3.298 \(\int \frac{\cot ^4(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=177 \[ -\frac{a \cot ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac{a \left (a^2-2 b^2\right ) \cot (c+d x)}{d \left (a^2-b^2\right )^2}+\frac{b \csc ^3(c+d x)}{3 d \left (a^2-b^2\right )}-\frac{b \left (a^2-2 b^2\right ) \csc (c+d x)}{d \left (a^2-b^2\right )^2}-\frac{2 b^5 \tanh ^{-1}\left (\frac{\sqrt{a^2-b^2} \tan \left (\frac{1}{2} (c+d x)\right )}{a+b}\right )}{a d \left (a^2-b^2\right )^{5/2}}+\frac{x}{a} \]

[Out]

x/a - (2*b^5*ArcTanh[(Sqrt[a^2 - b^2]*Tan[(c + d*x)/2])/(a + b)])/(a*(a^2 - b^2)^(5/2)*d) + (a*(a^2 - 2*b^2)*C
ot[c + d*x])/((a^2 - b^2)^2*d) - (a*Cot[c + d*x]^3)/(3*(a^2 - b^2)*d) - (b*(a^2 - 2*b^2)*Csc[c + d*x])/((a^2 -
b^2)^2*d) + (b*Csc[c + d*x]^3)/(3*(a^2 - b^2)*d)

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Rubi [A]  time = 0.386905, antiderivative size = 256, normalized size of antiderivative = 1.45, number of steps used = 15, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3898, 2902, 2606, 3473, 8, 2735, 2659, 208} \[ -\frac{a \cot ^3(c+d x)}{3 d \left (a^2-b^2\right )}-\frac{a b^2 \cot (c+d x)}{d \left (a^2-b^2\right )^2}+\frac{a \cot (c+d x)}{d \left (a^2-b^2\right )}+\frac{b \csc ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac{b^3 \csc (c+d x)}{d \left (a^2-b^2\right )^2}-\frac{b \csc (c+d x)}{d \left (a^2-b^2\right )}+\frac{b^4 x}{a \left (a^2-b^2\right )^2}-\frac{a b^2 x}{\left (a^2-b^2\right )^2}+\frac{a x}{a^2-b^2}-\frac{2 b^5 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d (a-b)^{5/2} (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4/(a + b*Sec[c + d*x]),x]

[Out]

-((a*b^2*x)/(a^2 - b^2)^2) + (b^4*x)/(a*(a^2 - b^2)^2) + (a*x)/(a^2 - b^2) - (2*b^5*ArcTanh[(Sqrt[a - b]*Tan[(
c + d*x)/2])/Sqrt[a + b]])/(a*(a - b)^(5/2)*(a + b)^(5/2)*d) - (a*b^2*Cot[c + d*x])/((a^2 - b^2)^2*d) + (a*Cot
[c + d*x])/((a^2 - b^2)*d) - (a*Cot[c + d*x]^3)/(3*(a^2 - b^2)*d) + (b^3*Csc[c + d*x])/((a^2 - b^2)^2*d) - (b*
Csc[c + d*x])/((a^2 - b^2)*d) + (b*Csc[c + d*x]^3)/(3*(a^2 - b^2)*d)

Rule 3898

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[(Cos[c + d*x]^
m*(b + a*Sin[c + d*x])^n)/Sin[c + d*x]^(m + n), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[
n] && IntegerQ[m] && (IntegerQ[m/2] || LeQ[m, 1])

Rule 2902

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[(a*d^2)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[(b*d)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a^2*d^2)/(g^2*(a^2 - b^
2)), Int[((g*Cos[e + f*x])^(p + 2)*(d*Sin[e + f*x])^(n - 2))/(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [B]  time = 6.16237, size = 416, normalized size = 2.35 \[ \frac{2 b^5 \sec (c+d x) (a \cos (c+d x)+b) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a d \sqrt{a^2-b^2} \left (b^2-a^2\right )^2 (a+b \sec (c+d x))}+\frac{(c+d x) \sec (c+d x) (a \cos (c+d x)+b)}{a d (a+b \sec (c+d x))}+\frac{\csc \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (8 a \cos \left (\frac{1}{2} (c+d x)\right )+11 b \cos \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)}{12 d (a+b)^2 (a+b \sec (c+d x))}-\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) (a \cos (c+d x)+b)}{24 d (b-a) (a+b \sec (c+d x))}+\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (11 b \sin \left (\frac{1}{2} (c+d x)\right )-8 a \sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)}{12 d (b-a)^2 (a+b \sec (c+d x))}-\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) (a \cos (c+d x)+b)}{24 d (a+b) (a+b \sec (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4/(a + b*Sec[c + d*x]),x]

[Out]

