3.315 \(\int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=97 \[ -\frac {3 \cos (c+d x)}{a^3 d}+\frac {\sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {19 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)}+\frac {2 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)^2}-\frac {11 x}{2 a^3} \]

[Out]

-11/2*x/a^3-3*cos(d*x+c)/a^3/d+1/2*cos(d*x+c)*sin(d*x+c)/a^3/d+2/3*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^2-19/3*cos(
d*x+c)/a^3/d/(1+sin(d*x+c))

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Rubi [A]  time = 0.26, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2874, 2966, 2638, 2635, 8, 2650, 2648} \[ -\frac {3 \cos (c+d x)}{a^3 d}+\frac {\sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {19 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)}+\frac {2 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)^2}-\frac {11 x}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]

[Out]

(-11*x)/(2*a^3) - (3*Cos[c + d*x])/(a^3*d) + (Cos[c + d*x]*Sin[c + d*x])/(2*a^3*d) + (2*Cos[c + d*x])/(3*a^3*d
*(1 + Sin[c + d*x])^2) - (19*Cos[c + d*x])/(3*a^3*d*(1 + Sin[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 2874

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 0])

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 {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\int \frac {\sin ^3(c+d x) (a-a \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx}{a^2}\\ &=\frac {\int \left (-\frac {5}{a}+\frac {3 \sin (c+d x)}{a}-\frac {\sin ^2(c+d x)}{a}-\frac {2}{a (1+\sin (c+d x))^2}+\frac {7}{a (1+\sin (c+d x))}\right ) \, dx}{a^2}\\ &=-\frac {5 x}{a^3}-\frac {\int \sin ^2(c+d x) \, dx}{a^3}-\frac {2 \int \frac {1}{(1+\sin (c+d x))^2} \, dx}{a^3}+\frac {3 \int \sin (c+d x) \, dx}{a^3}+\frac {7 \int \frac {1}{1+\sin (c+d x)} \, dx}{a^3}\\ &=-\frac {5 x}{a^3}-\frac {3 \cos (c+d x)}{a^3 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac {2 \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))^2}-\frac {7 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}-\frac {\int 1 \, dx}{2 a^3}-\frac {2 \int \frac {1}{1+\sin (c+d x)} \, dx}{3 a^3}\\ &=-\frac {11 x}{2 a^3}-\frac {3 \cos (c+d x)}{a^3 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac {2 \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))^2}-\frac {19 \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 1.09, size = 197, normalized size = 2.03 \[ -\frac {1980 d x \sin \left (c+\frac {d x}{2}\right )+660 d x \sin \left (c+\frac {3 d x}{2}\right )+498 \sin \left (2 c+\frac {3 d x}{2}\right )+135 \sin \left (2 c+\frac {5 d x}{2}\right )+15 \sin \left (4 c+\frac {7 d x}{2}\right )-1326 \cos \left (c+\frac {d x}{2}\right )+2012 \cos \left (c+\frac {3 d x}{2}\right )-660 d x \cos \left (2 c+\frac {3 d x}{2}\right )-135 \cos \left (3 c+\frac {5 d x}{2}\right )+15 \cos \left (3 c+\frac {7 d x}{2}\right )-3216 \sin \left (\frac {d x}{2}\right )+1980 d x \cos \left (\frac {d x}{2}\right )}{240 a^3 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]

[Out]

-1/240*(1980*d*x*Cos[(d*x)/2] - 1326*Cos[c + (d*x)/2] + 2012*Cos[c + (3*d*x)/2] - 660*d*x*Cos[2*c + (3*d*x)/2]
- 135*Cos[3*c + (5*d*x)/2] + 15*Cos[3*c + (7*d*x)/2] - 3216*Sin[(d*x)/2] + 1980*d*x*Sin[c + (d*x)/2] + 660*d*
x*Sin[c + (3*d*x)/2] + 498*Sin[2*c + (3*d*x)/2] + 135*Sin[2*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])^3)

