3.270 \(\int \frac{\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=228 \[ -\frac{a b \left (11 a^2+b^2\right )}{2 d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}+\frac{a \left (2 a^2+b^2\right )}{2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}-\frac{2 a^3 \left (a^2+5 b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^4}-\frac{\csc ^2(c+d x) (a-b \cos (c+d x))}{2 d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac{(4 a+b) \log (1-\cos (c+d x))}{4 d (a+b)^4}+\frac{(4 a-b) \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

[Out]

(a*(2*a^2 + b^2))/(2*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x])^2) - (a*b*(11*a^2 + b^2))/(2*(a^2 - b^2)^3*d*(b + a*
Cos[c + d*x])) - ((a - b*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 - b^2)*d*(b + a*Cos[c + d*x])^2) + ((4*a + b)*L
og[1 - Cos[c + d*x]])/(4*(a + b)^4*d) + ((4*a - b)*Log[1 + Cos[c + d*x]])/(4*(a - b)^4*d) - (2*a^3*(a^2 + 5*b^
2)*Log[b + a*Cos[c + d*x]])/((a^2 - b^2)^4*d)

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Rubi [A]  time = 0.40793, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4397, 2668, 741, 801} \[ -\frac{a b \left (11 a^2+b^2\right )}{2 d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}+\frac{a \left (2 a^2+b^2\right )}{2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}-\frac{2 a^3 \left (a^2+5 b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^4}-\frac{\csc ^2(c+d x) (a-b \cos (c+d x))}{2 d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac{(4 a+b) \log (1-\cos (c+d x))}{4 d (a+b)^4}+\frac{(4 a-b) \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]

[Out]

(a*(2*a^2 + b^2))/(2*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x])^2) - (a*b*(11*a^2 + b^2))/(2*(a^2 - b^2)^3*d*(b + a*
Cos[c + d*x])) - ((a - b*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 - b^2)*d*(b + a*Cos[c + d*x])^2) + ((4*a + b)*L
og[1 - Cos[c + d*x]])/(4*(a + b)^4*d) + ((4*a - b)*Log[1 + Cos[c + d*x]])/(4*(a - b)^4*d) - (2*a^3*(a^2 + 5*b^
2)*Log[b + a*Cos[c + d*x]])/((a^2 - b^2)^4*d)

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

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{\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx &=\int \frac{\csc ^3(c+d x)}{(b+a \cos (c+d x))^3} \, dx\\ &=-\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{(b+x)^3 \left (a^2-x^2\right )^2} \, dx,x,a \cos (c+d x)\right )}{d}\\ &=-\frac{(a-b \cos (c+d x)) \csc ^2(c+d x)}{2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac{a \operatorname{Subst}\left (\int \frac{-4 a^2+b^2+3 b x}{(b+x)^3 \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=-\frac{(a-b \cos (c+d x)) \csc ^2(c+d x)}{2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac{a \operatorname{Subst}\left (\int \left (\frac{(-4 a-b) (a-b)}{2 a (a+b)^3 (a-x)}+\frac{(4 a-b) (a+b)}{2 a (a-b)^3 (a+x)}-\frac{2 \left (2 a^2+b^2\right )}{(a-b) (a+b) (b+x)^3}+\frac{b \left (11 a^2+b^2\right )}{(a-b)^2 (a+b)^2 (b+x)^2}-\frac{4 \left (a^4+5 a^2 b^2\right )}{(a-b)^3 (a+b)^3 (b+x)}\right ) \, dx,x,a \cos (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=\frac{a \left (2 a^2+b^2\right )}{2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}-\frac{a b \left (11 a^2+b^2\right )}{2 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}-\frac{(a-b \cos (c+d x)) \csc ^2(c+d x)}{2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac{(4 a+b) \log (1-\cos (c+d x))}{4 (a+b)^4 d}+\frac{(4 a-b) \log (1+\cos (c+d x))}{4 (a-b)^4 d}-\frac{2 a^3 \left (a^2+5 b^2\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^4 d}\\ \end{align*}

Mathematica [A]  time = 6.26157, size = 217, normalized size = 0.95 \[ -\frac{2 \left (5 a^3 b^2+a^5\right ) \log (a \cos (c+d x)+b)}{d \left (b^2-a^2\right )^4}+\frac{4 a^3 b}{d (b-a)^3 (a+b)^3 (a \cos (c+d x)+b)}+\frac{a^3}{2 d (b-a)^2 (a+b)^2 (a \cos (c+d x)+b)^2}-\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 d (a+b)^3}+\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 d (b-a)^3}+\frac{(4 a+b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d (a+b)^4}+\frac{(4 a-b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d (b-a)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]

[Out]

a^3/(2*(-a + b)^2*(a + b)^2*d*(b + a*Cos[c + d*x])^2) + (4*a^3*b)/((-a + b)^3*(a + b)^3*d*(b + a*Cos[c + d*x])
) - Csc[(c + d*x)/2]^2/(8*(a + b)^3*d) + ((4*a - b)*Log[Cos[(c + d*x)/2]])/(2*(-a + b)^4*d) - (2*(a^5 + 5*a^3*
b^2)*Log[b + a*Cos[c + d*x]])/((-a^2 + b^2)^4*d) + ((4*a + b)*Log[Sin[(c + d*x)/2]])/(2*(a + b)^4*d) + Sec[(c
+ d*x)/2]^2/(8*(-a + b)^3*d)

