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

Optimal. Leaf size=77 \[ \frac{3 d \sin (2 a+2 b x)}{128 b^2}-\frac{d \sin (6 a+6 b x)}{1152 b^2}-\frac{3 (c+d x) \cos (2 a+2 b x)}{64 b}+\frac{(c+d x) \cos (6 a+6 b x)}{192 b} \]

[Out]

(-3*(c + d*x)*Cos[2*a + 2*b*x])/(64*b) + ((c + d*x)*Cos[6*a + 6*b*x])/(192*b) + (3*d*Sin[2*a + 2*b*x])/(128*b^
2) - (d*Sin[6*a + 6*b*x])/(1152*b^2)

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Rubi [A]  time = 0.0744242, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {4406, 3296, 2637} \[ \frac{3 d \sin (2 a+2 b x)}{128 b^2}-\frac{d \sin (6 a+6 b x)}{1152 b^2}-\frac{3 (c+d x) \cos (2 a+2 b x)}{64 b}+\frac{(c+d x) \cos (6 a+6 b x)}{192 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(c + d*x)*Cos[2*a + 2*b*x])/(64*b) + ((c + d*x)*Cos[6*a + 6*b*x])/(192*b) + (3*d*Sin[2*a + 2*b*x])/(128*b^
2) - (d*Sin[6*a + 6*b*x])/(1152*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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int (c+d x) \cos ^3(a+b x) \sin ^3(a+b x) \, dx &=\int \left (\frac{3}{32} (c+d x) \sin (2 a+2 b x)-\frac{1}{32} (c+d x) \sin (6 a+6 b x)\right ) \, dx\\ &=-\left (\frac{1}{32} \int (c+d x) \sin (6 a+6 b x) \, dx\right )+\frac{3}{32} \int (c+d x) \sin (2 a+2 b x) \, dx\\ &=-\frac{3 (c+d x) \cos (2 a+2 b x)}{64 b}+\frac{(c+d x) \cos (6 a+6 b x)}{192 b}-\frac{d \int \cos (6 a+6 b x) \, dx}{192 b}+\frac{(3 d) \int \cos (2 a+2 b x) \, dx}{64 b}\\ &=-\frac{3 (c+d x) \cos (2 a+2 b x)}{64 b}+\frac{(c+d x) \cos (6 a+6 b x)}{192 b}+\frac{3 d \sin (2 a+2 b x)}{128 b^2}-\frac{d \sin (6 a+6 b x)}{1152 b^2}\\ \end{align*}

Mathematica [A]  time = 0.222521, size = 63, normalized size = 0.82 \[ \frac{-54 b (c+d x) \cos (2 (a+b x))+6 b (c+d x) \cos (6 (a+b x))+d (27 \sin (2 (a+b x))-\sin (6 (a+b x)))}{1152 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-54*b*(c + d*x)*Cos[2*(a + b*x)] + 6*b*(c + d*x)*Cos[6*(a + b*x)] + d*(27*Sin[2*(a + b*x)] - Sin[6*(a + b*x)]
))/(1152*b^2)

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Maple [B]  time = 0.022, size = 176, normalized size = 2.3 \begin{align*}{\frac{1}{b} \left ({\frac{d}{b} \left ({\frac{ \left ( bx+a \right ) \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{4}}+{\frac{\cos \left ( bx+a \right ) }{16} \left ( \left ( \sin \left ( bx+a \right ) \right ) ^{3}+{\frac{3\,\sin \left ( bx+a \right ) }{2}} \right ) }-{\frac{bx}{24}}-{\frac{a}{24}}-{\frac{ \left ( bx+a \right ) \left ( \sin \left ( bx+a \right ) \right ) ^{6}}{6}}-{\frac{\cos \left ( bx+a \right ) }{36} \left ( \left ( \sin \left ( bx+a \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( bx+a \right ) }{8}} \right ) } \right ) }-{\frac{ad}{b} \left ( -{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{2} \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{6}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{12}} \right ) }+c \left ( -{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{2} \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{6}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{12}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.17966, size = 161, normalized size = 2.09 \begin{align*} -\frac{96 \,{\left (2 \, \sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4}\right )} c - \frac{96 \,{\left (2 \, \sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4}\right )} a d}{b} - \frac{{\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 )} d}{b}}{1152 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1152*(96*(2*sin(b*x + a)^6 - 3*sin(b*x + a)^4)*c - 96*(2*sin(b*x + a)^6 - 3*sin(b*x + a)^4)*a*d/b - (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))*d/b)/b

