3.2 \(\int (c+d x)^4 \cos (a+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=156 \[ -\frac{3 d^2 (c+d x)^2 \sin ^2(a+b x)}{2 b^3}-\frac{3 d^3 (c+d x) \sin (a+b x) \cos (a+b x)}{2 b^4}+\frac{d (c+d x)^3 \sin (a+b x) \cos (a+b x)}{b^2}+\frac{3 d^4 \sin ^2(a+b x)}{4 b^5}+\frac{(c+d x)^4 \sin ^2(a+b x)}{2 b}+\frac{3 c d^3 x}{2 b^3}+\frac{3 d^4 x^2}{4 b^3}-\frac{(c+d x)^4}{4 b} \]

[Out]

(3*c*d^3*x)/(2*b^3) + (3*d^4*x^2)/(4*b^3) - (c + d*x)^4/(4*b) - (3*d^3*(c + d*x)*Cos[a + b*x]*Sin[a + b*x])/(2
*b^4) + (d*(c + d*x)^3*Cos[a + b*x]*Sin[a + b*x])/b^2 + (3*d^4*Sin[a + b*x]^2)/(4*b^5) - (3*d^2*(c + d*x)^2*Si
n[a + b*x]^2)/(2*b^3) + ((c + d*x)^4*Sin[a + b*x]^2)/(2*b)

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Rubi [A]  time = 0.106998, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4404, 3311, 32, 3310} \[ -\frac{3 d^2 (c+d x)^2 \sin ^2(a+b x)}{2 b^3}-\frac{3 d^3 (c+d x) \sin (a+b x) \cos (a+b x)}{2 b^4}+\frac{d (c+d x)^3 \sin (a+b x) \cos (a+b x)}{b^2}+\frac{3 d^4 \sin ^2(a+b x)}{4 b^5}+\frac{(c+d x)^4 \sin ^2(a+b x)}{2 b}+\frac{3 c d^3 x}{2 b^3}+\frac{3 d^4 x^2}{4 b^3}-\frac{(c+d x)^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^4*Cos[a + b*x]*Sin[a + b*x],x]

[Out]

(3*c*d^3*x)/(2*b^3) + (3*d^4*x^2)/(4*b^3) - (c + d*x)^4/(4*b) - (3*d^3*(c + d*x)*Cos[a + b*x]*Sin[a + b*x])/(2
*b^4) + (d*(c + d*x)^3*Cos[a + b*x]*Sin[a + b*x])/b^2 + (3*d^4*Sin[a + b*x]^2)/(4*b^5) - (3*d^2*(c + d*x)^2*Si
n[a + b*x]^2)/(2*b^3) + ((c + d*x)^4*Sin[a + b*x]^2)/(2*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 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)^4 \cos (a+b x) \sin (a+b x) \, dx &=\frac{(c+d x)^4 \sin ^2(a+b x)}{2 b}-\frac{(2 d) \int (c+d x)^3 \sin ^2(a+b x) \, dx}{b}\\ &=\frac{d (c+d x)^3 \cos (a+b x) \sin (a+b x)}{b^2}-\frac{3 d^2 (c+d x)^2 \sin ^2(a+b x)}{2 b^3}+\frac{(c+d x)^4 \sin ^2(a+b x)}{2 b}-\frac{d \int (c+d x)^3 \, dx}{b}+\frac{\left (3 d^3\right ) \int (c+d x) \sin ^2(a+b x) \, dx}{b^3}\\ &=-\frac{(c+d x)^4}{4 b}-\frac{3 d^3 (c+d x) \cos (a+b x) \sin (a+b x)}{2 b^4}+\frac{d (c+d x)^3 \cos (a+b x) \sin (a+b x)}{b^2}+\frac{3 d^4 \sin ^2(a+b x)}{4 b^5}-\frac{3 d^2 (c+d x)^2 \sin ^2(a+b x)}{2 b^3}+\frac{(c+d x)^4 \sin ^2(a+b x)}{2 b}+\frac{\left (3 d^3\right ) \int (c+d x) \, dx}{2 b^3}\\ &=\frac{3 c d^3 x}{2 b^3}+\frac{3 d^4 x^2}{4 b^3}-\frac{(c+d x)^4}{4 b}-\frac{3 d^3 (c+d x) \cos (a+b x) \sin (a+b x)}{2 b^4}+\frac{d (c+d x)^3 \cos (a+b x) \sin (a+b x)}{b^2}+\frac{3 d^4 \sin ^2(a+b x)}{4 b^5}-\frac{3 d^2 (c+d x)^2 \sin ^2(a+b x)}{2 b^3}+\frac{(c+d x)^4 \sin ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.512725, size = 86, normalized size = 0.55 \[ \frac{4 b d (c+d x) \sin (2 (a+b x)) \left (2 b^2 (c+d x)^2-3 d^2\right )-2 \cos (2 (a+b x)) \left (-6 b^2 d^2 (c+d x)^2+2 b^4 (c+d x)^4+3 d^4\right )}{16 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^4*Cos[a + b*x]*Sin[a + b*x],x]

