\(\int \cot ^5(c+d x) (a+a \sin (c+d x))^4 \, dx\) [531]

Optimal result

Integrand size = 21, antiderivative size = 148 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {4 a^4 \csc (c+d x)}{d}-\frac {2 a^4 \csc ^2(c+d x)}{d}-\frac {4 a^4 \csc ^3(c+d x)}{3 d}-\frac {a^4 \csc ^4(c+d x)}{4 d}-\frac {10 a^4 \log (\sin (c+d x))}{d}-\frac {4 a^4 \sin (c+d x)}{d}+\frac {2 a^4 \sin ^2(c+d x)}{d}+\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {a^4 \sin ^4(c+d x)}{4 d} \]

[Out]

4*a^4*csc(d*x+c)/d-2*a^4*csc(d*x+c)^2/d-4/3*a^4*csc(d*x+c)^3/d-1/4*a^4*csc(d*x+c)^4/d-10*a^4*ln(sin(d*x+c))/d-
4*a^4*sin(d*x+c)/d+2*a^4*sin(d*x+c)^2/d+4/3*a^4*sin(d*x+c)^3/d+1/4*a^4*sin(d*x+c)^4/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2786, 90} \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \sin ^4(c+d x)}{4 d}+\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {2 a^4 \sin ^2(c+d x)}{d}-\frac {4 a^4 \sin (c+d x)}{d}-\frac {a^4 \csc ^4(c+d x)}{4 d}-\frac {4 a^4 \csc ^3(c+d x)}{3 d}-\frac {2 a^4 \csc ^2(c+d x)}{d}+\frac {4 a^4 \csc (c+d x)}{d}-\frac {10 a^4 \log (\sin (c+d x))}{d} \]

[In]

Int[Cot[c + d*x]^5*(a + a*Sin[c + d*x])^4,x]

[Out]

(4*a^4*Csc[c + d*x])/d - (2*a^4*Csc[c + d*x]^2)/d - (4*a^4*Csc[c + d*x]^3)/(3*d) - (a^4*Csc[c + d*x]^4)/(4*d)
- (10*a^4*Log[Sin[c + d*x]])/d - (4*a^4*Sin[c + d*x])/d + (2*a^4*Sin[c + d*x]^2)/d + (4*a^4*Sin[c + d*x]^3)/(3
*d) + (a^4*Sin[c + d*x]^4)/(4*d)

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2786

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[x^p*((a + x)^(m - (p + 1)/2)/(a - x)^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x)^2 (a+x)^6}{x^5} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (-4 a^3+\frac {a^8}{x^5}+\frac {4 a^7}{x^4}+\frac {4 a^6}{x^3}-\frac {4 a^5}{x^2}-\frac {10 a^4}{x}+4 a^2 x+4 a x^2+x^3\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {4 a^4 \csc (c+d x)}{d}-\frac {2 a^4 \csc ^2(c+d x)}{d}-\frac {4 a^4 \csc ^3(c+d x)}{3 d}-\frac {a^4 \csc ^4(c+d x)}{4 d}-\frac {10 a^4 \log (\sin (c+d x))}{d}-\frac {4 a^4 \sin (c+d x)}{d}+\frac {2 a^4 \sin ^2(c+d x)}{d}+\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {a^4 \sin ^4(c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.65 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \left (48 \csc (c+d x)-24 \csc ^2(c+d x)-16 \csc ^3(c+d x)-3 \csc ^4(c+d x)-120 \log (\sin (c+d x))-48 \sin (c+d x)+24 \sin ^2(c+d x)+16 \sin ^3(c+d x)+3 \sin ^4(c+d x)\right )}{12 d} \]

[In]

Integrate[Cot[c + d*x]^5*(a + a*Sin[c + d*x])^4,x]

[Out]

