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

Optimal. Leaf size=196 \[ -\frac{3 d^2 (c+d x) \sin ^4(a+b x)}{32 b^3}-\frac{9 d^2 (c+d x) \sin ^2(a+b x)}{32 b^3}+\frac{3 d (c+d x)^2 \sin ^3(a+b x) \cos (a+b x)}{16 b^2}+\frac{9 d (c+d x)^2 \sin (a+b x) \cos (a+b x)}{32 b^2}-\frac{3 d^3 \sin ^3(a+b x) \cos (a+b x)}{128 b^4}-\frac{45 d^3 \sin (a+b x) \cos (a+b x)}{256 b^4}+\frac{(c+d x)^3 \sin ^4(a+b x)}{4 b}+\frac{45 d^3 x}{256 b^3}-\frac{3 (c+d x)^3}{32 b} \]

[Out]

(45*d^3*x)/(256*b^3) - (3*(c + d*x)^3)/(32*b) - (45*d^3*Cos[a + b*x]*Sin[a + b*x])/(256*b^4) + (9*d*(c + d*x)^
2*Cos[a + b*x]*Sin[a + b*x])/(32*b^2) - (9*d^2*(c + d*x)*Sin[a + b*x]^2)/(32*b^3) - (3*d^3*Cos[a + b*x]*Sin[a
+ b*x]^3)/(128*b^4) + (3*d*(c + d*x)^2*Cos[a + b*x]*Sin[a + b*x]^3)/(16*b^2) - (3*d^2*(c + d*x)*Sin[a + b*x]^4
)/(32*b^3) + ((c + d*x)^3*Sin[a + b*x]^4)/(4*b)

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Rubi [A]  time = 0.165402, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {4404, 3311, 32, 2635, 8} \[ -\frac{3 d^2 (c+d x) \sin ^4(a+b x)}{32 b^3}-\frac{9 d^2 (c+d x) \sin ^2(a+b x)}{32 b^3}+\frac{3 d (c+d x)^2 \sin ^3(a+b x) \cos (a+b x)}{16 b^2}+\frac{9 d (c+d x)^2 \sin (a+b x) \cos (a+b x)}{32 b^2}-\frac{3 d^3 \sin ^3(a+b x) \cos (a+b x)}{128 b^4}-\frac{45 d^3 \sin (a+b x) \cos (a+b x)}{256 b^4}+\frac{(c+d x)^3 \sin ^4(a+b x)}{4 b}+\frac{45 d^3 x}{256 b^3}-\frac{3 (c+d x)^3}{32 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(45*d^3*x)/(256*b^3) - (3*(c + d*x)^3)/(32*b) - (45*d^3*Cos[a + b*x]*Sin[a + b*x])/(256*b^4) + (9*d*(c + d*x)^
2*Cos[a + b*x]*Sin[a + b*x])/(32*b^2) - (9*d^2*(c + d*x)*Sin[a + b*x]^2)/(32*b^3) - (3*d^3*Cos[a + b*x]*Sin[a
+ b*x]^3)/(128*b^4) + (3*d*(c + d*x)^2*Cos[a + b*x]*Sin[a + b*x]^3)/(16*b^2) - (3*d^2*(c + d*x)*Sin[a + b*x]^4
)/(32*b^3) + ((c + d*x)^3*Sin[a + b*x]^4)/(4*b)

Rule 4404

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]

