\(\int \sin ^2(c+d x) (a+b \sin ^2(c+d x)) \, dx\) [69]

Optimal result

Integrand size = 21, antiderivative size = 61 \[ \int \sin ^2(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=\frac {1}{8} (4 a+3 b) x-\frac {(4 a+3 b) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b \cos (c+d x) \sin ^3(c+d x)}{4 d} \]

[Out]

1/8*(4*a+3*b)*x-1/8*(4*a+3*b)*cos(d*x+c)*sin(d*x+c)/d-1/4*b*cos(d*x+c)*sin(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3093, 2715, 8} \[ \int \sin ^2(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=-\frac {(4 a+3 b) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x (4 a+3 b)-\frac {b \sin ^3(c+d x) \cos (c+d x)}{4 d} \]

[In]

Int[Sin[c + d*x]^2*(a + b*Sin[c + d*x]^2),x]

[Out]

((4*a + 3*b)*x)/8 - ((4*a + 3*b)*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (b*Cos[c + d*x]*Sin[c + d*x]^3)/(4*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]

Rule 3093

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos
[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e +
f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {b \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{4} (4 a+3 b) \int \sin ^2(c+d x) \, dx \\ & = -\frac {(4 a+3 b) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{8} (4 a+3 b) \int 1 \, dx \\ & = \frac {1}{8} (4 a+3 b) x-\frac {(4 a+3 b) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b \cos (c+d x) \sin ^3(c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74 \[ \int \sin ^2(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=\frac {4 (4 a+3 b) (c+d x)-8 (a+b) \sin (2 (c+d x))+b \sin (4 (c+d x))}{32 d} \]

[In]

Integrate[Sin[c + d*x]^2*(a + b*Sin[c + d*x]^2),x]

[Out]

(4*(4*a + 3*b)*(c + d*x) - 8*(a + b)*Sin[2*(c + d*x)] + b*Sin[4*(c + d*x)])/(32*d)

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72

method	result	size
parallelrisch	\(\frac {\left (-8 a -8 b \right ) \sin \left (2 d x +2 c \right )+\sin \left (4 d x +4 c \right ) b +16 d \left (a +\frac {3 b}{4}\right ) x}{32 d}\)	\(44\)
risch	\(\frac {a x}{2}+\frac {3 b x}{8}+\frac {b \sin \left (4 d x +4 c \right )}{32 d}-\frac {\sin \left (2 d x +2 c \right ) a}{4 d}-\frac {\sin \left (2 d x +2 c \right ) b}{4 d}\)	\(55\)
derivativedivides	\(\frac {b \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(65\)
default	\(\frac {b \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(65\)
parts	\(\frac {a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(67\)
norman	\(\frac {\left (\frac {a}{2}+\frac {3 b}{8}\right ) x +\left (2 a +\frac {3 b}{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 a +\frac {3 b}{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a +\frac {9 b}{4}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {a}{2}+\frac {3 b}{8}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\left (4 a +3 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (4 a +3 b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {\left (4 a +11 b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (4 a +11 b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(197\)

[In]

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

[Out]

1/32*((-8*a-8*b)*sin(2*d*x+2*c)+sin(4*d*x+4*c)*b+16*d*(a+3/4*b)*x)/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.82 \[ \int \sin ^2(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=\frac {{\left (4 \, a + 3 \, b\right )} d x + {\left (2 \, b \cos \left (d x + c\right )^{3} - {\left (4 \, a + 5 \, b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

[In]

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

[Out]

1/8*((4*a + 3*b)*d*x + (2*b*cos(d*x + c)^3 - (4*a + 5*b)*cos(d*x + c))*sin(d*x + c))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (53) = 106\).

Time = 0.16 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.59 \[ \int \sin ^2(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=\begin {cases} \frac {a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a x \cos ^{2}{\left (c + d x \right )}}{2} - \frac {a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {3 b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 b x \cos ^{4}{\left (c + d x \right )}}{8} - \frac {5 b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {3 b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin ^{2}{\left (c \right )}\right ) \sin ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*x*sin(c + d*x)**2/2 + a*x*cos(c + d*x)**2/2 - a*sin(c + d*x)*cos(c + d*x)/(2*d) + 3*b*x*sin(c + d
*x)**4/8 + 3*b*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*b*x*cos(c + d*x)**4/8 - 5*b*sin(c + d*x)**3*cos(c + d*x
)/(8*d) - 3*b*sin(c + d*x)*cos(c + d*x)**3/(8*d), Ne(d, 0)), (x*(a + b*sin(c)**2)*sin(c)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.21 \[ \int \sin ^2(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=\frac {{\left (d x + c\right )} {\left (4 \, a + 3 \, b\right )} - \frac {{\left (4 \, a + 5 \, b\right )} \tan \left (d x + c\right )^{3} + {\left (4 \, a + 3 \, b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{8 \, d} \]

[In]

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

[Out]

1/8*((d*x + c)*(4*a + 3*b) - ((4*a + 5*b)*tan(d*x + c)^3 + (4*a + 3*b)*tan(d*x + c))/(tan(d*x + c)^4 + 2*tan(d
*x + c)^2 + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.70 \[ \int \sin ^2(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=\frac {1}{8} \, {\left (4 \, a + 3 \, b\right )} x + \frac {b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {{\left (a + b\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

[In]

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

[Out]

1/8*(4*a + 3*b)*x + 1/32*b*sin(4*d*x + 4*c)/d - 1/4*(a + b)*sin(2*d*x + 2*c)/d

Mupad [B] (verification not implemented)

Time = 13.83 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11 \[ \int \sin ^2(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx=x\,\left (\frac {a}{2}+\frac {3\,b}{8}\right )-\frac {\left (\frac {a}{2}+\frac {5\,b}{8}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (\frac {a}{2}+\frac {3\,b}{8}\right )\,\mathrm {tan}\left (c+d\,x\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4+2\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )} \]

[In]

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

[Out]

x*(a/2 + (3*b)/8) - (tan(c + d*x)^3*(a/2 + (5*b)/8) + tan(c + d*x)*(a/2 + (3*b)/8))/(d*(2*tan(c + d*x)^2 + tan
(c + d*x)^4 + 1))