(a^4*(48*Csc[c + d*x] - 24*Csc[c + d*x]^2 - 16*Csc[c + d*x]^3 - 3*Csc[c + d*x]^4 - 120*Log[Sin[c + d*x]] - 48*
Sin[c + d*x] + 24*Sin[c + d*x]^2 + 16*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4))/(12*d)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.27

method	result	size
parallelrisch	\(-\frac {3 \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\left (\frac {80 \cos \left (2 d x +2 c \right )}{3}-\frac {20 \cos \left (4 d x +4 c \right )}{3}-20\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {80 \cos \left (2 d x +2 c \right )}{3}+\frac {20 \cos \left (4 d x +4 c \right )}{3}+20\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin \left (3 d x +3 c \right )+\frac {5 \sin \left (5 d x +5 c \right )}{9}+\frac {\sin \left (7 d x +7 c \right )}{9}+\frac {113 \cos \left (2 d x +2 c \right )}{12}-\frac {55 \cos \left (4 d x +4 c \right )}{12}+\frac {5 \cos \left (6 d x +6 c \right )}{12}-\frac {\cos \left (8 d x +8 c \right )}{96}+\frac {5 \sin \left (d x +c \right )}{9}-\frac {125}{32}\right ) a^{4}}{256 d}\)	\(188\)
derivativedivides	\(\frac {a^{4} \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+4 a^{4} \left (-\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+6 a^{4} \left (-\frac {\cos ^{6}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{2}-\left (\cos ^{2}\left (d x +c \right )\right )-2 \ln \left (\sin \left (d x +c \right )\right )\right )+4 a^{4} \left (-\frac {\cos ^{6}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+a^{4} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(243\)
default	\(\frac {a^{4} \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+4 a^{4} \left (-\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+6 a^{4} \left (-\frac {\cos ^{6}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{2}-\left (\cos ^{2}\left (d x +c \right )\right )-2 \ln \left (\sin \left (d x +c \right )\right )\right )+4 a^{4} \left (-\frac {\cos ^{6}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+a^{4} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(243\)
risch	\(10 i a^{4} x +\frac {a^{4} {\mathrm e}^{4 i \left (d x +c \right )}}{64 d}+\frac {i a^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{6 d}-\frac {9 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {3 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {3 i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {9 a^{4} {\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}-\frac {i a^{4} {\mathrm e}^{-3 i \left (d x +c \right )}}{6 d}+\frac {a^{4} {\mathrm e}^{-4 i \left (d x +c \right )}}{64 d}+\frac {20 i a^{4} c}{d}+\frac {4 i a^{4} \left (-6 i {\mathrm e}^{6 i \left (d x +c \right )}+6 \,{\mathrm e}^{7 i \left (d x +c \right )}+15 i {\mathrm e}^{4 i \left (d x +c \right )}-10 \,{\mathrm e}^{5 i \left (d x +c \right )}-6 i {\mathrm e}^{2 i \left (d x +c \right )}+10 \,{\mathrm e}^{3 i \left (d x +c \right )}-6 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}-\frac {10 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(282\)
norman	\(\frac {-\frac {a^{4}}{64 d}-\frac {a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d}-\frac {5 a^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {5 a^{4} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {3 a^{4} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a^{4} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {5 a^{4} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {3 a^{4} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a^{4} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {5 a^{4} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {a^{4} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {a^{4} \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}+\frac {107 a^{4} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {107 a^{4} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {945 a^{4} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {10 a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {10 a^{4} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(342\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^5*(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

-3/256*sec(1/2*d*x+1/2*c)^4*csc(1/2*d*x+1/2*c)^4*((80/3*cos(2*d*x+2*c)-20/3*cos(4*d*x+4*c)-20)*ln(sec(1/2*d*x+
1/2*c)^2)+(-80/3*cos(2*d*x+2*c)+20/3*cos(4*d*x+4*c)+20)*ln(tan(1/2*d*x+1/2*c))+sin(3*d*x+3*c)+5/9*sin(5*d*x+5*
c)+1/9*sin(7*d*x+7*c)+113/12*cos(2*d*x+2*c)-55/12*cos(4*d*x+4*c)+5/12*cos(6*d*x+6*c)-1/96*cos(8*d*x+8*c)+5/9*s
in(d*x+c)-125/32)*a^4/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.97 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {24 \, a^{4} \cos \left (d x + c\right )^{8} - 128 \, a^{4} \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 288 \, a^{4} \cos \left (d x + c\right )^{6} + 615 \, a^{4} \cos \left (d x + c\right )^{4} - 270 \, a^{4} \cos \left (d x + c\right )^{2} - 105 \, a^{4} - 960 \, {\left (a^{4} \cos \left (d x + c\right )^{4} - 2 \, a^{4} \cos \left (d x + c\right )^{2} + a^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right )}{96 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5*(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/96*(24*a^4*cos(d*x + c)^8 - 128*a^4*cos(d*x + c)^6*sin(d*x + c) - 288*a^4*cos(d*x + c)^6 + 615*a^4*cos(d*x +
c)^4 - 270*a^4*cos(d*x + c)^2 - 105*a^4 - 960*(a^4*cos(d*x + c)^4 - 2*a^4*cos(d*x + c)^2 + a^4)*log(1/2*sin(d
*x + c)))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

