3.29 \(\int \frac {1}{(-5+3 \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=108 \[ -\frac {311 \cos (c+d x)}{8192 d (5-3 \sin (c+d x))}-\frac {25 \cos (c+d x)}{512 d (5-3 \sin (c+d x))^2}-\frac {\cos (c+d x)}{16 d (5-3 \sin (c+d x))^3}-\frac {385 \tan ^{-1}\left (\frac {\cos (c+d x)}{3-\sin (c+d x)}\right )}{16384 d}+\frac {385 x}{32768} \]

[Out]

385/32768*x-385/16384*arctan(cos(d*x+c)/(3-sin(d*x+c)))/d-1/16*cos(d*x+c)/d/(5-3*sin(d*x+c))^3-25/512*cos(d*x+
c)/d/(5-3*sin(d*x+c))^2-311/8192*cos(d*x+c)/d/(5-3*sin(d*x+c))

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Rubi [A]  time = 0.09, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2664, 2754, 12, 2658} \[ -\frac {311 \cos (c+d x)}{8192 d (5-3 \sin (c+d x))}-\frac {25 \cos (c+d x)}{512 d (5-3 \sin (c+d x))^2}-\frac {\cos (c+d x)}{16 d (5-3 \sin (c+d x))^3}-\frac {385 \tan ^{-1}\left (\frac {\cos (c+d x)}{3-\sin (c+d x)}\right )}{16384 d}+\frac {385 x}{32768} \]

Antiderivative was successfully verified.

[In]

Int[(-5 + 3*Sin[c + d*x])^(-4),x]

[Out]

(385*x)/32768 - (385*ArcTan[Cos[c + d*x]/(3 - Sin[c + d*x])])/(16384*d) - Cos[c + d*x]/(16*d*(5 - 3*Sin[c + d*
x])^3) - (25*Cos[c + d*x])/(512*d*(5 - 3*Sin[c + d*x])^2) - (311*Cos[c + d*x])/(8192*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 2658

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*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] && NegQ[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]

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {1}{(-5+3 \sin (c+d x))^4} \, dx &=-\frac {\cos (c+d x)}{16 d (5-3 \sin (c+d x))^3}-\frac {1}{48} \int \frac {15+6 \sin (c+d x)}{(-5+3 \sin (c+d x))^3} \, dx\\ &=-\frac {\cos (c+d x)}{16 d (5-3 \sin (c+d x))^3}-\frac {25 \cos (c+d x)}{512 d (5-3 \sin (c+d x))^2}+\frac {\int \frac {186+75 \sin (c+d x)}{(-5+3 \sin (c+d x))^2} \, dx}{1536}\\ &=-\frac {\cos (c+d x)}{16 d (5-3 \sin (c+d x))^3}-\frac {25 \cos (c+d x)}{512 d (5-3 \sin (c+d x))^2}-\frac {311 \cos (c+d x)}{8192 d (5-3 \sin (c+d x))}-\frac {\int \frac {1155}{-5+3 \sin (c+d x)} \, dx}{24576}\\ &=-\frac {\cos (c+d x)}{16 d (5-3 \sin (c+d x))^3}-\frac {25 \cos (c+d x)}{512 d (5-3 \sin (c+d x))^2}-\frac {311 \cos (c+d x)}{8192 d (5-3 \sin (c+d x))}-\frac {385 \int \frac {1}{-5+3 \sin (c+d x)} \, dx}{8192}\\ &=\frac {385 x}{32768}-\frac {385 \tan ^{-1}\left (\frac {\cos (c+d x)}{3-\sin (c+d x)}\right )}{16384 d}-\frac {\cos (c+d x)}{16 d (5-3 \sin (c+d x))^3}-\frac {25 \cos (c+d x)}{512 d (5-3 \sin (c+d x))^2}-\frac {311 \cos (c+d x)}{8192 d (5-3 \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 133, normalized size = 1.23 \[ \frac {\frac {305091 \sin (c+d x)-105300 \sin (2 (c+d x))-8397 \sin (3 (c+d x))+219735 \cos (c+d x)+83970 \cos (2 (c+d x))-13995 \cos (3 (c+d x))-239470}{2 (3 \sin (c+d x)-5)^3}-1925 \tan ^{-1}\left (\frac {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}\right )}{81920 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 3*Sin[c + d*x])^(-4),x]

[Out]

