3.189 \(\int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=162 \[ -\frac{b \left (3 a^2+2 b^2\right ) \log (\cos (c+d x))}{d}-\frac{a^2 \csc ^2(c+d x) \left (a \left (\frac{3 b^2}{a^2}+1\right ) \cos (c+d x)+b \left (\frac{b^2}{a^2}+3\right )\right )}{2 d}+\frac{3 a b^2 \sec (c+d x)}{d}+\frac{(a+b)^2 (a+4 b) \log (1-\cos (c+d x))}{4 d}-\frac{(a-4 b) (a-b)^2 \log (\cos (c+d x)+1)}{4 d}+\frac{b^3 \sec ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.348948, antiderivative size = 162, 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, 1805, 1802} \[ -\frac{b \left (3 a^2+2 b^2\right ) \log (\cos (c+d x))}{d}-\frac{a^2 \csc ^2(c+d x) \left (a \left (\frac{3 b^2}{a^2}+1\right ) \cos (c+d x)+b \left (\frac{b^2}{a^2}+3\right )\right )}{2 d}+\frac{3 a b^2 \sec (c+d x)}{d}+\frac{(a+b)^2 (a+4 b) \log (1-\cos (c+d x))}{4 d}-\frac{(a-4 b) (a-b)^2 \log (\cos (c+d x)+1)}{4 d}+\frac{b^3 \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(b*(3 + b^2/a^2) + a*(1 + (3*b^2)/a^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(2*d) + ((a + b)^2*(a + 4*b)*Log[1
- Cos[c + d*x]])/(4*d) - (b*(3*a^2 + 2*b^2)*Log[Cos[c + d*x]])/d - ((a - 4*b)*(a - b)^2*Log[1 + Cos[c + d*x]])
/(4*d) + (3*a*b^2*Sec[c + d*x])/d + (b^3*Sec[c + d*x]^2)/(2*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 1805

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

Rule 1802

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

Rubi steps

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

Mathematica [B]  time = 6.19619, size = 669, normalized size = 4.13 \[ \frac{\left (-3 a^2 b+a^3+3 a b^2-b^3\right ) \cos ^3(c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a+b \sec (c+d x))^3}{8 d (a \cos (c+d x)+b)^3}+\frac{\left (6 a^2 b-a^3-9 a b^2+4 b^3\right ) \cos ^3(c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3}{2 d (a \cos (c+d x)+b)^3}+\frac{\left (-3 a^2 b-2 b^3\right ) \cos ^3(c+d x) \log (\cos (c+d x)) (a+b \sec (c+d x))^3}{d (a \cos (c+d x)+b)^3}+\frac{\left (-3 a^2 b-a^3-3 a b^2-b^3\right ) \cos ^3(c+d x) \csc ^2\left (\frac{1}{2} (c+d x)\right ) (a+b \sec (c+d x))^3}{8 d (a \cos (c+d x)+b)^3}+\frac{\left (6 a^2 b+a^3+9 a b^2+4 b^3\right ) \cos ^3(c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3}{2 d (a \cos (c+d x)+b)^3}+\frac{3 a b^2 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d (a \cos (c+d x)+b)^3}+\frac{3 a b^2 \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^3}-\frac{3 a b^2 \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^3}+\frac{b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b)^3}+\frac{b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*b^2*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3)/(d*(b + a*Cos[c + d*x])^3) + ((-a^3 - 3*a^2*b - 3*a*b^2 - b^3)
*Cos[c + d*x]^3*Csc[(c + d*x)/2]^2*(a + b*Sec[c + d*x])^3)/(8*d*(b + a*Cos[c + d*x])^3) + ((-a^3 + 6*a^2*b - 9
*a*b^2 + 4*b^3)*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2]]*(a + b*Sec[c + d*x])^3)/(2*d*(b + a*Cos[c + d*x])^3) + ((
-3*a^2*b - 2*b^3)*Cos[c + d*x]^3*Log[Cos[c + d*x]]*(a + b*Sec[c + d*x])^3)/(d*(b + a*Cos[c + d*x])^3) + ((a^3
+ 6*a^2*b + 9*a*b^2 + 4*b^3)*Cos[c + d*x]^3*Log[Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^3)/(2*d*(b + a*Cos[c +
d*x])^3) + ((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*Cos[c + d*x]^3*Sec[(c + d*x)/2]^2*(a + b*Sec[c + d*x])^3)/(8*d*(b
+ a*Cos[c + d*x])^3) + (b^3*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3)/(4*d*(b + a*Cos[c + d*x])^3*(Cos[(c + d*x)/
2] - Sin[(c + d*x)/2])^2) + (3*a*b^2*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[(c + d*x)/2])/(d*(b + a*Cos[c +
d*x])^3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (b^3*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3)/(4*d*(b + a*Cos[
c + d*x])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) - (3*a*b^2*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[(c +
d*x)/2])/(d*(b + a*Cos[c + d*x])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [A]  time = 0.053, size = 201, normalized size = 1.2 \begin{align*} -{\frac{{a}^{3}\csc \left ( dx+c \right ) \cot \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-{\frac{3\,{a}^{2}b}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{{a}^{2}b\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}-{\frac{3\,a{b}^{2}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }}+{\frac{9\,a{b}^{2}}{2\,d\cos \left ( dx+c \right ) }}+{\frac{9\,a{b}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{b}^{3}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{3}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{{b}^{3}\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}} \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))^3,x)

