3.1324 \(\int \frac{\cos ^4(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=183 \[ \frac{\left (a^2-3 b^2\right ) \cos (c+d x)}{b^3 d}-\frac{2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 b^4 d}+\frac{a x \left (a^2-3 b^2\right )}{b^4}+\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{a \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{a x}{2 b^2}-\frac{\cot (c+d x)}{a d}-\frac{\cos ^3(c+d x)}{3 b d}+\frac{\cos (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.255159, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.345, Rules used = {2897, 3770, 3767, 8, 2638, 2635, 2633, 2660, 618, 204} \[ \frac{\left (a^2-3 b^2\right ) \cos (c+d x)}{b^3 d}-\frac{2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 b^4 d}+\frac{a x \left (a^2-3 b^2\right )}{b^4}+\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{a \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{a x}{2 b^2}-\frac{\cot (c+d x)}{a d}-\frac{\cos ^3(c+d x)}{3 b d}+\frac{\cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

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

Rule 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 1.38042, size = 208, normalized size = 1.14 \[ -\frac{3 a^3 b^2 \sin (2 (c+d x))-3 a^2 b \left (4 a^2-9 b^2\right ) \cos (c+d x)+a^2 b^3 \cos (3 (c+d x))+24 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )+30 a^3 b^2 c+30 a^3 b^2 d x-12 a^5 c-12 a^5 d x-6 a b^4 \tan \left (\frac{1}{2} (c+d x)\right )+6 a b^4 \cot \left (\frac{1}{2} (c+d x)\right )+12 b^5 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-12 b^5 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{12 a^2 b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

-(-12*a^5*c + 30*a^3*b^2*c - 12*a^5*d*x + 30*a^3*b^2*d*x + 24*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*x)/2]
)/Sqrt[a^2 - b^2]] - 3*a^2*b*(4*a^2 - 9*b^2)*Cos[c + d*x] + a^2*b^3*Cos[3*(c + d*x)] + 6*a*b^4*Cot[(c + d*x)/2
] - 12*b^5*Log[Cos[(c + d*x)/2]] + 12*b^5*Log[Sin[(c + d*x)/2]] + 3*a^3*b^2*Sin[2*(c + d*x)] - 6*a*b^4*Tan[(c
+ d*x)/2])/(12*a^2*b^4*d)

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Maple [B]  time = 0.119, size = 557, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^2/(a+b*sin(d*x+c)),x)

[Out]

1/2/d/a*tan(1/2*d*x+1/2*c)+1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*a*tan(1/2*d*x+1/2*c)^5+2/d/b^3/(1+tan(1/2*d*x+1/
2*c)^2)^3*tan(1/2*d*x+1/2*c)^4*a^2-6/d/b/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^4+4/d/b^3/(1+tan(1/2*d*
x+1/2*c)^2)^3*a^2*tan(1/2*d*x+1/2*c)^2-8/d/b/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2-1/d/b^2/(1+tan(1/
2*d*x+1/2*c)^2)^3*a*tan(1/2*d*x+1/2*c)+2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*a^2-14/3/d/b/(1+tan(1/2*d*x+1/2*c)^2
)^3+2/d/b^4*arctan(tan(1/2*d*x+1/2*c))*a^3-5/d/b^2*a*arctan(tan(1/2*d*x+1/2*c))-2/d*a^4/b^4/(a^2-b^2)^(1/2)*ar
ctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+6/d/b^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2
*c)+2*b)/(a^2-b^2)^(1/2))*a^2-6/d/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+2/d
/a^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))*b^2-1/2/d/a/tan(1/2*d*x+1/2*c)-1
/d/a^2*b*ln(tan(1/2*d*x+1/2*c))

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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(cos(d*x+c)^6*csc(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.04288, size = 1311, normalized size = 7.16 \begin{align*} \left [\frac{3 \, a^{3} b^{2} \cos \left (d x + c\right )^{3} + 3 \, b^{5} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 3 \, b^{5} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 3 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) \sin \left (d x + c\right ) - 3 \,{\left (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right ) -{\left (2 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} - 3 \,{\left (2 \, a^{5} - 5 \, a^{3} b^{2}\right )} d x - 6 \,{\left (a^{4} b - 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, a^{2} b^{4} d \sin \left (d x + c\right )}, \frac{3 \, a^{3} b^{2} \cos \left (d x + c\right )^{3} + 3 \, b^{5} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 3 \, b^{5} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 6 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 3 \,{\left (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right ) -{\left (2 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} - 3 \,{\left (2 \, a^{5} - 5 \, a^{3} b^{2}\right )} d x - 6 \,{\left (a^{4} b - 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, a^{2} b^{4} d \sin \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

[1/6*(3*a^3*b^2*cos(d*x + c)^3 + 3*b^5*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*b^5*log(-1/2*cos(d*x + c)
+ 1/2)*sin(d*x + c) + 3*(a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin
(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2
- 2*a*b*sin(d*x + c) - a^2 - b^2))*sin(d*x + c) - 3*(a^3*b^2 + 2*a*b^4)*cos(d*x + c) - (2*a^2*b^3*cos(d*x + c
)^3 - 3*(2*a^5 - 5*a^3*b^2)*d*x - 6*(a^4*b - 2*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/(a^2*b^4*d*sin(d*x + c)),
1/6*(3*a^3*b^2*cos(d*x + c)^3 + 3*b^5*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*b^5*log(-1/2*cos(d*x + c) +
1/2)*sin(d*x + c) + 6*(a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*c
os(d*x + c)))*sin(d*x + c) - 3*(a^3*b^2 + 2*a*b^4)*cos(d*x + c) - (2*a^2*b^3*cos(d*x + c)^3 - 3*(2*a^5 - 5*a^3
*b^2)*d*x - 6*(a^4*b - 2*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/(a^2*b^4*d*sin(d*x + c))]

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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(cos(d*x+c)**6*csc(d*x+c)**2/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.25942, size = 408, normalized size = 2.23 \begin{align*} -\frac{\frac{6 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} - \frac{3 \,{\left (2 \, a^{3} - 5 \, a b^{2}\right )}{\left (d x + c\right )}}{b^{4}} - \frac{3 \,{\left (2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} + \frac{12 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} a^{2} b^{4}} - \frac{2 \,{\left (3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 18 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 24 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a^{2} - 14 \, b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 3*tan(1/2*d*x + 1/2*c)/a - 3*(2*a^3 - 5*a*b^2)*(d*x + c)/b^4 -
3*(2*b*tan(1/2*d*x + 1/2*c) - a)/(a^2*tan(1/2*d*x + 1/2*c)) + 12*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(pi*floor
(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^2*b
^4) - 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*a^2*tan(1/2*d*x + 1/2*c)^4 - 18*b^2*tan(1/2*d*x + 1/2*c)^4 + 12*a^2*
tan(1/2*d*x + 1/2*c)^2 - 24*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c) + 6*a^2 - 14*b^2)/((tan(1/
2*d*x + 1/2*c)^2 + 1)^3*b^3))/d