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Maple [A]  time = 0.183, size = 256, normalized size = 1.1 \begin{align*} -{\frac{1}{4\,d \left ( a-b \right ) ^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) }}+{\frac{a\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{ \left ( a-b \right ) ^{4}d}}-{\frac{b\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{4\, \left ( a-b \right ) ^{4}d}}+{\frac{1}{4\,d \left ( a+b \right ) ^{3} \left ( -1+\cos \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) a}{d \left ( a+b \right ) ^{4}}}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) b}{4\,d \left ( a+b \right ) ^{4}}}+{\frac{{a}^{3}}{2\,d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2} \left ( b+a\cos \left ( dx+c \right ) \right ) ^{2}}}-4\,{\frac{{a}^{3}b}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3} \left ( b+a\cos \left ( dx+c \right ) \right ) }}-2\,{\frac{{a}^{5}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}}}-10\,{\frac{{a}^{3}{b}^{2}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^3,x)

[Out]

-1/4/d/(a-b)^3/(cos(d*x+c)+1)+a*ln(cos(d*x+c)+1)/(a-b)^4/d-1/4*b*ln(cos(d*x+c)+1)/(a-b)^4/d+1/4/d/(a+b)^3/(-1+
cos(d*x+c))+1/d/(a+b)^4*ln(-1+cos(d*x+c))*a+1/4/d/(a+b)^4*ln(-1+cos(d*x+c))*b+1/2/d*a^3/(a+b)^2/(a-b)^2/(b+a*c
os(d*x+c))^2-4/d*a^3*b/(a+b)^3/(a-b)^3/(b+a*cos(d*x+c))-2/d*a^5/(a+b)^4/(a-b)^4*ln(b+a*cos(d*x+c))-10/d*a^3*b^
2/(a+b)^4/(a-b)^4*ln(b+a*cos(d*x+c))

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Maxima [B]  time = 1.23796, size = 798, normalized size = 3.5 \begin{align*} -\frac{\frac{16 \,{\left (a^{5} + 5 \, a^{3} b^{2}\right )} \log \left (a + b - \frac{{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} - \frac{4 \,{\left (4 \, a + b\right )} \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac{a^{6} - 2 \, a^{5} b - a^{4} b^{2} + 4 \, a^{3} b^{3} - a^{2} b^{4} - 2 \, a b^{5} + b^{6} - \frac{2 \,{\left (a^{6} - 44 \, a^{5} b - 35 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + 4 \, a b^{5} - b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{{\left (15 \, a^{6} + 70 \, a^{5} b - 95 \, a^{4} b^{2} + 20 \, a^{3} b^{3} - 15 \, a^{2} b^{4} + 6 \, a b^{5} - b^{6}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{\frac{{\left (a^{9} + a^{8} b - 4 \, a^{7} b^{2} - 4 \, a^{6} b^{3} + 6 \, a^{5} b^{4} + 6 \, a^{4} b^{5} - 4 \, a^{3} b^{6} - 4 \, a^{2} b^{7} + a b^{8} + b^{9}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{2 \,{\left (a^{9} - a^{8} b - 4 \, a^{7} b^{2} + 4 \, a^{6} b^{3} + 6 \, a^{5} b^{4} - 6 \, a^{4} b^{5} - 4 \, a^{3} b^{6} + 4 \, a^{2} b^{7} + a b^{8} - b^{9}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{{\left (a^{9} - 3 \, a^{8} b + 8 \, a^{6} b^{3} - 6 \, a^{5} b^{4} - 6 \, a^{4} b^{5} + 8 \, a^{3} b^{6} - 3 \, a b^{8} + b^{9}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{\sin \left (d x + c\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}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/8*(16*(a^5 + 5*a^3*b^2)*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^8 - 4*a^6*b^2 + 6*a^4*b
^4 - 4*a^2*b^6 + b^8) - 4*(4*a + b)*log(sin(d*x + c)/(cos(d*x + c) + 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3
+ b^4) + (a^6 - 2*a^5*b - a^4*b^2 + 4*a^3*b^3 - a^2*b^4 - 2*a*b^5 + b^6 - 2*(a^6 - 44*a^5*b - 35*a^4*b^2 - 5*a
^2*b^4 + 4*a*b^5 - b^6)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - (15*a^6 + 70*a^5*b - 95*a^4*b^2 + 20*a^3*b^3 - 1
5*a^2*b^4 + 6*a*b^5 - b^6)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/((a^9 + a^8*b - 4*a^7*b^2 - 4*a^6*b^3 + 6*a^5*
b^4 + 6*a^4*b^5 - 4*a^3*b^6 - 4*a^2*b^7 + a*b^8 + b^9)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*(a^9 - a^8*b -
4*a^7*b^2 + 4*a^6*b^3 + 6*a^5*b^4 - 6*a^4*b^5 - 4*a^3*b^6 + 4*a^2*b^7 + a*b^8 - b^9)*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4 + (a^9 - 3*a^8*b + 8*a^6*b^3 - 6*a^5*b^4 - 6*a^4*b^5 + 8*a^3*b^6 - 3*a*b^8 + b^9)*sin(d*x + c)^6/(c
os(d*x + c) + 1)^6) + sin(d*x + c)^2/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(cos(d*x + c) + 1)^2))/d

