b*x)/b**2 + 7*c**2*d*cos(a + b*x)**3/(3*b**2) + 4*c*d**2*x*sin(a + b*x)**2*cos(a + b*x)/b**2
+ 14*c*d**2*x*cos(a + b*x)**3/(3*b**2) + 2*d**3*x**2*sin(a + b*x)**2*cos(a + b*x)/b**2 + 7*d**3*x**2*cos(a +
b*x)**3/(3*b**2) - 40*c*d**2*sin(a + b*x)**3/(9*b**3) - 14*c*d**2*sin(a + b*x)*cos(a + b*x)**2/(3*b**3) - 40*d
**3*x*sin(a + b*x)**3/(9*b**3) - 14*d**3*x*sin(a + b*x)*cos(a + b*x)**2/(3*b**3) - 40*d**3*sin(a + b*x)**2*cos
(a + b*x)/(9*b**4) - 122*d**3*cos(a + b*x)**3/(27*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)*cos(a)**3, True))

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Giac [A]  time = 1.1231, size = 312, normalized size = 1.78 \begin{align*} \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 )} \cos \left (3 \, b x + 3 \, a\right )}{108 \, b^{4}} + \frac{9 \,{\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d - 2 \, d^{3}\right )} \cos \left (b x + a\right )}{4 \, b^{4}} + \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 )} \sin \left (3 \, b x + 3 \, a\right )}{36 \, b^{4}} + \frac{3 \,{\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 )} \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)^3,x, algorithm="giac")

[Out]

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)*cos(3*b*x + 3*a)/b^4 + 9/4*(b^2*d^3*x^2 + 2*b^2*c
*d^2*x + b^2*c^2*d - 2*d^3)*cos(b*x + a)/b^4 + 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)*sin(3*b*x + 3*a)/b^4 + 3/4*(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)*sin(b*x + a)/b^4