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

Optimal. Leaf size=53 \[ -\frac{d \cos (4 a+4 b x)}{128 b^2}-\frac{(c+d x) \sin (4 a+4 b x)}{32 b}+\frac{(c+d x)^2}{16 d} \]

[Out]

(c + d*x)^2/(16*d) - (d*Cos[4*a + 4*b*x])/(128*b^2) - ((c + d*x)*Sin[4*a + 4*b*x])/(32*b)

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Rubi [A]  time = 0.0537823, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {4406, 3296, 2638} \[ -\frac{d \cos (4 a+4 b x)}{128 b^2}-\frac{(c+d x) \sin (4 a+4 b x)}{32 b}+\frac{(c+d x)^2}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)^2/(16*d) - (d*Cos[4*a + 4*b*x])/(128*b^2) - ((c + d*x)*Sin[4*a + 4*b*x])/(32*b)

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) \cos ^2(a+b x) \sin ^2(a+b x) \, dx &=\int \left (\frac{1}{8} (c+d x)-\frac{1}{8} (c+d x) \cos (4 a+4 b x)\right ) \, dx\\ &=\frac{(c+d x)^2}{16 d}-\frac{1}{8} \int (c+d x) \cos (4 a+4 b x) \, dx\\ &=\frac{(c+d x)^2}{16 d}-\frac{(c+d x) \sin (4 a+4 b x)}{32 b}+\frac{d \int \sin (4 a+4 b x) \, dx}{32 b}\\ &=\frac{(c+d x)^2}{16 d}-\frac{d \cos (4 a+4 b x)}{128 b^2}-\frac{(c+d x) \sin (4 a+4 b x)}{32 b}\\ \end{align*}

Mathematica [A]  time = 0.297987, size = 54, normalized size = 1.02 \[ -\frac{8 (a+b x) (a d-2 b c-b d x)+4 b (c+d x) \sin (4 (a+b x))+d \cos (4 (a+b x))}{128 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(8*(a + b*x)*(-2*b*c + a*d - b*d*x) + d*Cos[4*(a + b*x)] + 4*b*(c + d*x)*Sin[4*(a + b*x)])/(128*b^2)

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Maple [B]  time = 0.02, size = 194, normalized size = 3.7 \begin{align*}{\frac{1}{b} \left ({\frac{d}{b} \left ( \left ( bx+a \right ) \left ( -{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{2}}+{\frac{bx}{2}}+{\frac{a}{2}} \right ) -{\frac{ \left ( bx+a \right ) ^{2}}{16}}+{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{16}}- \left ( bx+a \right ) \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{ \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{16}} \right ) }-{\frac{ad}{b} \left ( -{\frac{\sin \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{4}}+{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{8}}+{\frac{bx}{8}}+{\frac{a}{8}} \right ) }+c \left ( -{\frac{\sin \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{4}}+{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{8}}+{\frac{bx}{8}}+{\frac{a}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.25609, size = 130, normalized size = 2.45 \begin{align*} \frac{4 \,{\left (4 \, b x + 4 \, a - \sin \left (4 \, b x + 4 \, a\right )\right )} c - \frac{4 \,{\left (4 \, b x + 4 \, a - \sin \left (4 \, b x + 4 \, a\right )\right )} a d}{b} + \frac{{\left (8 \,{\left (b x + a\right )}^{2} - 4 \,{\left (b x + a\right )} \sin \left (4 \, b x + 4 \, a\right ) - \cos \left (4 \, b x + 4 \, a\right )\right )} d}{b}}{128 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(4*(4*b*x + 4*a - sin(4*b*x + 4*a))*c - 4*(4*b*x + 4*a - sin(4*b*x + 4*a))*a*d/b + (8*(b*x + a)^2 - 4*(b
*x + a)*sin(4*b*x + 4*a) - cos(4*b*x + 4*a))*d/b)/b

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Fricas [A]  time = 0.481737, size = 204, normalized size = 3.85 \begin{align*} \frac{b^{2} d x^{2} - d \cos \left (b x + a\right )^{4} + 2 \, b^{2} c x + d \cos \left (b x + a\right )^{2} - 2 \,{\left (2 \,{\left (b d x + b c\right )} \cos \left (b x + a\right )^{3} -{\left (b d x + b c\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{16 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.355, size = 231, normalized size = 4.36 \begin{align*} \begin{cases} \frac{c x \sin ^{4}{\left (a + b x \right )}}{8} + \frac{c x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac{c x \cos ^{4}{\left (a + b x \right )}}{8} + \frac{d x^{2} \sin ^{4}{\left (a + b x \right )}}{16} + \frac{d x^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{8} + \frac{d x^{2} \cos ^{4}{\left (a + b x \right )}}{16} + \frac{c \sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{8 b} - \frac{c \sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} + \frac{d x \sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{8 b} - \frac{d x \sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} + \frac{d \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{16 b^{2}} & \text{for}\: b \neq 0 \\\left (c x + \frac{d x^{2}}{2}\right ) \sin ^{2}{\left (a \right )} \cos ^{2}{\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)**2*sin(b*x+a)**2,x)

[Out]

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

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Giac [A]  time = 1.08702, size = 65, normalized size = 1.23 \begin{align*} \frac{1}{16} \, d x^{2} + \frac{1}{8} \, c x - \frac{d \cos \left (4 \, b x + 4 \, a\right )}{128 \, b^{2}} - \frac{{\left (b d x + b c\right )} \sin \left (4 \, b x + 4 \, a\right )}{32 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*d*x^2 + 1/8*c*x - 1/128*d*cos(4*b*x + 4*a)/b^2 - 1/32*(b*d*x + b*c)*sin(4*b*x + 4*a)/b^2