3.202 \(\int \frac{\csc ^5(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=179 \[ \frac{a^4 b \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^3}+\frac{\csc ^4(c+d x) (b-a \cos (c+d x))}{4 d \left (a^2-b^2\right )}+\frac{\csc ^2(c+d x) \left (4 a^2 b-a \left (3 a^2+b^2\right ) \cos (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac{a (3 a+b) \log (1-\cos (c+d x))}{16 d (a+b)^3}-\frac{a (3 a-b) \log (\cos (c+d x)+1)}{16 d (a-b)^3} \]

[Out]

((4*a^2*b - a*(3*a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(8*(a^2 - b^2)^2*d) + ((b - a*Cos[c + d*x])*Csc[c +
d*x]^4)/(4*(a^2 - b^2)*d) + (a*(3*a + b)*Log[1 - Cos[c + d*x]])/(16*(a + b)^3*d) - (a*(3*a - b)*Log[1 + Cos[c
+ d*x]])/(16*(a - b)^3*d) + (a^4*b*Log[b + a*Cos[c + d*x]])/((a^2 - b^2)^3*d)

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Rubi [A]  time = 0.30139, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3872, 2837, 12, 823, 801} \[ \frac{a^4 b \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^3}+\frac{\csc ^4(c+d x) (b-a \cos (c+d x))}{4 d \left (a^2-b^2\right )}+\frac{\csc ^2(c+d x) \left (4 a^2 b-a \left (3 a^2+b^2\right ) \cos (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac{a (3 a+b) \log (1-\cos (c+d x))}{16 d (a+b)^3}-\frac{a (3 a-b) \log (\cos (c+d x)+1)}{16 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^5/(a + b*Sec[c + d*x]),x]

[Out]

((4*a^2*b - a*(3*a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(8*(a^2 - b^2)^2*d) + ((b - a*Cos[c + d*x])*Csc[c +
d*x]^4)/(4*(a^2 - b^2)*d) + (a*(3*a + b)*Log[1 - Cos[c + d*x]])/(16*(a + b)^3*d) - (a*(3*a - b)*Log[1 + Cos[c
+ d*x]])/(16*(a - b)^3*d) + (a^4*b*Log[b + a*Cos[c + d*x]])/((a^2 - b^2)^3*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2837

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

\begin{align*} \int \frac{\csc ^5(c+d x)}{a+b \sec (c+d x)} \, dx &=-\int \frac{\cot (c+d x) \csc ^4(c+d x)}{-b-a \cos (c+d x)} \, dx\\ &=\frac{a^5 \operatorname{Subst}\left (\int \frac{x}{a (-b+x) \left (a^2-x^2\right )^3} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac{a^4 \operatorname{Subst}\left (\int \frac{x}{(-b+x) \left (a^2-x^2\right )^3} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac{(b-a \cos (c+d x)) \csc ^4(c+d x)}{4 \left (a^2-b^2\right ) d}+\frac{a^2 \operatorname{Subst}\left (\int \frac{a^2 b+3 a^2 x}{(-b+x) \left (a^2-x^2\right )^2} \, dx,x,-a \cos (c+d x)\right )}{4 \left (a^2-b^2\right ) d}\\ &=\frac{\left (4 a^2 b-a \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{8 \left (a^2-b^2\right )^2 d}+\frac{(b-a \cos (c+d x)) \csc ^4(c+d x)}{4 \left (a^2-b^2\right ) d}+\frac{\operatorname{Subst}\left (\int \frac{a^2 b \left (5 a^2-b^2\right )+a^2 \left (3 a^2+b^2\right ) x}{(-b+x) \left (a^2-x^2\right )} \, dx,x,-a \cos (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=\frac{\left (4 a^2 b-a \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{8 \left (a^2-b^2\right )^2 d}+\frac{(b-a \cos (c+d x)) \csc ^4(c+d x)}{4 \left (a^2-b^2\right ) d}+\frac{\operatorname{Subst}\left (\int \left (\frac{a (3 a-b) (a+b)^2}{2 (a-b) (a-x)}-\frac{8 a^4 b}{(a-b) (a+b) (b-x)}+\frac{a (a-b)^2 (3 a+b)}{2 (a+b) (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=\frac{\left (4 a^2 b-a \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{8 \left (a^2-b^2\right )^2 d}+\frac{(b-a \cos (c+d x)) \csc ^4(c+d x)}{4 \left (a^2-b^2\right ) d}+\frac{a (3 a+b) \log (1-\cos (c+d x))}{16 (a+b)^3 d}-\frac{a (3 a-b) \log (1+\cos (c+d x))}{16 (a-b)^3 d}+\frac{a^4 b \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^3 d}\\ \end{align*}