((c + d*x)*(b + a*Cos[c + d*x])*Sec[c + d*x])/(a*d*(a + b*Sec[c + d*x])) + (2*b^5*ArcTanh[((-a + b)*Tan[(c + d
*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x])*Sec[c + d*x])/(a*Sqrt[a^2 - b^2]*(-a^2 + b^2)^2*d*(a + b*Sec[c +
d*x])) + ((8*a*Cos[(c + d*x)/2] + 11*b*Cos[(c + d*x)/2])*(b + a*Cos[c + d*x])*Csc[(c + d*x)/2]*Sec[c + d*x])/
(12*(a + b)^2*d*(a + b*Sec[c + d*x])) - ((b + a*Cos[c + d*x])*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2*Sec[c + d*x]
)/(24*(a + b)*d*(a + b*Sec[c + d*x])) + ((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]*Sec[c + d*x]*(-8*a*Sin[(c + d*x
)/2] + 11*b*Sin[(c + d*x)/2]))/(12*(-a + b)^2*d*(a + b*Sec[c + d*x])) - ((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]
^2*Sec[c + d*x]*Tan[(c + d*x)/2])/(24*(-a + b)*d*(a + b*Sec[c + d*x]))

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Maple [A]  time = 0.085, size = 238, normalized size = 1.3 \begin{align*}{\frac{a}{24\,d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{b}{24\,d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{5\,a}{8\,d \left ( a-b \right ) ^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{7\,b}{8\,d \left ( a-b \right ) ^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{ad}}-2\,{\frac{{b}^{5}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}a\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{24\,d \left ( a+b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{5\,a}{8\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{7\,b}{8\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4/(a+b*sec(d*x+c)),x)

[Out]

1/24/d/(a-b)^2*tan(1/2*d*x+1/2*c)^3*a-1/24/d/(a-b)^2*b*tan(1/2*d*x+1/2*c)^3-5/8/d/(a-b)^2*a*tan(1/2*d*x+1/2*c)
+7/8/d/(a-b)^2*b*tan(1/2*d*x+1/2*c)+2/d/a*arctan(tan(1/2*d*x+1/2*c))-2/d/(a+b)^2/(a-b)^2*b^5/a/((a+b)*(a-b))^(
1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-1/24/d/(a+b)/tan(1/2*d*x+1/2*c)^3+5/8/d/(a+b)^2/tan
(1/2*d*x+1/2*c)*a+7/8/d/(a+b)^2/tan(1/2*d*x+1/2*c)*b

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.974179, size = 1623, normalized size = 9.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/6*(4*a^5*b - 14*a^3*b^3 + 10*a*b^5 + 2*(4*a^6 - 11*a^4*b^2 + 7*a^2*b^4)*cos(d*x + c)^3 + 3*(b^5*cos(d*x + c
)^2 - b^5)*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d
*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2))*sin(d*x + c) - 6*(a^
5*b - 3*a^3*b^3 + 2*a*b^5)*cos(d*x + c)^2 - 6*(a^6 - 3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c) + 6*((a^6 - 3*a^4*b^2
+ 3*a^2*b^4 - b^6)*d*x*cos(d*x + c)^2 - (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*x)*sin(d*x + c))/(((a^7 - 3*a^5
*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2 - (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d)*sin(d*x + c)), 1/3*(2*a^
5*b - 7*a^3*b^3 + 5*a*b^5 + (4*a^6 - 11*a^4*b^2 + 7*a^2*b^4)*cos(d*x + c)^3 - 3*(b^5*cos(d*x + c)^2 - b^5)*sqr
t(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*sin(d*x + c) - 3*(a^5*
b - 3*a^3*b^3 + 2*a*b^5)*cos(d*x + c)^2 - 3*(a^6 - 3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c) + 3*((a^6 - 3*a^4*b^2 +
3*a^2*b^4 - b^6)*d*x*cos(d*x + c)^2 - (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*x)*sin(d*x + c))/(((a^7 - 3*a^5*b
^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2 - (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d)*sin(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{4}{\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(cot(d*x+c)**4/(a+b*sec(d*x+c)),x)

[Out]

Integral(cot(c + d*x)**4/(a + b*sec(c + d*x)), x)

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Giac [A]  time = 1.34975, size = 386, normalized size = 2.18 \begin{align*} -\frac{\frac{48 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )} b^{5}}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 36 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 21 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{24 \,{\left (d x + c\right )}}{a} - \frac{15 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 21 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/24*(48*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x
+ 1/2*c))/sqrt(-a^2 + b^2)))*b^5/((a^5 - 2*a^3*b^2 + a*b^4)*sqrt(-a^2 + b^2)) - (a^2*tan(1/2*d*x + 1/2*c)^3 -
2*a*b*tan(1/2*d*x + 1/2*c)^3 + b^2*tan(1/2*d*x + 1/2*c)^3 - 15*a^2*tan(1/2*d*x + 1/2*c) + 36*a*b*tan(1/2*d*x +
1/2*c) - 21*b^2*tan(1/2*d*x + 1/2*c))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 24*(d*x + c)/a - (15*a*tan(1/2*d*x +
1/2*c)^2 + 21*b*tan(1/2*d*x + 1/2*c)^2 - a - b)/((a^2 + 2*a*b + b^2)*tan(1/2*d*x + 1/2*c)^3))/d