Integrand size = 29, antiderivative size = 98 \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {11 a^3 \text {arctanh}(\cos (c+d x))}{2 d}-\frac {5 a^3 \cot (c+d x)}{d}-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))} \] Output:
-11/2*a^3*arctanh(cos(d*x+c))/d-5*a^3*cot(d*x+c)/d-1/3*a^3*cot(d*x+c)^3/d- 3/2*a^3*cot(d*x+c)*csc(d*x+c)/d+4*a^3*cos(d*x+c)/d/(1-sin(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(211\) vs. \(2(98)=196\).
Time = 6.70 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.15 \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=a^3 \left (-\frac {7 \cot \left (\frac {1}{2} (c+d x)\right )}{3 d}-\frac {3 \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 d}-\frac {11 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {11 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {3 \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {8 \sin \left (\frac {1}{2} (c+d x)\right )}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {7 \tan \left (\frac {1}{2} (c+d x)\right )}{3 d}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{24 d}\right ) \] Input:
Integrate[Csc[c + d*x]^4*Sec[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]
Output:
a^3*((-7*Cot[(c + d*x)/2])/(3*d) - (3*Csc[(c + d*x)/2]^2)/(8*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*d) - (11*Log[Cos[(c + d*x)/2]])/(2*d) + (11*Log[Sin[(c + d*x)/2]])/(2*d) + (3*Sec[(c + d*x)/2]^2)/(8*d) + (8*Sin[( c + d*x)/2])/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (7*Tan[(c + d*x)/ 2])/(3*d) + (Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(24*d))
Time = 0.39 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3351, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \csc ^4(c+d x) \sec ^2(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^3}{\sin (c+d x)^4 \cos (c+d x)^2}dx\) |
\(\Big \downarrow \) 3351 |
\(\displaystyle a^2 \int \left (a \csc ^4(c+d x)+3 a \csc ^3(c+d x)+4 a \csc ^2(c+d x)+4 a \csc (c+d x)+\frac {4 a}{1-\sin (c+d x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^2 \left (-\frac {11 a \text {arctanh}(\cos (c+d x))}{2 d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {5 a \cot (c+d x)}{d}+\frac {4 a \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {3 a \cot (c+d x) \csc (c+d x)}{2 d}\right )\) |
Input:
Int[Csc[c + d*x]^4*Sec[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]
Output:
a^2*((-11*a*ArcTanh[Cos[c + d*x]])/(2*d) - (5*a*Cot[c + d*x])/d - (a*Cot[c + d*x]^3)/(3*d) - (3*a*Cot[c + d*x]*Csc[c + d*x])/(2*d) + (4*a*Cos[c + d* x])/(d*(1 - Sin[c + d*x])))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/a^p Int[Expan dTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x])^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && In tegersQ[m, n, p/2] && ((GtQ[m, 0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (G tQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))
Time = 0.79 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.21
method | result | size |
parallelrisch | \(\frac {\left (-306+132 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+8 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+48 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{24 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\) | \(119\) |
derivativedivides | \(\frac {a^{3} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+3 a^{3} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+a^{3} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )}{d}\) | \(167\) |
default | \(\frac {a^{3} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+3 a^{3} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+a^{3} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )}{d}\) | \(167\) |
risch | \(\frac {-33 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+33 a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+60 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-96 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-19 i a^{3} {\mathrm e}^{i \left (d x +c \right )}+123 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-52 a^{3}}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) d}-\frac {11 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}+\frac {11 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}\) | \(167\) |
norman | \(\frac {\frac {a^{3}}{24 d}+\frac {59 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{24 d}-\frac {231 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8 d}+\frac {3 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {45 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8 d}-\frac {51 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 d}-\frac {231 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8 d}-\frac {45 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8 d}+\frac {59 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{24 d}+\frac {3 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{8 d}+\frac {a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{24 d}-\frac {13 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 d}-\frac {79 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{8 d}-\frac {183 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {11 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}\) | \(316\) |
Input:
int(csc(d*x+c)^4*sec(d*x+c)^2*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/24*(-306+132*ln(tan(1/2*d*x+1/2*c))*(tan(1/2*d*x+1/2*c)-1)+tan(1/2*d*x+1 /2*c)^4+cot(1/2*d*x+1/2*c)^3+8*tan(1/2*d*x+1/2*c)^3+8*cot(1/2*d*x+1/2*c)^2 +48*tan(1/2*d*x+1/2*c)^2+48*cot(1/2*d*x+1/2*c))*a^3/d/(tan(1/2*d*x+1/2*c)- 1)
Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (90) = 180\).
