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]

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

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Rubi [A]  time = 0.11, antiderivative size = 58, normalized size of antiderivative = 1.00, 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 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]

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 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]))

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*}

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Mathematica [A]  time = 0.24, 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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fricas [B]  time = 0.41, size = 154, normalized size = 2.66 \[ \frac {A \cos \left (d x + c\right )^{3} - 2 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) - {\left (A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + 6 \, A\right )} \sin \left (d x + c\right ) + 6 \, A}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

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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giac [A]  time = 0.17, size = 79, normalized size = 1.36 \[ -\frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 25 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, A\right )}}{15 \, a^{3} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}} \]

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)

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maple [A]  time = 0.41, size = 86, normalized size = 1.48 \[ \frac {2 A \left (-\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {3}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {14}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\right )}{d \,a^{3}} \]

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+3/(tan(1/2*d*x+1/2*c)+1)^2-1/(tan(1/2*d*x+1/2*c)+1)+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 = 0.53, size = 387, normalized size = 6.67 \[ -\frac {2 \, {\left (\frac {A {\left (\frac {20 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 7\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} - \frac {3 \, A {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}\right )}}{15 \, d} \]

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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mupad [B]  time = 13.31, size = 134, normalized size = 2.31 \[ -\frac {2\,A\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+25\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+15\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+15\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{15\,a^3\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*A*cos(c/2 + (d*x)/2)*(4*cos(c/2 + (d*x)/2)^4 + 15*sin(c/2 + (d*x)/2)^4 + 15*cos(c/2 + (d*x)/2)*sin(c/2 + (
d*x)/2)^3 + 5*cos(c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/2) + 25*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2))/(15*a^3
*d*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^5)

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sympy [A]  time = 9.11, size = 573, normalized size = 9.88 \[ \begin {cases} - \frac {30 A \tan ^{4}{\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 {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 {50 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 {8 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 {\relax (c )} + A\right )}{\left (a \sin {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

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((-30*A*tan(c/2 + d*x/2)**4/(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) - 50*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*t
an(c/2 + d*x/2) + 15*a**3*d) - 8*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)/(a*sin(c) + a)**3, True))

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