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fricas [A]  time = 0.50, size = 163, normalized size = 1.68 \[ \frac {3 \, \cos \left (d x + c\right )^{4} - {\left (33 \, d x - 53\right )} \cos \left (d x + c\right )^{2} - 12 \, \cos \left (d x + c\right )^{3} + 66 \, d x + {\left (33 \, d x + 64\right )} \cos \left (d x + c\right ) + {\left (3 \, \cos \left (d x + c\right )^{3} + 66 \, d x + {\left (33 \, d x + 68\right )} \cos \left (d x + c\right ) + 15 \, \cos \left (d x + c\right )^{2} + 4\right )} \sin \left (d x + c\right ) - 4}{6 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d \cos \left (d x + c\right ) - 2 \, a^{3} d - {\left (a^{3} d \cos \left (d x + c\right ) + 2 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/6*(3*cos(d*x + c)^4 - (33*d*x - 53)*cos(d*x + c)^2 - 12*cos(d*x + c)^3 + 66*d*x + (33*d*x + 64)*cos(d*x + c)
+ (3*cos(d*x + c)^3 + 66*d*x + (33*d*x + 68)*cos(d*x + c) + 15*cos(d*x + c)^2 + 4)*sin(d*x + c) - 4)/(a^3*d*c
os(d*x + c)^2 - a^3*d*cos(d*x + c) - 2*a^3*d - (a^3*d*cos(d*x + c) + 2*a^3*d)*sin(d*x + c))

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giac [A]  time = 0.22, size = 117, normalized size = 1.21 \[ -\frac {\frac {33 \, {\left (d x + c\right )}}{a^{3}} + \frac {6 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} + \frac {4 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 17\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/6*(33*(d*x + c)/a^3 + 6*(tan(1/2*d*x + 1/2*c)^3 + 6*tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c) + 6)/((ta
n(1/2*d*x + 1/2*c)^2 + 1)^2*a^3) + 4*(15*tan(1/2*d*x + 1/2*c)^2 + 36*tan(1/2*d*x + 1/2*c) + 17)/(a^3*(tan(1/2*
d*x + 1/2*c) + 1)^3))/d

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maple [B]  time = 0.41, size = 205, normalized size = 2.11 \[ -\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {6}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {11 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {8}{3 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {4}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {10}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x)

[Out]

-1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-6/d/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^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-6/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2-11/d/a^3*arctan(ta
n(1/2*d*x+1/2*c))+8/3/d/a^3/(tan(1/2*d*x+1/2*c)+1)^3-4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2-10/d/a^3/(tan(1/2*d*x+1/
2*c)+1)

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maxima [B]  time = 0.45, size = 314, normalized size = 3.24 \[ -\frac {\frac {\frac {123 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {161 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {210 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {154 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {99 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {33 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 52}{a^{3} + \frac {3 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {7 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {7 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} + \frac {33 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/3*((123*sin(d*x + c)/(cos(d*x + c) + 1) + 161*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 210*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 154*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 99*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 33*sin(d
*x + c)^6/(cos(d*x + c) + 1)^6 + 52)/(a^3 + 3*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*a^3*sin(d*x + c)^2/(cos(
d*x + c) + 1)^2 + 7*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 7*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 5*a^
3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3*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) + 33*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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mupad [B]  time = 12.07, size = 121, normalized size = 1.25 \[ -\frac {11\,x}{2\,a^3}-\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+33\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {154\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+70\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {161\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+41\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {52}{3}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^3\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*sin(c + d*x)^3)/(a + a*sin(c + d*x))^3,x)

[Out]

- (11*x)/(2*a^3) - (41*tan(c/2 + (d*x)/2) + (161*tan(c/2 + (d*x)/2)^2)/3 + 70*tan(c/2 + (d*x)/2)^3 + (154*tan(
c/2 + (d*x)/2)^4)/3 + 33*tan(c/2 + (d*x)/2)^5 + 11*tan(c/2 + (d*x)/2)^6 + 52/3)/(a^3*d*(tan(c/2 + (d*x)/2) + 1
)^3*(tan(c/2 + (d*x)/2)^2 + 1)^2)

