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

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

[Out]

-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 = 0.077704, 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 \[ \frac{\sin (a+b x) \cos (a+b x)}{4 b^3}+\frac{x \sin ^2(a+b x)}{2 b^2}-\frac{x^2 \sin (a+b x) \cos (a+b x)}{2 b}-\frac{x}{4 b^2}+\frac{x^3}{6} \]

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)

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Rubi in Sympy [A]  time = 2.62825, size = 65, normalized size = 0.89 \[ \frac{x^{3}}{6} - \frac{x^{2} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b} + \frac{x \sin ^{2}{\left (a + b x \right )}}{2 b^{2}} - \frac{x}{4 b^{2}} + \frac{\sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{4 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**2*sin(b*x+a)**2,x)

[Out]

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

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Mathematica [A]  time = 0.133366, 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 = 0.004, size = 158, normalized size = 2.2 \[{\frac{1}{{b}^{3}} \left ( \left ( bx+a \right ) ^{2} \left ( -{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{2}}+{\frac{bx}{2}}+{\frac{a}{2}} \right ) -{\frac{ \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{2}}+{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{4}}+{\frac{bx}{4}}+{\frac{a}{4}}-{\frac{ \left ( bx+a \right ) ^{3}}{3}}-2\,a \left ( \left ( bx+a \right ) \left ( -1/2\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) +1/2\,bx+a/2 \right ) -1/4\, \left ( bx+a \right ) ^{2}+1/4\, \left ( \sin \left ( bx+a \right ) \right ) ^{2} \right ) +{a}^{2} \left ( -{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{2}}+{\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) } \]

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*co
s(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 = 1.40111, size = 158, normalized size = 2.16 \[ \frac{4 \,{\left (b x + a\right )}^{3} + 6 \,{\left (2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )\right )} a^{2} - 6 \,{\left (2 \,{\left (b x + a\right )}^{2} - 2 \,{\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) - \cos \left (2 \, b x + 2 \, a\right )\right )} a - 6 \,{\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - 3 \,{\left (2 \,{\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )}{24 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^2*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 = 0.240634, size = 73, normalized size = 1. \[ \frac{2 \, b^{3} x^{3} - 6 \, b x \cos \left (b x + a\right )^{2} - 3 \,{\left (2 \, b^{2} x^{2} - 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 3 \, b x}{12 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^2*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 = 1.52803, size = 105, normalized size = 1.44 \[ \begin{cases} \frac{x^{3} \sin ^{2}{\left (a + b x \right )}}{6} + \frac{x^{3} \cos ^{2}{\left (a + b x \right )}}{6} - \frac{x^{2} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b} + \frac{x \sin ^{2}{\left (a + b x \right )}}{4 b^{2}} - \frac{x \cos ^{2}{\left (a + b x \right )}}{4 b^{2}} + \frac{\sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{4 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \sin ^{2}{\left (a \right )}}{3} & \text{otherwise} \end{cases} \]

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)*c
os(a + b*x)/(2*b) + x*sin(a + b*x)**2/(4*b**2) - x*cos(a + b*x)**2/(4*b**2) + si
n(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/XCAS [A]  time = 0.199986, size = 61, normalized size = 0.84 \[ \frac{1}{6} \, x^{3} - \frac{x \cos \left (2 \, b x + 2 \, a\right )}{4 \, b^{2}} - \frac{{\left (2 \, b^{2} x^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^2*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