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

Optimal. Leaf size=88 \[ -\frac {\sin (2 (a-c)+2 x (b-d))}{16 (b-d)}-\frac {\sin (2 (a+c)+2 x (b+d))}{16 (b+d)}-\frac {\sin (2 a+2 b x)}{8 b}+\frac {\sin (2 c+2 d x)}{8 d}+\frac {x}{4} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4574, 2637} \[ -\frac {\sin (2 (a-c)+2 x (b-d))}{16 (b-d)}-\frac {\sin (2 (a+c)+2 x (b+d))}{16 (b+d)}-\frac {\sin (2 a+2 b x)}{8 b}+\frac {\sin (2 c+2 d x)}{8 d}+\frac {x}{4} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Sin[a + b*x]^2,x]

[Out]

x/4 - Sin[2*a + 2*b*x]/(8*b) - Sin[2*(a - c) + 2*(b - d)*x]/(16*(b - d)) + Sin[2*c + 2*d*x]/(8*d) - Sin[2*(a +
c) + 2*(b + d)*x]/(16*(b + d))

Rule 2637

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

Rule 4574

Int[Cos[w_]^(q_.)*Sin[v_]^(p_.), x_Symbol] :> Int[ExpandTrigReduce[Sin[v]^p*Cos[w]^q, x], x] /; IGtQ[p, 0] &&
IGtQ[q, 0] && ((PolynomialQ[v, x] && PolynomialQ[w, x]) || (BinomialQ[{v, w}, x] && IndependentQ[Cancel[v/w],
x]))

Rubi steps

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

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Mathematica [A]  time = 0.73, size = 108, normalized size = 1.23 \[ \frac {\left (2 d^3-2 b^2 d\right ) \sin (2 (a+b x))-b d (b+d) \sin (2 (a+x (b-d)-c))+b (b-d) (-d \sin (2 (a+x (b+d)+c))+2 (b+d) \sin (2 (c+d x))+4 d x (b+d))}{16 b d (b-d) (b+d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b^2*d + 2*d^3)*Sin[2*(a + b*x)] - b*d*(b + d)*Sin[2*(a - c + (b - d)*x)] + b*(b - d)*(4*d*(b + d)*x + 2*(
b + d)*Sin[2*(c + d*x)] - d*Sin[2*(a + c + (b + d)*x)]))/(16*b*(b - d)*d*(b + d))