[Out]

(-2*(3*d^4 - 6*b^2*d^2*(c + d*x)^2 + 2*b^4*(c + d*x)^4)*Cos[2*(a + b*x)] + 4*b*d*(c + d*x)*(-3*d^2 + 2*b^2*(c
+ d*x)^2)*Sin[2*(a + b*x)])/(16*b^5)

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Maple [B]  time = 0.034, size = 853, normalized size = 5.5 \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)^4*cos(b*x+a)*sin(b*x+a),x)

[Out]

1/b*(1/b^4*d^4*(-1/2*(b*x+a)^4*cos(b*x+a)^2+2*(b*x+a)^3*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)+3/2*(b*x+a)^
2*cos(b*x+a)^2-3*(b*x+a)*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)+3/4*(b*x+a)^2+3/4*sin(b*x+a)^2-3/4*(b*x+a)^
4)-4/b^4*a*d^4*(-1/2*(b*x+a)^3*cos(b*x+a)^2+3/2*(b*x+a)^2*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)+3/4*(b*x+a
)*cos(b*x+a)^2-3/8*cos(b*x+a)*sin(b*x+a)-3/8*b*x-3/8*a-1/2*(b*x+a)^3)+4/b^3*c*d^3*(-1/2*(b*x+a)^3*cos(b*x+a)^2
+3/2*(b*x+a)^2*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)+3/4*(b*x+a)*cos(b*x+a)^2-3/8*cos(b*x+a)*sin(b*x+a)-3/
8*b*x-3/8*a-1/2*(b*x+a)^3)+6/b^4*a^2*d^4*(-1/2*(b*x+a)^2*cos(b*x+a)^2+(b*x+a)*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b
*x+1/2*a)-1/4*(b*x+a)^2-1/4*sin(b*x+a)^2)-12/b^3*a*c*d^3*(-1/2*(b*x+a)^2*cos(b*x+a)^2+(b*x+a)*(1/2*cos(b*x+a)*
sin(b*x+a)+1/2*b*x+1/2*a)-1/4*(b*x+a)^2-1/4*sin(b*x+a)^2)+6/b^2*c^2*d^2*(-1/2*(b*x+a)^2*cos(b*x+a)^2+(b*x+a)*(
1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)-1/4*(b*x+a)^2-1/4*sin(b*x+a)^2)-4/b^4*a^3*d^4*(-1/2*(b*x+a)*cos(b*x+a
)^2+1/4*cos(b*x+a)*sin(b*x+a)+1/4*b*x+1/4*a)+12/b^3*a^2*c*d^3*(-1/2*(b*x+a)*cos(b*x+a)^2+1/4*cos(b*x+a)*sin(b*
x+a)+1/4*b*x+1/4*a)-12/b^2*a*c^2*d^2*(-1/2*(b*x+a)*cos(b*x+a)^2+1/4*cos(b*x+a)*sin(b*x+a)+1/4*b*x+1/4*a)+4/b*c
^3*d*(-1/2*(b*x+a)*cos(b*x+a)^2+1/4*cos(b*x+a)*sin(b*x+a)+1/4*b*x+1/4*a)-1/2/b^4*a^4*d^4*cos(b*x+a)^2+2/b^3*a^
3*c*d^3*cos(b*x+a)^2-3/b^2*a^2*c^2*d^2*cos(b*x+a)^2+2/b*a*c^3*d*cos(b*x+a)^2-1/2*c^4*cos(b*x+a)^2)

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Maxima [B]  time = 1.22132, size = 791, normalized size = 5.07 \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)^4*cos(b*x+a)*sin(b*x+a),x, algorithm="maxima")

[Out]

