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(d*x+c)/a^3/d+2/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^3-31/15*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^2
+104/15*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))

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Rubi [A]  time = 0.19, antiderivative size = 103, normalized size of antiderivative = 1.00, 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 2638

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

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]

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 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]

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*}

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Mathematica [B]  time = 0.80, 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]

-1/120*(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]))/(a^3*d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin
[(c + d*x)/2])^5)

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fricas [B]  time = 0.45, size = 225, normalized size = 2.18 \[ \frac {15 \, A \cos \left (d x + c\right )^{4} + {\left (60 \, A d x + 149 \, A\right )} \cos \left (d x + c\right )^{3} - 240 \, A d x + {\left (180 \, A d x - 103 \, A\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (40 \, A d x + 81 \, A\right )} \cos \left (d x + c\right ) + {\left (15 \, A \cos \left (d x + c\right )^{3} - 240 \, A d x + 2 \, {\left (30 \, A d x - 67 \, A\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (40 \, A d x + 79 \, A\right )} \cos \left (d x + c\right ) + 6 \, A\right )} \sin \left (d x + c\right ) - 6 \, A}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

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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giac [A]  time = 0.18, size = 113, normalized size = 1.10 \[ \frac {2 \, {\left (\frac {30 \, {\left (d x + c\right )} A}{a^{3}} + \frac {15 \, A}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} + \frac {60 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 285 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 505 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 335 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 79 \, A}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}\right )}}{15 \, d} \]

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

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maple [A]  time = 0.43, size = 155, normalized size = 1.50 \[ \frac {2 A}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {8 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {16 A}{5 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {8 A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {4 A}{3 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {6 A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8 A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

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 = 0.56, size = 543, normalized size = 5.27 \[ \frac {2 \, {\left (3 \, A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {189 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {200 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {160 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {75 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 24}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {11 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {11 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} + \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + A {\left (\frac {\frac {95 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {145 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {75 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 22}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}\right )}}{15 \, d} \]

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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mupad [B]  time = 16.93, size = 261, normalized size = 2.53 \[ \frac {4\,A\,x}{a^3}-\frac {\left (20\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (75\,c+75\,d\,x+30\right )}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (44\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (165\,c+165\,d\,x+150\right )}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (60\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (225\,c+225\,d\,x+320\right )}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (60\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (225\,c+225\,d\,x+385\right )}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (44\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (165\,c+165\,d\,x+367\right )}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (20\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (75\,c+75\,d\,x+205\right )}{15}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+4\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (15\,c+15\,d\,x+47\right )}{15}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 - (tan(c/2 + (d*x)/2)*(20*A*(c + d*x) - (4*A*(75*c + 75*d*x + 205))/15) + tan(c/2 + (d*x)/2)^6*(20
*A*(c + d*x) - (4*A*(75*c + 75*d*x + 30))/15) + tan(c/2 + (d*x)/2)^5*(44*A*(c + d*x) - (4*A*(165*c + 165*d*x +
150))/15) + tan(c/2 + (d*x)/2)^2*(44*A*(c + d*x) - (4*A*(165*c + 165*d*x + 367))/15) + tan(c/2 + (d*x)/2)^4*(
60*A*(c + d*x) - (4*A*(225*c + 225*d*x + 320))/15) + tan(c/2 + (d*x)/2)^3*(60*A*(c + d*x) - (4*A*(225*c + 225*
d*x + 385))/15) + 4*A*(c + d*x) - (4*A*(15*c + 15*d*x + 47))/15)/(a^3*d*(tan(c/2 + (d*x)/2) + 1)^5*(tan(c/2 +
(d*x)/2)^2 + 1))

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sympy [A]  time = 45.41, size = 2290, normalized size = 22.23 \[ \text {result too large to display} \]

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]

Piecewise((60*A*d*x*tan(c/2 + d*x/2)**7/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a
**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*tan(c/2 + d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c
/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 300*A*d*x*tan(c/2 + d*x/2)**6/(15*a**3*d*tan(c/2 +
d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*tan(c/2 + d*x/2)**4 +
225*a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 66
0*A*d*x*tan(c/2 + d*x/2)**5/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/
2 + d*x/2)**5 + 225*a**3*d*tan(c/2 + d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)*
*2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 900*A*d*x*tan(c/2 + d*x/2)**4/(15*a**3*d*tan(c/2 + d*x/2)**7 +
75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*tan(c/2 + d*x/2)**4 + 225*a**3*d*t
an(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 900*A*d*x*tan(
c/2 + d*x/2)**3/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)**
5 + 225*a**3*d*tan(c/2 + d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3
*d*tan(c/2 + d*x/2) + 15*a**3*d) + 660*A*d*x*tan(c/2 + d*x/2)**2/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*a**3*d*ta
n(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*tan(c/2 + d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x
/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 300*A*d*x*tan(c/2 + d*x/2)
/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*
tan(c/2 + d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*
x/2) + 15*a**3*d) + 60*A*d*x/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c
/2 + d*x/2)**5 + 225*a**3*d*tan(c/2 + d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)
**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 120*A*tan(c/2 + d*x/2)**6/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*
a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*tan(c/2 + d*x/2)**4 + 225*a**3*d*tan(
c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 600*A*tan(c/2 + d
*x/2)**5/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)**5 + 225
*a**3*d*tan(c/2 + d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(
c/2 + d*x/2) + 15*a**3*d) + 1280*A*tan(c/2 + d*x/2)**4/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*
x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*tan(c/2 + d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x/2)**3 + 1
65*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 1540*A*tan(c/2 + d*x/2)**3/(15*a**3*
d*tan(c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*tan(c/2 +
d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*
a**3*d) + 1468*A*tan(c/2 + d*x/2)**2/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3
*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*tan(c/2 + d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2
+ d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 820*A*tan(c/2 + d*x/2)/(15*a**3*d*tan(c/2 + d*x/2)**7
+ 75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*tan(c/2 + d*x/2)**4 + 225*a**3*d
*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 188*A/(15*a*
*3*d*tan(c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*tan(c/2
+ d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) +
15*a**3*d), Ne(d, 0)), (x*(-A*sin(c) + A)*sin(c)**3/(a*sin(c) + a)**3, True))

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