3.60 \(\int \cos ^2(a+b x) \sin ^2(a+b x) \, dx\)

Optimal. Leaf size=46 \[ -\frac {\sin (a+b x) \cos ^3(a+b x)}{4 b}+\frac {\sin (a+b x) \cos (a+b x)}{8 b}+\frac {x}{8} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2568, 2635, 8} \[ -\frac {\sin (a+b x) \cos ^3(a+b x)}{4 b}+\frac {\sin (a+b x) \cos (a+b x)}{8 b}+\frac {x}{8} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Sin[a + b*x]^2,x]

[Out]

x/8 + (Cos[a + b*x]*Sin[a + b*x])/(8*b) - (Cos[a + b*x]^3*Sin[a + b*x])/(4*b)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rubi steps

\begin {align*} \int \cos ^2(a+b x) \sin ^2(a+b x) \, dx &=-\frac {\cos ^3(a+b x) \sin (a+b x)}{4 b}+\frac {1}{4} \int \cos ^2(a+b x) \, dx\\ &=\frac {\cos (a+b x) \sin (a+b x)}{8 b}-\frac {\cos ^3(a+b x) \sin (a+b x)}{4 b}+\frac {\int 1 \, dx}{8}\\ &=\frac {x}{8}+\frac {\cos (a+b x) \sin (a+b x)}{8 b}-\frac {\cos ^3(a+b x) \sin (a+b x)}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 23, normalized size = 0.50 \[ -\frac {\sin (4 (a+b x))-4 (a+b x)}{32 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*(-4*(a + b*x) + Sin[4*(a + b*x)])/b

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fricas [A]  time = 0.43, size = 36, normalized size = 0.78 \[ \frac {b x - {\left (2 \, \cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x - (2*cos(b*x + a)^3 - cos(b*x + a))*sin(b*x + a))/b

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giac [A]  time = 0.18, size = 18, normalized size = 0.39 \[ \frac {1}{8} \, x - \frac {\sin \left (4 \, b x + 4 \, a\right )}{32 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x - 1/32*sin(4*b*x + 4*a)/b

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maple [A]  time = 0.02, size = 43, normalized size = 0.93 \[ \frac {-\frac {\left (\cos ^{3}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{4}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{8}+\frac {b x}{8}+\frac {a}{8}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-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 [A]  time = 0.31, size = 24, normalized size = 0.52 \[ \frac {4 \, b x + 4 \, a - \sin \left (4 \, b x + 4 \, a\right )}{32 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(4*b*x + 4*a - sin(4*b*x + 4*a))/b

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mupad [B]  time = 0.46, size = 50, normalized size = 1.09 \[ \frac {x}{8}-\frac {\frac {\mathrm {tan}\left (a+b\,x\right )}{8}-\frac {{\mathrm {tan}\left (a+b\,x\right )}^3}{8}}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^4+2\,{\mathrm {tan}\left (a+b\,x\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2*sin(a + b*x)^2,x)

[Out]

x/8 - (tan(a + b*x)/8 - tan(a + b*x)^3/8)/(b*(2*tan(a + b*x)^2 + tan(a + b*x)^4 + 1))

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sympy [A]  time = 1.45, size = 92, normalized size = 2.00 \[ \begin {cases} \frac {x \sin ^{4}{\left (a + b x \right )}}{8} + \frac {x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac {x \cos ^{4}{\left (a + b x \right )}}{8} + \frac {\sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{8 b} - \frac {\sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\relax (a )} \cos ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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