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

Optimal. Leaf size=129 \[ \frac{3 d (c+d x) \sin (2 a+2 b x)}{64 b^2}-\frac{d (c+d x) \sin (6 a+6 b x)}{576 b^2}+\frac{3 d^2 \cos (2 a+2 b x)}{128 b^3}-\frac{d^2 \cos (6 a+6 b x)}{3456 b^3}-\frac{3 (c+d x)^2 \cos (2 a+2 b x)}{64 b}+\frac{(c+d x)^2 \cos (6 a+6 b x)}{192 b} \]

[Out]

(3*d^2*Cos[2*a + 2*b*x])/(128*b^3) - (3*(c + d*x)^2*Cos[2*a + 2*b*x])/(64*b) - (d^2*Cos[6*a + 6*b*x])/(3456*b^
3) + ((c + d*x)^2*Cos[6*a + 6*b*x])/(192*b) + (3*d*(c + d*x)*Sin[2*a + 2*b*x])/(64*b^2) - (d*(c + d*x)*Sin[6*a
+ 6*b*x])/(576*b^2)

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Rubi [A]  time = 0.143795, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4406, 3296, 2638} \[ \frac{3 d (c+d x) \sin (2 a+2 b x)}{64 b^2}-\frac{d (c+d x) \sin (6 a+6 b x)}{576 b^2}+\frac{3 d^2 \cos (2 a+2 b x)}{128 b^3}-\frac{d^2 \cos (6 a+6 b x)}{3456 b^3}-\frac{3 (c+d x)^2 \cos (2 a+2 b x)}{64 b}+\frac{(c+d x)^2 \cos (6 a+6 b x)}{192 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d^2*Cos[2*a + 2*b*x])/(128*b^3) - (3*(c + d*x)^2*Cos[2*a + 2*b*x])/(64*b) - (d^2*Cos[6*a + 6*b*x])/(3456*b^
3) + ((c + d*x)^2*Cos[6*a + 6*b*x])/(192*b) + (3*d*(c + d*x)*Sin[2*a + 2*b*x])/(64*b^2) - (d*(c + d*x)*Sin[6*a
+ 6*b*x])/(576*b^2)

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

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]

Rubi steps

\begin{align*} \int (c+d x)^2 \cos ^3(a+b x) \sin ^3(a+b x) \, dx &=\int \left (\frac{3}{32} (c+d x)^2 \sin (2 a+2 b x)-\frac{1}{32} (c+d x)^2 \sin (6 a+6 b x)\right ) \, dx\\ &=-\left (\frac{1}{32} \int (c+d x)^2 \sin (6 a+6 b x) \, dx\right )+\frac{3}{32} \int (c+d x)^2 \sin (2 a+2 b x) \, dx\\ &=-\frac{3 (c+d x)^2 \cos (2 a+2 b x)}{64 b}+\frac{(c+d x)^2 \cos (6 a+6 b x)}{192 b}-\frac{d \int (c+d x) \cos (6 a+6 b x) \, dx}{96 b}+\frac{(3 d) \int (c+d x) \cos (2 a+2 b x) \, dx}{32 b}\\ &=-\frac{3 (c+d x)^2 \cos (2 a+2 b x)}{64 b}+\frac{(c+d x)^2 \cos (6 a+6 b x)}{192 b}+\frac{3 d (c+d x) \sin (2 a+2 b x)}{64 b^2}-\frac{d (c+d x) \sin (6 a+6 b x)}{576 b^2}+\frac{d^2 \int \sin (6 a+6 b x) \, dx}{576 b^2}-\frac{\left (3 d^2\right ) \int \sin (2 a+2 b x) \, dx}{64 b^2}\\ &=\frac{3 d^2 \cos (2 a+2 b x)}{128 b^3}-\frac{3 (c+d x)^2 \cos (2 a+2 b x)}{64 b}-\frac{d^2 \cos (6 a+6 b x)}{3456 b^3}+\frac{(c+d x)^2 \cos (6 a+6 b x)}{192 b}+\frac{3 d (c+d x) \sin (2 a+2 b x)}{64 b^2}-\frac{d (c+d x) \sin (6 a+6 b x)}{576 b^2}\\ \end{align*}

Mathematica [A]  time = 0.544526, size = 91, normalized size = 0.71 \[ \frac{-81 \cos (2 (a+b x)) \left (2 b^2 (c+d x)^2-d^2\right )+\cos (6 (a+b x)) \left (18 b^2 (c+d x)^2-d^2\right )-6 b d (c+d x) (\sin (6 (a+b x))-27 \sin (2 (a+b x)))}{3456 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-81*(-d^2 + 2*b^2*(c + d*x)^2)*Cos[2*(a + b*x)] + (-d^2 + 18*b^2*(c + d*x)^2)*Cos[6*(a + b*x)] - 6*b*d*(c + d
*x)*(-27*Sin[2*(a + b*x)] + Sin[6*(a + b*x)]))/(3456*b^3)

