3.235 \(\int \frac{\sin ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=129 \[ -\frac{4 A \cos (c+d x)}{a^3 d}+\frac{A \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{199 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)}+\frac{41 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)^2}-\frac{2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}-\frac{19 A x}{2 a^3} \]

[Out]

(-19*A*x)/(2*a^3) - (4*A*Cos[c + d*x])/(a^3*d) + (A*Cos[c + d*x]*Sin[c + d*x])/(2*a^3*d) - (2*A*Cos[c + d*x])/
(5*a^3*d*(1 + Sin[c + d*x])^3) + (41*A*Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x])^2) - (199*A*Cos[c + d*x])/(1
5*a^3*d*(1 + Sin[c + d*x]))

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Rubi [A]  time = 0.208172, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {2966, 2638, 2635, 8, 2650, 2648} \[ -\frac{4 A \cos (c+d x)}{a^3 d}+\frac{A \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{199 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)}+\frac{41 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)^2}-\frac{2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}-\frac{19 A x}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^4*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

(-19*A*x)/(2*a^3) - (4*A*Cos[c + d*x])/(a^3*d) + (A*Cos[c + d*x]*Sin[c + d*x])/(2*a^3*d) - (2*A*Cos[c + d*x])/
(5*a^3*d*(1 + Sin[c + d*x])^3) + (41*A*Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x])^2) - (199*A*Cos[c + d*x])/(1
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 2638

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx &=\int \left (-\frac{9 A}{a^3}+\frac{4 A \sin (c+d x)}{a^3}-\frac{A \sin ^2(c+d x)}{a^3}+\frac{2 A}{a^3 (1+\sin (c+d x))^3}-\frac{9 A}{a^3 (1+\sin (c+d x))^2}+\frac{16 A}{a^3 (1+\sin (c+d x))}\right ) \, dx\\ &=-\frac{9 A x}{a^3}-\frac{A \int \sin ^2(c+d x) \, dx}{a^3}+\frac{(2 A) \int \frac{1}{(1+\sin (c+d x))^3} \, dx}{a^3}+\frac{(4 A) \int \sin (c+d x) \, dx}{a^3}-\frac{(9 A) \int \frac{1}{(1+\sin (c+d x))^2} \, dx}{a^3}+\frac{(16 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{a^3}\\ &=-\frac{9 A x}{a^3}-\frac{4 A \cos (c+d x)}{a^3 d}+\frac{A \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac{3 A \cos (c+d x)}{a^3 d (1+\sin (c+d x))^2}-\frac{16 A \cos (c+d x)}{a^3 d (1+\sin (c+d x))}-\frac{A \int 1 \, dx}{2 a^3}+\frac{(4 A) \int \frac{1}{(1+\sin (c+d x))^2} \, dx}{5 a^3}-\frac{(3 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{a^3}\\ &=-\frac{19 A x}{2 a^3}-\frac{4 A \cos (c+d x)}{a^3 d}+\frac{A \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac{41 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))^2}-\frac{13 A \cos (c+d x)}{a^3 d (1+\sin (c+d x))}+\frac{(4 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{15 a^3}\\ &=-\frac{19 A x}{2 a^3}-\frac{4 A \cos (c+d x)}{a^3 d}+\frac{A \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac{41 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))^2}-\frac{199 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.927193, size = 254, normalized size = 1.97 \[ \frac{A \left (-11400 d x \sin \left (c+\frac{d x}{2}\right )-5700 d x \sin \left (c+\frac{3 d x}{2}\right )+1830 \sin \left (2 c+\frac{3 d x}{2}\right )-4234 \sin \left (2 c+\frac{5 d x}{2}\right )+1140 d x \sin \left (3 c+\frac{5 d x}{2}\right )+165 \sin \left (4 c+\frac{7 d x}{2}\right )-15 \sin \left (4 c+\frac{9 d x}{2}\right )+12060 \cos \left (c+\frac{d x}{2}\right )-14090 \cos \left (c+\frac{3 d x}{2}\right )+5700 d x \cos \left (2 c+\frac{3 d x}{2}\right )+1140 d x \cos \left (2 c+\frac{5 d x}{2}\right )+1050 \cos \left (3 c+\frac{5 d x}{2}\right )+165 \cos \left (3 c+\frac{7 d x}{2}\right )+15 \cos \left (5 c+\frac{9 d x}{2}\right )+19780 \sin \left (\frac{d x}{2}\right )-11400 d x \cos \left (\frac{d x}{2}\right )\right )}{480 a^3 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^4*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

