3.69 \(\int \sin ^2(c+d x) (a+b \sin ^2(c+d x)) \, dx\)

Optimal. Leaf size=61 \[ -\frac{(4 a+3 b) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x (4 a+3 b)-\frac{b \sin ^3(c+d x) \cos (c+d x)}{4 d} \]

[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)

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Rubi [A]  time = 0.0404287, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3014, 2635, 8} \[ -\frac{(4 a+3 b) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x (4 a+3 b)-\frac{b \sin ^3(c+d x) \cos (c+d x)}{4 d} \]

Antiderivative was successfully verified.

[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 3014

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]

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(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx &=-\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]  time = 0.0938271, size = 45, normalized size = 0.74 \[ \frac{4 (4 a+3 b) (c+d x)-8 (a+b) \sin (2 (c+d x))+b \sin (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.027, size = 65, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( b \left ( -{\frac{\cos \left ( dx+c \right ) }{4} \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\sin \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +a \left ( -{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.43521, size = 100, normalized size = 1.64 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.64944, size = 119, normalized size = 1.95 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 2.12638, size = 158, normalized size = 2.59 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.11987, size = 58, normalized size = 0.95 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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