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

Optimal. Leaf size=134 \[ \frac{d (c+d x) \sin ^3(a+b x) \cos (a+b x)}{8 b^2}+\frac{3 d (c+d x) \sin (a+b x) \cos (a+b x)}{16 b^2}-\frac{d^2 \sin ^4(a+b x)}{32 b^3}-\frac{3 d^2 \sin ^2(a+b x)}{32 b^3}+\frac{(c+d x)^2 \sin ^4(a+b x)}{4 b}-\frac{3 c d x}{16 b}-\frac{3 d^2 x^2}{32 b} \]

[Out]

(-3*c*d*x)/(16*b) - (3*d^2*x^2)/(32*b) + (3*d*(c + d*x)*Cos[a + b*x]*Sin[a + b*x])/(16*b^2) - (3*d^2*Sin[a + b
*x]^2)/(32*b^3) + (d*(c + d*x)*Cos[a + b*x]*Sin[a + b*x]^3)/(8*b^2) - (d^2*Sin[a + b*x]^4)/(32*b^3) + ((c + d*
x)^2*Sin[a + b*x]^4)/(4*b)

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Rubi [A]  time = 0.0924045, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {4404, 3310} \[ \frac{d (c+d x) \sin ^3(a+b x) \cos (a+b x)}{8 b^2}+\frac{3 d (c+d x) \sin (a+b x) \cos (a+b x)}{16 b^2}-\frac{d^2 \sin ^4(a+b x)}{32 b^3}-\frac{3 d^2 \sin ^2(a+b x)}{32 b^3}+\frac{(c+d x)^2 \sin ^4(a+b x)}{4 b}-\frac{3 c d x}{16 b}-\frac{3 d^2 x^2}{32 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*c*d*x)/(16*b) - (3*d^2*x^2)/(32*b) + (3*d*(c + d*x)*Cos[a + b*x]*Sin[a + b*x])/(16*b^2) - (3*d^2*Sin[a + b
*x]^2)/(32*b^3) + (d*(c + d*x)*Cos[a + b*x]*Sin[a + b*x]^3)/(8*b^2) - (d^2*Sin[a + b*x]^4)/(32*b^3) + ((c + d*
x)^2*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 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)^2 \cos (a+b x) \sin ^3(a+b x) \, dx &=\frac{(c+d x)^2 \sin ^4(a+b x)}{4 b}-\frac{d \int (c+d x) \sin ^4(a+b x) \, dx}{2 b}\\ &=\frac{d (c+d x) \cos (a+b x) \sin ^3(a+b x)}{8 b^2}-\frac{d^2 \sin ^4(a+b x)}{32 b^3}+\frac{(c+d x)^2 \sin ^4(a+b x)}{4 b}-\frac{(3 d) \int (c+d x) \sin ^2(a+b x) \, dx}{8 b}\\ &=\frac{3 d (c+d x) \cos (a+b x) \sin (a+b x)}{16 b^2}-\frac{3 d^2 \sin ^2(a+b x)}{32 b^3}+\frac{d (c+d x) \cos (a+b x) \sin ^3(a+b x)}{8 b^2}-\frac{d^2 \sin ^4(a+b x)}{32 b^3}+\frac{(c+d x)^2 \sin ^4(a+b x)}{4 b}-\frac{(3 d) \int (c+d x) \, dx}{16 b}\\ &=-\frac{3 c d x}{16 b}-\frac{3 d^2 x^2}{32 b}+\frac{3 d (c+d x) \cos (a+b x) \sin (a+b x)}{16 b^2}-\frac{3 d^2 \sin ^2(a+b x)}{32 b^3}+\frac{d (c+d x) \cos (a+b x) \sin ^3(a+b x)}{8 b^2}-\frac{d^2 \sin ^4(a+b x)}{32 b^3}+\frac{(c+d x)^2 \sin ^4(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.524018, size = 91, normalized size = 0.68 \[ \frac{-16 \cos (2 (a+b x)) \left (2 b^2 (c+d x)^2-d^2\right )+\cos (4 (a+b x)) \left (8 b^2 (c+d x)^2-d^2\right )-4 b d (c+d x) (\sin (4 (a+b x))-8 \sin (2 (a+b x)))}{256 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*(-d^2 + 2*b^2*(c + d*x)^2)*Cos[2*(a + b*x)] + (-d^2 + 8*b^2*(c + d*x)^2)*Cos[4*(a + b*x)] - 4*b*d*(c + d*
x)*(-8*Sin[2*(a + b*x)] + Sin[4*(a + b*x)]))/(256*b^3)

