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

Optimal. Leaf size=211 \[ \frac{b^4}{2 a d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}-\frac{4 a b^3}{d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac{6 a b^2 \left (a^2+b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^4}-\frac{\csc ^2(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}-\frac{3 b \log (1-\cos (c+d x))}{4 d (a+b)^4}+\frac{3 b \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

[Out]

b^4/(2*a*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x])^2) - (4*a*b^3)/((a^2 - b^2)^3*d*(b + a*Cos[c + d*x])) - ((a*(a^2
+ 3*b^2) - b*(3*a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 - b^2)^3*d) - (3*b*Log[1 - Cos[c + d*x]])/(4
*(a + b)^4*d) + (3*b*Log[1 + Cos[c + d*x]])/(4*(a - b)^4*d) - (6*a*b^2*(a^2 + b^2)*Log[b + a*Cos[c + d*x]])/((
a^2 - b^2)^4*d)

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Rubi [A]  time = 0.667958, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {4397, 2837, 12, 1647, 1629} \[ \frac{b^4}{2 a d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}-\frac{4 a b^3}{d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac{6 a b^2 \left (a^2+b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^4}-\frac{\csc ^2(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}-\frac{3 b \log (1-\cos (c+d x))}{4 d (a+b)^4}+\frac{3 b \log (\cos (c+d x)+1)}{4 d (a-b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

b^4/(2*a*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x])^2) - (4*a*b^3)/((a^2 - b^2)^3*d*(b + a*Cos[c + d*x])) - ((a*(a^2
+ 3*b^2) - b*(3*a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 - b^2)^3*d) - (3*b*Log[1 - Cos[c + d*x]])/(4
*(a + b)^4*d) + (3*b*Log[1 + Cos[c + d*x]])/(4*(a - b)^4*d) - (6*a*b^2*(a^2 + 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 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 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx &=\int \frac{\cos (c+d x) \cot ^3(c+d x)}{(b+a \cos (c+d x))^3} \, dx\\ &=-\frac{a^3 \operatorname{Subst}\left (\int \frac{x^4}{a^4 (b+x)^3 \left (a^2-x^2\right )^2} \, dx,x,a \cos (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^4}{(b+x)^3 \left (a^2-x^2\right )^2} \, dx,x,a \cos (c+d x)\right )}{a d}\\ &=-\frac{\left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}-\frac{\operatorname{Subst}\left (\int \frac{\frac{a^4 b^4 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^3}+\frac{a^4 b^3 \left (7 a^2-3 b^2\right ) x}{\left (a^2-b^2\right )^3}+\frac{a^2 b^2 \left (3 a^4-9 a^2 b^2+2 b^4\right ) x^2}{\left (a^2-b^2\right )^3}-\frac{a^4 b \left (3 a^2+b^2\right ) x^3}{\left (a^2-b^2\right )^3}}{(b+x)^3 \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{2 a^3 d}\\ &=-\frac{\left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}-\frac{\operatorname{Subst}\left (\int \left (-\frac{3 a^3 b}{2 (a+b)^4 (a-x)}-\frac{3 a^3 b}{2 (a-b)^4 (a+x)}+\frac{2 a^2 b^4}{\left (a^2-b^2\right )^2 (b+x)^3}-\frac{8 a^4 b^3}{\left (a^2-b^2\right )^3 (b+x)^2}+\frac{12 a^4 b^2 \left (a^2+b^2\right )}{\left (a^2-b^2\right )^4 (b+x)}\right ) \, dx,x,a \cos (c+d x)\right )}{2 a^3 d}\\ &=\frac{b^4}{2 a \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}-\frac{4 a b^3}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}-\frac{\left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}-\frac{3 b \log (1-\cos (c+d x))}{4 (a+b)^4 d}+\frac{3 b \log (1+\cos (c+d x))}{4 (a-b)^4 d}-\frac{6 a b^2 \left (a^2+b^2\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^4 d}\\ \end{align*}

Mathematica [A]  time = 5.47795, size = 184, normalized size = 0.87 \[ \frac{-\frac{48 a b^2 \left (a^2+b^2\right ) \log (a \cos (c+d x)+b)}{\left (a^2-b^2\right )^4}+\frac{4 b^4}{a (a-b)^2 (a+b)^2 (a \cos (c+d x)+b)^2}+\frac{32 a b^3}{(b-a)^3 (a+b)^3 (a \cos (c+d x)+b)}-\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right )}{(a+b)^3}+\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right )}{(b-a)^3}-\frac{12 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{(a+b)^4}+\frac{12 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{(a-b)^4}}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*b^4)/(a*(a - b)^2*(a + b)^2*(b + a*Cos[c + d*x])^2) + (32*a*b^3)/((-a + b)^3*(a + b)^3*(b + a*Cos[c + d*x]
)) - Csc[(c + d*x)/2]^2/(a + b)^3 + (12*b*Log[Cos[(c + d*x)/2]])/(a - b)^4 - (48*a*b^2*(a^2 + b^2)*Log[b + a*C
os[c + d*x]])/(a^2 - b^2)^4 - (12*b*Log[Sin[(c + d*x)/2]])/(a + b)^4 + Sec[(c + d*x)/2]^2/(-a + b)^3)/(8*d)

