\(\int (a+b \sin ^2(c+d x)) \, dx\) [70]

Optimal result

Integrand size = 12, antiderivative size = 30 \[ \int \left (a+b \sin ^2(c+d x)\right ) \, dx=a x+\frac {b x}{2}-\frac {b \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2715, 8} \[ \int \left (a+b \sin ^2(c+d x)\right ) \, dx=a x-\frac {b \sin (c+d x) \cos (c+d x)}{2 d}+\frac {b x}{2} \]

[In]

Int[a + b*Sin[c + d*x]^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \left (a+b \sin ^2(c+d x)\right ) \, dx=a x+\frac {b (c+d x)}{2 d}-\frac {b \sin (2 (c+d x))}{4 d} \]

[In]

Integrate[a + b*Sin[c + d*x]^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80

method	result	size
risch	\(a x +\frac {b x}{2}-\frac {\sin \left (2 d x +2 c \right ) b}{4 d}\)	\(24\)
parallelrisch	\(\frac {b \left (2 d x -\sin \left (2 d x +2 c \right )\right )}{4 d}+a x\)	\(27\)
default	\(a x +\frac {b \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(32\)
parts	\(a x +\frac {b \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(32\)
derivativedivides	\(\frac {\left (d x +c \right ) a +b \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(37\)
norman	\(\frac {\left (a +\frac {b}{2}\right ) x +\frac {b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\left (a +\frac {b}{2}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 a +b \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(92\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \left (a+b \sin ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, a + b\right )} d x - b \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \left (a+b \sin ^2(c+d x)\right ) \, dx=a x + b \left (\begin {cases} \frac {x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {x \cos ^{2}{\left (c + d x \right )}}{2} - \frac {\sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \sin ^{2}{\left (c \right )} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \left (a+b \sin ^2(c+d x)\right ) \, dx=a x + \frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} b}{4 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \left (a+b \sin ^2(c+d x)\right ) \, dx=\frac {1}{4} \, b {\left (2 \, x - \frac {\sin \left (2 \, d x + 2 \, c\right )}{d}\right )} + a x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.51 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \left (a+b \sin ^2(c+d x)\right ) \, dx=-\frac {\frac {b\,\sin \left (2\,c+2\,d\,x\right )}{4}-d\,x\,\left (a+\frac {b}{2}\right )}{d} \]

[In]

int(a + b*sin(c + d*x)^2,x)

[Out]

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