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Maple [B]  time = 0.02, size = 260, normalized size = 1.9 \begin{align*}{\frac{1}{b} \left ({\frac{{d}^{2}}{{b}^{2}} \left ({\frac{ \left ( bx+a \right ) ^{2} \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{4}}-{\frac{bx+a}{2} \left ( -{\frac{\cos \left ( bx+a \right ) }{4} \left ( \left ( \sin \left ( bx+a \right ) \right ) ^{3}+{\frac{3\,\sin \left ( bx+a \right ) }{2}} \right ) }+{\frac{3\,bx}{8}}+{\frac{3\,a}{8}} \right ) }+{\frac{3\, \left ( bx+a \right ) ^{2}}{32}}-{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{32}}-{\frac{3\, \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{32}} \right ) }-2\,{\frac{a{d}^{2}}{{b}^{2}} \left ( 1/4\, \left ( bx+a \right ) \left ( \sin \left ( bx+a \right ) \right ) ^{4}+1/16\, \left ( \left ( \sin \left ( bx+a \right ) \right ) ^{3}+3/2\,\sin \left ( bx+a \right ) \right ) \cos \left ( bx+a \right ) -{\frac{3\,bx}{32}}-{\frac{3\,a}{32}} \right ) }+2\,{\frac{cd}{b} \left ( 1/4\, \left ( bx+a \right ) \left ( \sin \left ( bx+a \right ) \right ) ^{4}+1/16\, \left ( \left ( \sin \left ( bx+a \right ) \right ) ^{3}+3/2\,\sin \left ( bx+a \right ) \right ) \cos \left ( bx+a \right ) -{\frac{3\,bx}{32}}-{\frac{3\,a}{32}} \right ) }+{\frac{{a}^{2}{d}^{2} \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{4\,{b}^{2}}}-{\frac{acd \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{2\,b}}+{\frac{{c}^{2} \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/b^2*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)-2/b^2*a*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)+2/b*c*d*(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)+1/4/b^2*a^2*d^2*sin(b*x+a)^4-1/2/b*a*c*d*sin(b*x+a)^4+1/4*c^2*sin(b*x+a)^
4)

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Maxima [B]  time = 1.1864, size = 355, normalized size = 2.65 \begin{align*} \frac{64 \, c^{2} \sin \left (b x + a\right )^{4} - \frac{128 \, a c d \sin \left (b x + a\right )^{4}}{b} + \frac{64 \, a^{2} d^{2} \sin \left (b x + a\right )^{4}}{b^{2}} + \frac{4 \,{\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 )} c d}{b} - \frac{4 \,{\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 )} a d^{2}}{b^{2}} + \frac{{\left ({\left (8 \,{\left (b x + a\right )}^{2} - 1\right )} \cos \left (4 \, b x + 4 \, a\right ) - 16 \,{\left (2 \,{\left (b x + a\right )}^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 4 \,{\left (b x + a\right )} \sin \left (4 \, b x + 4 \, a\right ) + 32 \,{\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{2}}{b^{2}}}{256 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(64*c^2*sin(b*x + a)^4 - 128*a*c*d*sin(b*x + a)^4/b + 64*a^2*d^2*sin(b*x + a)^4/b^2 + 4*(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*d/b - 4*(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*d^2/b^2 + ((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))*d^2/b^2)/b

