3.1110 \(\int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=189 \[ -\frac{\left (4 a^2-23 b^2\right ) \cos (c+d x)}{6 d}-\frac{\left (2 a^2-3 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{6 a^2 d}-\frac{b \left (a^2-3 b^2\right ) \sin (c+d x) \cos (c+d x)}{3 a d}+\frac{\left (3 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^3}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 a d}-3 a b x \]

[Out]

-3*a*b*x + ((3*a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(2*d) - ((4*a^2 - 23*b^2)*Cos[c + d*x])/(6*d) - (b*(a^2 - 3
*b^2)*Cos[c + d*x]*Sin[c + d*x])/(3*a*d) - ((2*a^2 - 3*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x])^2)/(6*a^2*d) - (
b*Cot[c + d*x]*(a + b*Sin[c + d*x])^3)/(2*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]*(a + b*Sin[c + d*x])^3)/(2*a*d)

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Rubi [A]  time = 0.483869, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2893, 3049, 3033, 3023, 2735, 3770} \[ -\frac{\left (4 a^2-23 b^2\right ) \cos (c+d x)}{6 d}-\frac{\left (2 a^2-3 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{6 a^2 d}-\frac{b \left (a^2-3 b^2\right ) \sin (c+d x) \cos (c+d x)}{3 a d}+\frac{\left (3 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^3}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 a d}-3 a b x \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*Cot[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

-3*a*b*x + ((3*a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(2*d) - ((4*a^2 - 23*b^2)*Cos[c + d*x])/(6*d) - (b*(a^2 - 3
*b^2)*Cos[c + d*x]*Sin[c + d*x])/(3*a*d) - ((2*a^2 - 3*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x])^2)/(6*a^2*d) - (
b*Cot[c + d*x]*(a + b*Sin[c + d*x])^3)/(2*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]*(a + b*Sin[c + d*x])^3)/(2*a*d)

Rule 2893

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*d*f*(n + 1)), x] +
(-Dist[1/(a^2*d^2*(n + 1)*(n + 2)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) -
b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4))*Sin[e +
f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/
(a^2*d^2*f*(n + 1)*(n + 2)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Intege
rsQ[2*m, 2*n]) &&  !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[m + n + 4, 0])

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 3770

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

Rubi steps

\begin{align*} \int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx &=-\frac{b \cot (c+d x) (a+b \sin (c+d x))^3}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 a d}-\frac{\int \csc (c+d x) (a+b \sin (c+d x))^2 \left (3 a^2-2 b^2+2 a b \sin (c+d x)-\left (2 a^2-3 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{2 a^2}\\ &=-\frac{\left (2 a^2-3 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{6 a^2 d}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^3}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 a d}-\frac{\int \csc (c+d x) (a+b \sin (c+d x)) \left (3 a \left (3 a^2-2 b^2\right )+11 a^2 b \sin (c+d x)-4 a \left (a^2-3 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{6 a^2}\\ &=-\frac{b \left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{3 a d}-\frac{\left (2 a^2-3 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{6 a^2 d}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^3}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 a d}-\frac{\int \csc (c+d x) \left (6 a^2 \left (3 a^2-2 b^2\right )+36 a^3 b \sin (c+d x)-2 a^2 \left (4 a^2-23 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{12 a^2}\\ &=-\frac{\left (4 a^2-23 b^2\right ) \cos (c+d x)}{6 d}-\frac{b \left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{3 a d}-\frac{\left (2 a^2-3 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{6 a^2 d}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^3}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 a d}-\frac{\int \csc (c+d x) \left (6 a^2 \left (3 a^2-2 b^2\right )+36 a^3 b \sin (c+d x)\right ) \, dx}{12 a^2}\\ &=-3 a b x-\frac{\left (4 a^2-23 b^2\right ) \cos (c+d x)}{6 d}-\frac{b \left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{3 a d}-\frac{\left (2 a^2-3 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{6 a^2 d}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^3}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 a d}-\frac{1}{2} \left (3 a^2-2 b^2\right ) \int \csc (c+d x) \, dx\\ &=-3 a b x+\frac{\left (3 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{\left (4 a^2-23 b^2\right ) \cos (c+d x)}{6 d}-\frac{b \left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{3 a d}-\frac{\left (2 a^2-3 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{6 a^2 d}-\frac{b \cot (c+d x) (a+b \sin (c+d x))^3}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 a d}\\ \end{align*}

Mathematica [A]  time = 3.56679, size = 191, normalized size = 1.01 \[ \frac{-6 \left (4 a^2-5 b^2\right ) \cos (c+d x)+3 \left (a^2 \left (-\csc ^2\left (\frac{1}{2} (c+d x)\right )\right )+a^2 \sec ^2\left (\frac{1}{2} (c+d x)\right )-12 a^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 a^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-4 a b \sin (2 (c+d x))+8 a b \tan \left (\frac{1}{2} (c+d x)\right )-8 a b \cot \left (\frac{1}{2} (c+d x)\right )-24 a b c-24 a b d x+8 b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-8 b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+2 b^2 \cos (3 (c+d x))}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*Cot[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