Sympy [F(-1)]

Timed out. \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^4 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**5*(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.81 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 24 \, a^{4} \sin \left (d x + c\right )^{2} - 120 \, a^{4} \log \left (\sin \left (d x + c\right )\right ) - 48 \, a^{4} \sin \left (d x + c\right ) + \frac {48 \, a^{4} \sin \left (d x + c\right )^{3} - 24 \, a^{4} \sin \left (d x + c\right )^{2} - 16 \, a^{4} \sin \left (d x + c\right ) - 3 \, a^{4}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5*(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 24*a^4*sin(d*x + c)^2 - 120*a^4*log(sin(d*x + c)) - 48*a^
4*sin(d*x + c) + (48*a^4*sin(d*x + c)^3 - 24*a^4*sin(d*x + c)^2 - 16*a^4*sin(d*x + c) - 3*a^4)/sin(d*x + c)^4)
/d

Giac [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 24 \, a^{4} \sin \left (d x + c\right )^{2} - 120 \, a^{4} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 48 \, a^{4} \sin \left (d x + c\right ) + \frac {250 \, a^{4} \sin \left (d x + c\right )^{4} + 48 \, a^{4} \sin \left (d x + c\right )^{3} - 24 \, a^{4} \sin \left (d x + c\right )^{2} - 16 \, a^{4} \sin \left (d x + c\right ) - 3 \, a^{4}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5*(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 24*a^4*sin(d*x + c)^2 - 120*a^4*log(abs(sin(d*x + c))) -
48*a^4*sin(d*x + c) + (250*a^4*sin(d*x + c)^4 + 48*a^4*sin(d*x + c)^3 - 24*a^4*sin(d*x + c)^2 - 16*a^4*sin(d*x
+ c) - 3*a^4)/sin(d*x + c)^4)/d

Mupad [B] (verification not implemented)

Time = 9.62 (sec) , antiderivative size = 368, normalized size of antiderivative = 2.49 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {3\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6\,d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {10\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {104\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-119\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+120\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-\frac {1135\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{4}+80\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-73\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+48\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {75\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}-\frac {40\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+10\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {8\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {a^4}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+96\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {9\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}+\frac {10\,a^4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x))^4)/sin(c + d*x)^5,x)

[Out]

(3*a^4*tan(c/2 + (d*x)/2))/(2*d) - (a^4*tan(c/2 + (d*x)/2)^3)/(6*d) - (a^4*tan(c/2 + (d*x)/2)^4)/(64*d) - (10*
a^4*log(tan(c/2 + (d*x)/2)))/d - (10*a^4*tan(c/2 + (d*x)/2)^2 - (40*a^4*tan(c/2 + (d*x)/2)^3)/3 + (75*a^4*tan(
c/2 + (d*x)/2)^4)/2 + 48*a^4*tan(c/2 + (d*x)/2)^5 - 73*a^4*tan(c/2 + (d*x)/2)^6 + 80*a^4*tan(c/2 + (d*x)/2)^7
- (1135*a^4*tan(c/2 + (d*x)/2)^8)/4 + 120*a^4*tan(c/2 + (d*x)/2)^9 - 119*a^4*tan(c/2 + (d*x)/2)^10 + 104*a^4*t
an(c/2 + (d*x)/2)^11 + a^4/4 + (8*a^4*tan(c/2 + (d*x)/2))/3)/(d*(16*tan(c/2 + (d*x)/2)^4 + 64*tan(c/2 + (d*x)/
2)^6 + 96*tan(c/2 + (d*x)/2)^8 + 64*tan(c/2 + (d*x)/2)^10 + 16*tan(c/2 + (d*x)/2)^12)) - (9*a^4*tan(c/2 + (d*x
)/2)^2)/(16*d) + (10*a^4*log(tan(c/2 + (d*x)/2)^2 + 1))/d