3.225 \(\int \sin ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx\)

Optimal. Leaf size=121 \[ \frac{2 a^3 A \cos ^5(c+d x)}{5 d}-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \sin ^5(c+d x) \cos (c+d x)}{6 d}+\frac{5 a^3 A \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac{3 a^3 A \sin (c+d x) \cos (c+d x)}{16 d}+\frac{3}{16} a^3 A x \]

[Out]

(3*a^3*A*x)/16 - (2*a^3*A*Cos[c + d*x]^3)/(3*d) + (2*a^3*A*Cos[c + d*x]^5)/(5*d) - (3*a^3*A*Cos[c + d*x]*Sin[c
+ d*x])/(16*d) + (5*a^3*A*Cos[c + d*x]*Sin[c + d*x]^3)/(24*d) + (a^3*A*Cos[c + d*x]*Sin[c + d*x]^5)/(6*d)

________________________________________________________________________________________

Rubi [A]  time = 0.168042, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2966, 2635, 8, 2633} \[ \frac{2 a^3 A \cos ^5(c+d x)}{5 d}-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \sin ^5(c+d x) \cos (c+d x)}{6 d}+\frac{5 a^3 A \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac{3 a^3 A \sin (c+d x) \cos (c+d x)}{16 d}+\frac{3}{16} a^3 A x \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

(3*a^3*A*x)/16 - (2*a^3*A*Cos[c + d*x]^3)/(3*d) + (2*a^3*A*Cos[c + d*x]^5)/(5*d) - (3*a^3*A*Cos[c + d*x]*Sin[c
+ d*x])/(16*d) + (5*a^3*A*Cos[c + d*x]*Sin[c + d*x]^3)/(24*d) + (a^3*A*Cos[c + d*x]*Sin[c + d*x]^5)/(6*d)

Rule 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

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]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \sin ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx &=\int \left (a^3 A \sin ^2(c+d x)+2 a^3 A \sin ^3(c+d x)-2 a^3 A \sin ^5(c+d x)-a^3 A \sin ^6(c+d x)\right ) \, dx\\ &=\left (a^3 A\right ) \int \sin ^2(c+d x) \, dx-\left (a^3 A\right ) \int \sin ^6(c+d x) \, dx+\left (2 a^3 A\right ) \int \sin ^3(c+d x) \, dx-\left (2 a^3 A\right ) \int \sin ^5(c+d x) \, dx\\ &=-\frac{a^3 A \cos (c+d x) \sin (c+d x)}{2 d}+\frac{a^3 A \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac{1}{2} \left (a^3 A\right ) \int 1 \, dx-\frac{1}{6} \left (5 a^3 A\right ) \int \sin ^4(c+d x) \, dx-\frac{\left (2 a^3 A\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (2 a^3 A\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{1}{2} a^3 A x-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{2 a^3 A \cos ^5(c+d x)}{5 d}-\frac{a^3 A \cos (c+d x) \sin (c+d x)}{2 d}+\frac{5 a^3 A \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac{a^3 A \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac{1}{8} \left (5 a^3 A\right ) \int \sin ^2(c+d x) \, dx\\ &=\frac{1}{2} a^3 A x-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{2 a^3 A \cos ^5(c+d x)}{5 d}-\frac{3 a^3 A \cos (c+d x) \sin (c+d x)}{16 d}+\frac{5 a^3 A \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac{a^3 A \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac{1}{16} \left (5 a^3 A\right ) \int 1 \, dx\\ &=\frac{3}{16} a^3 A x-\frac{2 a^3 A \cos ^3(c+d x)}{3 d}+\frac{2 a^3 A \cos ^5(c+d x)}{5 d}-\frac{3 a^3 A \cos (c+d x) \sin (c+d x)}{16 d}+\frac{5 a^3 A \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac{a^3 A \cos (c+d x) \sin ^5(c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.107905, size = 77, normalized size = 0.64 \[ \frac{a^3 A (-15 \sin (2 (c+d x))-45 \sin (4 (c+d x))+5 \sin (6 (c+d x))-240 \cos (c+d x)-40 \cos (3 (c+d x))+24 \cos (5 (c+d x))+180 c+180 d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

