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

Optimal. Leaf size=25 \[ \frac{x}{2}-\frac{\sin (a+b x) \cos (a+b x)}{2 b} \]

[Out]

x/2 - (Cos[a + b*x]*Sin[a + b*x])/(2*b)

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Rubi [A]  time = 0.0086649, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2635, 8} \[ \frac{x}{2}-\frac{\sin (a+b x) \cos (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2,x]

[Out]

x/2 - (Cos[a + b*x]*Sin[a + b*x])/(2*b)

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

Mathematica [A]  time = 0.0256389, size = 23, normalized size = 0.92 \[ -\frac{\sin (2 (a+b x))-2 (a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^2,x]

[Out]

-(-2*(a + b*x) + Sin[2*(a + b*x)])/(4*b)

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Maple [A]  time = 0.005, size = 27, normalized size = 1.1 \begin{align*}{\frac{1}{b} \left ( -{\frac{\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{2}}+{\frac{bx}{2}}+{\frac{a}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.94602, size = 32, normalized size = 1.28 \begin{align*} \frac{2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.69346, size = 55, normalized size = 2.2 \begin{align*} \frac{b x - \cos \left (b x + a\right ) \sin \left (b x + a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.227452, size = 46, normalized size = 1.84 \begin{align*} \begin{cases} \frac{x \sin ^{2}{\left (a + b x \right )}}{2} + \frac{x \cos ^{2}{\left (a + b x \right )}}{2} - \frac{\sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b} & \text{for}\: b \neq 0 \\x \sin ^{2}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.06503, size = 24, normalized size = 0.96 \begin{align*} \frac{1}{2} \, x - \frac{\sin \left (2 \, b x + 2 \, a\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x - 1/4*sin(2*b*x + 2*a)/b