(-1925*ArcTan[(2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])] + (-239470 + 21
9735*Cos[c + d*x] + 83970*Cos[2*(c + d*x)] - 13995*Cos[3*(c + d*x)] + 305091*Sin[c + d*x] - 105300*Sin[2*(c +
d*x)] - 8397*Sin[3*(c + d*x)])/(2*(-5 + 3*Sin[c + d*x])^3))/(81920*d)

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fricas [A]  time = 0.43, size = 130, normalized size = 1.20 \[ -\frac {11196 \, \cos \left (d x + c\right )^{3} - 385 \, {\left (135 \, \cos \left (d x + c\right )^{2} - 9 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 28\right )} \sin \left (d x + c\right ) - 260\right )} \arctan \left (\frac {5 \, \sin \left (d x + c\right ) - 3}{4 \, \cos \left (d x + c\right )}\right ) + 42120 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 52344 \, \cos \left (d x + c\right )}{32768 \, {\left (135 \, d \cos \left (d x + c\right )^{2} - 9 \, {\left (3 \, d \cos \left (d x + c\right )^{2} - 28 \, d\right )} \sin \left (d x + c\right ) - 260 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+3*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/32768*(11196*cos(d*x + c)^3 - 385*(135*cos(d*x + c)^2 - 9*(3*cos(d*x + c)^2 - 28)*sin(d*x + c) - 260)*arcta
n(1/4*(5*sin(d*x + c) - 3)/cos(d*x + c)) + 42120*cos(d*x + c)*sin(d*x + c) - 52344*cos(d*x + c))/(135*d*cos(d*
x + c)^2 - 9*(3*d*cos(d*x + c)^2 - 28*d)*sin(d*x + c) - 260*d)

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giac [A]  time = 0.39, size = 148, normalized size = 1.37 \[ \frac {48125 \, d x + 48125 \, c + \frac {72 \, {\left (110925 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 373735 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 637794 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 672110 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 403425 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 142875\right )}}{{\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}^{3}} + 96250 \, \arctan \left (\frac {3 \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 3}{\cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) - 9}\right )}{4096000 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+3*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/4096000*(48125*d*x + 48125*c + 72*(110925*tan(1/2*d*x + 1/2*c)^5 - 373735*tan(1/2*d*x + 1/2*c)^4 + 637794*ta
n(1/2*d*x + 1/2*c)^3 - 672110*tan(1/2*d*x + 1/2*c)^2 + 403425*tan(1/2*d*x + 1/2*c) - 142875)/(5*tan(1/2*d*x +
1/2*c)^2 - 6*tan(1/2*d*x + 1/2*c) + 5)^3 + 96250*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 [B]  time = 0.09, size = 272, normalized size = 2.52 \[ \frac {39933 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20480 d \left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )^{3}}-\frac {672723 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{102400 d \left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )^{3}}+\frac {2870073 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{256000 d \left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )^{3}}-\frac {604899 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{51200 d \left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )^{3}}+\frac {145233 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{20480 d \left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )^{3}}-\frac {10287}{4096 d \left (5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5\right )^{3}}+\frac {385 \arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}-\frac {3}{4}\right )}{16384 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-5+3*sin(d*x+c))^4,x)

[Out]

39933/20480/d/(5*tan(1/2*d*x+1/2*c)^2-6*tan(1/2*d*x+1/2*c)+5)^3*tan(1/2*d*x+1/2*c)^5-672723/102400/d/(5*tan(1/
2*d*x+1/2*c)^2-6*tan(1/2*d*x+1/2*c)+5)^3*tan(1/2*d*x+1/2*c)^4+2870073/256000/d/(5*tan(1/2*d*x+1/2*c)^2-6*tan(1
/2*d*x+1/2*c)+5)^3*tan(1/2*d*x+1/2*c)^3-604899/51200/d/(5*tan(1/2*d*x+1/2*c)^2-6*tan(1/2*d*x+1/2*c)+5)^3*tan(1
/2*d*x+1/2*c)^2+145233/20480/d/(5*tan(1/2*d*x+1/2*c)^2-6*tan(1/2*d*x+1/2*c)+5)^3*tan(1/2*d*x+1/2*c)-10287/4096
/d/(5*tan(1/2*d*x+1/2*c)^2-6*tan(1/2*d*x+1/2*c)+5)^3+385/16384/d*arctan(5/4*tan(1/2*d*x+1/2*c)-3/4)

