3.276 \(\int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=103 \[ -\frac{a^2 \cos ^5(c+d x)}{10 d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac{3 a^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{3 a^2 x}{16}-\frac{\cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 a d} \]

[Out]

(3*a^2*x)/16 - (a^2*Cos[c + d*x]^5)/(10*d) + (3*a^2*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (a^2*Cos[c + d*x]^3*Si
n[c + d*x])/(8*d) - (Cos[c + d*x]^3*(a + a*Sin[c + d*x])^3)/(6*a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.167851, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2870, 2669, 2635, 8} \[ -\frac{a^2 \cos ^5(c+d x)}{10 d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac{3 a^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{3 a^2 x}{16}-\frac{\cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*x)/16 - (a^2*Cos[c + d*x]^5)/(10*d) + (3*a^2*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (a^2*Cos[c + d*x]^3*Si
n[c + d*x])/(8*d) - (Cos[c + d*x]^3*(a + a*Sin[c + d*x])^3)/(6*a*d)

Rule 2870

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> -Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(2*b*f*g*(m + 1)), x] + Dist[a/(2
*g^2), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] && EqQ[m - p, 0]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

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 \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx &=-\frac{\cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 a d}+\frac{1}{2} a \int \cos ^4(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac{a^2 \cos ^5(c+d x)}{10 d}-\frac{\cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 a d}+\frac{1}{2} a^2 \int \cos ^4(c+d x) \, dx\\ &=-\frac{a^2 \cos ^5(c+d x)}{10 d}+\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac{\cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 a d}+\frac{1}{8} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{a^2 \cos ^5(c+d x)}{10 d}+\frac{3 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac{\cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 a d}+\frac{1}{16} \left (3 a^2\right ) \int 1 \, dx\\ &=\frac{3 a^2 x}{16}-\frac{a^2 \cos ^5(c+d x)}{10 d}+\frac{3 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac{\cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 a d}\\ \end{align*}

Mathematica [A]  time = 0.429726, size = 76, normalized size = 0.74 \[ \frac{a^2 (-15 \sin (2 (c+d x))-45 \sin (4 (c+d x))+5 \sin (6 (c+d x))-240 \cos (c+d x)-40 \cos (3 (c+d x))+24 \cos (5 (c+d x))+180 c+180 d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(180*c + 180*d*x - 240*Cos[c + d*x] - 40*Cos[3*(c + d*x)] + 24*Cos[5*(c + d*x)] - 15*Sin[2*(c + d*x)] - 4
5*Sin[4*(c + d*x)] + 5*Sin[6*(c + d*x)]))/(960*d)

________________________________________________________________________________________

Maple [A]  time = 0.034, size = 142, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{6}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{8}}+{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{16}}+{\frac{dx}{16}}+{\frac{c}{16}} \right ) +2\,{a}^{2} \left ( -1/5\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-2/15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3} \right ) +{a}^{2} \left ( -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{4}}+{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8}}+{\frac{dx}{8}}+{\frac{c}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*cos(d*x+c)^3*sin(d*x+c)+1/16*cos(d*x+c)*sin(d*x+c)+1/16*d*x+1/16*
c)+2*a^2*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)+a^2*(-1/4*cos(d*x+c)^3*sin(d*x+c)+1/8*cos(d*x+c)*s
in(d*x+c)+1/8*d*x+1/8*c))

________________________________________________________________________________________

Maxima [A]  time = 1.13309, size = 126, normalized size = 1.22 \begin{align*} \frac{128 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} - 5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} + 30 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(128*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^2 - 5*(4*sin(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x +
4*c))*a^2 + 30*(4*d*x + 4*c - sin(4*d*x + 4*c))*a^2)/d

________________________________________________________________________________________

Fricas [A]  time = 1.65558, size = 211, normalized size = 2.05 \begin{align*} \frac{96 \, a^{2} \cos \left (d x + c\right )^{5} - 160 \, a^{2} \cos \left (d x + c\right )^{3} + 45 \, a^{2} d x + 5 \,{\left (8 \, a^{2} \cos \left (d x + c\right )^{5} - 26 \, a^{2} \cos \left (d x + c\right )^{3} + 9 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(96*a^2*cos(d*x + c)^5 - 160*a^2*cos(d*x + c)^3 + 45*a^2*d*x + 5*(8*a^2*cos(d*x + c)^5 - 26*a^2*cos(d*x
+ c)^3 + 9*a^2*cos(d*x + c))*sin(d*x + c))/d

________________________________________________________________________________________

Sympy [A]  time = 11.8154, size = 309, normalized size = 3. \begin{align*} \begin{cases} \frac{a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{3 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{a^{2} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} - \frac{a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac{a^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{2 a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{4 a^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{2} \sin ^{2}{\left (c \right )} \cos ^{2}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x*sin(c + d*x)**6/16 + 3*a**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + a**2*x*sin(c + d*x)**4/8
+ 3*a**2*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + a**2*x*cos(c + d*x)
**6/16 + a**2*x*cos(c + d*x)**4/8 + a**2*sin(c + d*x)**5*cos(c + d*x)/(16*d) - a**2*sin(c + d*x)**3*cos(c + d*
x)**3/(6*d) + a**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) - 2*a**2*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - a**2*si
n(c + d*x)*cos(c + d*x)**5/(16*d) - a**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) - 4*a**2*cos(c + d*x)**5/(15*d), N
e(d, 0)), (x*(a*sin(c) + a)**2*sin(c)**2*cos(c)**2, True))

________________________________________________________________________________________

Giac [A]  time = 1.34954, size = 143, normalized size = 1.39 \begin{align*} \frac{3}{16} \, a^{2} x + \frac{a^{2} \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac{a^{2} \cos \left (3 \, d x + 3 \, c\right )}{24 \, d} - \frac{a^{2} \cos \left (d x + c\right )}{4 \, d} + \frac{a^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{3 \, a^{2} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac{a^{2} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*a^2*x + 1/40*a^2*cos(5*d*x + 5*c)/d - 1/24*a^2*cos(3*d*x + 3*c)/d - 1/4*a^2*cos(d*x + c)/d + 1/192*a^2*si
n(6*d*x + 6*c)/d - 3/64*a^2*sin(4*d*x + 4*c)/d - 1/64*a^2*sin(2*d*x + 2*c)/d