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Fricas [B]  time = 1.16053, size = 2142, normalized size = 9.39 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*(2*a^7 - 22*a^5*b^2 + 14*a^3*b^4 + 6*a*b^6 + 2*(11*a^6*b - 10*a^4*b^3 - a^2*b^5)*cos(d*x + c)^3 - 4*(a^7
- 7*a^5*b^2 + 5*a^3*b^4 + a*b^6)*cos(d*x + c)^2 - 2*(10*a^6*b - 7*a^4*b^3 - 4*a^2*b^5 + b^7)*cos(d*x + c) - 8*
(a^5*b^2 + 5*a^3*b^4 - (a^7 + 5*a^5*b^2)*cos(d*x + c)^4 - 2*(a^6*b + 5*a^4*b^3)*cos(d*x + c)^3 + (a^7 + 4*a^5*
b^2 - 5*a^3*b^4)*cos(d*x + c)^2 + 2*(a^6*b + 5*a^4*b^3)*cos(d*x + c))*log(a*cos(d*x + c) + b) + (4*a^5*b^2 + 1
5*a^4*b^3 + 20*a^3*b^4 + 10*a^2*b^5 - b^7 - (4*a^7 + 15*a^6*b + 20*a^5*b^2 + 10*a^4*b^3 - a^2*b^5)*cos(d*x + c
)^4 - 2*(4*a^6*b + 15*a^5*b^2 + 20*a^4*b^3 + 10*a^3*b^4 - a*b^6)*cos(d*x + c)^3 + (4*a^7 + 15*a^6*b + 16*a^5*b
^2 - 5*a^4*b^3 - 20*a^3*b^4 - 11*a^2*b^5 + b^7)*cos(d*x + c)^2 + 2*(4*a^6*b + 15*a^5*b^2 + 20*a^4*b^3 + 10*a^3
*b^4 - a*b^6)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (4*a^5*b^2 - 15*a^4*b^3 + 20*a^3*b^4 - 10*a^2*b^5 +
b^7 - (4*a^7 - 15*a^6*b + 20*a^5*b^2 - 10*a^4*b^3 + a^2*b^5)*cos(d*x + c)^4 - 2*(4*a^6*b - 15*a^5*b^2 + 20*a^4
*b^3 - 10*a^3*b^4 + a*b^6)*cos(d*x + c)^3 + (4*a^7 - 15*a^6*b + 16*a^5*b^2 + 5*a^4*b^3 - 20*a^3*b^4 + 11*a^2*b
^5 - b^7)*cos(d*x + c)^2 + 2*(4*a^6*b - 15*a^5*b^2 + 20*a^4*b^3 - 10*a^3*b^4 + a*b^6)*cos(d*x + c))*log(-1/2*c
os(d*x + c) + 1/2))/((a^10 - 4*a^8*b^2 + 6*a^6*b^4 - 4*a^4*b^6 + a^2*b^8)*d*cos(d*x + c)^4 + 2*(a^9*b - 4*a^7*
b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)^3 - (a^10 - 5*a^8*b^2 + 10*a^6*b^4 - 10*a^4*b^6 + 5*a^2*b^
8 - b^10)*d*cos(d*x + c)^2 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c) - (a^8*b^2 -
4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (c + d x \right )}}{\left (a \sin{\left (c + d x \right )} + b \tan{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a*sin(d*x+c)+b*tan(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**3/(a*sin(c + d*x) + b*tan(c + d*x))**3, x)

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Giac [B]  time = 1.39256, size = 911, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/8*(2*(4*a + b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)
- 16*(a^5 + 5*a^3*b^2)*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^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8) + (a + b - 8*a*(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) - 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 +
b^4)*(cos(d*x + c) - 1)) + (cos(d*x + c) - 1)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(cos(d*x + c) + 1)) + 8*(3*a^7
- 4*a^6*b - 2*a^5*b^2 + 20*a^4*b^3 + 15*a^3*b^4 + 4*a^7*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 10*a^6*b*(cos(
d*x + c) - 1)/(cos(d*x + c) + 1) + 26*a^5*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 10*a^4*b^3*(cos(d*x + c)
- 1)/(cos(d*x + c) + 1) - 30*a^3*b^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a^7*(cos(d*x + c) - 1)^2/(cos(
d*x + c) + 1)^2 - 6*a^6*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 18*a^5*b^2*(cos(d*x + c) - 1)^2/(cos(d*x
+ c) + 1)^2 - 30*a^4*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 15*a^3*b^4*(cos(d*x + c) - 1)^2/(cos(d*x
+ c) + 1)^2)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(a + b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1
) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))^2))/d