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

Optimal result

Integrand size = 8, antiderivative size = 25 \[ \int \sin ^2(a+b x) \, dx=\frac {x}{2}-\frac {\cos (a+b x) \sin (a+b x)}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2715, 8} \[ \int \sin ^2(a+b x) \, dx=\frac {x}{2}-\frac {\sin (a+b x) \cos (a+b x)}{2 b} \]

[In]

Int[Sin[a + b*x]^2,x]

[Out]

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 2715

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] && Integ
erQ[2*n]

Rubi steps \begin{align*} \text {integral}& = -\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] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \sin ^2(a+b x) \, dx=-\frac {-2 (a+b x)+\sin (2 (a+b x))}{4 b} \]

[In]

Integrate[Sin[a + b*x]^2,x]

[Out]

-1/4*(-2*(a + b*x) + Sin[2*(a + b*x)])/b

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {x}{2}-\frac {\sin \left (2 b x +2 a \right )}{4 b}\)	\(19\)
parallelrisch	\(\frac {2 b x -\sin \left (2 b x +2 a \right )}{4 b}\)	\(22\)
derivativedivides	\(\frac {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}}{b}\)	\(27\)
default	\(\frac {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}}{b}\)	\(27\)
norman	\(\frac {\frac {\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )}{b}+x \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\frac {x}{2}-\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{b}+\frac {x \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{2}}\)	\(77\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \sin ^2(a+b x) \, dx=\frac {b x - \cos \left (b x + a\right ) \sin \left (b x + a\right )}{2 \, b} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (19) = 38\).

Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \sin ^2(a+b x) \, dx=\begin {cases} \frac {x \sin ^{2}{\left (a + b x \right )}}{2} + \frac {x \cos ^{2}{\left (a + b x \right )}}{2} - \frac {\sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \sin ^2(a+b x) \, dx=\frac {2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )}{4 \, b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \sin ^2(a+b x) \, dx=\frac {1}{2} \, x - \frac {\sin \left (2 \, b x + 2 \, a\right )}{4 \, b} \]

[In]

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

[Out]

1/2*x - 1/4*sin(2*b*x + 2*a)/b

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \sin ^2(a+b x) \, dx=\frac {x}{2}-\frac {\sin \left (2\,a+2\,b\,x\right )}{4\,b} \]

[In]

int(sin(a + b*x)^2,x)

[Out]

x/2 - sin(2*a + 2*b*x)/(4*b)