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Fricas [A]  time = 0.486378, size = 221, normalized size = 2.87 \begin{align*} \frac{12 \,{\left (b d x + b c\right )} \cos \left (b x + a\right )^{6} - 18 \,{\left (b d x + b c\right )} \cos \left (b x + a\right )^{4} + 3 \, b d x -{\left (2 \, d \cos \left (b x + a\right )^{5} - 2 \, d \cos \left (b x + a\right )^{3} - 3 \, d \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{72 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(12*(b*d*x + b*c)*cos(b*x + a)^6 - 18*(b*d*x + b*c)*cos(b*x + a)^4 + 3*b*d*x - (2*d*cos(b*x + a)^5 - 2*d*
cos(b*x + a)^3 - 3*d*cos(b*x + a))*sin(b*x + a))/b^2

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Sympy [A]  time = 11.1778, size = 201, normalized size = 2.61 \begin{align*} \begin{cases} \frac{c \sin ^{6}{\left (a + b x \right )}}{12 b} + \frac{c \sin ^{4}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac{d x \sin ^{6}{\left (a + b x \right )}}{24 b} + \frac{d x \sin ^{4}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{8 b} - \frac{d x \sin ^{2}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{8 b} - \frac{d x \cos ^{6}{\left (a + b x \right )}}{24 b} + \frac{d \sin ^{5}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{24 b^{2}} + \frac{d \sin ^{3}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac{d \sin{\left (a + b x \right )} \cos ^{5}{\left (a + b x \right )}}{24 b^{2}} & \text{for}\: b \neq 0 \\\left (c x + \frac{d x^{2}}{2}\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)*cos(b*x+a)**3*sin(b*x+a)**3,x)

[Out]

Piecewise((c*sin(a + b*x)**6/(12*b) + c*sin(a + b*x)**4*cos(a + b*x)**2/(4*b) + d*x*sin(a + b*x)**6/(24*b) + d
*x*sin(a + b*x)**4*cos(a + b*x)**2/(8*b) - d*x*sin(a + b*x)**2*cos(a + b*x)**4/(8*b) - d*x*cos(a + b*x)**6/(24
*b) + d*sin(a + b*x)**5*cos(a + b*x)/(24*b**2) + d*sin(a + b*x)**3*cos(a + b*x)**3/(9*b**2) + d*sin(a + b*x)*c
os(a + b*x)**5/(24*b**2), Ne(b, 0)), ((c*x + d*x**2/2)*sin(a)**3*cos(a)**3, True))

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Giac [A]  time = 1.12175, size = 101, normalized size = 1.31 \begin{align*} \frac{{\left (b d x + b c\right )} \cos \left (6 \, b x + 6 \, a\right )}{192 \, b^{2}} - \frac{3 \,{\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right )}{64 \, b^{2}} - \frac{d \sin \left (6 \, b x + 6 \, a\right )}{1152 \, b^{2}} + \frac{3 \, d \sin \left (2 \, b x + 2 \, a\right )}{128 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(b*d*x + b*c)*cos(6*b*x + 6*a)/b^2 - 3/64*(b*d*x + b*c)*cos(2*b*x + 2*a)/b^2 - 1/1152*d*sin(6*b*x + 6*a)
/b^2 + 3/128*d*sin(2*b*x + 2*a)/b^2