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -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)^3 \cos (a+b x) \sin ^3(a+b x) \, dx &=\frac{(c+d x)^3 \sin ^4(a+b x)}{4 b}-\frac{(3 d) \int (c+d x)^2 \sin ^4(a+b x) \, dx}{4 b}\\ &=\frac{3 d (c+d x)^2 \cos (a+b x) \sin ^3(a+b x)}{16 b^2}-\frac{3 d^2 (c+d x) \sin ^4(a+b x)}{32 b^3}+\frac{(c+d x)^3 \sin ^4(a+b x)}{4 b}-\frac{(9 d) \int (c+d x)^2 \sin ^2(a+b x) \, dx}{16 b}+\frac{\left (3 d^3\right ) \int \sin ^4(a+b x) \, dx}{32 b^3}\\ &=\frac{9 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{32 b^2}-\frac{9 d^2 (c+d x) \sin ^2(a+b x)}{32 b^3}-\frac{3 d^3 \cos (a+b x) \sin ^3(a+b x)}{128 b^4}+\frac{3 d (c+d x)^2 \cos (a+b x) \sin ^3(a+b x)}{16 b^2}-\frac{3 d^2 (c+d x) \sin ^4(a+b x)}{32 b^3}+\frac{(c+d x)^3 \sin ^4(a+b x)}{4 b}-\frac{(9 d) \int (c+d x)^2 \, dx}{32 b}+\frac{\left (9 d^3\right ) \int \sin ^2(a+b x) \, dx}{128 b^3}+\frac{\left (9 d^3\right ) \int \sin ^2(a+b x) \, dx}{32 b^3}\\ &=-\frac{3 (c+d x)^3}{32 b}-\frac{45 d^3 \cos (a+b x) \sin (a+b x)}{256 b^4}+\frac{9 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{32 b^2}-\frac{9 d^2 (c+d x) \sin ^2(a+b x)}{32 b^3}-\frac{3 d^3 \cos (a+b x) \sin ^3(a+b x)}{128 b^4}+\frac{3 d (c+d x)^2 \cos (a+b x) \sin ^3(a+b x)}{16 b^2}-\frac{3 d^2 (c+d x) \sin ^4(a+b x)}{32 b^3}+\frac{(c+d x)^3 \sin ^4(a+b x)}{4 b}+\frac{\left (9 d^3\right ) \int 1 \, dx}{256 b^3}+\frac{\left (9 d^3\right ) \int 1 \, dx}{64 b^3}\\ &=\frac{45 d^3 x}{256 b^3}-\frac{3 (c+d x)^3}{32 b}-\frac{45 d^3 \cos (a+b x) \sin (a+b x)}{256 b^4}+\frac{9 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{32 b^2}-\frac{9 d^2 (c+d x) \sin ^2(a+b x)}{32 b^3}-\frac{3 d^3 \cos (a+b x) \sin ^3(a+b x)}{128 b^4}+\frac{3 d (c+d x)^2 \cos (a+b x) \sin ^3(a+b x)}{16 b^2}-\frac{3 d^2 (c+d x) \sin ^4(a+b x)}{32 b^3}+\frac{(c+d x)^3 \sin ^4(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.969691, size = 135, normalized size = 0.69 \[ \frac{-64 b (c+d x) \cos (2 (a+b x)) \left (2 b^2 (c+d x)^2-3 d^2\right )+4 b (c+d x) \cos (4 (a+b x)) \left (8 b^2 (c+d x)^2-3 d^2\right )-6 d \sin (2 (a+b x)) \left (\cos (2 (a+b x)) \left (8 b^2 (c+d x)^2-d^2\right )-16 \left (2 b^2 (c+d x)^2-d^2\right )\right )}{1024 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*b*(c + d*x)*(-3*d^2 + 2*b^2*(c + d*x)^2)*Cos[2*(a + b*x)] + 4*b*(c + d*x)*(-3*d^2 + 8*b^2*(c + d*x)^2)*Co
s[4*(a + b*x)] - 6*d*(-16*(-d^2 + 2*b^2*(c + d*x)^2) + (-d^2 + 8*b^2*(c + d*x)^2)*Cos[2*(a + b*x)])*Sin[2*(a +
b*x)])/(1024*b^4)

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Maple [B]  time = 0.019, size = 594, 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)*sin(b*x+a)^3,x)

[Out]

1/b*(1/b^3*d^3*(1/4*(b*x+a)^3*sin(b*x+a)^4-3/4*(b*x+a)^2*(-1/4*(sin(b*x+a)^3+3/2*sin(b*x+a))*cos(b*x+a)+3/8*b*
x+3/8*a)-3/32*(b*x+a)*sin(b*x+a)^4-3/128*(sin(b*x+a)^3+3/2*sin(b*x+a))*cos(b*x+a)-27/256*b*x-27/256*a+9/32*(b*
x+a)*cos(b*x+a)^2-9/64*cos(b*x+a)*sin(b*x+a)+3/16*(b*x+a)^3)-3/b^3*a*d^3*(1/4*(b*x+a)^2*sin(b*x+a)^4-1/2*(b*x+
a)*(-1/4*(sin(b*x+a)^3+3/2*sin(b*x+a))*cos(b*x+a)+3/8*b*x+3/8*a)+3/32*(b*x+a)^2-1/32*sin(b*x+a)^4-3/32*sin(b*x
+a)^2)+3/b^2*c*d^2*(1/4*(b*x+a)^2*sin(b*x+a)^4-1/2*(b*x+a)*(-1/4*(sin(b*x+a)^3+3/2*sin(b*x+a))*cos(b*x+a)+3/8*
b*x+3/8*a)+3/32*(b*x+a)^2-1/32*sin(b*x+a)^4-3/32*sin(b*x+a)^2)+3/b^3*a^2*d^3*(1/4*(b*x+a)*sin(b*x+a)^4+1/16*(s
in(b*x+a)^3+3/2*sin(b*x+a))*cos(b*x+a)-3/32*b*x-3/32*a)-6/b^2*a*c*d^2*(1/4*(b*x+a)*sin(b*x+a)^4+1/16*(sin(b*x+
a)^3+3/2*sin(b*x+a))*cos(b*x+a)-3/32*b*x-3/32*a)+3/b*c^2*d*(1/4*(b*x+a)*sin(b*x+a)^4+1/16*(sin(b*x+a)^3+3/2*si
n(b*x+a))*cos(b*x+a)-3/32*b*x-3/32*a)-1/4/b^3*a^3*d^3*sin(b*x+a)^4+3/4/b^2*a^2*c*d^2*sin(b*x+a)^4-3/4/b*a*c^2*
d*sin(b*x+a)^4+1/4*c^3*sin(b*x+a)^4)