Mathematica [A]  time = 5.16327, size = 207, normalized size = 1.16 \[ \frac{-2 (a-b)^3 \left (3 a^2+4 a b+b^2\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )+2 (a+b)^3 \left (3 a^2-4 a b+b^2\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )+8 a \left (8 a^3 b \log (a \cos (c+d x)+b)+(a-b)^3 (3 a+b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-(3 a-b) (a+b)^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-(a-b)^3 (a+b)^2 \csc ^4\left (\frac{1}{2} (c+d x)\right )+(a-b)^2 (a+b)^3 \sec ^4\left (\frac{1}{2} (c+d x)\right )}{64 d (a-b)^3 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^5/(a + b*Sec[c + d*x]),x]

[Out]

(-2*(a - b)^3*(3*a^2 + 4*a*b + b^2)*Csc[(c + d*x)/2]^2 - (a - b)^3*(a + b)^2*Csc[(c + d*x)/2]^4 + 8*a*(-((3*a
- b)*(a + b)^3*Log[Cos[(c + d*x)/2]]) + 8*a^3*b*Log[b + a*Cos[c + d*x]] + (a - b)^3*(3*a + b)*Log[Sin[(c + d*x
)/2]]) + 2*(a + b)^3*(3*a^2 - 4*a*b + b^2)*Sec[(c + d*x)/2]^2 + (a - b)^2*(a + b)^3*Sec[(c + d*x)/2]^4)/(64*(a
- b)^3*(a + b)^3*d)

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Maple [A]  time = 0.067, size = 259, normalized size = 1.5 \begin{align*}{\frac{{a}^{4}b\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}+{\frac{1}{2\,d \left ( 8\,a-8\,b \right ) \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}}+{\frac{3\,a}{16\,d \left ( a-b \right ) ^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) }}-{\frac{b}{16\,d \left ( a-b \right ) ^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) }}-{\frac{3\,{a}^{2}\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{16\,d \left ( a-b \right ) ^{3}}}+{\frac{a\ln \left ( \cos \left ( dx+c \right ) +1 \right ) b}{16\,d \left ( a-b \right ) ^{3}}}-{\frac{1}{2\,d \left ( 8\,a+8\,b \right ) \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,a}{16\,d \left ( a+b \right ) ^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) }}+{\frac{b}{16\,d \left ( a+b \right ) ^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) }}+{\frac{3\,{a}^{2}\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{16\,d \left ( a+b \right ) ^{3}}}+{\frac{a\ln \left ( -1+\cos \left ( dx+c \right ) \right ) b}{16\,d \left ( a+b \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^5/(a+b*sec(d*x+c)),x)

[Out]

1/d*b*a^4/(a+b)^3/(a-b)^3*ln(b+a*cos(d*x+c))+1/2/d/(8*a-8*b)/(cos(d*x+c)+1)^2+3/16/d/(a-b)^2/(cos(d*x+c)+1)*a-
1/16/d/(a-b)^2/(cos(d*x+c)+1)*b-3/16/d*a^2/(a-b)^3*ln(cos(d*x+c)+1)+1/16/d*a/(a-b)^3*ln(cos(d*x+c)+1)*b-1/2/d/
(8*a+8*b)/(-1+cos(d*x+c))^2+3/16/d/(a+b)^2/(-1+cos(d*x+c))*a+1/16/d/(a+b)^2/(-1+cos(d*x+c))*b+3/16/d/(a+b)^3*a
^2*ln(-1+cos(d*x+c))+1/16/d/(a+b)^3*a*ln(-1+cos(d*x+c))*b