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Maple [B]  time = 0.024, size = 498, normalized size = 3.9 \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)^2*cos(b*x+a)^3*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)+1/24*(b*x+a)^2-1/72*sin(b*x+a)^4-1/24*sin(b*x+a)^2-1/6*(b*x+a)^2*sin(b*x+a)^6+1/3*(b*x+a)*(-1/6*(sin(b*
x+a)^5+5/4*sin(b*x+a)^3+15/8*sin(b*x+a))*cos(b*x+a)+5/16*b*x+5/16*a)+1/108*sin(b*x+a)^6)-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)-1/24*b*x-1/24*a-1/6*(b*x+a)*sin(b*x+a)^6-1/36*(
sin(b*x+a)^5+5/4*sin(b*x+a)^3+15/8*sin(b*x+a))*cos(b*x+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)-1/24*b*x-1/24*a-1/6*(b*x+a)*sin(b*x+a)^6-1/36*(sin(b*x+a)^5+5/4*sin(b*x+a)^3+15/8
*sin(b*x+a))*cos(b*x+a))+d^2/b^2*a^2*(-1/6*sin(b*x+a)^2*cos(b*x+a)^4-1/12*cos(b*x+a)^4)-2*c*d/b*a*(-1/6*sin(b*
x+a)^2*cos(b*x+a)^4-1/12*cos(b*x+a)^4)+c^2*(-1/6*sin(b*x+a)^2*cos(b*x+a)^4-1/12*cos(b*x+a)^4))

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Maxima [B]  time = 1.23834, size = 409, normalized size = 3.17 \begin{align*} -\frac{288 \,{\left (2 \, \sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4}\right )} c^{2} - \frac{576 \,{\left (2 \, \sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4}\right )} a c d}{b} + \frac{288 \,{\left (2 \, \sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4}\right )} a^{2} d^{2}}{b^{2}} - \frac{6 \,{\left (6 \,{\left (b x + a\right )} \cos \left (6 \, b x + 6 \, a\right ) - 54 \,{\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (6 \, b x + 6 \, a\right ) + 27 \, \sin \left (2 \, b x + 2 \, a\right )\right )} c d}{b} + \frac{6 \,{\left (6 \,{\left (b x + a\right )} \cos \left (6 \, b x + 6 \, a\right ) - 54 \,{\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (6 \, b x + 6 \, a\right ) + 27 \, \sin \left (2 \, b x + 2 \, a\right )\right )} a d^{2}}{b^{2}} - \frac{{\left ({\left (18 \,{\left (b x + a\right )}^{2} - 1\right )} \cos \left (6 \, b x + 6 \, a\right ) - 81 \,{\left (2 \,{\left (b x + a\right )}^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 6 \,{\left (b x + a\right )} \sin \left (6 \, b x + 6 \, a\right ) + 162 \,{\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{2}}{b^{2}}}{3456 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3456*(288*(2*sin(b*x + a)^6 - 3*sin(b*x + a)^4)*c^2 - 576*(2*sin(b*x + a)^6 - 3*sin(b*x + a)^4)*a*c*d/b + 2
88*(2*sin(b*x + a)^6 - 3*sin(b*x + a)^4)*a^2*d^2/b^2 - 6*(6*(b*x + a)*cos(6*b*x + 6*a) - 54*(b*x + a)*cos(2*b*
x + 2*a) - sin(6*b*x + 6*a) + 27*sin(2*b*x + 2*a))*c*d/b + 6*(6*(b*x + a)*cos(6*b*x + 6*a) - 54*(b*x + a)*cos(
2*b*x + 2*a) - sin(6*b*x + 6*a) + 27*sin(2*b*x + 2*a))*a*d^2/b^2 - ((18*(b*x + a)^2 - 1)*cos(6*b*x + 6*a) - 81
*(2*(b*x + a)^2 - 1)*cos(2*b*x + 2*a) - 6*(b*x + a)*sin(6*b*x + 6*a) + 162*(b*x + a)*sin(2*b*x + 2*a))*d^2/b^2
)/b

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Fricas [A]  time = 0.503551, size = 447, normalized size = 3.47 \begin{align*} \frac{2 \,{\left (18 \, b^{2} d^{2} x^{2} + 36 \, b^{2} c d x + 18 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )^{6} + 9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x - 3 \,{\left (18 \, b^{2} d^{2} x^{2} + 36 \, b^{2} c d x + 18 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )^{4} + 9 \, d^{2} \cos \left (b x + a\right )^{2} - 6 \,{\left (2 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{5} - 2 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{3} - 3 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{216 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/216*(2*(18*b^2*d^2*x^2 + 36*b^2*c*d*x + 18*b^2*c^2 - d^2)*cos(b*x + a)^6 + 9*b^2*d^2*x^2 + 18*b^2*c*d*x - 3*
(18*b^2*d^2*x^2 + 36*b^2*c*d*x + 18*b^2*c^2 - d^2)*cos(b*x + a)^4 + 9*d^2*cos(b*x + a)^2 - 6*(2*(b*d^2*x + b*c
*d)*cos(b*x + a)^5 - 2*(b*d^2*x + b*c*d)*cos(b*x + a)^3 - 3*(b*d^2*x + b*c*d)*cos(b*x + a))*sin(b*x + a))/b^3