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Maxima [B]  time = 1.21312, size = 741, normalized size = 3.78 \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)*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

1/1024*(256*c^3*sin(b*x + a)^4 - 768*a*c^2*d*sin(b*x + a)^4/b + 768*a^2*c*d^2*sin(b*x + a)^4/b^2 - 256*a^3*d^3
*sin(b*x + a)^4/b^3 + 24*(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))*c^2*d/b - 48*(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))*a*c*d^2/b^2 + 24*(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))*a^2*d^3/b^3 + 12*((8*(b*x + a)^2 - 1)*cos(4*b*x + 4*a) - 16*(2*(b*x + a)^2 -
1)*cos(2*b*x + 2*a) - 4*(b*x + a)*sin(4*b*x + 4*a) + 32*(b*x + a)*sin(2*b*x + 2*a))*c*d^2/b^2 - 12*((8*(b*x +
a)^2 - 1)*cos(4*b*x + 4*a) - 16*(2*(b*x + a)^2 - 1)*cos(2*b*x + 2*a) - 4*(b*x + a)*sin(4*b*x + 4*a) + 32*(b*x
+ a)*sin(2*b*x + 2*a))*a*d^3/b^3 + (4*(8*(b*x + a)^3 - 3*b*x - 3*a)*cos(4*b*x + 4*a) - 64*(2*(b*x + a)^3 - 3*
b*x - 3*a)*cos(2*b*x + 2*a) - 3*(8*(b*x + a)^2 - 1)*sin(4*b*x + 4*a) + 96*(2*(b*x + a)^2 - 1)*sin(2*b*x + 2*a)
)*d^3/b^3)/b

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Fricas [A]  time = 0.517854, size = 621, normalized size = 3.17 \begin{align*} \frac{40 \, b^{3} d^{3} x^{3} + 120 \, b^{3} c d^{2} x^{2} + 8 \,{\left (8 \, b^{3} d^{3} x^{3} + 24 \, b^{3} c d^{2} x^{2} + 8 \, b^{3} c^{3} - 3 \, b c d^{2} + 3 \,{\left (8 \, b^{3} c^{2} d - b d^{3}\right )} x\right )} \cos \left (b x + a\right )^{4} - 8 \,{\left (16 \, b^{3} d^{3} x^{3} + 48 \, b^{3} c d^{2} x^{2} + 16 \, b^{3} c^{3} - 15 \, b c d^{2} + 3 \,{\left (16 \, b^{3} c^{2} d - 5 \, b d^{3}\right )} x\right )} \cos \left (b x + a\right )^{2} + 3 \,{\left (40 \, b^{3} c^{2} d - 17 \, b d^{3}\right )} x - 3 \,{\left (2 \,{\left (8 \, b^{2} d^{3} x^{2} + 16 \, b^{2} c d^{2} x + 8 \, b^{2} c^{2} d - d^{3}\right )} \cos \left (b x + a\right )^{3} -{\left (40 \, b^{2} d^{3} x^{2} + 80 \, b^{2} c d^{2} x + 40 \, b^{2} c^{2} d - 17 \, d^{3}\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{256 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(40*b^3*d^3*x^3 + 120*b^3*c*d^2*x^2 + 8*(8*b^3*d^3*x^3 + 24*b^3*c*d^2*x^2 + 8*b^3*c^3 - 3*b*c*d^2 + 3*(8
*b^3*c^2*d - b*d^3)*x)*cos(b*x + a)^4 - 8*(16*b^3*d^3*x^3 + 48*b^3*c*d^2*x^2 + 16*b^3*c^3 - 15*b*c*d^2 + 3*(16
*b^3*c^2*d - 5*b*d^3)*x)*cos(b*x + a)^2 + 3*(40*b^3*c^2*d - 17*b*d^3)*x - 3*(2*(8*b^2*d^3*x^2 + 16*b^2*c*d^2*x
+ 8*b^2*c^2*d - d^3)*cos(b*x + a)^3 - (40*b^2*d^3*x^2 + 80*b^2*c*d^2*x + 40*b^2*c^2*d - 17*d^3)*cos(b*x + a))
*sin(b*x + a))/b^4

