3.23 \(\int \frac {1}{(5-3 \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=58 \[ -\frac {3 \cos (c+d x)}{16 d (5-3 \sin (c+d x))}-\frac {5 \tan ^{-1}\left (\frac {\cos (c+d x)}{3-\sin (c+d x)}\right )}{32 d}+\frac {5 x}{64} \]

[Out]

5/64*x-5/32*arctan(cos(d*x+c)/(3-sin(d*x+c)))/d-3/16*cos(d*x+c)/d/(5-3*sin(d*x+c))

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Rubi [A]  time = 0.03, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2664, 12, 2657} \[ -\frac {3 \cos (c+d x)}{16 d (5-3 \sin (c+d x))}-\frac {5 \tan ^{-1}\left (\frac {\cos (c+d x)}{3-\sin (c+d x)}\right )}{32 d}+\frac {5 x}{64} \]

Antiderivative was successfully verified.

[In]

Int[(5 - 3*Sin[c + d*x])^(-2),x]

[Out]

(5*x)/64 - (5*ArcTan[Cos[c + d*x]/(3 - Sin[c + d*x])])/(32*d) - (3*Cos[c + d*x])/(16*d*(5 - 3*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2657

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[a^2 - b^2, 2]}, Simp[x/q, x] + Simp
[(2*ArcTan[(b*Cos[c + d*x])/(a + q + b*Sin[c + d*x])])/(d*q), x]] /; FreeQ[{a, b, c, d}, x] && GtQ[a^2 - b^2,
0] && PosQ[a]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rubi steps

\begin {align*} \int \frac {1}{(5-3 \sin (c+d x))^2} \, dx &=-\frac {3 \cos (c+d x)}{16 d (5-3 \sin (c+d x))}-\frac {1}{16} \int -\frac {5}{5-3 \sin (c+d x)} \, dx\\ &=-\frac {3 \cos (c+d x)}{16 d (5-3 \sin (c+d x))}+\frac {5}{16} \int \frac {1}{5-3 \sin (c+d x)} \, dx\\ &=\frac {5 x}{64}-\frac {5 \tan ^{-1}\left (\frac {\cos (c+d x)}{3-\sin (c+d x)}\right )}{32 d}-\frac {3 \cos (c+d x)}{16 d (5-3 \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 91, normalized size = 1.57 \[ \frac {\frac {6 (3 \sin (c+d x)+5 \cos (c+d x)-5)}{3 \sin (c+d x)-5}-25 \tan ^{-1}\left (\frac {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}\right )}{160 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(5 - 3*Sin[c + d*x])^(-2),x]

[Out]

(-25*ArcTan[(2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])] + (6*(-5 + 5*Cos[
c + d*x] + 3*Sin[c + d*x]))/(-5 + 3*Sin[c + d*x]))/(160*d)

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fricas [A]  time = 0.44, size = 59, normalized size = 1.02 \[ \frac {5 \, {\left (3 \, \sin \left (d x + c\right ) - 5\right )} \arctan \left (\frac {5 \, \sin \left (d x + c\right ) - 3}{4 \, \cos \left (d x + c\right )}\right ) + 12 \, \cos \left (d x + c\right )}{64 \, {\left (3 \, d \sin \left (d x + c\right ) - 5 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5-3*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/64*(5*(3*sin(d*x + c) - 5)*arctan(1/4*(5*sin(d*x + c) - 3)/cos(d*x + c)) + 12*cos(d*x + c))/(3*d*sin(d*x + c
) - 5*d)

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giac [A]  time = 0.77, size = 96, normalized size = 1.66 \[ \frac {25 \, d x + 25 \, c + \frac {24 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5\right )}}{5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5} + 50 \, \arctan \left (\frac {3 \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 3}{\cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) - 9}\right )}{320 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5-3*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/320*(25*d*x + 25*c + 24*(3*tan(1/2*d*x + 1/2*c) - 5)/(5*tan(1/2*d*x + 1/2*c)^2 - 6*tan(1/2*d*x + 1/2*c) + 5)
+ 50*arctan((3*cos(d*x + c) - sin(d*x + c) + 3)/(cos(d*x + c) + 3*sin(d*x + c) - 9)))/d

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maple [A]  time = 0.09, size = 92, normalized size = 1.59 \[ \frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{200 d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+1\right )}-\frac {3}{40 d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+1\right )}+\frac {5 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}-\frac {3}{4}\right )}{32 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5-3*sin(d*x+c))^2,x)

[Out]