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Maple [A]  time = 0.142, size = 220, normalized size = 1. \begin{align*} -{\frac{1}{4\,d \left ( a-b \right ) ^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) }}+{\frac{3\,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{3\,\ln \left ( -1+\cos \left ( dx+c \right ) \right ) b}{4\,d \left ( a+b \right ) ^{4}}}+{\frac{{b}^{4}}{2\,d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}a \left ( b+a\cos \left ( dx+c \right ) \right ) ^{2}}}-4\,{\frac{a{b}^{3}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3} \left ( b+a\cos \left ( dx+c \right ) \right ) }}-6\,{\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}}}-6\,{\frac{a{b}^{4}\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(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))^3,x)

[Out]

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

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Maxima [B]  time = 1.30881, size = 801, normalized size = 3.8 \begin{align*} -\frac{\frac{48 \,{\left (a^{3} b^{2} + a b^{4}\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{12 \, b \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} - 4 \, a^{5} b + 5 \, a^{4} b^{2} - 32 \, a^{3} b^{3} - 37 \, a^{2} b^{4} - 4 \, a b^{5} - 9 \, b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{{\left (a^{6} - 6 \, a^{5} b + 15 \, a^{4} b^{2} - 84 \, a^{3} b^{3} + 63 \, a^{2} b^{4} - 6 \, a b^{5} + 17 \, 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(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/8*(48*(a^3*b^2 + a*b^4)*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) + 12*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 - 4*a^5*b + 5*a^4*b^2 - 32*a^3*b^3 -
37*a^2*b^4 - 4*a*b^5 - 9*b^6)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + (a^6 - 6*a^5*b + 15*a^4*b^2 - 84*a^3*b^3
+ 63*a^2*b^4 - 6*a*b^5 + 17*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/(cos(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 = 1.10824, size = 2115, normalized size = 10.02 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a^6*b^2 + 18*a^4*b^4 - 18*a^2*b^6 - 2*b^8 - 6*(a^7*b + 2*a^5*b^3 - 3*a^3*b^5)*cos(d*x + c)^3 + 2*(a^8 -
4*a^6*b^2 - 6*a^4*b^4 + 8*a^2*b^6 + b^8)*cos(d*x + c)^2 + 2*(2*a^7*b + 9*a^5*b^3 - 12*a^3*b^5 + a*b^7)*cos(d*
x + c) + 24*(a^4*b^4 + a^2*b^6 - (a^6*b^2 + a^4*b^4)*cos(d*x + c)^4 - 2*(a^5*b^3 + a^3*b^5)*cos(d*x + c)^3 + (
a^6*b^2 - a^2*b^6)*cos(d*x + c)^2 + 2*(a^5*b^3 + a^3*b^5)*cos(d*x + c))*log(a*cos(d*x + c) + b) - 3*(a^5*b^3 +
4*a^4*b^4 + 6*a^3*b^5 + 4*a^2*b^6 + a*b^7 - (a^7*b + 4*a^6*b^2 + 6*a^5*b^3 + 4*a^4*b^4 + a^3*b^5)*cos(d*x + c
)^4 - 2*(a^6*b^2 + 4*a^5*b^3 + 6*a^4*b^4 + 4*a^3*b^5 + a^2*b^6)*cos(d*x + c)^3 + (a^7*b + 4*a^6*b^2 + 5*a^5*b^
3 - 5*a^3*b^5 - 4*a^2*b^6 - a*b^7)*cos(d*x + c)^2 + 2*(a^6*b^2 + 4*a^5*b^3 + 6*a^4*b^4 + 4*a^3*b^5 + a^2*b^6)*
cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 3*(a^5*b^3 - 4*a^4*b^4 + 6*a^3*b^5 - 4*a^2*b^6 + a*b^7 - (a^7*b -
4*a^6*b^2 + 6*a^5*b^3 - 4*a^4*b^4 + a^3*b^5)*cos(d*x + c)^4 - 2*(a^6*b^2 - 4*a^5*b^3 + 6*a^4*b^4 - 4*a^3*b^5 +
a^2*b^6)*cos(d*x + c)^3 + (a^7*b - 4*a^6*b^2 + 5*a^5*b^3 - 5*a^3*b^5 + 4*a^2*b^6 - a*b^7)*cos(d*x + c)^2 + 2*
(a^6*b^2 - 4*a^5*b^3 + 6*a^4*b^4 - 4*a^3*b^5 + a^2*b^6)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^11 - 4
*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d*cos(d*x + c)^4 + 2*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 +
a^2*b^9)*d*cos(d*x + c)^3 - (a^11 - 5*a^9*b^2 + 10*a^7*b^4 - 10*a^5*b^6 + 5*a^3*b^8 - a*b^10)*d*cos(d*x + c)^
2 - 2*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c) - (a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6
- 4*a^3*b^8 + a*b^10)*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\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(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))**3,x)

[Out]

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

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Giac [B]  time = 1.3658, size = 932, normalized size = 4.42 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(6*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) + 48*(
a^3*b^2 + a*b^4)*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 + 6*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*(9*a^5*b^2 + 10*a^4*b^3 + 2*a^3*b^4 + 8*a^2*b^5 + 5*a*b^6
- 2*b^7 + 18*a^5*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 8*a^4*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) -
2*a^3*b^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 6*a^2*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 16*a*b^6*
(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*b^7*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 9*a^5*b^2*(cos(d*x + c)
- 1)^2/(cos(d*x + c) + 1)^2 - 18*a^4*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 18*a^3*b^4*(cos(d*x + c)
- 1)^2/(cos(d*x + c) + 1)^2 - 18*a^2*b^5*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 9*a*b^6*(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