(-6*(4*a^2 - 5*b^2)*Cos[c + d*x] + 2*b^2*Cos[3*(c + d*x)] + 3*(-24*a*b*c - 24*a*b*d*x - 8*a*b*Cot[(c + d*x)/2]
- a^2*Csc[(c + d*x)/2]^2 + 12*a^2*Log[Cos[(c + d*x)/2]] - 8*b^2*Log[Cos[(c + d*x)/2]] - 12*a^2*Log[Sin[(c + d
*x)/2]] + 8*b^2*Log[Sin[(c + d*x)/2]] + a^2*Sec[(c + d*x)/2]^2 - 4*a*b*Sin[2*(c + d*x)] + 8*a*b*Tan[(c + d*x)/
2]))/(24*d)

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Maple [A]  time = 0.089, size = 208, normalized size = 1.1 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{2\,d}}-{\frac{3\,{a}^{2}\cos \left ( dx+c \right ) }{2\,d}}-{\frac{3\,{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-2\,{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{d\sin \left ( dx+c \right ) }}-2\,{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{d}}-3\,{\frac{ab\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}-3\,abx-3\,{\frac{abc}{d}}+{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{b}^{2}\cos \left ( dx+c \right ) }{d}}+{\frac{{b}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^3*(a+b*sin(d*x+c))^2,x)

[Out]

-1/2/d*a^2/sin(d*x+c)^2*cos(d*x+c)^5-1/2*a^2*cos(d*x+c)^3/d-3/2*a^2*cos(d*x+c)/d-3/2/d*a^2*ln(csc(d*x+c)-cot(d
*x+c))-2/d*a*b/sin(d*x+c)*cos(d*x+c)^5-2*a*b*cos(d*x+c)^3*sin(d*x+c)/d-3*a*b*cos(d*x+c)*sin(d*x+c)/d-3*a*b*x-3
/d*a*b*c+1/3*b^2*cos(d*x+c)^3/d+b^2*cos(d*x+c)/d+1/d*b^2*ln(csc(d*x+c)-cot(d*x+c))

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Maxima [A]  time = 1.49058, size = 203, normalized size = 1.07 \begin{align*} -\frac{12 \,{\left (3 \, d x + 3 \, c + \frac{3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a b - 2 \,{\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} b^{2} - 3 \, a^{2}{\left (\frac{2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.85399, size = 518, normalized size = 2.74 \begin{align*} \frac{4 \, b^{2} \cos \left (d x + c\right )^{5} - 36 \, a b d x \cos \left (d x + c\right )^{2} + 36 \, a b d x - 4 \,{\left (3 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (3 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right ) + 3 \,{\left ({\left (3 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} + 2 \, b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 3 \,{\left ({\left (3 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} + 2 \, b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 12 \,{\left (a b \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \,{\left (d \cos \left (d x + c\right )^{2} - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/12*(4*b^2*cos(d*x + c)^5 - 36*a*b*d*x*cos(d*x + c)^2 + 36*a*b*d*x - 4*(3*a^2 - 2*b^2)*cos(d*x + c)^3 + 6*(3*
a^2 - 2*b^2)*cos(d*x + c) + 3*((3*a^2 - 2*b^2)*cos(d*x + c)^2 - 3*a^2 + 2*b^2)*log(1/2*cos(d*x + c) + 1/2) - 3
*((3*a^2 - 2*b^2)*cos(d*x + c)^2 - 3*a^2 + 2*b^2)*log(-1/2*cos(d*x + c) + 1/2) - 12*(a*b*cos(d*x + c)^3 - 3*a*
b*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 - d)

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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)**4*csc(d*x+c)**3*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.33913, size = 340, normalized size = 1.8 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 72 \,{\left (d x + c\right )} a b + 24 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 12 \,{\left (3 \, a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{3 \,{\left (18 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{2}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}} + \frac{16 \,{\left (3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 6 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 6 \, 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 ) - 3 \, a^{2} + 4 \, b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(3*a^2*tan(1/2*d*x + 1/2*c)^2 - 72*(d*x + c)*a*b + 24*a*b*tan(1/2*d*x + 1/2*c) - 12*(3*a^2 - 2*b^2)*log(a
bs(tan(1/2*d*x + 1/2*c))) + 3*(18*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d*x + 1/2*c)^2 - 8*a*b*tan(1/2*d
*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^2 + 16*(3*a*b*tan(1/2*d*x + 1/2*c)^5 - 3*a^2*tan(1/2*d*x + 1/2*c)^4 +
6*b^2*tan(1/2*d*x + 1/2*c)^4 - 6*a^2*tan(1/2*d*x + 1/2*c)^2 + 6*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x
+ 1/2*c) - 3*a^2 + 4*b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d