3.6.64 \(\int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx\) [564]

3.6.64.1 Optimal result
3.6.64.2 Mathematica [A] (verified)
3.6.64.3 Rubi [A] (verified)
3.6.64.4 Maple [A] (verified)
3.6.64.5 Fricas [B] (verification not implemented)
3.6.64.6 Sympy [F(-1)]
3.6.64.7 Maxima [A] (verification not implemented)
3.6.64.8 Giac [A] (verification not implemented)
3.6.64.9 Mupad [B] (verification not implemented)

3.6.64.1 Optimal result

Integrand size = 27, antiderivative size = 135 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {16 \csc (c+d x)}{a^4 d}+\frac {6 \csc ^2(c+d x)}{a^4 d}-\frac {8 \csc ^3(c+d x)}{3 a^4 d}+\frac {\csc ^4(c+d x)}{a^4 d}-\frac {\csc ^5(c+d x)}{5 a^4 d}-\frac {20 \log (\sin (c+d x))}{a^4 d}+\frac {20 \log (1+\sin (c+d x))}{a^4 d}-\frac {4}{d \left (a^4+a^4 \sin (c+d x)\right )} \]

output
-16*csc(d*x+c)/a^4/d+6*csc(d*x+c)^2/a^4/d-8/3*csc(d*x+c)^3/a^4/d+csc(d*x+c 
)^4/a^4/d-1/5*csc(d*x+c)^5/a^4/d-20*ln(sin(d*x+c))/a^4/d+20*ln(1+sin(d*x+c 
))/a^4/d-4/d/(a^4+a^4*sin(d*x+c))
 
3.6.64.2 Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.67 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {240 \csc (c+d x)-90 \csc ^2(c+d x)+40 \csc ^3(c+d x)-15 \csc ^4(c+d x)+3 \csc ^5(c+d x)+300 \log (\sin (c+d x))-300 \log (1+\sin (c+d x))+\frac {60}{1+\sin (c+d x)}}{15 a^4 d} \]

input
Integrate[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x])^4,x]
 
output
-1/15*(240*Csc[c + d*x] - 90*Csc[c + d*x]^2 + 40*Csc[c + d*x]^3 - 15*Csc[c 
 + d*x]^4 + 3*Csc[c + d*x]^5 + 300*Log[Sin[c + d*x]] - 300*Log[1 + Sin[c + 
 d*x]] + 60/(1 + Sin[c + d*x]))/(a^4*d)
 
3.6.64.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3315, 27, 99, 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.

\(\displaystyle \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a \sin (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^5}{\sin (c+d x)^6 (a \sin (c+d x)+a)^4}dx\)

\(\Big \downarrow \) 3315

\(\displaystyle \frac {\int \frac {\csc ^6(c+d x) (a-a \sin (c+d x))^2}{(\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a^5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a \int \frac {\csc ^6(c+d x) (a-a \sin (c+d x))^2}{a^6 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {a \int \left (\frac {\csc ^6(c+d x)}{a^6}-\frac {4 \csc ^5(c+d x)}{a^6}+\frac {8 \csc ^4(c+d x)}{a^6}-\frac {12 \csc ^3(c+d x)}{a^6}+\frac {16 \csc ^2(c+d x)}{a^6}-\frac {20 \csc (c+d x)}{a^6}+\frac {20}{a^5 (\sin (c+d x) a+a)}+\frac {4}{a^4 (\sin (c+d x) a+a)^2}\right )d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a \left (-\frac {\csc ^5(c+d x)}{5 a^5}+\frac {\csc ^4(c+d x)}{a^5}-\frac {8 \csc ^3(c+d x)}{3 a^5}+\frac {6 \csc ^2(c+d x)}{a^5}-\frac {16 \csc (c+d x)}{a^5}-\frac {20 \log (a \sin (c+d x))}{a^5}+\frac {20 \log (a \sin (c+d x)+a)}{a^5}-\frac {4}{a^4 (a \sin (c+d x)+a)}\right )}{d}\)

input
Int[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x])^4,x]
 
output
(a*((-16*Csc[c + d*x])/a^5 + (6*Csc[c + d*x]^2)/a^5 - (8*Csc[c + d*x]^3)/( 
3*a^5) + Csc[c + d*x]^4/a^5 - Csc[c + d*x]^5/(5*a^5) - (20*Log[a*Sin[c + d 
*x]])/a^5 + (20*Log[a + a*Sin[c + d*x]])/a^5 - 4/(a^4*(a + a*Sin[c + d*x]) 
)))/d
 