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sympy [A]  time = 63.55, size = 2264, normalized size = 23.34 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*sin(d*x+c)**3/(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((-33*d*x*tan(c/2 + d*x/2)**7/(6*a**3*d*tan(c/2 + d*x/2)**7 + 18*a**3*d*tan(c/2 + d*x/2)**6 + 30*a**3
*d*tan(c/2 + d*x/2)**5 + 42*a**3*d*tan(c/2 + d*x/2)**4 + 42*a**3*d*tan(c/2 + d*x/2)**3 + 30*a**3*d*tan(c/2 + d
*x/2)**2 + 18*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 99*d*x*tan(c/2 + d*x/2)**6/(6*a**3*d*tan(c/2 + d*x/2)**7 +
18*a**3*d*tan(c/2 + d*x/2)**6 + 30*a**3*d*tan(c/2 + d*x/2)**5 + 42*a**3*d*tan(c/2 + d*x/2)**4 + 42*a**3*d*tan
(c/2 + d*x/2)**3 + 30*a**3*d*tan(c/2 + d*x/2)**2 + 18*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 165*d*x*tan(c/2 +
d*x/2)**5/(6*a**3*d*tan(c/2 + d*x/2)**7 + 18*a**3*d*tan(c/2 + d*x/2)**6 + 30*a**3*d*tan(c/2 + d*x/2)**5 + 42*a
**3*d*tan(c/2 + d*x/2)**4 + 42*a**3*d*tan(c/2 + d*x/2)**3 + 30*a**3*d*tan(c/2 + d*x/2)**2 + 18*a**3*d*tan(c/2
+ d*x/2) + 6*a**3*d) - 231*d*x*tan(c/2 + d*x/2)**4/(6*a**3*d*tan(c/2 + d*x/2)**7 + 18*a**3*d*tan(c/2 + d*x/2)*
*6 + 30*a**3*d*tan(c/2 + d*x/2)**5 + 42*a**3*d*tan(c/2 + d*x/2)**4 + 42*a**3*d*tan(c/2 + d*x/2)**3 + 30*a**3*d
*tan(c/2 + d*x/2)**2 + 18*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 231*d*x*tan(c/2 + d*x/2)**3/(6*a**3*d*tan(c/2
+ d*x/2)**7 + 18*a**3*d*tan(c/2 + d*x/2)**6 + 30*a**3*d*tan(c/2 + d*x/2)**5 + 42*a**3*d*tan(c/2 + d*x/2)**4 +
42*a**3*d*tan(c/2 + d*x/2)**3 + 30*a**3*d*tan(c/2 + d*x/2)**2 + 18*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 165*d
*x*tan(c/2 + d*x/2)**2/(6*a**3*d*tan(c/2 + d*x/2)**7 + 18*a**3*d*tan(c/2 + d*x/2)**6 + 30*a**3*d*tan(c/2 + d*x
/2)**5 + 42*a**3*d*tan(c/2 + d*x/2)**4 + 42*a**3*d*tan(c/2 + d*x/2)**3 + 30*a**3*d*tan(c/2 + d*x/2)**2 + 18*a*
*3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 99*d*x*tan(c/2 + d*x/2)/(6*a**3*d*tan(c/2 + d*x/2)**7 + 18*a**3*d*tan(c/2
+ d*x/2)**6 + 30*a**3*d*tan(c/2 + d*x/2)**5 + 42*a**3*d*tan(c/2 + d*x/2)**4 + 42*a**3*d*tan(c/2 + d*x/2)**3 +
30*a**3*d*tan(c/2 + d*x/2)**2 + 18*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 33*d*x/(6*a**3*d*tan(c/2 + d*x/2)**7
