\(\int \cot ^2(c+d x) (a+a \sin (c+d x))^4 \, dx\) [296]

Optimal result

Integrand size = 21, antiderivative size = 116 \[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {17 a^4 x}{8}-\frac {4 a^4 \text {arctanh}(\cos (c+d x))}{d}+\frac {4 a^4 \cos (c+d x)}{d}-\frac {4 a^4 \cos ^3(c+d x)}{3 d}-\frac {a^4 \cot (c+d x)}{d}+\frac {23 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d} \]

[Out]

17/8*a^4*x-4*a^4*arctanh(cos(d*x+c))/d+4*a^4*cos(d*x+c)/d-4/3*a^4*cos(d*x+c)^3/d-a^4*cot(d*x+c)/d+23/8*a^4*cos
(d*x+c)*sin(d*x+c)/d+1/4*a^4*cos(d*x+c)*sin(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2788, 3855, 3852, 8, 2715, 2713} \[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {4 a^4 \text {arctanh}(\cos (c+d x))}{d}-\frac {4 a^4 \cos ^3(c+d x)}{3 d}+\frac {4 a^4 \cos (c+d x)}{d}-\frac {a^4 \cot (c+d x)}{d}+\frac {a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac {23 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {17 a^4 x}{8} \]

[In]

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

[Out]

(17*a^4*x)/8 - (4*a^4*ArcTanh[Cos[c + d*x]])/d + (4*a^4*Cos[c + d*x])/d - (4*a^4*Cos[c + d*x]^3)/(3*d) - (a^4*
Cot[c + d*x])/d + (23*a^4*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^4*Cos[c + d*x]*Sin[c + d*x]^3)/(4*d)

Rule 8

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

Rule 2713

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 2715

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] && Integ
erQ[2*n]

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 3852

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 3855

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (5 a^6+4 a^6 \csc (c+d x)+a^6 \csc ^2(c+d x)-5 a^6 \sin ^2(c+d x)-4 a^6 \sin ^3(c+d x)-a^6 \sin ^4(c+d x)\right ) \, dx}{a^2} \\ & = 5 a^4 x+a^4 \int \csc ^2(c+d x) \, dx-a^4 \int \sin ^4(c+d x) \, dx+\left (4 a^4\right ) \int \csc (c+d x) \, dx-\left (4 a^4\right ) \int \sin ^3(c+d x) \, dx-\left (5 a^4\right ) \int \sin ^2(c+d x) \, dx \\ & = 5 a^4 x-\frac {4 a^4 \text {arctanh}(\cos (c+d x))}{d}+\frac {5 a^4 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac {1}{4} \left (3 a^4\right ) \int \sin ^2(c+d x) \, dx-\frac {1}{2} \left (5 a^4\right ) \int 1 \, dx-\frac {a^4 \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}+\frac {\left (4 a^4\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = \frac {5 a^4 x}{2}-\frac {4 a^4 \text {arctanh}(\cos (c+d x))}{d}+\frac {4 a^4 \cos (c+d x)}{d}-\frac {4 a^4 \cos ^3(c+d x)}{3 d}-\frac {a^4 \cot (c+d x)}{d}+\frac {23 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac {1}{8} \left (3 a^4\right ) \int 1 \, dx \\ & = \frac {17 a^4 x}{8}-\frac {4 a^4 \text {arctanh}(\cos (c+d x))}{d}+\frac {4 a^4 \cos (c+d x)}{d}-\frac {4 a^4 \cos ^3(c+d x)}{3 d}-\frac {a^4 \cot (c+d x)}{d}+\frac {23 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.17 \[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \csc \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (-48 \cos (c+d x)-147 \cos (3 (c+d x))+3 \cos (5 (c+d x))+408 c \sin (c+d x)+408 d x \sin (c+d x)-768 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+768 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+320 \sin (2 (c+d x))-32 \sin (4 (c+d x))\right )}{384 d} \]

[In]

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

[Out]

(a^4*Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(-48*Cos[c + d*x] - 147*Cos[3*(c + d*x)] + 3*Cos[5*(c + d*x)] + 408*c*S
in[c + d*x] + 408*d*x*Sin[c + d*x] - 768*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] + 768*Log[Sin[(c + d*x)/2]]*Sin[c
+ d*x] + 320*Sin[2*(c + d*x)] - 32*Sin[4*(c + d*x)]))/(384*d)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.03

method	result	size
parallelrisch	\(\frac {\left (\frac {16}{3}+8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (-\frac {79}{2}+47 \cos \left (d x +c \right )-23 \cos \left (2 d x +2 c \right )-\cos \left (3 d x +3 c \right )+\frac {\cos \left (4 d x +4 c \right )}{2}\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}+\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {17 d x}{4}+6 \cos \left (d x +c \right )-\frac {2 \cos \left (3 d x +3 c \right )}{3}\right ) a^{4}}{2 d}\)	\(119\)
derivativedivides	\(\frac {a^{4} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {4 a^{4} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+6 a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{4} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(136\)
default	\(\frac {a^{4} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {4 a^{4} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+6 a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{4} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(136\)
risch	\(\frac {17 a^{4} x}{8}-\frac {3 i a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{4 d}+\frac {3 a^{4} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {3 a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {3 i a^{4} {\mathrm e}^{-2 i \left (d x +c \right )}}{4 d}-\frac {2 i a^{4}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a^{4} \sin \left (4 d x +4 c \right )}{32 d}-\frac {a^{4} \cos \left (3 d x +3 c \right )}{3 d}\)	\(174\)
norman	\(\frac {-\frac {a^{4}}{2 d}+\frac {17 a^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {27 a^{4} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {27 a^{4} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {17 a^{4} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a^{4} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {17 a^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}+\frac {17 a^{4} x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {51 a^{4} x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {17 a^{4} x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {17 a^{4} x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {16 a^{4} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {16 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d}+\frac {64 a^{4} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {4 a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(289\)