3.6.64.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 99
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(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 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3315
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, 
 x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege 
rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
 
3.6.64.4 Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.58

method result size
derivativedivides \(\frac {-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\csc ^{4}\left (d x +c \right )-\frac {8 \left (\csc ^{3}\left (d x +c \right )\right )}{3}+6 \left (\csc ^{2}\left (d x +c \right )\right )-16 \csc \left (d x +c \right )+\frac {4}{\csc \left (d x +c \right )+1}+20 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{4}}\) \(78\)
default \(\frac {-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\csc ^{4}\left (d x +c \right )-\frac {8 \left (\csc ^{3}\left (d x +c \right )\right )}{3}+6 \left (\csc ^{2}\left (d x +c \right )\right )-16 \csc \left (d x +c \right )+\frac {4}{\csc \left (d x +c \right )+1}+20 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{4}}\) \(78\)
risch \(-\frac {8 i \left (75 i {\mathrm e}^{10 i \left (d x +c \right )}+75 \,{\mathrm e}^{11 i \left (d x +c \right )}-350 i {\mathrm e}^{8 i \left (d x +c \right )}-325 \,{\mathrm e}^{9 i \left (d x +c \right )}+574 i {\mathrm e}^{6 i \left (d x +c \right )}+552 \,{\mathrm e}^{7 i \left (d x +c \right )}-350 i {\mathrm e}^{4 i \left (d x +c \right )}-552 \,{\mathrm e}^{5 i \left (d x +c \right )}+75 i {\mathrm e}^{2 i \left (d x +c \right )}+325 \,{\mathrm e}^{3 i \left (d x +c \right )}-75 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{15 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d \,a^{4}}+\frac {40 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}-\frac {20 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{4}}\) \(206\)
parallelrisch \(\frac {19200 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-9600 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 \left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-118 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+520 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-118 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2845 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+520 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2845 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+10860 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{480 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}\) \(214\)

input
int(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
 
output
1/d/a^4*(-1/5*csc(d*x+c)^5+csc(d*x+c)^4-8/3*csc(d*x+c)^3+6*csc(d*x+c)^2-16 
*csc(d*x+c)+4/(csc(d*x+c)+1)+20*ln(csc(d*x+c)+1))
 
3.6.64.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (131) = 262\).

Time = 0.27 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.10 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {150 \, \cos \left (d x + c\right )^{4} - 325 \, \cos \left (d x + c\right )^{2} - 300 \, {\left (\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 300 \, {\left (\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (150 \, \cos \left (d x + c\right )^{4} - 275 \, \cos \left (d x + c\right )^{2} + 119\right )} \sin \left (d x + c\right ) + 178}{15 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} - 3 \, a^{4} d \cos \left (d x + c\right )^{4} + 3 \, a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d - {\left (a^{4} d \cos \left (d x + c\right )^{4} - 2 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

input
integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^4,x, algorithm="frica 
s")
 
output
1/15*(150*cos(d*x + c)^4 - 325*cos(d*x + c)^2 - 300*(cos(d*x + c)^6 - 3*co 
s(d*x + c)^4 + 3*cos(d*x + c)^2 - (cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)* 
sin(d*x + c) - 1)*log(1/2*sin(d*x + c)) + 300*(cos(d*x + c)^6 - 3*cos(d*x 
+ c)^4 + 3*cos(d*x + c)^2 - (cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*sin(d* 
x + c) - 1)*log(sin(d*x + c) + 1) + 2*(150*cos(d*x + c)^4 - 275*cos(d*x + 
c)^2 + 119)*sin(d*x + c) + 178)/(a^4*d*cos(d*x + c)^6 - 3*a^4*d*cos(d*x + 
c)^4 + 3*a^4*d*cos(d*x + c)^2 - a^4*d - (a^4*d*cos(d*x + c)^4 - 2*a^4*d*co 
s(d*x + c)^2 + a^4*d)*sin(d*x + c))
 
