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

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

[Out]

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

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Rubi [A]  time = 0.0432089, antiderivative size = 73, normalized size of antiderivative = 1., 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} \[ \frac{x \sin ^2(a+b x)}{2 b^2}+\frac{\sin (a+b x) \cos (a+b x)}{4 b^3}-\frac{x^2 \sin (a+b x) \cos (a+b x)}{2 b}-\frac{x}{4 b^2}+\frac{x^3}{6} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sin[a + b*x]^2,x]

[Out]

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

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]

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 8

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

Rubi steps

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

Mathematica [A]  time = 0.128293, size = 47, normalized size = 0.64 \[ \frac{\left (3-6 b^2 x^2\right ) \sin (2 (a+b x))-6 b x \cos (2 (a+b x))+4 b^3 x^3}{24 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sin[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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Maple [B]  time = 0.005, size = 158, normalized size = 2.2 \begin{align*}{\frac{1}{{b}^{3}} \left ( \left ( bx+a \right ) ^{2} \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 ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{2}}+{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{4}}+{\frac{bx}{4}}+{\frac{a}{4}}-{\frac{ \left ( bx+a \right ) ^{3}}{3}}-2\,a \left ( \left ( bx+a \right ) \left ( -1/2\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) +1/2\,bx+a/2 \right ) -1/4\, \left ( bx+a \right ) ^{2}+1/4\, \left ( \sin \left ( bx+a \right ) \right ) ^{2} \right ) +{a}^{2} \left ( -{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{2}}+{\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.965575, size = 158, normalized size = 2.16 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sin(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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Fricas [A]  time = 1.77741, size = 134, normalized size = 1.84 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sin(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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Sympy [A]  time = 1.10156, size = 105, normalized size = 1.44 \begin{align*} \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} \sin ^{2}{\left (a \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*sin(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*sin(a)**2
/3, True))

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Giac [A]  time = 1.10103, size = 61, normalized size = 0.84 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sin(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