\(\int \frac {\sin (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx\) [238]

Optimal result

Integrand size = 30, antiderivative size = 82 \[ \int \frac {\sin (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac {11 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))^2}+\frac {4 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))} \]

[Out]

2/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^3-11/15*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^2+4/15*A*cos(d*x+c)/a^3/d/(1+s
in(d*x+c))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3045, 2729, 2727} \[ \int \frac {\sin (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {4 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)}-\frac {11 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)^2}+\frac {2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3} \]

[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 2727

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]

Rule 2729

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 3045

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 2.53 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.30 \[ \int \frac {\sin (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {A \left (15 \cos \left (c+\frac {d x}{2}\right )-5 \cos \left (c+\frac {3 d x}{2}\right )+25 \sin \left (\frac {d x}{2}\right )+15 \sin \left (2 c+\frac {3 d x}{2}\right )-4 \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{30 a^3 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

[In]

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

[Out]

-1/30*(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]))/(a^3*d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.76

method	result	size
parallelrisch	\(-\frac {2 A \left (15 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{15 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(62\)
derivativedivides	\(\frac {4 A \left (\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {5}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\right )}{d \,a^{3}}\)	\(71\)
default	\(\frac {4 A \left (\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {5}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\right )}{d \,a^{3}}\)	\(71\)
risch	\(\frac {2 A \left (15 i {\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{4 i \left (d x +c \right )}-5 i {\mathrm e}^{i \left (d x +c \right )}-25 \,{\mathrm e}^{2 i \left (d x +c \right )}+4\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5}}\)	\(72\)
norman	\(\frac {-\frac {2 A}{15 a d}-\frac {14 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 a d}-\frac {10 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {2 A \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 A \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}+\frac {2 A \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {6 A \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(181\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (76) = 152\).

Time = 0.26 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.90 \[ \int \frac {\sin (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \, A \cos \left (d x + c\right )^{3} + 7 \, A \cos \left (d x + c\right )^{2} - 3 \, A \cos \left (d x + c\right ) - {\left (4 \, A \cos \left (d x + c\right )^{2} - 3 \, A \cos \left (d x + c\right ) - 6 \, A\right )} \sin \left (d x + c\right ) - 6 \, A}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

[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))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 461 vs. \(2 (78) = 156\).

Time = 4.17 (sec) , antiderivative size = 461, normalized size of antiderivative = 5.62 \[ \int \frac {\sin (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\begin {cases} - \frac {30 A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{3} d} + \frac {10 A \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{3} d} - \frac {10 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{3} d} - \frac {2 A}{15 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (- A \sin {\left (c \right )} + A\right ) \sin {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (76) = 152\).

Time = 0.25 (sec) , antiderivative size = 348, normalized size of antiderivative = 4.24 \[ \int \frac {\sin (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (\frac {2 \, A {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} - \frac {3 \, A {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}\right )}}{15 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.77 \[ \int \frac {\sin (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A\right )}}{15 \, a^{3} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}} \]

[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)

Mupad [B] (verification not implemented)

Time = 12.42 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.34 \[ \int \frac {\sin (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {2\,A\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left ({\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+15\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}{15\,a^3\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^5} \]

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