(A*(-11400*d*x*Cos[(d*x)/2] + 12060*Cos[c + (d*x)/2] - 14090*Cos[c + (3*d*x)/2] + 5700*d*x*Cos[2*c + (3*d*x)/2
] + 1140*d*x*Cos[2*c + (5*d*x)/2] + 1050*Cos[3*c + (5*d*x)/2] + 165*Cos[3*c + (7*d*x)/2] + 15*Cos[5*c + (9*d*x
)/2] + 19780*Sin[(d*x)/2] - 11400*d*x*Sin[c + (d*x)/2] - 5700*d*x*Sin[c + (3*d*x)/2] + 1830*Sin[2*c + (3*d*x)/
2] - 4234*Sin[2*c + (5*d*x)/2] + 1140*d*x*Sin[3*c + (5*d*x)/2] + 165*Sin[4*c + (7*d*x)/2] - 15*Sin[4*c + (9*d*
x)/2]))/(480*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 [B]  time = 0.112, size = 257, normalized size = 2. \begin{align*} -{\frac{A}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-8\,{\frac{A \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{A}{d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-8\,{\frac{A}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-19\,{\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}}-10\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}-18\,{\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)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x)

[Out]

-1/d*A/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-8/d*A/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/
2*c)^2+1/d*A/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-8/d*A/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2-19/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)^3-10/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^2-18/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)

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Maxima [B]  time = 1.49411, size = 965, normalized size = 7.48 \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)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/15*(A*((1325*sin(d*x + c)/(cos(d*x + c) + 1) + 2673*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3805*sin(d*x + c)
^3/(cos(d*x + c) + 1)^3 + 4329*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 3575*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
+ 2275*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 975*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 195*sin(d*x + c)^8/(cos
(d*x + c) + 1)^8 + 304)/(a^3 + 5*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 12*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1
)^2 + 20*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 26*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 26*a^3*sin(d*x
+ c)^5/(cos(d*x + c) + 1)^5 + 20*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 12*a^3*sin(d*x + c)^7/(cos(d*x + c
) + 1)^7 + 5*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + a^3*sin(d*x + c)^9/(cos(d*x + c) + 1)^9) + 195*arctan(s
in(d*x + c)/(cos(d*x + c) + 1))/a^3) + 6*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*s
in(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))/d

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Fricas [B]  time = 2.07252, size = 656, normalized size = 5.09 \begin{align*} -\frac{15 \, A \cos \left (d x + c\right )^{5} + 90 \, A \cos \left (d x + c\right )^{4} +{\left (285 \, A d x + 683 \, A\right )} \cos \left (d x + c\right )^{3} - 1140 \, A d x +{\left (855 \, A d x - 526 \, A\right )} \cos \left (d x + c\right )^{2} - 6 \,{\left (95 \, A d x + 191 \, A\right )} \cos \left (d x + c\right ) -{\left (15 \, A \cos \left (d x + c\right )^{4} - 75 \, A \cos \left (d x + c\right )^{3} + 1140 \, A d x - 19 \,{\left (15 \, A d x - 32 \, A\right )} \cos \left (d x + c\right )^{2} + 6 \,{\left (95 \, A d x + 189 \, A\right )} \cos \left (d x + c\right ) - 12 \, A\right )} \sin \left (d x + c\right ) - 12 \, A}{30 \,{\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)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/30*(15*A*cos(d*x + c)^5 + 90*A*cos(d*x + c)^4 + (285*A*d*x + 683*A)*cos(d*x + c)^3 - 1140*A*d*x + (855*A*d*
x - 526*A)*cos(d*x + c)^2 - 6*(95*A*d*x + 191*A)*cos(d*x + c) - (15*A*cos(d*x + c)^4 - 75*A*cos(d*x + c)^3 + 1
140*A*d*x - 19*(15*A*d*x - 32*A)*cos(d*x + c)^2 + 6*(95*A*d*x + 189*A)*cos(d*x + c) - 12*A)*sin(d*x + c) - 12*
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)**4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.1601, size = 211, normalized size = 1.64 \begin{align*} -\frac{\frac{285 \,{\left (d x + c\right )} A}{a^{3}} + \frac{30 \,{\left (A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8 \, A\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} + \frac{4 \,{\left (135 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 615 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1025 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 685 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 164 \, A\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/30*(285*(d*x + c)*A/a^3 + 30*(A*tan(1/2*d*x + 1/2*c)^3 + 8*A*tan(1/2*d*x + 1/2*c)^2 - A*tan(1/2*d*x + 1/2*c
) + 8*A)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^3) + 4*(135*A*tan(1/2*d*x + 1/2*c)^4 + 615*A*tan(1/2*d*x + 1/2*c)^3
+ 1025*A*tan(1/2*d*x + 1/2*c)^2 + 685*A*tan(1/2*d*x + 1/2*c) + 164*A)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^5))/d