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]

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

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Rubi [A]  time = 0.07, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 2637

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

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 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]

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*}

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Mathematica [A]  time = 0.22, 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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fricas [A]  time = 0.44, size = 87, normalized size = 1.13 \[ \frac {12 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{6} - 18 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{4} + 3 \, b d x - {\left (2 \, d \cos \left (b x + a\right )^{5} - 2 \, d \cos \left (b x + a\right )^{3} - 3 \, d \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{72 \, b^{2}} \]

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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giac [A]  time = 0.24, size = 75, normalized size = 0.97 \[ \frac {{\left (b d x + b c\right )} \cos \left (6 \, b x + 6 \, a\right )}{192 \, b^{2}} - \frac {3 \, {\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right )}{64 \, b^{2}} - \frac {d \sin \left (6 \, b x + 6 \, a\right )}{1152 \, b^{2}} + \frac {3 \, d \sin \left (2 \, b x + 2 \, a\right )}{128 \, b^{2}} \]

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

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maple [B]  time = 0.02, size = 176, normalized size = 2.29 \[ \frac {\frac {d \left (\frac {\left (b x +a \right ) \left (\sin ^{4}\left (b x +a \right )\right )}{4}+\frac {\left (\sin ^{3}\left (b x +a \right )+\frac {3 \sin \left (b x +a \right )}{2}\right ) \cos \left (b x +a \right )}{16}-\frac {b x}{24}-\frac {a}{24}-\frac {\left (b x +a \right ) \left (\sin ^{6}\left (b x +a \right )\right )}{6}-\frac {\left (\sin ^{5}\left (b x +a \right )+\frac {5 \left (\sin ^{3}\left (b x +a \right )\right )}{4}+\frac {15 \sin \left (b x +a \right )}{8}\right ) \cos \left (b x +a \right )}{36}\right )}{b}-\frac {d a \left (-\frac {\left (\sin ^{2}\left (b x +a \right )\right ) \left (\cos ^{4}\left (b x +a \right )\right )}{6}-\frac {\left (\cos ^{4}\left (b x +a \right )\right )}{12}\right )}{b}+c \left (-\frac {\left (\sin ^{2}\left (b x +a \right )\right ) \left (\cos ^{4}\left (b x +a \right )\right )}{6}-\frac {\left (\cos ^{4}\left (b x +a \right )\right )}{12}\right )}{b} \]

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*(1/b*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))-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 = 0.80, size = 119, normalized size = 1.55 \[ -\frac {96 \, {\left (2 \, \sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4}\right )} c - \frac {96 \, {\left (2 \, \sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4}\right )} a d}{b} - \frac {{\left (6 \, {\left (b x + a\right )} \cos \left (6 \, b x + 6 \, a\right ) - 54 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (6 \, b x + 6 \, a\right ) + 27 \, \sin \left (2 \, b x + 2 \, a\right )\right )} d}{b}}{1152 \, b} \]

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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mupad [B]  time = 0.71, size = 84, normalized size = 1.09 \[ \frac {\frac {27\,d\,\sin \left (2\,a+2\,b\,x\right )}{4}-\frac {d\,\sin \left (6\,a+6\,b\,x\right )}{4}-\frac {27\,b\,c\,\cos \left (2\,a+2\,b\,x\right )}{2}+\frac {3\,b\,c\,\cos \left (6\,a+6\,b\,x\right )}{2}-\frac {27\,b\,d\,x\,\cos \left (2\,a+2\,b\,x\right )}{2}+\frac {3\,b\,d\,x\,\cos \left (6\,a+6\,b\,x\right )}{2}}{288\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((27*d*sin(2*a + 2*b*x))/4 - (d*sin(6*a + 6*b*x))/4 - (27*b*c*cos(2*a + 2*b*x))/2 + (3*b*c*cos(6*a + 6*b*x))/2
- (27*b*d*x*cos(2*a + 2*b*x))/2 + (3*b*d*x*cos(6*a + 6*b*x))/2)/(288*b^2)

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sympy [A]  time = 5.33, size = 201, normalized size = 2.61 \[ \begin {cases} \frac {c \sin ^{6}{\left (a + b x \right )}}{12 b} + \frac {c \sin ^{4}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac {d x \sin ^{6}{\left (a + b x \right )}}{24 b} + \frac {d x \sin ^{4}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{8 b} - \frac {d x \sin ^{2}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{8 b} - \frac {d x \cos ^{6}{\left (a + b x \right )}}{24 b} + \frac {d \sin ^{5}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{24 b^{2}} + \frac {d \sin ^{3}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac {d \sin {\left (a + b x \right )} \cos ^{5}{\left (a + b x \right )}}{24 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin ^{3}{\relax (a )} \cos ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

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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