-1/8*(4*c^4*cos(b*x + a)^2 - 16*a*c^3*d*cos(b*x + a)^2/b + 24*a^2*c^2*d^2*cos(b*x + a)^2/b^2 - 16*a^3*c*d^3*co
s(b*x + a)^2/b^3 + 4*a^4*d^4*cos(b*x + a)^2/b^4 + 4*(2*(b*x + a)*cos(2*b*x + 2*a) - sin(2*b*x + 2*a))*c^3*d/b
- 12*(2*(b*x + a)*cos(2*b*x + 2*a) - sin(2*b*x + 2*a))*a*c^2*d^2/b^2 + 12*(2*(b*x + a)*cos(2*b*x + 2*a) - sin(
2*b*x + 2*a))*a^2*c*d^3/b^3 - 4*(2*(b*x + a)*cos(2*b*x + 2*a) - sin(2*b*x + 2*a))*a^3*d^4/b^4 + 6*((2*(b*x + a
)^2 - 1)*cos(2*b*x + 2*a) - 2*(b*x + a)*sin(2*b*x + 2*a))*c^2*d^2/b^2 - 12*((2*(b*x + a)^2 - 1)*cos(2*b*x + 2*
a) - 2*(b*x + a)*sin(2*b*x + 2*a))*a*c*d^3/b^3 + 6*((2*(b*x + a)^2 - 1)*cos(2*b*x + 2*a) - 2*(b*x + a)*sin(2*b
*x + 2*a))*a^2*d^4/b^4 + 2*(2*(2*(b*x + a)^3 - 3*b*x - 3*a)*cos(2*b*x + 2*a) - 3*(2*(b*x + a)^2 - 1)*sin(2*b*x
+ 2*a))*c*d^3/b^3 - 2*(2*(2*(b*x + a)^3 - 3*b*x - 3*a)*cos(2*b*x + 2*a) - 3*(2*(b*x + a)^2 - 1)*sin(2*b*x + 2
*a))*a*d^4/b^4 + ((2*(b*x + a)^4 - 6*(b*x + a)^2 + 3)*cos(2*b*x + 2*a) - 2*(2*(b*x + a)^3 - 3*b*x - 3*a)*sin(2
*b*x + 2*a))*d^4/b^4)/b

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Fricas [A]  time = 0.49188, size = 518, normalized size = 3.32 \begin{align*} \frac{b^{4} d^{4} x^{4} + 4 \, b^{4} c d^{3} x^{3} + 3 \,{\left (2 \, b^{4} c^{2} d^{2} - b^{2} d^{4}\right )} x^{2} -{\left (2 \, b^{4} d^{4} x^{4} + 8 \, b^{4} c d^{3} x^{3} + 2 \, b^{4} c^{4} - 6 \, b^{2} c^{2} d^{2} + 3 \, d^{4} + 6 \,{\left (2 \, b^{4} c^{2} d^{2} - b^{2} d^{4}\right )} x^{2} + 4 \,{\left (2 \, b^{4} c^{3} d - 3 \, b^{2} c d^{3}\right )} x\right )} \cos \left (b x + a\right )^{2} + 2 \,{\left (2 \, b^{3} d^{4} x^{3} + 6 \, b^{3} c d^{3} x^{2} + 2 \, b^{3} c^{3} d - 3 \, b c d^{3} + 3 \,{\left (2 \, b^{3} c^{2} d^{2} - b d^{4}\right )} x\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \,{\left (2 \, b^{4} c^{3} d - 3 \, b^{2} c d^{3}\right )} x}{4 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*cos(b*x+a)*sin(b*x+a),x, algorithm="fricas")

[Out]

1/4*(b^4*d^4*x^4 + 4*b^4*c*d^3*x^3 + 3*(2*b^4*c^2*d^2 - b^2*d^4)*x^2 - (2*b^4*d^4*x^4 + 8*b^4*c*d^3*x^3 + 2*b^
4*c^4 - 6*b^2*c^2*d^2 + 3*d^4 + 6*(2*b^4*c^2*d^2 - b^2*d^4)*x^2 + 4*(2*b^4*c^3*d - 3*b^2*c*d^3)*x)*cos(b*x + a
)^2 + 2*(2*b^3*d^4*x^3 + 6*b^3*c*d^3*x^2 + 2*b^3*c^3*d - 3*b*c*d^3 + 3*(2*b^3*c^2*d^2 - b*d^4)*x)*cos(b*x + a)
*sin(b*x + a) + 2*(2*b^4*c^3*d - 3*b^2*c*d^3)*x)/b^5

