3.237 \(\int \frac{\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=89 \[ -\frac{13 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)}+\frac{7 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^2}-\frac{2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}-\frac{A x}{a^3} \]

[Out]

-((A*x)/a^3) - (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) + (7*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*
x])^2) - (13*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x]))

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Rubi [A]  time = 0.173079, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {2966, 2650, 2648} \[ -\frac{13 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)}+\frac{7 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^2}-\frac{2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}-\frac{A x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*x)/a^3) - (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) + (7*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*
x])^2) - (13*A*Cos[c + d*x])/(5*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 ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx &=\int \left (-\frac{A}{a^3}+\frac{2 A}{a^3 (1+\sin (c+d x))^3}-\frac{5 A}{a^3 (1+\sin (c+d x))^2}+\frac{4 A}{a^3 (1+\sin (c+d x))}\right ) \, dx\\ &=-\frac{A x}{a^3}+\frac{(2 A) \int \frac{1}{(1+\sin (c+d x))^3} \, dx}{a^3}+\frac{(4 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{a^3}-\frac{(5 A) \int \frac{1}{(1+\sin (c+d x))^2} \, dx}{a^3}\\ &=-\frac{A x}{a^3}-\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac{5 A \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))^2}-\frac{4 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{(5 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{3 a^3}\\ &=-\frac{A x}{a^3}-\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac{7 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac{7 A \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))}+\frac{(4 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{15 a^3}\\ &=-\frac{A x}{a^3}-\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac{7 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac{13 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [B]  time = 0.762797, size = 189, normalized size = 2.12 \[ \frac{A \left (-50 d x \sin \left (c+\frac{d x}{2}\right )-25 d x \sin \left (c+\frac{3 d x}{2}\right )+40 \sin \left (2 c+\frac{3 d x}{2}\right )-26 \sin \left (2 c+\frac{5 d x}{2}\right )+5 d x \sin \left (3 c+\frac{5 d x}{2}\right )+110 \cos \left (c+\frac{d x}{2}\right )-90 \cos \left (c+\frac{3 d x}{2}\right )+25 d x \cos \left (2 c+\frac{3 d x}{2}\right )+5 d x \cos \left (2 c+\frac{5 d x}{2}\right )+150 \sin \left (\frac{d x}{2}\right )-50 d x \cos \left (\frac{d x}{2}\right )\right )}{20 a^3 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*(-50*d*x*Cos[(d*x)/2] + 110*Cos[c + (d*x)/2] - 90*Cos[c + (3*d*x)/2] + 25*d*x*Cos[2*c + (3*d*x)/2] + 5*d*x*
Cos[2*c + (5*d*x)/2] + 150*Sin[(d*x)/2] - 50*d*x*Sin[c + (d*x)/2] - 25*d*x*Sin[c + (3*d*x)/2] + 40*Sin[2*c + (
3*d*x)/2] - 26*Sin[2*c + (5*d*x)/2] + 5*d*x*Sin[3*c + (5*d*x)/2]))/(20*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 = 0.099, size = 131, normalized size = 1.5 \begin{align*} -2\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}-{\frac{16\,A}{5\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}+8\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}-4\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}-2\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}-2\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \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))/(a+a*sin(d*x+c))^3,x)

[Out]

-2/d*A/a^3*arctan(tan(1/2*d*x+1/2*c))-16/5/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^5+8/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^4
-4/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^3-2/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^2-2/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)

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Maxima [B]  time = 1.48544, size = 529, normalized size = 5.94 \begin{align*} -\frac{2 \,{\left (A{\left (\frac{\frac{95 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{145 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{75 \, \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}} + 22}{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{15 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + \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}}}\right )}}{15 \, 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))/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-2/15*(A*((95*sin(d*x + c)/(cos(d*x + c) + 1) + 145*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 75*sin(d*x + c)^3/(c
os(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 22)/(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) + 15*arctan(sin(d*x + c)/(cos(d*x + c) + 1
))/a^3) + 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*si
n(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 = 1.76422, size = 508, normalized size = 5.71 \begin{align*} -\frac{{\left (5 \, A d x + 13 \, A\right )} \cos \left (d x + c\right )^{3} - 20 \, A d x + 3 \,{\left (5 \, A d x - 2 \, A\right )} \cos \left (d x + c\right )^{2} -{\left (10 \, A d x + 21 \, A\right )} \cos \left (d x + c\right ) -{\left (20 \, A d x -{\left (5 \, A d x - 13 \, A\right )} \cos \left (d x + c\right )^{2} +{\left (10 \, A d x + 19 \, A\right )} \cos \left (d x + c\right ) - 2 \, A\right )} \sin \left (d x + c\right ) - 2 \, A}{5 \,{\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)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/5*((5*A*d*x + 13*A)*cos(d*x + c)^3 - 20*A*d*x + 3*(5*A*d*x - 2*A)*cos(d*x + c)^2 - (10*A*d*x + 21*A)*cos(d*
x + c) - (20*A*d*x - (5*A*d*x - 13*A)*cos(d*x + c)^2 + (10*A*d*x + 19*A)*cos(d*x + c) - 2*A)*sin(d*x + c) - 2*
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 [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.16008, size = 126, normalized size = 1.42 \begin{align*} -\frac{\frac{5 \,{\left (d x + c\right )} A}{a^{3}} + \frac{2 \,{\left (5 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 25 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 55 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 35 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8 \, A\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}}}{5 \, 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))/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/5*(5*(d*x + c)*A/a^3 + 2*(5*A*tan(1/2*d*x + 1/2*c)^4 + 25*A*tan(1/2*d*x + 1/2*c)^3 + 55*A*tan(1/2*d*x + 1/2
*c)^2 + 35*A*tan(1/2*d*x + 1/2*c) + 8*A)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^5))/d