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]

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

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Rubi [A]  time = 0.05, antiderivative size = 53, normalized size of antiderivative = 1.00, 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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 ^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*}

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Mathematica [A]  time = 0.29, 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]

-1/128*(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)])/b^2

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fricas [A]  time = 0.59, size = 85, normalized size = 1.60 \[ \frac {b^{2} d x^{2} - d \cos \left (b x + a\right )^{4} + 2 \, b^{2} c x + d \cos \left (b x + a\right )^{2} - 2 \, {\left (2 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{3} - {\left (b d x + b c\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{16 \, b^{2}} \]

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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giac [A]  time = 1.36, size = 48, normalized size = 0.91 \[ \frac {1}{16} \, d x^{2} + \frac {1}{8} \, c x - \frac {d \cos \left (4 \, b x + 4 \, a\right )}{128 \, b^{2}} - \frac {{\left (b d x + b c\right )} \sin \left (4 \, b x + 4 \, a\right )}{32 \, b^{2}} \]

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

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maple [B]  time = 0.02, size = 194, normalized size = 3.66 \[ \frac {\frac {d \left (\left (b x +a \right ) \left (-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{16}+\frac {\left (\sin ^{2}\left (b x +a \right )\right )}{16}-\left (b x +a \right ) \left (-\frac {\left (\sin ^{3}\left (b x +a \right )+\frac {3 \sin \left (b x +a \right )}{2}\right ) \cos \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}\right )-\frac {\left (\sin ^{4}\left (b x +a \right )\right )}{16}\right )}{b}-\frac {d a \left (-\frac {\left (\cos ^{3}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{4}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{8}+\frac {b x}{8}+\frac {a}{8}\right )}{b}+c \left (-\frac {\left (\cos ^{3}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{4}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{8}+\frac {b x}{8}+\frac {a}{8}\right )}{b} \]

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

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maxima [B]  time = 0.33, size = 96, normalized size = 1.81 \[ \frac {4 \, {\left (4 \, b x + 4 \, a - \sin \left (4 \, b x + 4 \, a\right )\right )} c - \frac {4 \, {\left (4 \, b x + 4 \, a - \sin \left (4 \, b x + 4 \, a\right )\right )} a d}{b} + \frac {{\left (8 \, {\left (b x + a\right )}^{2} - 4 \, {\left (b x + a\right )} \sin \left (4 \, b x + 4 \, a\right ) - \cos \left (4 \, b x + 4 \, a\right )\right )} d}{b}}{128 \, b} \]

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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mupad [B]  time = 1.04, size = 57, normalized size = 1.08 \[ \frac {c\,x}{8}+\frac {d\,x^2}{16}-\frac {d\,\cos \left (4\,a+4\,b\,x\right )}{128\,b^2}-\frac {c\,\sin \left (4\,a+4\,b\,x\right )}{32\,b}-\frac {d\,x\,\sin \left (4\,a+4\,b\,x\right )}{32\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.00, size = 238, normalized size = 4.49 \[ \begin {cases} \frac {c x \sin ^{4}{\left (a + b x \right )}}{8} + \frac {c x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac {c x \cos ^{4}{\left (a + b x \right )}}{8} + \frac {d x^{2} \sin ^{4}{\left (a + b x \right )}}{16} + \frac {d x^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{8} + \frac {d x^{2} \cos ^{4}{\left (a + b x \right )}}{16} + \frac {c \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{8 b} - \frac {c \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} + \frac {d x \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{8 b} - \frac {d x \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} - \frac {d \sin ^{4}{\left (a + b x \right )}}{32 b^{2}} - \frac {d \cos ^{4}{\left (a + b x \right )}}{32 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin ^{2}{\relax (a )} \cos ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*cos(b*x+a)**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)**4/(32*b**2) - d*cos(a + b*x)**4/(32*b**2), Ne(b, 0)), ((c*x + d*x*
*2/2)*sin(a)**2*cos(a)**2, True))

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