+ 18*a**3*d*tan(c/2 + d*x/2)**6 + 30*a**3*d*tan(c/2 + d*x/2)**5 + 42*a**3*d*tan(c/2 + d*x/2)**4 + 42*a**3*d*ta
n(c/2 + d*x/2)**3 + 30*a**3*d*tan(c/2 + d*x/2)**2 + 18*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 66*tan(c/2 + d*x/
2)**6/(6*a**3*d*tan(c/2 + d*x/2)**7 + 18*a**3*d*tan(c/2 + d*x/2)**6 + 30*a**3*d*tan(c/2 + d*x/2)**5 + 42*a**3*
d*tan(c/2 + d*x/2)**4 + 42*a**3*d*tan(c/2 + d*x/2)**3 + 30*a**3*d*tan(c/2 + d*x/2)**2 + 18*a**3*d*tan(c/2 + d*
x/2) + 6*a**3*d) - 198*tan(c/2 + d*x/2)**5/(6*a**3*d*tan(c/2 + d*x/2)**7 + 18*a**3*d*tan(c/2 + d*x/2)**6 + 30*
a**3*d*tan(c/2 + d*x/2)**5 + 42*a**3*d*tan(c/2 + d*x/2)**4 + 42*a**3*d*tan(c/2 + d*x/2)**3 + 30*a**3*d*tan(c/2
+ d*x/2)**2 + 18*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 308*tan(c/2 + d*x/2)**4/(6*a**3*d*tan(c/2 + d*x/2)**7
+ 18*a**3*d*tan(c/2 + d*x/2)**6 + 30*a**3*d*tan(c/2 + d*x/2)**5 + 42*a**3*d*tan(c/2 + d*x/2)**4 + 42*a**3*d*ta
n(c/2 + d*x/2)**3 + 30*a**3*d*tan(c/2 + d*x/2)**2 + 18*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 420*tan(c/2 + d*x
/2)**3/(6*a**3*d*tan(c/2 + d*x/2)**7 + 18*a**3*d*tan(c/2 + d*x/2)**6 + 30*a**3*d*tan(c/2 + d*x/2)**5 + 42*a**3
*d*tan(c/2 + d*x/2)**4 + 42*a**3*d*tan(c/2 + d*x/2)**3 + 30*a**3*d*tan(c/2 + d*x/2)**2 + 18*a**3*d*tan(c/2 + d
*x/2) + 6*a**3*d) - 322*tan(c/2 + d*x/2)**2/(6*a**3*d*tan(c/2 + d*x/2)**7 + 18*a**3*d*tan(c/2 + d*x/2)**6 + 30
*a**3*d*tan(c/2 + d*x/2)**5 + 42*a**3*d*tan(c/2 + d*x/2)**4 + 42*a**3*d*tan(c/2 + d*x/2)**3 + 30*a**3*d*tan(c/
2 + d*x/2)**2 + 18*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 246*tan(c/2 + d*x/2)/(6*a**3*d*tan(c/2 + d*x/2)**7 +
18*a**3*d*tan(c/2 + d*x/2)**6 + 30*a**3*d*tan(c/2 + d*x/2)**5 + 42*a**3*d*tan(c/2 + d*x/2)**4 + 42*a**3*d*tan(
c/2 + d*x/2)**3 + 30*a**3*d*tan(c/2 + d*x/2)**2 + 18*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 104/(6*a**3*d*tan(c
/2 + d*x/2)**7 + 18*a**3*d*tan(c/2 + d*x/2)**6 + 30*a**3*d*tan(c/2 + d*x/2)**5 + 42*a**3*d*tan(c/2 + d*x/2)**4
+ 42*a**3*d*tan(c/2 + d*x/2)**3 + 30*a**3*d*tan(c/2 + d*x/2)**2 + 18*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d), Ne(
d, 0)), (x*sin(c)**3*cos(c)**2/(a*sin(c) + a)**3, True))

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