[Out]

-1/2/d*a^3*csc(d*x+c)*cot(d*x+c)+1/2/d*a^3*ln(csc(d*x+c)-cot(d*x+c))-3/2/d*a^2*b/sin(d*x+c)^2+3/d*a^2*b*ln(tan
(d*x+c))-3/2/d*a*b^2/sin(d*x+c)^2/cos(d*x+c)+9/2/d*a*b^2/cos(d*x+c)+9/2/d*a*b^2*ln(csc(d*x+c)-cot(d*x+c))+1/2/
d*b^3/sin(d*x+c)^2/cos(d*x+c)^2-1/d*b^3/sin(d*x+c)^2+2/d*b^3*ln(tan(d*x+c))

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Maxima [A]  time = 1.02559, size = 231, normalized size = 1.43 \begin{align*} -\frac{{\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) -{\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) + 4 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (\cos \left (d x + c\right )\right ) + \frac{2 \,{\left (6 \, a b^{2} \cos \left (d x + c\right ) -{\left (a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + b^{3} -{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )}}{\cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{2}}}{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))^3,x, algorithm="maxima")

[Out]

-1/4*((a^3 - 6*a^2*b + 9*a*b^2 - 4*b^3)*log(cos(d*x + c) + 1) - (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*log(cos(d*x
+ c) - 1) + 4*(3*a^2*b + 2*b^3)*log(cos(d*x + c)) + 2*(6*a*b^2*cos(d*x + c) - (a^3 + 9*a*b^2)*cos(d*x + c)^3 +
b^3 - (3*a^2*b + 2*b^3)*cos(d*x + c)^2)/(cos(d*x + c)^4 - cos(d*x + c)^2))/d

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Fricas [A]  time = 1.88695, size = 683, normalized size = 4.22 \begin{align*} -\frac{12 \, a b^{2} \cos \left (d x + c\right ) - 2 \,{\left (a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 2 \, b^{3} - 2 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left ({\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{4} -{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\cos \left (d x + c\right )\right ) +{\left ({\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} -{\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\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 ({\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} -{\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{2}\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))^3,x, algorithm="fricas")

[Out]

-1/4*(12*a*b^2*cos(d*x + c) - 2*(a^3 + 9*a*b^2)*cos(d*x + c)^3 + 2*b^3 - 2*(3*a^2*b + 2*b^3)*cos(d*x + c)^2 +
4*((3*a^2*b + 2*b^3)*cos(d*x + c)^4 - (3*a^2*b + 2*b^3)*cos(d*x + c)^2)*log(-cos(d*x + c)) + ((a^3 - 6*a^2*b +
9*a*b^2 - 4*b^3)*cos(d*x + c)^4 - (a^3 - 6*a^2*b + 9*a*b^2 - 4*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/
2) - ((a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*cos(d*x + c)^4 - (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*cos(d*x + c)^2)*log
(-1/2*cos(d*x + c) + 1/2))/(d*cos(d*x + c)^4 - d*cos(d*x + c)^2)

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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)**3*(a+b*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.5433, size = 651, normalized size = 4.02 \begin{align*} -\frac{\frac{a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{3 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{3 \, a b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 2 \,{\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) + 8 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) - \frac{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3} - \frac{2 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{12 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{18 \, a b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{8 \, b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}{\cos \left (d x + c\right ) - 1} - \frac{4 \,{\left (9 \, a^{2} b + 12 \, a b^{2} + 6 \, b^{3} + \frac{18 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{12 \, a b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{8 \, b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{9 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{6 \, b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}}}{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))^3,x, algorithm="giac")

[Out]

-1/8*(a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 3*a^2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a*b^2*(cos
(d*x + c) - 1)/(cos(d*x + c) + 1) - b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*(a^3 + 6*a^2*b + 9*a*b^2 + 4
*b^3)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) + 8*(3*a^2*b + 2*b^3)*log(abs(-(cos(d*x + c) - 1)/(cos
(d*x + c) + 1) - 1)) - (a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 12*a^2*b
*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 18*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 8*b^3*(cos(d*x + c)
- 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/(cos(d*x + c) - 1) - 4*(9*a^2*b + 12*a*b^2 + 6*b^3 + 18*a^2*b*(cos
(d*x + c) - 1)/(cos(d*x + c) + 1) + 12*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8*b^3*(cos(d*x + c) - 1)/
(cos(d*x + c) + 1) + 9*a^2*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 6*b^3*(cos(d*x + c) - 1)^2/(cos(d*x +
c) + 1)^2)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2)/d