3.243 \(\int \frac{\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=153 \[ -\frac{A \cot ^3(c+d x)}{3 a^3 d}-\frac{10 A \cot (c+d x)}{a^3 d}-\frac{93 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)}-\frac{13 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^2}-\frac{2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}+\frac{18 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{2 A \cot (c+d x) \csc (c+d x)}{a^3 d} \]

[Out]

(18*A*ArcTanh[Cos[c + d*x]])/(a^3*d) - (10*A*Cot[c + d*x])/(a^3*d) - (A*Cot[c + d*x]^3)/(3*a^3*d) + (2*A*Cot[c
+ d*x]*Csc[c + d*x])/(a^3*d) - (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) - (13*A*Cos[c + d*x])/(5*a^3
*d*(1 + Sin[c + d*x])^2) - (93*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x]))

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Rubi [A]  time = 0.246471, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 7, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.219, Rules used = {2966, 3770, 3767, 8, 3768, 2650, 2648} \[ -\frac{A \cot ^3(c+d x)}{3 a^3 d}-\frac{10 A \cot (c+d x)}{a^3 d}-\frac{93 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)}-\frac{13 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^2}-\frac{2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}+\frac{18 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{2 A \cot (c+d x) \csc (c+d x)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^4*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

(18*A*ArcTanh[Cos[c + d*x]])/(a^3*d) - (10*A*Cot[c + d*x])/(a^3*d) - (A*Cot[c + d*x]^3)/(3*a^3*d) + (2*A*Cot[c
+ d*x]*Csc[c + d*x])/(a^3*d) - (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) - (13*A*Cos[c + d*x])/(5*a^3
*d*(1 + Sin[c + d*x])^2) - (93*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x]))

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]

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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 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]

Rubi steps

\begin{align*} \int \frac{\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx &=\int \left (-\frac{16 A \csc (c+d x)}{a^3}+\frac{9 A \csc ^2(c+d x)}{a^3}-\frac{4 A \csc ^3(c+d x)}{a^3}+\frac{A \csc ^4(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{16 A}{a^3 (1+\sin (c+d x))}\right ) \, dx\\ &=\frac{A \int \csc ^4(c+d x) \, dx}{a^3}+\frac{(2 A) \int \frac{1}{(1+\sin (c+d x))^3} \, dx}{a^3}-\frac{(4 A) \int \csc ^3(c+d x) \, dx}{a^3}+\frac{(7 A) \int \frac{1}{(1+\sin (c+d x))^2} \, dx}{a^3}+\frac{(9 A) \int \csc ^2(c+d x) \, dx}{a^3}-\frac{(16 A) \int \csc (c+d x) \, dx}{a^3}+\frac{(16 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{a^3}\\ &=\frac{16 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{2 A \cot (c+d x) \csc (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{16 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{(2 A) \int \csc (c+d x) \, dx}{a^3}+\frac{(7 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{3 a^3}-\frac{A \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^3 d}-\frac{(9 A) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=\frac{18 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{10 A \cot (c+d x)}{a^3 d}-\frac{A \cot ^3(c+d x)}{3 a^3 d}+\frac{2 A \cot (c+d x) \csc (c+d x)}{a^3 d}-\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac{13 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac{55 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{18 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{10 A \cot (c+d x)}{a^3 d}-\frac{A \cot ^3(c+d x)}{3 a^3 d}+\frac{2 A \cot (c+d x) \csc (c+d x)}{a^3 d}-\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac{13 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac{93 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [B]  time = 6.2232, size = 348, normalized size = 2.27 \[ \frac{A \left (\frac{29 \tan \left (\frac{1}{2} (c+d x)\right )}{6 d}-\frac{29 \cot \left (\frac{1}{2} (c+d x)\right )}{6 d}+\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right )}{2 d}-\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right )}{2 d}-\frac{18 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{18 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{186 \sin \left (\frac{1}{2} (c+d x)\right )}{5 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{13}{5 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{26 \sin \left (\frac{1}{2} (c+d x)\right )}{5 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}-\frac{2}{5 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}+\frac{4 \sin \left (\frac{1}{2} (c+d x)\right )}{5 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5}-\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{24 d}+\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{24 d}\right )}{a^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Csc[c + d*x]^4*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

(A*((-29*Cot[(c + d*x)/2])/(6*d) + Csc[(c + d*x)/2]^2/(2*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*d) + (
18*Log[Cos[(c + d*x)/2]])/d - (18*Log[Sin[(c + d*x)/2]])/d - Sec[(c + d*x)/2]^2/(2*d) + (4*Sin[(c + d*x)/2])/(
5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5) - 2/(5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4) + (26*Sin[(c +
d*x)/2])/(5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - 13/(5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (1
86*Sin[(c + d*x)/2])/(5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (29*Tan[(c + d*x)/2])/(6*d) + (Sec[(c + d*x
)/2]^2*Tan[(c + d*x)/2])/(24*d)))/a^3

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Maple [A]  time = 0.208, size = 249, normalized size = 1.6 \begin{align*}{\frac{A}{24\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{A}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+{\frac{39\,A}{8\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{16\,A}{5\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}+8\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}-20\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}+22\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}-50\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-{\frac{A}{24\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{A}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}-{\frac{39\,A}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-18\,{\frac{A\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x)

