3.214 \(\int \frac{\csc ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=168 \[ \frac{a b^2}{d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac{2 a b \left (a^2+b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^3}+\frac{\csc ^2(c+d x) \left (2 a b-\left (a^2+b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^2}+\frac{(a-b) \log (1-\cos (c+d x))}{4 d (a+b)^3}-\frac{(a+b) \log (\cos (c+d x)+1)}{4 d (a-b)^3} \]

[Out]

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

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Rubi [A]  time = 0.432896, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3872, 2837, 12, 1647, 1629} \[ \frac{a b^2}{d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac{2 a b \left (a^2+b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^3}+\frac{\csc ^2(c+d x) \left (2 a b-\left (a^2+b^2\right ) \cos (c+d x)\right )}{2 d \left (a^2-b^2\right )^2}+\frac{(a-b) \log (1-\cos (c+d x))}{4 d (a+b)^3}-\frac{(a+b) \log (\cos (c+d x)+1)}{4 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3/(a + b*Sec[c + d*x])^2,x]

[Out]

(a*b^2)/((a^2 - b^2)^2*d*(b + a*Cos[c + d*x])) + ((2*a*b - (a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 -
b^2)^2*d) + ((a - b)*Log[1 - Cos[c + d*x]])/(4*(a + b)^3*d) - ((a + b)*Log[1 + Cos[c + d*x]])/(4*(a - b)^3*d)
+ (2*a*b*(a^2 + b^2)*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 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{\csc ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\int \frac{\cot ^2(c+d x) \csc (c+d x)}{(-b-a \cos (c+d x))^2} \, dx\\ &=\frac{a^3 \operatorname{Subst}\left (\int \frac{x^2}{a^2 (-b+x)^2 \left (a^2-x^2\right )^2} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac{a \operatorname{Subst}\left (\int \frac{x^2}{(-b+x)^2 \left (a^2-x^2\right )^2} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac{\left (2 a b-\left (a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^2 d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{a^2 b^2 \left (a^2+b^2\right )}{\left (a^2-b^2\right )^2}+\frac{2 a^2 b x}{a^2-b^2}+\frac{a^2 \left (a^2+b^2\right ) x^2}{\left (a^2-b^2\right )^2}}{(-b+x)^2 \left (a^2-x^2\right )} \, dx,x,-a \cos (c+d x)\right )}{2 a d}\\ &=\frac{\left (2 a b-\left (a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^2 d}+\frac{\operatorname{Subst}\left (\int \left (\frac{a (a+b)}{2 (a-b)^3 (a-x)}+\frac{2 a^2 b^2}{(a-b)^2 (a+b)^2 (b-x)^2}-\frac{4 a^2 b \left (a^2+b^2\right )}{(a-b)^3 (a+b)^3 (b-x)}+\frac{a (a-b)}{2 (a+b)^3 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{2 a d}\\ &=\frac{a b^2}{\left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac{\left (2 a b-\left (a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^2 d}+\frac{(a-b) \log (1-\cos (c+d x))}{4 (a+b)^3 d}-\frac{(a+b) \log (1+\cos (c+d x))}{4 (a-b)^3 d}+\frac{2 a b \left (a^2+b^2\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^3 d}\\ \end{align*}

Mathematica [A]  time = 1.30005, size = 224, normalized size = 1.33 \[ \frac{\sec ^2(c+d x) (a \cos (c+d x)+b) \left (\frac{16 a b \left (a^2+b^2\right ) (a \cos (c+d x)+b) \log (a \cos (c+d x)+b)}{\left (a^2-b^2\right )^3}+\frac{8 a b^2}{(a-b)^2 (a+b)^2}+\frac{4 (a+b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)}{(b-a)^3}-\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a+b)^2}+\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a-b)^2}+\frac{4 (a-b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)}{(a+b)^3}\right )}{8 d (a+b \sec (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3/(a + b*Sec[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.078, size = 224, normalized size = 1.3 \begin{align*}{\frac{a{b}^{2}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2} \left ( b+a\cos \left ( dx+c \right ) \right ) }}+2\,{\frac{{a}^{3}b\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}+2\,{\frac{a{b}^{3}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}+{\frac{1}{4\,d \left ( a-b \right ) ^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) }}-{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ) a}{4\,d \left ( a-b \right ) ^{3}}}-{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ) b}{4\,d \left ( a-b \right ) ^{3}}}+{\frac{1}{4\,d \left ( a+b \right ) ^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) a}{4\,d \left ( a+b \right ) ^{3}}}-{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) b}{4\,d \left ( a+b \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3/(a+b*sec(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.10275, size = 370, normalized size = 2.2 \begin{align*} \frac{\frac{8 \,{\left (a^{3} b + a b^{3}\right )} \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 (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 (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 b^{2} -{\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )}}{a^{4} b - 2 \, a^{2} b^{3} + b^{5} -{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.81496, size = 1378, normalized size = 8.2 \begin{align*} -\frac{8 \, a^{3} b^{2} - 8 \, a b^{4} - 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) + 8 \,{\left (a^{3} b^{2} + a b^{4} -{\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2} +{\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) -{\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5} -{\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} \cos \left (d x + c\right )^{2} +{\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5} -{\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} \cos \left (d x + c\right )^{2} +{\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{3} +{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) -{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/4*(8*a^3*b^2 - 8*a*b^4 - 2*(a^5 + 2*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 + 2*(a^4*b - 2*a^2*b^3 + b^5)*cos(d*x
+ c) + 8*(a^3*b^2 + a*b^4 - (a^4*b + a^2*b^3)*cos(d*x + c)^3 - (a^3*b^2 + a*b^4)*cos(d*x + c)^2 + (a^4*b + a^
2*b^3)*cos(d*x + c))*log(a*cos(d*x + c) + b) - (a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5 - (a^5 + 4*a^4*b
+ 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*cos(d*x + c)^3 - (a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*cos(d*x + c
)^2 + (a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (a^4*b - 4*a
^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5 - (a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*cos(d*x + c)^3 - (a^4*b -
4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*cos(d*x + c)^2 + (a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*cos(d
*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^3 + (a^6*b - 3*a^
4*b^3 + 3*a^2*b^5 - b^7)*d*cos(d*x + c)^2 - (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c) - (a^6*b - 3*
a^4*b^3 + 3*a^2*b^5 - b^7)*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{3}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**3/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(csc(c + d*x)**3/(a + b*sec(c + d*x))**2, x)

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Giac [B]  time = 1.43979, size = 616, normalized size = 3.67 \begin{align*} \frac{\frac{2 \,{\left (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{16 \,{\left (a^{3} b + a b^{3}\right )} \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{a^{3} - a^{2} b - a b^{2} + b^{3} - \frac{8 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{8 \, a b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{3 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{3 \, a b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\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} + \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}}\right )}} - \frac{\cos \left (d x + c\right ) - 1}{{\left (a^{2} - 2 \, a b + b^{2}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

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