[In]

int(cos(d*x+c)^2*csc(d*x+c)^2*(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/2*(16/3+8*ln(tan(1/2*d*x+1/2*c))+1/8*(-79/2+47*cos(d*x+c)-23*cos(2*d*x+2*c)-cos(3*d*x+3*c)+1/2*cos(4*d*x+4*c
))*cot(1/2*d*x+1/2*c)+sec(1/2*d*x+1/2*c)*csc(1/2*d*x+1/2*c)+17/4*d*x+6*cos(d*x+c)-2/3*cos(3*d*x+3*c))*a^4/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.16 \[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {6 \, a^{4} \cos \left (d x + c\right )^{5} - 81 \, a^{4} \cos \left (d x + c\right )^{3} - 48 \, a^{4} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 48 \, a^{4} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 51 \, a^{4} \cos \left (d x + c\right ) - {\left (32 \, a^{4} \cos \left (d x + c\right )^{3} - 51 \, a^{4} d x - 96 \, a^{4} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d \sin \left (d x + c\right )} \]

[In]

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

[Out]

1/24*(6*a^4*cos(d*x + c)^5 - 81*a^4*cos(d*x + c)^3 - 48*a^4*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 48*a^4*
log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 51*a^4*cos(d*x + c) - (32*a^4*cos(d*x + c)^3 - 51*a^4*d*x - 96*a^4
*cos(d*x + c))*sin(d*x + c))/(d*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**2*(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.01 \[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {128 \, a^{4} \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{4} - 144 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} + 96 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a^{4} - 192 \, a^{4} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{96 \, d} \]

[In]

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

[Out]

-1/96*(128*a^4*cos(d*x + c)^3 - 3*(4*d*x + 4*c - sin(4*d*x + 4*c))*a^4 - 144*(2*d*x + 2*c + sin(2*d*x + 2*c))*
a^4 + 96*(d*x + c + 1/tan(d*x + c))*a^4 - 192*a^4*(2*cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) -
1)))/d

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.67 \[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {51 \, {\left (d x + c\right )} a^{4} + 96 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 12 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {12 \, {\left (8 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{4}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (69 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 93 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 192 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 93 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 256 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 69 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 64 \, a^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

[In]

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

[Out]

1/24*(51*(d*x + c)*a^4 + 96*a^4*log(abs(tan(1/2*d*x + 1/2*c))) + 12*a^4*tan(1/2*d*x + 1/2*c) - 12*(8*a^4*tan(1
/2*d*x + 1/2*c) + a^4)/tan(1/2*d*x + 1/2*c) - 2*(69*a^4*tan(1/2*d*x + 1/2*c)^7 + 93*a^4*tan(1/2*d*x + 1/2*c)^5
- 192*a^4*tan(1/2*d*x + 1/2*c)^4 - 93*a^4*tan(1/2*d*x + 1/2*c)^3 - 256*a^4*tan(1/2*d*x + 1/2*c)^2 - 69*a^4*ta
n(1/2*d*x + 1/2*c) - 64*a^4)/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

Mupad [B] (verification not implemented)

Time = 9.12 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.54 \[ \int \cot ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {4\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {17\,a^4\,\mathrm {atan}\left (\frac {289\,a^8}{16\,\left (34\,a^8-\frac {289\,a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}+\frac {34\,a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{34\,a^8-\frac {289\,a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}}\right )}{4\,d}+\frac {-\frac {25\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}-\frac {39\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {19\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+\frac {128\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {15\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {32\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-a^4}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

[In]

int((cos(c + d*x)^2*(a + a*sin(c + d*x))^4)/sin(c + d*x)^2,x)

[Out]

(4*a^4*log(tan(c/2 + (d*x)/2)))/d + (17*a^4*atan((289*a^8)/(16*(34*a^8 - (289*a^8*tan(c/2 + (d*x)/2))/16)) + (
34*a^8*tan(c/2 + (d*x)/2))/(34*a^8 - (289*a^8*tan(c/2 + (d*x)/2))/16)))/(4*d) + ((15*a^4*tan(c/2 + (d*x)/2)^2)
/2 + (128*a^4*tan(c/2 + (d*x)/2)^3)/3 + (19*a^4*tan(c/2 + (d*x)/2)^4)/2 + 32*a^4*tan(c/2 + (d*x)/2)^5 - (39*a^
4*tan(c/2 + (d*x)/2)^6)/2 - (25*a^4*tan(c/2 + (d*x)/2)^8)/2 - a^4 + (32*a^4*tan(c/2 + (d*x)/2))/3)/(d*(2*tan(c
/2 + (d*x)/2) + 8*tan(c/2 + (d*x)/2)^3 + 12*tan(c/2 + (d*x)/2)^5 + 8*tan(c/2 + (d*x)/2)^7 + 2*tan(c/2 + (d*x)/
2)^9)) + (a^4*tan(c/2 + (d*x)/2))/(2*d)