3.6.64.6 Sympy [F(-1)]

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

input
integrate(cos(d*x+c)**5*csc(d*x+c)**6/(a+a*sin(d*x+c))**4,x)
 
output
Timed out
 
3.6.64.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.81 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\frac {300 \, \sin \left (d x + c\right )^{5} + 150 \, \sin \left (d x + c\right )^{4} - 50 \, \sin \left (d x + c\right )^{3} + 25 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right ) + 3}{a^{4} \sin \left (d x + c\right )^{6} + a^{4} \sin \left (d x + c\right )^{5}} - \frac {300 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {300 \, \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{15 \, d} \]

input
integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^4,x, algorithm="maxim 
a")
 
output
-1/15*((300*sin(d*x + c)^5 + 150*sin(d*x + c)^4 - 50*sin(d*x + c)^3 + 25*s 
in(d*x + c)^2 - 12*sin(d*x + c) + 3)/(a^4*sin(d*x + c)^6 + a^4*sin(d*x + c 
)^5) - 300*log(sin(d*x + c) + 1)/a^4 + 300*log(sin(d*x + c))/a^4)/d
 
3.6.64.8 Giac [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.84 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {19200 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {9600 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac {1920 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 28 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15\right )}}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{2}} + \frac {21920 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4350 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 840 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 175 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} - \frac {3 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 175 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 840 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4350 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{20}}}{480 \, d} \]

input
integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^4,x, algorithm="giac" 
)
 
output
1/480*(19200*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 9600*log(abs(tan(1/2 
*d*x + 1/2*c)))/a^4 - 1920*(15*tan(1/2*d*x + 1/2*c)^2 + 28*tan(1/2*d*x + 1 
/2*c) + 15)/(a^4*(tan(1/2*d*x + 1/2*c) + 1)^2) + (21920*tan(1/2*d*x + 1/2* 
c)^5 - 4350*tan(1/2*d*x + 1/2*c)^4 + 840*tan(1/2*d*x + 1/2*c)^3 - 175*tan( 
1/2*d*x + 1/2*c)^2 + 30*tan(1/2*d*x + 1/2*c) - 3)/(a^4*tan(1/2*d*x + 1/2*c 
)^5) - (3*a^16*tan(1/2*d*x + 1/2*c)^5 - 30*a^16*tan(1/2*d*x + 1/2*c)^4 + 1 
75*a^16*tan(1/2*d*x + 1/2*c)^3 - 840*a^16*tan(1/2*d*x + 1/2*c)^2 + 4350*a^ 
16*tan(1/2*d*x + 1/2*c))/a^20)/d
 
3.6.64.9 Mupad [B] (verification not implemented)

Time = 11.28 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.97 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^4\,d}-\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,a^4\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{16\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a^4\,d}-\frac {20\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {34\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+524\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {569\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-\frac {104\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {118\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-\frac {8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+\frac {1}{5}}{d\,\left (32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+64\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {40\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^4\,d}-\frac {145\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a^4\,d} \]

input
int(cos(c + d*x)^5/(sin(c + d*x)^6*(a + a*sin(c + d*x))^4),x)
 
output
(7*tan(c/2 + (d*x)/2)^2)/(4*a^4*d) - (35*tan(c/2 + (d*x)/2)^3)/(96*a^4*d) 
+ tan(c/2 + (d*x)/2)^4/(16*a^4*d) - tan(c/2 + (d*x)/2)^5/(160*a^4*d) - (20 
*log(tan(c/2 + (d*x)/2)))/(a^4*d) - ((118*tan(c/2 + (d*x)/2)^2)/15 - (8*ta 
n(c/2 + (d*x)/2))/5 - (104*tan(c/2 + (d*x)/2)^3)/3 + (569*tan(c/2 + (d*x)/ 
2)^4)/3 + 524*tan(c/2 + (d*x)/2)^5 + 34*tan(c/2 + (d*x)/2)^6 + 1/5)/(d*(32 
*a^4*tan(c/2 + (d*x)/2)^5 + 64*a^4*tan(c/2 + (d*x)/2)^6 + 32*a^4*tan(c/2 + 
 (d*x)/2)^7)) + (40*log(tan(c/2 + (d*x)/2) + 1))/(a^4*d) - (145*tan(c/2 + 
(d*x)/2))/(16*a^4*d)