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Fricas [A]  time = 0.486331, size = 359, normalized size = 2.68 \begin{align*} \frac{5 \, b^{2} d^{2} x^{2} + 10 \, b^{2} c d x +{\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )^{4} -{\left (16 \, b^{2} d^{2} x^{2} + 32 \, b^{2} c d x + 16 \, b^{2} c^{2} - 5 \, d^{2}\right )} \cos \left (b x + a\right )^{2} - 2 \,{\left (2 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{3} - 5 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{32 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(5*b^2*d^2*x^2 + 10*b^2*c*d*x + (8*b^2*d^2*x^2 + 16*b^2*c*d*x + 8*b^2*c^2 - d^2)*cos(b*x + a)^4 - (16*b^2
*d^2*x^2 + 32*b^2*c*d*x + 16*b^2*c^2 - 5*d^2)*cos(b*x + a)^2 - 2*(2*(b*d^2*x + b*c*d)*cos(b*x + a)^3 - 5*(b*d^
2*x + b*c*d)*cos(b*x + a))*sin(b*x + a))/b^3

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Sympy [A]  time = 5.11579, size = 350, normalized size = 2.61 \begin{align*} \begin{cases} - \frac{c^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{2 b} - \frac{c^{2} \cos ^{4}{\left (a + b x \right )}}{4 b} + \frac{5 c d x \sin ^{4}{\left (a + b x \right )}}{16 b} - \frac{3 c d x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{8 b} - \frac{3 c d x \cos ^{4}{\left (a + b x \right )}}{16 b} + \frac{5 d^{2} x^{2} \sin ^{4}{\left (a + b x \right )}}{32 b} - \frac{3 d^{2} x^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{16 b} - \frac{3 d^{2} x^{2} \cos ^{4}{\left (a + b x \right )}}{32 b} + \frac{5 c d \sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{16 b^{2}} + \frac{3 c d \sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{16 b^{2}} + \frac{5 d^{2} x \sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{16 b^{2}} + \frac{3 d^{2} x \sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{16 b^{2}} + \frac{5 d^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{32 b^{3}} + \frac{d^{2} \cos ^{4}{\left (a + b x \right )}}{8 b^{3}} & \text{for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac{d^{2} x^{3}}{3}\right ) \sin ^{3}{\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)**2*cos(b*x+a)*sin(b*x+a)**3,x)

[Out]

Piecewise((-c**2*sin(a + b*x)**2*cos(a + b*x)**2/(2*b) - c**2*cos(a + b*x)**4/(4*b) + 5*c*d*x*sin(a + b*x)**4/
(16*b) - 3*c*d*x*sin(a + b*x)**2*cos(a + b*x)**2/(8*b) - 3*c*d*x*cos(a + b*x)**4/(16*b) + 5*d**2*x**2*sin(a +
b*x)**4/(32*b) - 3*d**2*x**2*sin(a + b*x)**2*cos(a + b*x)**2/(16*b) - 3*d**2*x**2*cos(a + b*x)**4/(32*b) + 5*c
*d*sin(a + b*x)**3*cos(a + b*x)/(16*b**2) + 3*c*d*sin(a + b*x)*cos(a + b*x)**3/(16*b**2) + 5*d**2*x*sin(a + b*
x)**3*cos(a + b*x)/(16*b**2) + 3*d**2*x*sin(a + b*x)*cos(a + b*x)**3/(16*b**2) + 5*d**2*sin(a + b*x)**2*cos(a
+ b*x)**2/(32*b**3) + d**2*cos(a + b*x)**4/(8*b**3), Ne(b, 0)), ((c**2*x + c*d*x**2 + d**2*x**3/3)*sin(a)**3*c
os(a), True))

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Giac [A]  time = 1.13904, size = 196, normalized size = 1.46 \begin{align*} \frac{{\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - d^{2}\right )} \cos \left (4 \, b x + 4 \, a\right )}{256 \, b^{3}} - \frac{{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right )}{16 \, b^{3}} - \frac{{\left (b d^{2} x + b c d\right )} \sin \left (4 \, b x + 4 \, a\right )}{64 \, b^{3}} + \frac{{\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(8*b^2*d^2*x^2 + 16*b^2*c*d*x + 8*b^2*c^2 - d^2)*cos(4*b*x + 4*a)/b^3 - 1/16*(2*b^2*d^2*x^2 + 4*b^2*c*d*
x + 2*b^2*c^2 - d^2)*cos(2*b*x + 2*a)/b^3 - 1/64*(b*d^2*x + b*c*d)*sin(4*b*x + 4*a)/b^3 + 1/8*(b*d^2*x + b*c*d
)*sin(2*b*x + 2*a)/b^3