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Sympy [A]  time = 7.63633, size = 502, normalized size = 3.22 \begin{align*} \begin{cases} \frac{c^{4} \sin ^{2}{\left (a + b x \right )}}{2 b} + \frac{c^{3} d x \sin ^{2}{\left (a + b x \right )}}{b} - \frac{c^{3} d x \cos ^{2}{\left (a + b x \right )}}{b} + \frac{3 c^{2} d^{2} x^{2} \sin ^{2}{\left (a + b x \right )}}{2 b} - \frac{3 c^{2} d^{2} x^{2} \cos ^{2}{\left (a + b x \right )}}{2 b} + \frac{c d^{3} x^{3} \sin ^{2}{\left (a + b x \right )}}{b} - \frac{c d^{3} x^{3} \cos ^{2}{\left (a + b x \right )}}{b} + \frac{d^{4} x^{4} \sin ^{2}{\left (a + b x \right )}}{4 b} - \frac{d^{4} x^{4} \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac{c^{3} d \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{b^{2}} + \frac{3 c^{2} d^{2} x \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{b^{2}} + \frac{3 c d^{3} x^{2} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{b^{2}} + \frac{d^{4} x^{3} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{b^{2}} - \frac{3 c^{2} d^{2} \sin ^{2}{\left (a + b x \right )}}{2 b^{3}} - \frac{3 c d^{3} x \sin ^{2}{\left (a + b x \right )}}{2 b^{3}} + \frac{3 c d^{3} x \cos ^{2}{\left (a + b x \right )}}{2 b^{3}} - \frac{3 d^{4} x^{2} \sin ^{2}{\left (a + b x \right )}}{4 b^{3}} + \frac{3 d^{4} x^{2} \cos ^{2}{\left (a + b x \right )}}{4 b^{3}} - \frac{3 c d^{3} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b^{4}} - \frac{3 d^{4} x \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b^{4}} + \frac{3 d^{4} \sin ^{2}{\left (a + b x \right )}}{4 b^{5}} & \text{for}\: b \neq 0 \\\left (c^{4} x + 2 c^{3} d x^{2} + 2 c^{2} d^{2} x^{3} + c d^{3} x^{4} + \frac{d^{4} x^{5}}{5}\right ) \sin{\left (a \right )} \cos{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**4*cos(b*x+a)*sin(b*x+a),x)

[Out]

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

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Giac [A]  time = 1.09295, size = 244, normalized size = 1.56 \begin{align*} -\frac{{\left (2 \, b^{4} d^{4} x^{4} + 8 \, b^{4} c d^{3} x^{3} + 12 \, b^{4} c^{2} d^{2} x^{2} + 8 \, b^{4} c^{3} d x + 2 \, b^{4} c^{4} - 6 \, b^{2} d^{4} x^{2} - 12 \, b^{2} c d^{3} x - 6 \, b^{2} c^{2} d^{2} + 3 \, d^{4}\right )} \cos \left (2 \, b x + 2 \, a\right )}{8 \, b^{5}} + \frac{{\left (2 \, b^{3} d^{4} x^{3} + 6 \, b^{3} c d^{3} x^{2} + 6 \, b^{3} c^{2} d^{2} x + 2 \, b^{3} c^{3} d - 3 \, b d^{4} x - 3 \, b c d^{3}\right )} \sin \left (2 \, b x + 2 \, a\right )}{4 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*cos(b*x+a)*sin(b*x+a),x, algorithm="giac")

[Out]

-1/8*(2*b^4*d^4*x^4 + 8*b^4*c*d^3*x^3 + 12*b^4*c^2*d^2*x^2 + 8*b^4*c^3*d*x + 2*b^4*c^4 - 6*b^2*d^4*x^2 - 12*b^
2*c*d^3*x - 6*b^2*c^2*d^2 + 3*d^4)*cos(2*b*x + 2*a)/b^5 + 1/4*(2*b^3*d^4*x^3 + 6*b^3*c*d^3*x^2 + 6*b^3*c^2*d^2
*x + 2*b^3*c^3*d - 3*b*d^4*x - 3*b*c*d^3)*sin(2*b*x + 2*a)/b^5