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maxima [B]  time = 0.74, size = 253, normalized size = 2.34 \[ -\frac {\frac {36 \, {\left (\frac {403425 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {672110 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {637794 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {373735 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {110925 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 142875\right )}}{\frac {450 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {915 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1116 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {915 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {450 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {125 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - 125} - 48125 \, \arctan \left (\frac {5 \, \sin \left (d x + c\right )}{4 \, {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {3}{4}\right )}{2048000 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+3*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/2048000*(36*(403425*sin(d*x + c)/(cos(d*x + c) + 1) - 672110*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 637794*s
in(d*x + c)^3/(cos(d*x + c) + 1)^3 - 373735*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 110925*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5 - 142875)/(450*sin(d*x + c)/(cos(d*x + c) + 1) - 915*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1116
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 915*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 450*sin(d*x + c)^5/(cos(d*x +
c) + 1)^5 - 125*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 125) - 48125*arctan(5/4*sin(d*x + c)/(cos(d*x + c) + 1)
- 3/4))/d

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mupad [B]  time = 0.00, size = 187, normalized size = 1.73 \[ \frac {385\,\mathrm {atan}\left (\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {3}{4}\right )}{16384\,d}-\frac {385\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{16384\,d}+\frac {\frac {39933\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2560000}-\frac {672723\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{12800000}+\frac {2870073\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32000000}-\frac {604899\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{6400000}+\frac {145233\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2560000}-\frac {10287}{512000}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+\frac {183\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{25}-\frac {1116\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{125}+\frac {183\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{25}-\frac {18\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*sin(c + d*x) - 5)^4,x)

[Out]

(385*atan((5*tan(c/2 + (d*x)/2))/4 - 3/4))/(16384*d) - (385*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(16384*d) +
((145233*tan(c/2 + (d*x)/2))/2560000 - (604899*tan(c/2 + (d*x)/2)^2)/6400000 + (2870073*tan(c/2 + (d*x)/2)^3)/
32000000 - (672723*tan(c/2 + (d*x)/2)^4)/12800000 + (39933*tan(c/2 + (d*x)/2)^5)/2560000 - 10287/512000)/(d*((
183*tan(c/2 + (d*x)/2)^2)/25 - (18*tan(c/2 + (d*x)/2))/5 - (1116*tan(c/2 + (d*x)/2)^3)/125 + (183*tan(c/2 + (d
*x)/2)^4)/25 - (18*tan(c/2 + (d*x)/2)^5)/5 + tan(c/2 + (d*x)/2)^6 + 1))

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sympy [A]  time = 10.78, size = 1690, normalized size = 15.65 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+3*sin(d*x+c))**4,x)

[Out]