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Sympy [A]  time = 22.629, size = 471, normalized size = 3.65 \begin{align*} \begin{cases} \frac{c^{2} \sin ^{6}{\left (a + b x \right )}}{12 b} + \frac{c^{2} \sin ^{4}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac{c d x \sin ^{6}{\left (a + b x \right )}}{12 b} + \frac{c d x \sin ^{4}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4 b} - \frac{c d x \sin ^{2}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{4 b} - \frac{c d x \cos ^{6}{\left (a + b x \right )}}{12 b} + \frac{d^{2} x^{2} \sin ^{6}{\left (a + b x \right )}}{24 b} + \frac{d^{2} x^{2} \sin ^{4}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{8 b} - \frac{d^{2} x^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{8 b} - \frac{d^{2} x^{2} \cos ^{6}{\left (a + b x \right )}}{24 b} + \frac{c d \sin ^{5}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{12 b^{2}} + \frac{2 c d \sin ^{3}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac{c d \sin{\left (a + b x \right )} \cos ^{5}{\left (a + b x \right )}}{12 b^{2}} + \frac{d^{2} x \sin ^{5}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{12 b^{2}} + \frac{2 d^{2} x \sin ^{3}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac{d^{2} x \sin{\left (a + b x \right )} \cos ^{5}{\left (a + b x \right )}}{12 b^{2}} - \frac{5 d^{2} \sin ^{6}{\left (a + b x \right )}}{108 b^{3}} - \frac{7 d^{2} \sin ^{4}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{72 b^{3}} - \frac{d^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{24 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 ^{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)**2*cos(b*x+a)**3*sin(b*x+a)**3,x)

[Out]

Piecewise((c**2*sin(a + b*x)**6/(12*b) + c**2*sin(a + b*x)**4*cos(a + b*x)**2/(4*b) + c*d*x*sin(a + b*x)**6/(1
2*b) + c*d*x*sin(a + b*x)**4*cos(a + b*x)**2/(4*b) - c*d*x*sin(a + b*x)**2*cos(a + b*x)**4/(4*b) - c*d*x*cos(a
+ b*x)**6/(12*b) + d**2*x**2*sin(a + b*x)**6/(24*b) + d**2*x**2*sin(a + b*x)**4*cos(a + b*x)**2/(8*b) - d**2*
x**2*sin(a + b*x)**2*cos(a + b*x)**4/(8*b) - d**2*x**2*cos(a + b*x)**6/(24*b) + c*d*sin(a + b*x)**5*cos(a + b*
x)/(12*b**2) + 2*c*d*sin(a + b*x)**3*cos(a + b*x)**3/(9*b**2) + c*d*sin(a + b*x)*cos(a + b*x)**5/(12*b**2) + d
**2*x*sin(a + b*x)**5*cos(a + b*x)/(12*b**2) + 2*d**2*x*sin(a + b*x)**3*cos(a + b*x)**3/(9*b**2) + d**2*x*sin(
a + b*x)*cos(a + b*x)**5/(12*b**2) - 5*d**2*sin(a + b*x)**6/(108*b**3) - 7*d**2*sin(a + b*x)**4*cos(a + b*x)**
2/(72*b**3) - d**2*sin(a + b*x)**2*cos(a + b*x)**4/(24*b**3), Ne(b, 0)), ((c**2*x + c*d*x**2 + d**2*x**3/3)*si
n(a)**3*cos(a)**3, True))

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Giac [A]  time = 1.14459, size = 196, normalized size = 1.52 \begin{align*} \frac{{\left (18 \, b^{2} d^{2} x^{2} + 36 \, b^{2} c d x + 18 \, b^{2} c^{2} - d^{2}\right )} \cos \left (6 \, b x + 6 \, a\right )}{3456 \, b^{3}} - \frac{3 \,{\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 )}{128 \, b^{3}} - \frac{{\left (b d^{2} x + b c d\right )} \sin \left (6 \, b x + 6 \, a\right )}{576 \, b^{3}} + \frac{3 \,{\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )}{64 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3456*(18*b^2*d^2*x^2 + 36*b^2*c*d*x + 18*b^2*c^2 - d^2)*cos(6*b*x + 6*a)/b^3 - 3/128*(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 - 1/576*(b*d^2*x + b*c*d)*sin(6*b*x + 6*a)/b^3 + 3/64*(b*d^2*x +
b*c*d)*sin(2*b*x + 2*a)/b^3