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Maxima [A]  time = 1.01762, size = 362, normalized size = 2.02 \begin{align*} \frac{\frac{16 \, a^{4} b \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac{{\left (3 \, a^{2} - a b\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{{\left (3 \, a^{2} + a b\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{2 \,{\left (4 \, a^{2} b \cos \left (d x + c\right )^{2} -{\left (3 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{3} - 6 \, a^{2} b + 2 \, b^{3} +{\left (5 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/16*(16*a^4*b*log(a*cos(d*x + c) + b)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - (3*a^2 - a*b)*log(cos(d*x + c) +
1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a^2 + a*b)*log(cos(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 2*(
4*a^2*b*cos(d*x + c)^2 - (3*a^3 + a*b^2)*cos(d*x + c)^3 - 6*a^2*b + 2*b^3 + (5*a^3 - a*b^2)*cos(d*x + c))/((a^
4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2))/d

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Fricas [B]  time = 3.13934, size = 1033, normalized size = 5.77 \begin{align*} \frac{12 \, a^{4} b - 16 \, a^{2} b^{3} + 4 \, b^{5} + 2 \,{\left (3 \, a^{5} - 2 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} - 8 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (5 \, a^{5} - 6 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right ) + 16 \,{\left (a^{4} b \cos \left (d x + c\right )^{4} - 2 \, a^{4} b \cos \left (d x + c\right )^{2} + a^{4} b\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) -{\left (3 \, a^{5} + 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4} +{\left (3 \, a^{5} + 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (3 \, a^{5} + 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (3 \, a^{5} - 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4} +{\left (3 \, a^{5} - 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (3 \, a^{5} - 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{16 \,{\left ({\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4} - 2 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(12*a^4*b - 16*a^2*b^3 + 4*b^5 + 2*(3*a^5 - 2*a^3*b^2 - a*b^4)*cos(d*x + c)^3 - 8*(a^4*b - a^2*b^3)*cos(d
*x + c)^2 - 2*(5*a^5 - 6*a^3*b^2 + a*b^4)*cos(d*x + c) + 16*(a^4*b*cos(d*x + c)^4 - 2*a^4*b*cos(d*x + c)^2 + a
^4*b)*log(a*cos(d*x + c) + b) - (3*a^5 + 8*a^4*b + 6*a^3*b^2 - a*b^4 + (3*a^5 + 8*a^4*b + 6*a^3*b^2 - a*b^4)*c
os(d*x + c)^4 - 2*(3*a^5 + 8*a^4*b + 6*a^3*b^2 - a*b^4)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) + (3*a^5 -
8*a^4*b + 6*a^3*b^2 - a*b^4 + (3*a^5 - 8*a^4*b + 6*a^3*b^2 - a*b^4)*cos(d*x + c)^4 - 2*(3*a^5 - 8*a^4*b + 6*a
^3*b^2 - a*b^4)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x +
c)^4 - 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c)^2 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{5}{\left (c + d x \right )}}{a + b \sec{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**5/(a+b*sec(d*x+c)),x)

[Out]

Integral(csc(c + d*x)**5/(a + b*sec(c + d*x)), x)

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Giac [B]  time = 1.44221, size = 566, normalized size = 3.16 \begin{align*} \frac{\frac{64 \, a^{4} b \log \left ({\left | -a - b - \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac{4 \,{\left (3 \, a^{2} + a b\right )} \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{\frac{8 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{4 \, b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2} - 2 \, a b + b^{2}} - \frac{{\left (a^{2} + 2 \, a b + b^{2} - \frac{8 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{12 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{4 \, b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{18 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{6 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

1/64*(64*a^4*b*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1)/(cos(d*x + c) +
1)))/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + 4*(3*a^2 + a*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/
(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - (8*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 4*b*(cos(d*x + c) - 1)/(cos(d*x
+ c) + 1) - a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/(a^2 -
2*a*b + b^2) - (a^2 + 2*a*b + b^2 - 8*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 12*a*b*(cos(d*x + c) - 1)/(
cos(d*x + c) + 1) - 4*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 18*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) +
1)^2 + 6*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)^2/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*
(cos(d*x + c) - 1)^2))/d