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fricas [A]  time = 0.44, size = 108, normalized size = 1.23 \[ \frac {{\left (2 \, b d^{2} \cos \left (b x + a\right )^{2} + b^{3} - 2 \, b d^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (b^{3} d - b d^{3}\right )} x - {\left (2 \, b^{2} d \cos \left (b x + a\right ) \cos \left (d x + c\right )^{2} - d^{3} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{4 \, {\left (b^{3} d - b d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 80, normalized size = 0.91 \[ \frac {1}{4} \, x - \frac {\sin \left (2 \, b x + 2 \, d x + 2 \, a + 2 \, c\right )}{16 \, {\left (b + d\right )}} - \frac {\sin \left (2 \, b x - 2 \, d x + 2 \, a - 2 \, c\right )}{16 \, {\left (b - d\right )}} - \frac {\sin \left (2 \, b x + 2 \, a\right )}{8 \, b} + \frac {\sin \left (2 \, d x + 2 \, c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.23, size = 83, normalized size = 0.94 \[ \frac {x}{4}-\frac {\sin \left (2 b x +2 a \right )}{8 b}+\frac {\sin \left (2 d x +2 c \right )}{8 d}-\frac {\sin \left (\left (2 b -2 d \right ) x +2 a -2 c \right )}{16 \left (b -d \right )}-\frac {\sin \left (\left (2 b +2 d \right ) x +2 a +2 c \right )}{16 \left (b +d \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.38, size = 620, normalized size = 7.05 \[ \frac {8 \, {\left ({\left (b \cos \left (2 \, c\right )^{2} + b \sin \left (2 \, c\right )^{2}\right )} d^{3} - {\left (b^{3} \cos \left (2 \, c\right )^{2} + b^{3} \sin \left (2 \, c\right )^{2}\right )} d\right )} x - {\left (b^{2} d \sin \left (2 \, c\right ) - b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (2 \, {\left (b + d\right )} x + 2 \, a + 4 \, c\right ) + {\left (b^{2} d \sin \left (2 \, c\right ) - b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (2 \, {\left (b + d\right )} x + 2 \, a\right ) + {\left (b^{2} d \sin \left (2 \, c\right ) + b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (-2 \, {\left (b - d\right )} x - 2 \, a + 4 \, c\right ) - {\left (b^{2} d \sin \left (2 \, c\right ) + b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (-2 \, {\left (b - d\right )} x - 2 \, a\right ) - 2 \, {\left (b^{2} d \sin \left (2 \, c\right ) - d^{3} \sin \left (2 \, c\right )\right )} \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) + 2 \, {\left (b^{2} d \sin \left (2 \, c\right ) - d^{3} \sin \left (2 \, c\right )\right )} \cos \left (2 \, b x + 2 \, a - 2 \, c\right ) - 2 \, {\left (b^{3} \sin \left (2 \, c\right ) - b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (2 \, d x\right ) + 2 \, {\left (b^{3} \sin \left (2 \, c\right ) - b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (2 \, d x + 4 \, c\right ) + {\left (b^{2} d \cos \left (2 \, c\right ) - b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (2 \, {\left (b + d\right )} x + 2 \, a + 4 \, c\right ) + {\left (b^{2} d \cos \left (2 \, c\right ) - b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (2 \, {\left (b + d\right )} x + 2 \, a\right ) - {\left (b^{2} d \cos \left (2 \, c\right ) + b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (-2 \, {\left (b - d\right )} x - 2 \, a + 4 \, c\right ) - {\left (b^{2} d \cos \left (2 \, c\right ) + b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (-2 \, {\left (b - d\right )} x - 2 \, a\right ) + 2 \, {\left (b^{2} d \cos \left (2 \, c\right ) - d^{3} \cos \left (2 \, c\right )\right )} \sin \left (2 \, b x + 2 \, a + 2 \, c\right ) + 2 \, {\left (b^{2} d \cos \left (2 \, c\right ) - d^{3} \cos \left (2 \, c\right )\right )} \sin \left (2 \, b x + 2 \, a - 2 \, c\right ) - 2 \, {\left (b^{3} \cos \left (2 \, c\right ) - b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (2 \, d x\right ) - 2 \, {\left (b^{3} \cos \left (2 \, c\right ) - b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (2 \, d x + 4 \, c\right )}{32 \, {\left ({\left (b \cos \left (2 \, c\right )^{2} + b \sin \left (2 \, c\right )^{2}\right )} d^{3} - {\left (b^{3} \cos \left (2 \, c\right )^{2} + b^{3} \sin \left (2 \, c\right )^{2}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(8*((b*cos(2*c)^2 + b*sin(2*c)^2)*d^3 - (b^3*cos(2*c)^2 + b^3*sin(2*c)^2)*d)*x - (b^2*d*sin(2*c) - b*d^2*
sin(2*c))*cos(2*(b + d)*x + 2*a + 4*c) + (b^2*d*sin(2*c) - b*d^2*sin(2*c))*cos(2*(b + d)*x + 2*a) + (b^2*d*sin
(2*c) + b*d^2*sin(2*c))*cos(-2*(b - d)*x - 2*a + 4*c) - (b^2*d*sin(2*c) + b*d^2*sin(2*c))*cos(-2*(b - d)*x - 2
*a) - 2*(b^2*d*sin(2*c) - d^3*sin(2*c))*cos(2*b*x + 2*a + 2*c) + 2*(b^2*d*sin(2*c) - d^3*sin(2*c))*cos(2*b*x +
2*a - 2*c) - 2*(b^3*sin(2*c) - b*d^2*sin(2*c))*cos(2*d*x) + 2*(b^3*sin(2*c) - b*d^2*sin(2*c))*cos(2*d*x + 4*c
) + (b^2*d*cos(2*c) - b*d^2*cos(2*c))*sin(2*(b + d)*x + 2*a + 4*c) + (b^2*d*cos(2*c) - b*d^2*cos(2*c))*sin(2*(
b + d)*x + 2*a) - (b^2*d*cos(2*c) + b*d^2*cos(2*c))*sin(-2*(b - d)*x - 2*a + 4*c) - (b^2*d*cos(2*c) + b*d^2*co
s(2*c))*sin(-2*(b - d)*x - 2*a) + 2*(b^2*d*cos(2*c) - d^3*cos(2*c))*sin(2*b*x + 2*a + 2*c) + 2*(b^2*d*cos(2*c)
- d^3*cos(2*c))*sin(2*b*x + 2*a - 2*c) - 2*(b^3*cos(2*c) - b*d^2*cos(2*c))*sin(2*d*x) - 2*(b^3*cos(2*c) - b*d
^2*cos(2*c))*sin(2*d*x + 4*c))/((b*cos(2*c)^2 + b*sin(2*c)^2)*d^3 - (b^3*cos(2*c)^2 + b^3*sin(2*c)^2)*d)