Piecewise((x/(-5 + 3*sin(2*atan(3/5 - 4*I/5)))**4, Eq(c, -d*x + 2*atan(3/5 - 4*I/5))), (x/(-5 + 3*sin(2*atan(3
/5 + 4*I/5)))**4, Eq(c, -d*x + 2*atan(3/5 + 4*I/5))), (x/(3*sin(c) - 5)**4, Eq(d, 0)), (6015625*(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)**6/(256000000*d*tan(c/2 + d*x/2)**6
- 921600000*d*tan(c/2 + d*x/2)**5 + 1873920000*d*tan(c/2 + d*x/2)**4 - 2285568000*d*tan(c/2 + d*x/2)**3 + 1873
920000*d*tan(c/2 + d*x/2)**2 - 921600000*d*tan(c/2 + d*x/2) + 256000000*d) - 21656250*(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)**5/(256000000*d*tan(c/2 + d*x/2)**6 - 92160000
0*d*tan(c/2 + d*x/2)**5 + 1873920000*d*tan(c/2 + d*x/2)**4 - 2285568000*d*tan(c/2 + d*x/2)**3 + 1873920000*d*t
an(c/2 + d*x/2)**2 - 921600000*d*tan(c/2 + d*x/2) + 256000000*d) + 44034375*(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)**4/(256000000*d*tan(c/2 + d*x/2)**6 - 921600000*d*tan(c/
2 + d*x/2)**5 + 1873920000*d*tan(c/2 + d*x/2)**4 - 2285568000*d*tan(c/2 + d*x/2)**3 + 1873920000*d*tan(c/2 + d
*x/2)**2 - 921600000*d*tan(c/2 + d*x/2) + 256000000*d) - 53707500*(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)**3/(256000000*d*tan(c/2 + d*x/2)**6 - 921600000*d*tan(c/2 + d*x/2)
**5 + 1873920000*d*tan(c/2 + d*x/2)**4 - 2285568000*d*tan(c/2 + d*x/2)**3 + 1873920000*d*tan(c/2 + d*x/2)**2 -
921600000*d*tan(c/2 + d*x/2) + 256000000*d) + 44034375*(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/(256000000*d*tan(c/2 + d*x/2)**6 - 921600000*d*tan(c/2 + d*x/2)**5 + 1873
920000*d*tan(c/2 + d*x/2)**4 - 2285568000*d*tan(c/2 + d*x/2)**3 + 1873920000*d*tan(c/2 + d*x/2)**2 - 921600000
*d*tan(c/2 + d*x/2) + 256000000*d) - 21656250*(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)/(256000000*d*tan(c/2 + d*x/2)**6 - 921600000*d*tan(c/2 + d*x/2)**5 + 1873920000*d*tan(
c/2 + d*x/2)**4 - 2285568000*d*tan(c/2 + d*x/2)**3 + 1873920000*d*tan(c/2 + d*x/2)**2 - 921600000*d*tan(c/2 +
d*x/2) + 256000000*d) + 6015625*(atan(5*tan(c/2 + d*x/2)/4 - 3/4) + pi*floor((c/2 + d*x/2 - pi/2)/pi))/(256000
000*d*tan(c/2 + d*x/2)**6 - 921600000*d*tan(c/2 + d*x/2)**5 + 1873920000*d*tan(c/2 + d*x/2)**4 - 2285568000*d*
tan(c/2 + d*x/2)**3 + 1873920000*d*tan(c/2 + d*x/2)**2 - 921600000*d*tan(c/2 + d*x/2) + 256000000*d) + 3993300
*tan(c/2 + d*x/2)**5/(256000000*d*tan(c/2 + d*x/2)**6 - 921600000*d*tan(c/2 + d*x/2)**5 + 1873920000*d*tan(c/2
+ d*x/2)**4 - 2285568000*d*tan(c/2 + d*x/2)**3 + 1873920000*d*tan(c/2 + d*x/2)**2 - 921600000*d*tan(c/2 + d*x
/2) + 256000000*d) - 13454460*tan(c/2 + d*x/2)**4/(256000000*d*tan(c/2 + d*x/2)**6 - 921600000*d*tan(c/2 + d*x
/2)**5 + 1873920000*d*tan(c/2 + d*x/2)**4 - 2285568000*d*tan(c/2 + d*x/2)**3 + 1873920000*d*tan(c/2 + d*x/2)**
2 - 921600000*d*tan(c/2 + d*x/2) + 256000000*d) + 22960584*tan(c/2 + d*x/2)**3/(256000000*d*tan(c/2 + d*x/2)**
6 - 921600000*d*tan(c/2 + d*x/2)**5 + 1873920000*d*tan(c/2 + d*x/2)**4 - 2285568000*d*tan(c/2 + d*x/2)**3 + 18
73920000*d*tan(c/2 + d*x/2)**2 - 921600000*d*tan(c/2 + d*x/2) + 256000000*d) - 24195960*tan(c/2 + d*x/2)**2/(2
56000000*d*tan(c/2 + d*x/2)**6 - 921600000*d*tan(c/2 + d*x/2)**5 + 1873920000*d*tan(c/2 + d*x/2)**4 - 22855680
00*d*tan(c/2 + d*x/2)**3 + 1873920000*d*tan(c/2 + d*x/2)**2 - 921600000*d*tan(c/2 + d*x/2) + 256000000*d) + 14
523300*tan(c/2 + d*x/2)/(256000000*d*tan(c/2 + d*x/2)**6 - 921600000*d*tan(c/2 + d*x/2)**5 + 1873920000*d*tan(
c/2 + d*x/2)**4 - 2285568000*d*tan(c/2 + d*x/2)**3 + 1873920000*d*tan(c/2 + d*x/2)**2 - 921600000*d*tan(c/2 +
d*x/2) + 256000000*d) - 5143500/(256000000*d*tan(c/2 + d*x/2)**6 - 921600000*d*tan(c/2 + d*x/2)**5 + 187392000
0*d*tan(c/2 + d*x/2)**4 - 2285568000*d*tan(c/2 + d*x/2)**3 + 1873920000*d*tan(c/2 + d*x/2)**2 - 921600000*d*ta
n(c/2 + d*x/2) + 256000000*d), True))

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