(a^3*A*(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)] -
45*Sin[4*(c + d*x)] + 5*Sin[6*(c + d*x)]))/(960*d)

________________________________________________________________________________________

Maple [A]  time = 0.032, size = 136, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( -{a}^{3}A \left ( -{\frac{\cos \left ( dx+c \right ) }{6} \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +{\frac{2\,{a}^{3}A\cos \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }-{\frac{2\,{a}^{3}A \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}}+{a}^{3}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+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.955482, size = 186, normalized size = 1.54 \begin{align*} \frac{128 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} A a^{3} + 640 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} A a^{3} - 5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 240 \,{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/960*(128*(3*cos(d*x + c)^5 - 10*cos(d*x + c)^3 + 15*cos(d*x + c))*A*a^3 + 640*(cos(d*x + c)^3 - 3*cos(d*x +
c))*A*a^3 - 5*(4*sin(2*d*x + 2*c)^3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*A*a^3 + 240*(2
*d*x + 2*c - sin(2*d*x + 2*c))*A*a^3)/d

________________________________________________________________________________________

Fricas [A]  time = 1.95713, size = 227, normalized size = 1.88 \begin{align*} \frac{96 \, A a^{3} \cos \left (d x + c\right )^{5} - 160 \, A a^{3} \cos \left (d x + c\right )^{3} + 45 \, A a^{3} d x + 5 \,{\left (8 \, A a^{3} \cos \left (d x + c\right )^{5} - 26 \, A a^{3} \cos \left (d x + c\right )^{3} + 9 \, A a^{3} \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(sin(d*x+c)^2*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 9.33865, size = 359, normalized size = 2.97 \begin{align*} \begin{cases} - \frac{5 A a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} - \frac{15 A a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} - \frac{15 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{A a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} - \frac{5 A a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{A a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{11 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{2 A a^{3} \sin ^{4}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{5 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac{8 A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{2 A a^{3} \sin ^{2}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{5 A a^{3} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac{A a^{3} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{16 A a^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac{4 A a^{3} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text{for}\: d \neq 0 \\x \left (- A \sin{\left (c \right )} + A\right ) \left (a \sin{\left (c \right )} + a\right )^{3} \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+a*sin(d*x+c))**3*(A-A*sin(d*x+c)),x)

[Out]

Piecewise((-5*A*a**3*x*sin(c + d*x)**6/16 - 15*A*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 - 15*A*a**3*x*sin(c
+ d*x)**2*cos(c + d*x)**4/16 + A*a**3*x*sin(c + d*x)**2/2 - 5*A*a**3*x*cos(c + d*x)**6/16 + A*a**3*x*cos(c +
d*x)**2/2 + 11*A*a**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 2*A*a**3*sin(c + d*x)**4*cos(c + d*x)/d + 5*A*a**3
*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 8*A*a**3*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - 2*A*a**3*sin(c + d*x
)**2*cos(c + d*x)/d + 5*A*a**3*sin(c + d*x)*cos(c + d*x)**5/(16*d) - A*a**3*sin(c + d*x)*cos(c + d*x)/(2*d) +
16*A*a**3*cos(c + d*x)**5/(15*d) - 4*A*a**3*cos(c + d*x)**3/(3*d), Ne(d, 0)), (x*(-A*sin(c) + A)*(a*sin(c) + a
)**3*sin(c)**2, True))

________________________________________________________________________________________

Giac [A]  time = 1.12209, size = 153, normalized size = 1.26 \begin{align*} \frac{3}{16} \, A a^{3} x + \frac{A a^{3} \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac{A a^{3} \cos \left (3 \, d x + 3 \, c\right )}{24 \, d} - \frac{A a^{3} \cos \left (d x + c\right )}{4 \, d} + \frac{A a^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{3 \, A a^{3} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac{A a^{3} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="giac")

[Out]

3/16*A*a^3*x + 1/40*A*a^3*cos(5*d*x + 5*c)/d - 1/24*A*a^3*cos(3*d*x + 3*c)/d - 1/4*A*a^3*cos(d*x + c)/d + 1/19
2*A*a^3*sin(6*d*x + 6*c)/d - 3/64*A*a^3*sin(4*d*x + 4*c)/d - 1/64*A*a^3*sin(2*d*x + 2*c)/d