\(\int \frac {1}{(-5+3 \sin (c+d x))^2} \, dx\) [27]

Optimal result

Integrand size = 12, antiderivative size = 58 \[ \int \frac {1}{(-5+3 \sin (c+d x))^2} \, dx=\frac {5 x}{64}-\frac {5 \arctan \left (\frac {\cos (c+d x)}{3-\sin (c+d x)}\right )}{32 d}-\frac {3 \cos (c+d x)}{16 d (5-3 \sin (c+d x))} \]

[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))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2743, 12, 2737} \[ \int \frac {1}{(-5+3 \sin (c+d x))^2} \, dx=-\frac {5 \arctan \left (\frac {\cos (c+d x)}{3-\sin (c+d x)}\right )}{32 d}-\frac {3 \cos (c+d x)}{16 d (5-3 \sin (c+d x))}+\frac {5 x}{64} \]

[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 2737

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

Rule 2743

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] && Integ
erQ[2*n]

Rubi steps \begin{align*} \text {integral}& = -\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 \arctan \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*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.57 \[ \int \frac {1}{(-5+3 \sin (c+d x))^2} \, dx=\frac {-25 \arctan \left (\frac {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}\right )+\frac {6 (-5+5 \cos (c+d x)+3 \sin (c+d x))}{-5+3 \sin (c+d x)}}{160 d} \]

[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)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {\frac {\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{200}-\frac {3}{40}}{\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+1}+\frac {5 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}-\frac {3}{4}\right )}{32}}{d}\)	\(63\)
default	\(\frac {\frac {\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{200}-\frac {3}{40}}{\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+1}+\frac {5 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}-\frac {3}{4}\right )}{32}}{d}\)	\(63\)
risch	\(-\frac {5 \,{\mathrm e}^{i \left (d x +c \right )}-3 i}{8 d \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-3-10 i {\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {5 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-3 i\right )}{64 d}-\frac {5 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i}{3}\right )}{64 d}\)	\(86\)
parallelrisch	\(\frac {\left (-75 i \sin \left (d x +c \right )+125 i\right ) \ln \left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3-4 i\right )+\left (75 i \sin \left (d x +c \right )-125 i\right ) \ln \left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3+4 i\right )+60 \cos \left (d x +c \right )-36 \sin \left (d x +c \right )+60}{960 d \sin \left (d x +c \right )-1600 d}\)	\(93\)

[In]

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

[Out]

1/d*(2*(9/400*tan(1/2*d*x+1/2*c)-3/80)/(tan(1/2*d*x+1/2*c)^2-6/5*tan(1/2*d*x+1/2*c)+1)+5/32*arctan(5/4*tan(1/2
*d*x+1/2*c)-3/4))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(-5+3 \sin (c+d x))^2} \, dx=\frac {5 \, {\left (3 \, \sin \left (d x + c\right ) - 5\right )} \arctan \left (\frac {5 \, \sin \left (d x + c\right ) - 3}{4 \, \cos \left (d x + c\right )}\right ) + 12 \, \cos \left (d x + c\right )}{64 \, {\left (3 \, d \sin \left (d x + c\right ) - 5 \, d\right )}} \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.91 (sec) , antiderivative size = 384, normalized size of antiderivative = 6.62 \[ \int \frac {1}{(-5+3 \sin (c+d x))^2} \, dx=\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 (3 \sin {\left (c \right )} - 5\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} \]

[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/(3*sin(c) - 5)**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*ta
n(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) +
pi*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*
tan(c/2 + d*x/2) + 800*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.60 \[ \int \frac {1}{(-5+3 \sin (c+d x))^2} \, dx=-\frac {\frac {12 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 5\right )}}{\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 5} - 25 \, \arctan \left (\frac {5 \, \sin \left (d x + c\right )}{4 \, {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {3}{4}\right )}{160 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.66 \[ \int \frac {1}{(-5+3 \sin (c+d x))^2} \, dx=\frac {25 \, d x + 25 \, c + \frac {24 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5\right )}}{5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5} + 50 \, \arctan \left (\frac {3 \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 3}{\cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) - 9}\right )}{320 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.43 \[ \int \frac {1}{(-5+3 \sin (c+d x))^2} \, dx=\frac {5\,\mathrm {atan}\left (\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {3}{4}\right )}{32\,d}-\frac {5\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{32\,d}+\frac {\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{200}-\frac {3}{40}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+1\right )} \]

[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))