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mupad [B]  time = 0.93, size = 177, normalized size = 2.01 \[ -\frac {b\,d^2\,\sin \left (2\,a-2\,c+2\,b\,x-2\,d\,x\right )-2\,b^3\,\sin \left (2\,c+2\,d\,x\right )-2\,d^3\,\sin \left (2\,a+2\,b\,x\right )-b\,d^2\,\sin \left (2\,a+2\,c+2\,b\,x+2\,d\,x\right )+b^2\,d\,\sin \left (2\,a-2\,c+2\,b\,x-2\,d\,x\right )+b^2\,d\,\sin \left (2\,a+2\,c+2\,b\,x+2\,d\,x\right )+2\,b^2\,d\,\sin \left (2\,a+2\,b\,x\right )+2\,b\,d^2\,\sin \left (2\,c+2\,d\,x\right )+4\,b\,d^3\,x-4\,b^3\,d\,x}{16\,b\,d\,\left (b^2-d^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*sin(a + b*x)^2,x)

[Out]

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

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sympy [A]  time = 22.98, size = 1027, normalized size = 11.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*sin(a)**2*cos(c)**2, Eq(b, 0) & Eq(d, 0)), ((x*sin(c + d*x)**2/2 + x*cos(c + d*x)**2/2 + sin(c +
d*x)*cos(c + d*x)/(2*d))*sin(a)**2, Eq(b, 0)), (x*sin(a - d*x)**2*sin(c + d*x)**2/8 + 3*x*sin(a - d*x)**2*cos(
c + d*x)**2/8 + x*sin(a - d*x)*sin(c + d*x)*cos(a - d*x)*cos(c + d*x)/2 + 3*x*sin(c + d*x)**2*cos(a - d*x)**2/
8 + x*cos(a - d*x)**2*cos(c + d*x)**2/8 + sin(a - d*x)**2*sin(c + d*x)*cos(c + d*x)/(8*d) + sin(a - d*x)*cos(a
- d*x)*cos(c + d*x)**2/(2*d) + 3*sin(c + d*x)*cos(a - d*x)**2*cos(c + d*x)/(8*d), Eq(b, -d)), (x*sin(a + d*x)
**2*sin(c + d*x)**2/8 + 3*x*sin(a + d*x)**2*cos(c + d*x)**2/8 - x*sin(a + d*x)*sin(c + d*x)*cos(a + d*x)*cos(c
+ d*x)/2 + 3*x*sin(c + d*x)**2*cos(a + d*x)**2/8 + x*cos(a + d*x)**2*cos(c + d*x)**2/8 + sin(a + d*x)*sin(c +
d*x)**2*cos(a + d*x)/(8*d) - 5*sin(a + d*x)*cos(a + d*x)*cos(c + d*x)**2/(8*d) + sin(c + d*x)*cos(a + d*x)**2
*cos(c + d*x)/(2*d), Eq(b, d)), ((x*sin(a + b*x)**2/2 + x*cos(a + b*x)**2/2 - sin(a + b*x)*cos(a + b*x)/(2*b))
*cos(c)**2, Eq(d, 0)), (b**3*d*x*sin(a + b*x)**2*sin(c + d*x)**2/(4*b**3*d - 4*b*d**3) + b**3*d*x*sin(a + b*x)
**2*cos(c + d*x)**2/(4*b**3*d - 4*b*d**3) + b**3*d*x*sin(c + d*x)**2*cos(a + b*x)**2/(4*b**3*d - 4*b*d**3) + b
**3*d*x*cos(a + b*x)**2*cos(c + d*x)**2/(4*b**3*d - 4*b*d**3) + b**3*sin(a + b*x)**2*sin(c + d*x)*cos(c + d*x)
/(4*b**3*d - 4*b*d**3) + b**3*sin(c + d*x)*cos(a + b*x)**2*cos(c + d*x)/(4*b**3*d - 4*b*d**3) - 2*b**2*d*sin(a
+ b*x)*cos(a + b*x)*cos(c + d*x)**2/(4*b**3*d - 4*b*d**3) - b*d**3*x*sin(a + b*x)**2*sin(c + d*x)**2/(4*b**3*
d - 4*b*d**3) - b*d**3*x*sin(a + b*x)**2*cos(c + d*x)**2/(4*b**3*d - 4*b*d**3) - b*d**3*x*sin(c + d*x)**2*cos(
a + b*x)**2/(4*b**3*d - 4*b*d**3) - b*d**3*x*cos(a + b*x)**2*cos(c + d*x)**2/(4*b**3*d - 4*b*d**3) - 2*b*d**2*
sin(a + b*x)**2*sin(c + d*x)*cos(c + d*x)/(4*b**3*d - 4*b*d**3) + d**3*sin(a + b*x)*sin(c + d*x)**2*cos(a + b*
x)/(4*b**3*d - 4*b*d**3) + d**3*sin(a + b*x)*cos(a + b*x)*cos(c + d*x)**2/(4*b**3*d - 4*b*d**3), True))

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