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

Optimal. Leaf size=103 \[ \frac{A \cos (c+d x)}{a^3 d}+\frac{104 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)}-\frac{31 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)^2}+\frac{2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}+\frac{4 A x}{a^3} \]

[Out]

(4*A*x)/a^3 + (A*Cos[c + d*x])/(a^3*d) + (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) - (31*A*Cos[c + d*x
])/(15*a^3*d*(1 + Sin[c + d*x])^2) + (104*A*Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x]))

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Rubi [A]  time = 0.18774, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2966, 2638, 2650, 2648} \[ \frac{A \cos (c+d x)}{a^3 d}+\frac{104 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)}-\frac{31 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)^2}+\frac{2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}+\frac{4 A x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*A*x)/a^3 + (A*Cos[c + d*x])/(a^3*d) + (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) - (31*A*Cos[c + d*x
])/(15*a^3*d*(1 + Sin[c + d*x])^2) + (104*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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 ^3(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx &=\int \left (\frac{4 A}{a^3}-\frac{A \sin (c+d x)}{a^3}-\frac{2 A}{a^3 (1+\sin (c+d x))^3}+\frac{7 A}{a^3 (1+\sin (c+d x))^2}-\frac{9 A}{a^3 (1+\sin (c+d x))}\right ) \, dx\\ &=\frac{4 A x}{a^3}-\frac{A \int \sin (c+d x) \, dx}{a^3}-\frac{(2 A) \int \frac{1}{(1+\sin (c+d x))^3} \, dx}{a^3}+\frac{(7 A) \int \frac{1}{(1+\sin (c+d x))^2} \, dx}{a^3}-\frac{(9 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{a^3}\\ &=\frac{4 A x}{a^3}+\frac{A \cos (c+d x)}{a^3 d}+\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac{7 A \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))^2}+\frac{9 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{(7 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{3 a^3}\\ &=\frac{4 A x}{a^3}+\frac{A \cos (c+d x)}{a^3 d}+\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac{31 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))^2}+\frac{20 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{4 A x}{a^3}+\frac{A \cos (c+d x)}{a^3 d}+\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac{31 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))^2}+\frac{104 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [B]  time = 0.789342, size = 228, normalized size = 2.21 \[ -\frac{A \left (-1200 d x \sin \left (c+\frac{d x}{2}\right )-600 d x \sin \left (c+\frac{3 d x}{2}\right )+405 \sin \left (2 c+\frac{3 d x}{2}\right )-491 \sin \left (2 c+\frac{5 d x}{2}\right )+120 d x \sin \left (3 c+\frac{5 d x}{2}\right )+15 \sin \left (4 c+\frac{7 d x}{2}\right )+1665 \cos \left (c+\frac{d x}{2}\right )-1675 \cos \left (c+\frac{3 d x}{2}\right )+600 d x \cos \left (2 c+\frac{3 d x}{2}\right )+120 d x \cos \left (2 c+\frac{5 d x}{2}\right )+75 \cos \left (3 c+\frac{5 d x}{2}\right )+15 \cos \left (3 c+\frac{7 d x}{2}\right )+2495 \sin \left (\frac{d x}{2}\right )-1200 d x \cos \left (\frac{d x}{2}\right )\right )}{120 a^3 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*(-1200*d*x*Cos[(d*x)/2] + 1665*Cos[c + (d*x)/2] - 1675*Cos[c + (3*d*x)/2] + 600*d*x*Cos[2*c + (3*d*x)/2] +
120*d*x*Cos[2*c + (5*d*x)/2] + 75*Cos[3*c + (5*d*x)/2] + 15*Cos[3*c + (7*d*x)/2] + 2495*Sin[(d*x)/2] - 1200*d
*x*Sin[c + (d*x)/2] - 600*d*x*Sin[c + (3*d*x)/2] + 405*Sin[2*c + (3*d*x)/2] - 491*Sin[2*c + (5*d*x)/2] + 120*d
*x*Sin[3*c + (5*d*x)/2] + 15*Sin[4*c + (7*d*x)/2]))/(120*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 = 0.102, size = 155, normalized size = 1.5 \begin{align*} 2\,{\frac{A}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+8\,{\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}}}+{\frac{4\,A}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+6\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}+8\,{\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)^3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x)

[Out]

2/d*A/a^3/(1+tan(1/2*d*x+1/2*c)^2)+8/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+4/3/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^3+6/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^2+8/d
*A/a^3/(tan(1/2*d*x+1/2*c)+1)

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Maxima [B]  time = 1.50767, size = 733, normalized size = 7.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*A*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 189*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 200*sin(d*x + c)^3
/(cos(d*x + c) + 1)^3 + 160*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 75*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*
sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 24)/(a^3 + 5*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 11*a^3*sin(d*x + c)^2
/(cos(d*x + c) + 1)^2 + 15*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^
4 + 11*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d*x + c)^
7/(cos(d*x + c) + 1)^7) + 15*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) + 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/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(c
os(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))/d

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Fricas [B]  time = 1.96004, size = 585, normalized size = 5.68 \begin{align*} \frac{15 \, A \cos \left (d x + c\right )^{4} +{\left (60 \, A d x + 149 \, A\right )} \cos \left (d x + c\right )^{3} - 240 \, A d x +{\left (180 \, A d x - 103 \, A\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (40 \, A d x + 81 \, A\right )} \cos \left (d x + c\right ) +{\left (15 \, A \cos \left (d x + c\right )^{3} - 240 \, A d x + 2 \,{\left (30 \, A d x - 67 \, A\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (40 \, A d x + 79 \, A\right )} \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)^3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(15*A*cos(d*x + c)^4 + (60*A*d*x + 149*A)*cos(d*x + c)^3 - 240*A*d*x + (180*A*d*x - 103*A)*cos(d*x + c)^2
- 3*(40*A*d*x + 81*A)*cos(d*x + c) + (15*A*cos(d*x + c)^3 - 240*A*d*x + 2*(30*A*d*x - 67*A)*cos(d*x + c)^2 -
3*(40*A*d*x + 79*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 [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)**3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.14841, size = 153, normalized size = 1.49 \begin{align*} \frac{2 \,{\left (\frac{30 \,{\left (d x + c\right )} A}{a^{3}} + \frac{15 \, A}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} a^{3}} + \frac{60 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 285 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 505 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 335 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 79 \, A}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, 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)^3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

2/15*(30*(d*x + c)*A/a^3 + 15*A/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) + (60*A*tan(1/2*d*x + 1/2*c)^4 + 285*A*tan(
1/2*d*x + 1/2*c)^3 + 505*A*tan(1/2*d*x + 1/2*c)^2 + 335*A*tan(1/2*d*x + 1/2*c) + 79*A)/(a^3*(tan(1/2*d*x + 1/2
*c) + 1)^5))/d