[Out]

1/24/d*A/a^3*tan(1/2*d*x+1/2*c)^3-1/2/d*A/a^3*tan(1/2*d*x+1/2*c)^2+39/8/d*A/a^3*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)^4-20/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^3+22/d*A/a^3/(
tan(1/2*d*x+1/2*c)+1)^2-50/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)-1/24/d*A/a^3/tan(1/2*d*x+1/2*c)^3+1/2/d*A/a^3/tan(1/
2*d*x+1/2*c)^2-39/8/d*A/a^3/tan(1/2*d*x+1/2*c)-18/d*A/a^3*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.0227, size = 953, normalized size = 6.23 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/120*(A*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 2782*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 9410*sin(d*x + c)
^3/(cos(d*x + c) + 1)^3 + 13645*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 9285*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
+ 2580*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 15)/(a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 5*a^3*sin(d*x + c
)^3/(cos(d*x + c) + 1)^3 + 10*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*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*(12*sin(d*x +
c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^3 + 780*log(sin(d*x + c)/(cos(d*x + c) + 1))/a
^3) - A*((20*sin(d*x + c)/(cos(d*x + c) + 1) - 230*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4777*sin(d*x + c)^3/(
cos(d*x + c) + 1)^3 - 15785*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 22390*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 -
14940*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 4005*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 5)/(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 + 10*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
+ 10*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + a^3*sin(d*x + c)^8/
(cos(d*x + c) + 1)^8) + 5*(81*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*
x + c)^3/(cos(d*x + c) + 1)^3)/a^3 - 1380*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3))/d

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Fricas [B]  time = 2.20219, size = 1563, normalized size = 10.22 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(424*A*cos(d*x + c)^6 + 1002*A*cos(d*x + c)^5 - 944*A*cos(d*x + c)^4 - 2074*A*cos(d*x + c)^3 + 531*A*cos(
d*x + c)^2 + 1077*A*cos(d*x + c) + 135*(A*cos(d*x + c)^6 - 2*A*cos(d*x + c)^5 - 6*A*cos(d*x + c)^4 + 4*A*cos(d
*x + c)^3 + 9*A*cos(d*x + c)^2 - 2*A*cos(d*x + c) - (A*cos(d*x + c)^5 + 3*A*cos(d*x + c)^4 - 3*A*cos(d*x + c)^
3 - 7*A*cos(d*x + c)^2 + 2*A*cos(d*x + c) + 4*A)*sin(d*x + c) - 4*A)*log(1/2*cos(d*x + c) + 1/2) - 135*(A*cos(
d*x + c)^6 - 2*A*cos(d*x + c)^5 - 6*A*cos(d*x + c)^4 + 4*A*cos(d*x + c)^3 + 9*A*cos(d*x + c)^2 - 2*A*cos(d*x +
c) - (A*cos(d*x + c)^5 + 3*A*cos(d*x + c)^4 - 3*A*cos(d*x + c)^3 - 7*A*cos(d*x + c)^2 + 2*A*cos(d*x + c) + 4*
A)*sin(d*x + c) - 4*A)*log(-1/2*cos(d*x + c) + 1/2) + (424*A*cos(d*x + c)^5 - 578*A*cos(d*x + c)^4 - 1522*A*co
s(d*x + c)^3 + 552*A*cos(d*x + c)^2 + 1083*A*cos(d*x + c) + 6*A)*sin(d*x + c) - 6*A)/(a^3*d*cos(d*x + c)^6 - 2
*a^3*d*cos(d*x + c)^5 - 6*a^3*d*cos(d*x + c)^4 + 4*a^3*d*cos(d*x + c)^3 + 9*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)^5 + 3*a^3*d*cos(d*x + c)^4 - 3*a^3*d*cos(d*x + c)^3 - 7*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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.1754, size = 288, normalized size = 1.88 \begin{align*} -\frac{\frac{2160 \, A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{5 \,{\left (792 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 117 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A\right )}}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}} + \frac{48 \,{\left (125 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 445 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 635 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 415 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 108 \, A\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}} - \frac{5 \,{\left (A a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, A a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 117 \, A a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a^{9}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/120*(2160*A*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 5*(792*A*tan(1/2*d*x + 1/2*c)^3 - 117*A*tan(1/2*d*x + 1/2*
c)^2 + 12*A*tan(1/2*d*x + 1/2*c) - A)/(a^3*tan(1/2*d*x + 1/2*c)^3) + 48*(125*A*tan(1/2*d*x + 1/2*c)^4 + 445*A*
tan(1/2*d*x + 1/2*c)^3 + 635*A*tan(1/2*d*x + 1/2*c)^2 + 415*A*tan(1/2*d*x + 1/2*c) + 108*A)/(a^3*(tan(1/2*d*x
+ 1/2*c) + 1)^5) - 5*(A*a^6*tan(1/2*d*x + 1/2*c)^3 - 12*A*a^6*tan(1/2*d*x + 1/2*c)^2 + 117*A*a^6*tan(1/2*d*x +
1/2*c))/a^9)/d