\(\int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\) [283]

Optimal result

Integrand size = 29, antiderivative size = 100 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \text {arctanh}(\cos (c+d x))}{4 d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {a^2 \cot (c+d x) \csc (c+d x)}{4 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d} \]

[Out]

1/4*a^2*arctanh(cos(d*x+c))/d-2/3*a^2*cot(d*x+c)^3/d-1/5*a^2*cot(d*x+c)^5/d+1/4*a^2*cot(d*x+c)*csc(d*x+c)/d-1/
2*a^2*cot(d*x+c)*csc(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2952, 2687, 30, 2691, 3853, 3855, 14} \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \text {arctanh}(\cos (c+d x))}{4 d}-\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {a^2 \cot (c+d x) \csc (c+d x)}{4 d} \]

[In]

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

[Out]

(a^2*ArcTanh[Cos[c + d*x]])/(4*d) - (2*a^2*Cot[c + d*x]^3)/(3*d) - (a^2*Cot[c + d*x]^5)/(5*d) + (a^2*Cot[c + d
*x]*Csc[c + d*x])/(4*d) - (a^2*Cot[c + d*x]*Csc[c + d*x]^3)/(2*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2952

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

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}& = \int \left (a^2 \cot ^2(c+d x) \csc ^2(c+d x)+2 a^2 \cot ^2(c+d x) \csc ^3(c+d x)+a^2 \cot ^2(c+d x) \csc ^4(c+d x)\right ) \, dx \\ & = a^2 \int \cot ^2(c+d x) \csc ^2(c+d x) \, dx+a^2 \int \cot ^2(c+d x) \csc ^4(c+d x) \, dx+\left (2 a^2\right ) \int \cot ^2(c+d x) \csc ^3(c+d x) \, dx \\ & = -\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}-\frac {1}{2} a^2 \int \csc ^3(c+d x) \, dx+\frac {a^2 \text {Subst}\left (\int x^2 \, dx,x,-\cot (c+d x)\right )}{d}+\frac {a^2 \text {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{d} \\ & = -\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot (c+d x) \csc (c+d x)}{4 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}-\frac {1}{4} a^2 \int \csc (c+d x) \, dx+\frac {a^2 \text {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,-\cot (c+d x)\right )}{d} \\ & = \frac {a^2 \text {arctanh}(\cos (c+d x))}{4 d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {a^2 \cot (c+d x) \csc (c+d x)}{4 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.21 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.89 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \csc ^5(c+d x) \left (200 \cos (c+d x)+20 \cos (3 (c+d x))-28 \cos (5 (c+d x))-150 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+150 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+180 \sin (2 (c+d x))+75 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-75 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+30 \sin (4 (c+d x))-15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))+15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{960 d} \]

[In]

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

[Out]

-1/960*(a^2*Csc[c + d*x]^5*(200*Cos[c + d*x] + 20*Cos[3*(c + d*x)] - 28*Cos[5*(c + d*x)] - 150*Log[Cos[(c + d*
x)/2]]*Sin[c + d*x] + 150*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] + 180*Sin[2*(c + d*x)] + 75*Log[Cos[(c + d*x)/2]]
*Sin[3*(c + d*x)] - 75*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 30*Sin[4*(c + d*x)] - 15*Log[Cos[(c + d*x)/2]]
*Sin[5*(c + d*x)] + 15*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/d

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.20

method	result	size
parallelrisch	\(-\frac {\left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )-\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {25 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {25 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-30 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+30 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+40 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) a^{2}}{160 d}\)	\(120\)
derivativedivides	\(\frac {-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 \sin \left (d x +c \right )^{3}}+2 a^{2} \left (-\frac {\cos ^{3}\left (d x +c \right )}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos ^{3}\left (d x +c \right )}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+a^{2} \left (-\frac {\cos ^{3}\left (d x +c \right )}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15 \sin \left (d x +c \right )^{3}}\right )}{d}\)	\(136\)
default	\(\frac {-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 \sin \left (d x +c \right )^{3}}+2 a^{2} \left (-\frac {\cos ^{3}\left (d x +c \right )}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos ^{3}\left (d x +c \right )}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+a^{2} \left (-\frac {\cos ^{3}\left (d x +c \right )}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15 \sin \left (d x +c \right )^{3}}\right )}{d}\)	\(136\)
risch	\(-\frac {a^{2} \left (-60 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}+240 i {\mathrm e}^{6 i \left (d x +c \right )}+90 \,{\mathrm e}^{7 i \left (d x +c \right )}-40 i {\mathrm e}^{4 i \left (d x +c \right )}+80 i {\mathrm e}^{2 i \left (d x +c \right )}-90 \,{\mathrm e}^{3 i \left (d x +c \right )}-28 i-15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d}\)	\(158\)
norman	\(\frac {-\frac {a^{2}}{160 d}-\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32 d}-\frac {31 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d}-\frac {a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}+\frac {37 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d}+\frac {13 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}-\frac {13 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}-\frac {37 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d}+\frac {a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}+\frac {31 a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d}+\frac {a^{2} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}+\frac {a^{2} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}-\frac {a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}-\frac {a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}\)	\(301\)

