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

Optimal. Leaf size=58 \[ -\frac{A \cos ^3(c+d x)}{15 d (a \sin (c+d x)+a)^3}-\frac{a A \cos ^3(c+d x)}{5 d (a \sin (c+d x)+a)^4} \]

[Out]

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

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Rubi [A]  time = 0.114368, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2736, 2672, 2671} \[ -\frac{A \cos ^3(c+d x)}{15 d (a \sin (c+d x)+a)^3}-\frac{a A \cos ^3(c+d x)}{5 d (a \sin (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rule 2672

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(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] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 2671

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

Rubi steps

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

Mathematica [A]  time = 0.238575, size = 92, normalized size = 1.59 \[ \frac{A \left (\sin \left (2 c+\frac{5 d x}{2}\right )-15 \cos \left (c+\frac{d x}{2}\right )+5 \cos \left (c+\frac{3 d x}{2}\right )+5 \sin \left (\frac{d x}{2}\right )\right )}{30 a^3 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(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] + 5*Sin[(d*x)/2] + 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 = 0.087, size = 86, normalized size = 1.5 \begin{align*} 2\,{\frac{A}{d{a}^{3}} \left ( -8/5\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-5}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-1}+3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-2}+4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4}-14/3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*A/a^3*(-8/5/(tan(1/2*d*x+1/2*c)+1)^5-1/(tan(1/2*d*x+1/2*c)+1)+3/(tan(1/2*d*x+1/2*c)+1)^2+4/(tan(1/2*d*x+1/
2*c)+1)^4-14/3/(tan(1/2*d*x+1/2*c)+1)^3)

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Maxima [B]  time = 1.01554, size = 522, normalized size = 9. \begin{align*} -\frac{2 \,{\left (\frac{A{\left (\frac{20 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{30 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 7\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((A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-2/15*(A*(20*sin(d*x + c)/(cos(d*x + c) + 1) + 40*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 30*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 7)/(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)/(co
s(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 = 1.86094, size = 381, normalized size = 6.57 \begin{align*} \frac{A \cos \left (d x + c\right )^{3} - 2 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) -{\left (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((A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(A*cos(d*x + c)^3 - 2*A*cos(d*x + c)^2 + 3*A*cos(d*x + c) - (A*cos(d*x + c)^2 + 3*A*cos(d*x + c) + 6*A)*s
in(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*co
s(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 = 14.1702, size = 571, normalized size = 9.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((6*A*tan(c/2 + d*x/2)**5/(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) + 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) + 20*A*tan(c/2 + d*x/2)/(15*a**3*d*tan(c/2 + d*x/2)**5 + 7
5*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*t
an(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)/(a*sin(c) + a)**3, True))

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Giac [A]  time = 1.12946, size = 107, normalized size = 1.84 \begin{align*} -\frac{2 \,{\left (15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 25 \, 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 ) + 4 \, 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((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)^4 + 15*A*tan(1/2*d*x + 1/2*c)^3 + 25*A*tan(1/2*d*x + 1/2*c)^2 + 5*A*tan(1/2*d
*x + 1/2*c) + 4*A)/(a^3*d*(tan(1/2*d*x + 1/2*c) + 1)^5)