3.238 \(\int \frac{\sin (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=82 \[ \frac{4 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)}-\frac{11 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)^2}+\frac{2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3} \]

[Out]

(2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) - (11*A*Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x])^2) + (4*A
*Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x]))

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Rubi [A]  time = 0.138005, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {2966, 2650, 2648} \[ \frac{4 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)}-\frac{11 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)^2}+\frac{2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) - (11*A*Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x])^2) + (4*A
*Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x]))

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 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\sin (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx &=\int \left (-\frac{2 A}{a^3 (1+\sin (c+d x))^3}+\frac{3 A}{a^3 (1+\sin (c+d x))^2}-\frac{A}{a^3 (1+\sin (c+d x))}\right ) \, dx\\ &=-\frac{A \int \frac{1}{1+\sin (c+d x)} \, dx}{a^3}-\frac{(2 A) \int \frac{1}{(1+\sin (c+d x))^3} \, dx}{a^3}+\frac{(3 A) \int \frac{1}{(1+\sin (c+d x))^2} \, dx}{a^3}\\ &=\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac{A \cos (c+d x)}{a^3 d (1+\sin (c+d x))^2}+\frac{A \cos (c+d x)}{a^3 d (1+\sin (c+d x))}-\frac{(4 A) \int \frac{1}{(1+\sin (c+d x))^2} \, dx}{5 a^3}+\frac{A \int \frac{1}{1+\sin (c+d x)} \, dx}{a^3}\\ &=\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac{11 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))^2}-\frac{(4 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{15 a^3}\\ &=\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac{11 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))^2}+\frac{4 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.459339, size = 107, normalized size = 1.3 \[ -\frac{A \left (15 \sin \left (2 c+\frac{3 d x}{2}\right )-4 \sin \left (2 c+\frac{5 d x}{2}\right )+15 \cos \left (c+\frac{d x}{2}\right )-5 \cos \left (c+\frac{3 d x}{2}\right )+25 \sin \left (\frac{d x}{2}\right )\right )}{30 a^3 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*(15*Cos[c + (d*x)/2] - 5*Cos[c + (3*d*x)/2] + 25*Sin[(d*x)/2] + 15*Sin[2*c + (3*d*x)/2] - 4*Sin[2*c + (5*d
*x)/2]))/(30*a^3*d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

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Maple [A]  time = 0.093, size = 71, normalized size = 0.9 \begin{align*} 4\,{\frac{A}{d{a}^{3}} \left ( -1/2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-2}+4/5\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-5}+5/3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-3}-2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x)

[Out]

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

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Maxima [B]  time = 1.00779, size = 470, normalized size = 5.73 \begin{align*} \frac{2 \,{\left (\frac{2 \, A{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}}{a^{3} + \frac{5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} - \frac{3 \, A{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 1\right )}}{a^{3} + \frac{5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}\right )}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(2*A*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)/(a^3 + 5*a^3*sin(d*
x + c)/(cos(d*x + c) + 1) + 10*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)^3/(cos(d*x + c) +
1)^3 + 5*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) - 3*A*(5*sin(d*x
+ c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 1)/(
a^3 + 5*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 10*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)
^3/(cos(d*x + c) + 1)^3 + 5*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)
)/d

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Fricas [B]  time = 1.85775, size = 386, normalized size = 4.71 \begin{align*} \frac{4 \, A \cos \left (d x + c\right )^{3} + 7 \, A \cos \left (d x + c\right )^{2} - 3 \, A \cos \left (d x + c\right ) -{\left (4 \, A \cos \left (d x + c\right )^{2} - 3 \, A \cos \left (d x + c\right ) - 6 \, A\right )} \sin \left (d x + c\right ) - 6 \, A}{15 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d +{\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(4*A*cos(d*x + c)^3 + 7*A*cos(d*x + c)^2 - 3*A*cos(d*x + c) - (4*A*cos(d*x + c)^2 - 3*A*cos(d*x + c) - 6*
A)*sin(d*x + c) - 6*A)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d + (a^3*
d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d)*sin(d*x + c))

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Sympy [A]  time = 34.0405, size = 461, normalized size = 5.62 \begin{align*} \begin{cases} - \frac{30 A \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{15 a^{3} d \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 75 a^{3} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 150 a^{3} d \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 150 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 75 a^{3} d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 15 a^{3} d} + \frac{10 A \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{15 a^{3} d \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 75 a^{3} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 150 a^{3} d \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 150 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 75 a^{3} d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 15 a^{3} d} - \frac{10 A \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{15 a^{3} d \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 75 a^{3} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 150 a^{3} d \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 150 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 75 a^{3} d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 15 a^{3} d} - \frac{2 A}{15 a^{3} d \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 75 a^{3} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 150 a^{3} d \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 150 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 75 a^{3} d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 15 a^{3} d} & \text{for}\: d \neq 0 \\\frac{x \left (- A \sin{\left (c \right )} + A\right ) \sin{\left (c \right )}}{\left (a \sin{\left (c \right )} + a\right )^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((-30*A*tan(c/2 + d*x/2)**3/(15*a**3*d*tan(c/2 + d*x/2)**5 + 75*a**3*d*tan(c/2 + d*x/2)**4 + 150*a**3
*d*tan(c/2 + d*x/2)**3 + 150*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 10*A*tan(c
/2 + d*x/2)**2/(15*a**3*d*tan(c/2 + d*x/2)**5 + 75*a**3*d*tan(c/2 + d*x/2)**4 + 150*a**3*d*tan(c/2 + d*x/2)**3
+ 150*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) - 10*A*tan(c/2 + d*x/2)/(15*a**3*d
*tan(c/2 + d*x/2)**5 + 75*a**3*d*tan(c/2 + d*x/2)**4 + 150*a**3*d*tan(c/2 + d*x/2)**3 + 150*a**3*d*tan(c/2 + d
*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) - 2*A/(15*a**3*d*tan(c/2 + d*x/2)**5 + 75*a**3*d*tan(c/2 +
d*x/2)**4 + 150*a**3*d*tan(c/2 + d*x/2)**3 + 150*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*
a**3*d), Ne(d, 0)), (x*(-A*sin(c) + A)*sin(c)/(a*sin(c) + a)**3, True))

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Giac [A]  time = 1.16554, size = 85, normalized size = 1.04 \begin{align*} -\frac{2 \,{\left (15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 5 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A\right )}}{15 \, a^{3} d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(15*A*tan(1/2*d*x + 1/2*c)^3 - 5*A*tan(1/2*d*x + 1/2*c)^2 + 5*A*tan(1/2*d*x + 1/2*c) + A)/(a^3*d*(tan(1/
2*d*x + 1/2*c) + 1)^5)