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Sympy [A]  time = 10.4897, size = 634, normalized size = 3.23 \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)*sin(b*x+a)**3,x)

[Out]

Piecewise((-c**3*sin(a + b*x)**2*cos(a + b*x)**2/(2*b) - c**3*cos(a + b*x)**4/(4*b) + 15*c**2*d*x*sin(a + b*x)
**4/(32*b) - 9*c**2*d*x*sin(a + b*x)**2*cos(a + b*x)**2/(16*b) - 9*c**2*d*x*cos(a + b*x)**4/(32*b) + 15*c*d**2
*x**2*sin(a + b*x)**4/(32*b) - 9*c*d**2*x**2*sin(a + b*x)**2*cos(a + b*x)**2/(16*b) - 9*c*d**2*x**2*cos(a + b*
x)**4/(32*b) + 5*d**3*x**3*sin(a + b*x)**4/(32*b) - 3*d**3*x**3*sin(a + b*x)**2*cos(a + b*x)**2/(16*b) - 3*d**
3*x**3*cos(a + b*x)**4/(32*b) + 15*c**2*d*sin(a + b*x)**3*cos(a + b*x)/(32*b**2) + 9*c**2*d*sin(a + b*x)*cos(a
+ b*x)**3/(32*b**2) + 15*c*d**2*x*sin(a + b*x)**3*cos(a + b*x)/(16*b**2) + 9*c*d**2*x*sin(a + b*x)*cos(a + b*
x)**3/(16*b**2) + 15*d**3*x**2*sin(a + b*x)**3*cos(a + b*x)/(32*b**2) + 9*d**3*x**2*sin(a + b*x)*cos(a + b*x)*
*3/(32*b**2) + 15*c*d**2*sin(a + b*x)**2*cos(a + b*x)**2/(32*b**3) + 3*c*d**2*cos(a + b*x)**4/(8*b**3) - 51*d*
*3*x*sin(a + b*x)**4/(256*b**3) + 9*d**3*x*sin(a + b*x)**2*cos(a + b*x)**2/(128*b**3) + 45*d**3*x*cos(a + b*x)
**4/(256*b**3) - 51*d**3*sin(a + b*x)**3*cos(a + b*x)/(256*b**4) - 45*d**3*sin(a + b*x)*cos(a + b*x)**3/(256*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)**3*cos(a), True))

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Giac [A]  time = 1.13294, size = 325, normalized size = 1.66 \begin{align*} \frac{{\left (8 \, b^{3} d^{3} x^{3} + 24 \, b^{3} c d^{2} x^{2} + 24 \, b^{3} c^{2} d x + 8 \, b^{3} c^{3} - 3 \, b d^{3} x - 3 \, b c d^{2}\right )} \cos \left (4 \, b x + 4 \, a\right )}{256 \, b^{4}} - \frac{{\left (2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} + 6 \, b^{3} c^{2} d x + 2 \, b^{3} c^{3} - 3 \, b d^{3} x - 3 \, b c d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right )}{16 \, b^{4}} - \frac{3 \,{\left (8 \, b^{2} d^{3} x^{2} + 16 \, b^{2} c d^{2} x + 8 \, b^{2} c^{2} d - d^{3}\right )} \sin \left (4 \, b x + 4 \, a\right )}{1024 \, b^{4}} + \frac{3 \,{\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \sin \left (2 \, b x + 2 \, a\right )}{32 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(8*b^3*d^3*x^3 + 24*b^3*c*d^2*x^2 + 24*b^3*c^2*d*x + 8*b^3*c^3 - 3*b*d^3*x - 3*b*c*d^2)*cos(4*b*x + 4*a)
/b^4 - 1/16*(2*b^3*d^3*x^3 + 6*b^3*c*d^2*x^2 + 6*b^3*c^2*d*x + 2*b^3*c^3 - 3*b*d^3*x - 3*b*c*d^2)*cos(2*b*x +
2*a)/b^4 - 3/1024*(8*b^2*d^3*x^2 + 16*b^2*c*d^2*x + 8*b^2*c^2*d - d^3)*sin(4*b*x + 4*a)/b^4 + 3/32*(2*b^2*d^3*
x^2 + 4*b^2*c*d^2*x + 2*b^2*c^2*d - d^3)*sin(2*b*x + 2*a)/b^4