[In]

int(cos(d*x+c)^2*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/160*(cot(1/2*d*x+1/2*c)^5-tan(1/2*d*x+1/2*c)^5+5*cot(1/2*d*x+1/2*c)^4-5*tan(1/2*d*x+1/2*c)^4+25/3*cot(1/2*d
*x+1/2*c)^3-25/3*tan(1/2*d*x+1/2*c)^3-30*cot(1/2*d*x+1/2*c)+30*tan(1/2*d*x+1/2*c)+40*ln(tan(1/2*d*x+1/2*c)))*a
^2/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (90) = 180\).

Time = 0.27 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.89 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {56 \, a^{2} \cos \left (d x + c\right )^{5} - 80 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/120*(56*a^2*cos(d*x + c)^5 - 80*a^2*cos(d*x + c)^3 + 15*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*lo
g(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 15*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*log(-1/2*cos(d*x
+ c) + 1/2)*sin(d*x + c) - 30*(a^2*cos(d*x + c)^3 + a^2*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*
cos(d*x + c)^2 + d)*sin(d*x + c))

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**6*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.09 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {15 \, a^{2} {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {40 \, a^{2}}{\tan \left (d x + c\right )^{3}} + \frac {8 \, {\left (5 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{2}}{\tan \left (d x + c\right )^{5}}}{120 \, d} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/120*(15*a^2*(2*(cos(d*x + c)^3 + cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - log(cos(d*x + c) +
1) + log(cos(d*x + c) - 1)) + 40*a^2/tan(d*x + c)^3 + 8*(5*tan(d*x + c)^2 + 3)*a^2/tan(d*x + c)^5)/d

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.64 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 90 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {274 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 90 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 25 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/480*(3*a^2*tan(1/2*d*x + 1/2*c)^5 + 15*a^2*tan(1/2*d*x + 1/2*c)^4 + 25*a^2*tan(1/2*d*x + 1/2*c)^3 - 120*a^2*
log(abs(tan(1/2*d*x + 1/2*c))) - 90*a^2*tan(1/2*d*x + 1/2*c) + (274*a^2*tan(1/2*d*x + 1/2*c)^5 + 90*a^2*tan(1/
2*d*x + 1/2*c)^4 - 25*a^2*tan(1/2*d*x + 1/2*c)^2 - 15*a^2*tan(1/2*d*x + 1/2*c) - 3*a^2)/tan(1/2*d*x + 1/2*c)^5
)/d

Mupad [B] (verification not implemented)

Time = 9.56 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.60 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}-\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (-6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^2}{5}\right )}{32\,d}-\frac {3\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d} \]

[In]

int((cos(c + d*x)^2*(a + a*sin(c + d*x))^2)/sin(c + d*x)^6,x)

[Out]

(5*a^2*tan(c/2 + (d*x)/2)^3)/(96*d) + (a^2*tan(c/2 + (d*x)/2)^4)/(32*d) + (a^2*tan(c/2 + (d*x)/2)^5)/(160*d) -
(a^2*log(tan(c/2 + (d*x)/2)))/(4*d) - (cot(c/2 + (d*x)/2)^5*((5*a^2*tan(c/2 + (d*x)/2)^2)/3 - 6*a^2*tan(c/2 +
(d*x)/2)^4 + a^2/5 + a^2*tan(c/2 + (d*x)/2)))/(32*d) - (3*a^2*tan(c/2 + (d*x)/2))/(16*d)