9/200/d/(tan(1/2*d*x+1/2*c)^2-6/5*tan(1/2*d*x+1/2*c)+1)*tan(1/2*d*x+1/2*c)-3/40/d/(tan(1/2*d*x+1/2*c)^2-6/5*ta
n(1/2*d*x+1/2*c)+1)+5/32/d*arctan(5/4*tan(1/2*d*x+1/2*c)-3/4)

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maxima [A]  time = 0.68, size = 93, normalized size = 1.60 \[ -\frac {\frac {12 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 5\right )}}{\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 5} - 25 \, \arctan \left (\frac {5 \, \sin \left (d x + c\right )}{4 \, {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {3}{4}\right )}{160 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5-3*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/160*(12*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5)/(6*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^2/(cos(
d*x + c) + 1)^2 - 5) - 25*arctan(5/4*sin(d*x + c)/(cos(d*x + c) + 1) - 3/4))/d

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mupad [B]  time = 4.32, size = 83, normalized size = 1.43 \[ \frac {5\,\mathrm {atan}\left (\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {3}{4}\right )}{32\,d}-\frac {5\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{32\,d}+\frac {\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{200}-\frac {3}{40}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*sin(c + d*x) - 5)^2,x)

[Out]

(5*atan((5*tan(c/2 + (d*x)/2))/4 - 3/4))/(32*d) - (5*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(32*d) + ((9*tan(c/
2 + (d*x)/2))/200 - 3/40)/(d*(tan(c/2 + (d*x)/2)^2 - (6*tan(c/2 + (d*x)/2))/5 + 1))

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sympy [A]  time = 2.05, size = 384, normalized size = 6.62 \[ \begin {cases} \frac {x}{\left (5 - 3 \sin {\left (2 \operatorname {atan}{\left (\frac {3}{5} - \frac {4 i}{5} \right )} \right )}\right )^{2}} & \text {for}\: c = - d x + 2 \operatorname {atan}{\left (\frac {3}{5} - \frac {4 i}{5} \right )} \\\frac {x}{\left (5 - 3 \sin {\left (2 \operatorname {atan}{\left (\frac {3}{5} + \frac {4 i}{5} \right )} \right )}\right )^{2}} & \text {for}\: c = - d x + 2 \operatorname {atan}{\left (\frac {3}{5} + \frac {4 i}{5} \right )} \\\frac {x}{\left (5 - 3 \sin {\relax (c )}\right )^{2}} & \text {for}\: d = 0 \\\frac {125 \left (\operatorname {atan}{\left (\frac {5 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4} - \frac {3}{4} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{800 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 960 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 800 d} - \frac {150 \left (\operatorname {atan}{\left (\frac {5 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4} - \frac {3}{4} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{800 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 960 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 800 d} + \frac {125 \left (\operatorname {atan}{\left (\frac {5 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4} - \frac {3}{4} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{800 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 960 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 800 d} + \frac {36 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{800 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 960 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 800 d} - \frac {60}{800 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 960 d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 800 d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5-3*sin(d*x+c))**2,x)

[Out]

Piecewise((x/(5 - 3*sin(2*atan(3/5 - 4*I/5)))**2, Eq(c, -d*x + 2*atan(3/5 - 4*I/5))), (x/(5 - 3*sin(2*atan(3/5
+ 4*I/5)))**2, Eq(c, -d*x + 2*atan(3/5 + 4*I/5))), (x/(5 - 3*sin(c))**2, Eq(d, 0)), (125*(atan(5*tan(c/2 + d*
x/2)/4 - 3/4) + pi*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d*x/2)**2/(800*d*tan(c/2 + d*x/2)**2 - 960*d*tan(
c/2 + d*x/2) + 800*d) - 150*(atan(5*tan(c/2 + d*x/2)/4 - 3/4) + pi*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d
*x/2)/(800*d*tan(c/2 + d*x/2)**2 - 960*d*tan(c/2 + d*x/2) + 800*d) + 125*(atan(5*tan(c/2 + d*x/2)/4 - 3/4) + p
i*floor((c/2 + d*x/2 - pi/2)/pi))/(800*d*tan(c/2 + d*x/2)**2 - 960*d*tan(c/2 + d*x/2) + 800*d) + 36*tan(c/2 +
d*x/2)/(800*d*tan(c/2 + d*x/2)**2 - 960*d*tan(c/2 + d*x/2) + 800*d) - 60/(800*d*tan(c/2 + d*x/2)**2 - 960*d*ta
n(c/2 + d*x/2) + 800*d), True))

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