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

Optimal. Leaf size=73 \[ -\frac {\sin (a+b x) \cos (a+b x)}{4 b^3}+\frac {x \cos ^2(a+b x)}{2 b^2}+\frac {x^2 \sin (a+b x) \cos (a+b x)}{2 b}-\frac {x}{4 b^2}+\frac {x^3}{6} \]

[Out]

-1/4*x/b^2+1/6*x^3+1/2*x*cos(b*x+a)^2/b^2-1/4*cos(b*x+a)*sin(b*x+a)/b^3+1/2*x^2*cos(b*x+a)*sin(b*x+a)/b

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Rubi [A]  time = 0.04, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3311, 30, 2635, 8} \[ \frac {x \cos ^2(a+b x)}{2 b^2}-\frac {\sin (a+b x) \cos (a+b x)}{4 b^3}+\frac {x^2 \sin (a+b x) \cos (a+b x)}{2 b}-\frac {x}{4 b^2}+\frac {x^3}{6} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cos[a + b*x]^2,x]

[Out]

-x/(4*b^2) + x^3/6 + (x*Cos[a + b*x]^2)/(2*b^2) - (Cos[a + b*x]*Sin[a + b*x])/(4*b^3) + (x^2*Cos[a + b*x]*Sin[
a + b*x])/(2*b)

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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]

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rubi steps

\begin {align*} \int x^2 \cos ^2(a+b x) \, dx &=\frac {x \cos ^2(a+b x)}{2 b^2}+\frac {x^2 \cos (a+b x) \sin (a+b x)}{2 b}+\frac {\int x^2 \, dx}{2}-\frac {\int \cos ^2(a+b x) \, dx}{2 b^2}\\ &=\frac {x^3}{6}+\frac {x \cos ^2(a+b x)}{2 b^2}-\frac {\cos (a+b x) \sin (a+b x)}{4 b^3}+\frac {x^2 \cos (a+b x) \sin (a+b x)}{2 b}-\frac {\int 1 \, dx}{4 b^2}\\ &=-\frac {x}{4 b^2}+\frac {x^3}{6}+\frac {x \cos ^2(a+b x)}{2 b^2}-\frac {\cos (a+b x) \sin (a+b x)}{4 b^3}+\frac {x^2 \cos (a+b x) \sin (a+b x)}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 47, normalized size = 0.64 \[ \frac {\left (6 b^2 x^2-3\right ) \sin (2 (a+b x))+6 b x \cos (2 (a+b x))+4 b^3 x^3}{24 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[a + b*x]^2,x]

[Out]

(4*b^3*x^3 + 6*b*x*Cos[2*(a + b*x)] + (-3 + 6*b^2*x^2)*Sin[2*(a + b*x)])/(24*b^3)

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fricas [A]  time = 0.44, size = 54, normalized size = 0.74 \[ \frac {2 \, b^{3} x^{3} + 6 \, b x \cos \left (b x + a\right )^{2} + 3 \, {\left (2 \, b^{2} x^{2} - 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 3 \, b x}{12 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*b^3*x^3 + 6*b*x*cos(b*x + a)^2 + 3*(2*b^2*x^2 - 1)*cos(b*x + a)*sin(b*x + a) - 3*b*x)/b^3

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giac [A]  time = 0.98, size = 45, normalized size = 0.62 \[ \frac {1}{6} \, x^{3} + \frac {x \cos \left (2 \, b x + 2 \, a\right )}{4 \, b^{2}} + \frac {{\left (2 \, b^{2} x^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 + 1/4*x*cos(2*b*x + 2*a)/b^2 + 1/8*(2*b^2*x^2 - 1)*sin(2*b*x + 2*a)/b^3

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maple [B]  time = 0.02, size = 158, normalized size = 2.16 \[ \frac {\left (\frac {b x}{2}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {a}{2}\right ) a^{2}-\frac {b x}{4}+\frac {\left (b x +a \right ) \left (\cos ^{2}\left (b x +a \right )\right )}{2}-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{4}-\frac {a}{4}-2 \left (-\frac {\left (\sin ^{2}\left (b x +a \right )\right )}{4}+\left (b x +a \right ) \left (\frac {b x}{2}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{4}\right ) a +\left (b x +a \right )^{2} \left (\frac {b x}{2}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{3}}{3}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 113, normalized size = 1.55 \[ \frac {4 \, {\left (b x + a\right )}^{3} + 6 \, {\left (2 \, b x + 2 \, a + \sin \left (2 \, b x + 2 \, a\right )\right )} a^{2} - 6 \, {\left (2 \, {\left (b x + a\right )}^{2} + 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right )\right )} a + 6 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) + 3 \, {\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )}{24 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(4*(b*x + a)^3 + 6*(2*b*x + 2*a + sin(2*b*x + 2*a))*a^2 - 6*(2*(b*x + a)^2 + 2*(b*x + a)*sin(2*b*x + 2*a)
+ cos(2*b*x + 2*a))*a + 6*(b*x + a)*cos(2*b*x + 2*a) + 3*(2*(b*x + a)^2 - 1)*sin(2*b*x + 2*a))/b^3

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mupad [B]  time = 0.10, size = 52, normalized size = 0.71 \[ \frac {x^3}{6}-\frac {\sin \left (2\,a+2\,b\,x\right )}{8\,b^3}+\frac {x\,\cos \left (2\,a+2\,b\,x\right )}{4\,b^2}+\frac {x^2\,\sin \left (2\,a+2\,b\,x\right )}{4\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos(a + b*x)^2,x)

[Out]

x^3/6 - sin(2*a + 2*b*x)/(8*b^3) + (x*cos(2*a + 2*b*x))/(4*b^2) + (x^2*sin(2*a + 2*b*x))/(4*b)

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sympy [A]  time = 1.11, size = 105, normalized size = 1.44 \[ \begin {cases} \frac {x^{3} \sin ^{2}{\left (a + b x \right )}}{6} + \frac {x^{3} \cos ^{2}{\left (a + b x \right )}}{6} + \frac {x^{2} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b} - \frac {x \sin ^{2}{\left (a + b x \right )}}{4 b^{2}} + \frac {x \cos ^{2}{\left (a + b x \right )}}{4 b^{2}} - \frac {\sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{4 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \cos ^{2}{\relax (a )}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**3*sin(a + b*x)**2/6 + x**3*cos(a + b*x)**2/6 + x**2*sin(a + b*x)*cos(a + b*x)/(2*b) - x*sin(a +
b*x)**2/(4*b**2) + x*cos(a + b*x)**2/(4*b**2) - sin(a + b*x)*cos(a + b*x)/(4*b**3), Ne(b, 0)), (x**3*cos(a)**2
/3, True))

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