\(\int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx\) [382]

Optimal result

Integrand size = 27, antiderivative size = 119 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 a^2 x}{4}-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2952, 2715, 8, 2672, 308, 212, 2645, 30} \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}-\frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac {3 a^2 x}{4} \]

[In]

Int[Cos[c + d*x]^3*Cot[c + d*x]*(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 8

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

Rule 30

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

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 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]

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

Mathematica [A] (verified)

Time = 5.75 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.81 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (270 \cos (c+d x)+5 \cos (3 (c+d x))-3 \cos (5 (c+d x))+15 \left (4 \left (3 c+3 d x-4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+8 \sin (2 (c+d x))+\sin (4 (c+d x))\right )\right )}{240 d} \]

[In]

Integrate[Cos[c + d*x]^3*Cot[c + d*x]*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(270*Cos[c + d*x] + 5*Cos[3*(c + d*x)] - 3*Cos[5*(c + d*x)] + 15*(4*(3*c + 3*d*x - 4*Log[Cos[(c + d*x)/2]
] + 4*Log[Sin[(c + d*x)/2]]) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)])))/(240*d)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.66

method	result	size
parallelrisch	\(\frac {a^{2} \left (180 d x +270 \cos \left (d x +c \right )+15 \sin \left (4 d x +4 c \right )-3 \cos \left (5 d x +5 c \right )+120 \sin \left (2 d x +2 c \right )+5 \cos \left (3 d x +3 c \right )+240 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+272\right )}{240 d}\)	\(79\)
derivativedivides	\(\frac {-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+2 a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(94\)
default	\(\frac {-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+2 a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(94\)
risch	\(\frac {3 a^{2} x}{4}+\frac {9 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{16 d}+\frac {9 a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{16 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a^{2} \cos \left (5 d x +5 c \right )}{80 d}+\frac {a^{2} \sin \left (4 d x +4 c \right )}{16 d}+\frac {a^{2} \cos \left (3 d x +3 c \right )}{48 d}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{2 d}\)	\(149\)
norman	\(\frac {\frac {a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 a^{2} x}{4}+\frac {34 a^{2}}{15 d}+\frac {5 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}-\frac {a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {15 a^{2} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a^{2} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a^{2} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a^{2} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {3 a^{2} x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {12 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {32 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {28 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(284\)

[In]

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

[Out]

1/240*a^2*(180*d*x+270*cos(d*x+c)+15*sin(4*d*x+4*c)-3*cos(5*d*x+5*c)+120*sin(2*d*x+2*c)+5*cos(3*d*x+3*c)+240*l
n(tan(1/2*d*x+1/2*c))+272)/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.97 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {12 \, a^{2} \cos \left (d x + c\right )^{5} - 20 \, a^{2} \cos \left (d x + c\right )^{3} - 45 \, a^{2} d x - 60 \, a^{2} \cos \left (d x + c\right ) + 30 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 30 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{3} + 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

-1/60*(12*a^2*cos(d*x + c)^5 - 20*a^2*cos(d*x + c)^3 - 45*a^2*d*x - 60*a^2*cos(d*x + c) + 30*a^2*log(1/2*cos(d
*x + c) + 1/2) - 30*a^2*log(-1/2*cos(d*x + c) + 1/2) - 15*(2*a^2*cos(d*x + c)^3 + 3*a^2*cos(d*x + c))*sin(d*x
+ c))/d

Sympy [F]

\[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cos ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a**2*(Integral(cos(c + d*x)**4*csc(c + d*x), x) + Integral(2*sin(c + d*x)*cos(c + d*x)**4*csc(c + d*x), x) + I
ntegral(sin(c + d*x)**2*cos(c + d*x)**4*csc(c + d*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.82 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {48 \, a^{2} \cos \left (d x + c\right )^{5} - 40 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{240 \, d} \]

[In]

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

[Out]

-1/240*(48*a^2*cos(d*x + c)^5 - 40*(2*cos(d*x + c)^3 + 6*cos(d*x + c) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*
x + c) - 1))*a^2 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.52 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {45 \, {\left (d x + c\right )} a^{2} + 60 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (75 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 360 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 320 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 280 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 75 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 68 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{60 \, d} \]

[In]

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

[Out]

1/60*(45*(d*x + c)*a^2 + 60*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(75*a^2*tan(1/2*d*x + 1/2*c)^9 - 60*a^2*tan
(1/2*d*x + 1/2*c)^8 + 30*a^2*tan(1/2*d*x + 1/2*c)^7 - 360*a^2*tan(1/2*d*x + 1/2*c)^6 - 320*a^2*tan(1/2*d*x + 1
/2*c)^4 - 30*a^2*tan(1/2*d*x + 1/2*c)^3 - 280*a^2*tan(1/2*d*x + 1/2*c)^2 - 75*a^2*tan(1/2*d*x + 1/2*c) - 68*a^
2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

Mupad [B] (verification not implemented)

Time = 11.05 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.46 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {3\,a^2\,\mathrm {atan}\left (\frac {9\,a^4}{4\,\left (3\,a^4-\frac {9\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )}+\frac {3\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3\,a^4-\frac {9\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}\right )}{2\,d}+\frac {-\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {32\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {28\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {34\,a^2}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

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

[Out]

(a^2*log(tan(c/2 + (d*x)/2)))/d + (3*a^2*atan((9*a^4)/(4*(3*a^4 - (9*a^4*tan(c/2 + (d*x)/2))/4)) + (3*a^4*tan(
c/2 + (d*x)/2))/(3*a^4 - (9*a^4*tan(c/2 + (d*x)/2))/4)))/(2*d) + ((28*a^2*tan(c/2 + (d*x)/2)^2)/3 + a^2*tan(c/
2 + (d*x)/2)^3 + (32*a^2*tan(c/2 + (d*x)/2)^4)/3 + 12*a^2*tan(c/2 + (d*x)/2)^6 - a^2*tan(c/2 + (d*x)/2)^7 + 2*
a^2*tan(c/2 + (d*x)/2)^8 - (5*a^2*tan(c/2 + (d*x)/2)^9)/2 + (34*a^2)/15 + (5*a^2*tan(c/2 + (d*x)/2))/2)/(d*(5*
tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 +
(d*x)/2)^10 + 1))