Time = 0.08 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.61 \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {104 \, a^{3} \cos \left (d x + c\right )^{4} + 38 \, a^{3} \cos \left (d x + c\right )^{3} - 156 \, a^{3} \cos \left (d x + c\right )^{2} - 42 \, a^{3} \cos \left (d x + c\right ) + 48 \, a^{3} + 33 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3} + {\left (a^{3} \cos \left (d x + c\right )^{3} + a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 33 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3} + {\left (a^{3} \cos \left (d x + c\right )^{3} + a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (52 \, a^{3} \cos \left (d x + c\right )^{3} + 33 \, a^{3} \cos \left (d x + c\right )^{2} - 45 \, a^{3} \cos \left (d x + c\right ) - 24 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right ) + d\right )}} \] Input:
integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
-1/12*(104*a^3*cos(d*x + c)^4 + 38*a^3*cos(d*x + c)^3 - 156*a^3*cos(d*x + c)^2 - 42*a^3*cos(d*x + c) + 48*a^3 + 33*(a^3*cos(d*x + c)^4 - 2*a^3*cos(d *x + c)^2 + a^3 + (a^3*cos(d*x + c)^3 + a^3*cos(d*x + c)^2 - a^3*cos(d*x + c) - a^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 33*(a^3*cos(d*x + c )^4 - 2*a^3*cos(d*x + c)^2 + a^3 + (a^3*cos(d*x + c)^3 + a^3*cos(d*x + c)^ 2 - a^3*cos(d*x + c) - a^3)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2 *(52*a^3*cos(d*x + c)^3 + 33*a^3*cos(d*x + c)^2 - 45*a^3*cos(d*x + c) - 24 *a^3)*sin(d*x + c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + (d*cos(d*x + c)^3 + d*cos(d*x + c)^2 - d*cos(d*x + c) - d)*sin(d*x + c) + d)
Timed out. \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \] Input:
integrate(csc(d*x+c)**4*sec(d*x+c)**2*(a+a*sin(d*x+c))**3,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.63 \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {9 \, a^{3} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 36 \, a^{3} {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} - 4 \, a^{3} {\left (\frac {6 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{3}} - 3 \, \tan \left (d x + c\right )\right )}}{12 \, d} \] Input:
integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
1/12*(9*a^3*(2*(3*cos(d*x + c)^2 - 2)/(cos(d*x + c)^3 - cos(d*x + c)) - 3* log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) + 6*a^3*(2/cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 36*a^3*(1/tan(d*x + c) - tan(d*x + c)) - 4*a^3*((6*tan(d*x + c)^2 + 1)/tan(d*x + c)^3 - 3*tan(d*x + c)))/d
Time = 0.19 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.51 \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 132 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 57 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {192 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} - \frac {242 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 57 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \] Input:
integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
1/24*(a^3*tan(1/2*d*x + 1/2*c)^3 + 9*a^3*tan(1/2*d*x + 1/2*c)^2 + 132*a^3* log(abs(tan(1/2*d*x + 1/2*c))) + 57*a^3*tan(1/2*d*x + 1/2*c) - 192*a^3/(ta n(1/2*d*x + 1/2*c) - 1) - (242*a^3*tan(1/2*d*x + 1/2*c)^3 + 57*a^3*tan(1/2 *d*x + 1/2*c)^2 + 9*a^3*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c)^3 )/d
Time = 31.56 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.63 \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {-83\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+16\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {8\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {a^3}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}+\frac {11\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}+\frac {19\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d} \] Input:
int((a + a*sin(c + d*x))^3/(cos(c + d*x)^2*sin(c + d*x)^4),x)
Output:
(3*a^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (a^3*tan(c/2 + (d*x)/2)^3)/(24*d) - ( 16*a^3*tan(c/2 + (d*x)/2)^2 - 83*a^3*tan(c/2 + (d*x)/2)^3 + a^3/3 + (8*a^3 *tan(c/2 + (d*x)/2))/3)/(d*(8*tan(c/2 + (d*x)/2)^3 - 8*tan(c/2 + (d*x)/2)^ 4)) + (11*a^3*log(tan(c/2 + (d*x)/2)))/(2*d) + (19*a^3*tan(c/2 + (d*x)/2)) /(8*d)
Time = 0.17 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.57 \[ \int \csc ^4(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \left (132 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-132 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-306 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{24 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )} \] Input:
int(csc(d*x+c)^4*sec(d*x+c)^2*(a+a*sin(d*x+c))^3,x)
Output:
(a**3*(132*log(tan((c + d*x)/2))*tan((c + d*x)/2)**4 - 132*log(tan((c + d* x)/2))*tan((c + d*x)/2)**3 + tan((c + d*x)/2)**7 + 8*tan((c + d*x)/2)**6 + 48*tan((c + d*x)/2)**5 - 306*tan((c + d*x)/2)**4 + 48*tan((c + d*x)/2)**2 + 8*tan((c + d*x)/2) + 1))/(24*tan((c + d*x)/2)**3*d*(tan((c + d*x)/2) - 1))