\(\int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx\) [628]

Optimal result

Integrand size = 27, antiderivative size = 101 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 x}{8 a}-\frac {\text {arctanh}(\cos (c+d x))}{a d}+\frac {\cos (c+d x)}{a d}+\frac {\cos ^3(c+d x)}{3 a d}-\frac {3 \cos (c+d x) \sin (c+d x)}{8 a d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2918, 2672, 308, 212, 2715, 8} \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{a d}+\frac {\cos ^3(c+d x)}{3 a d}+\frac {\cos (c+d x)}{a d}-\frac {\sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {3 \sin (c+d x) \cos (c+d x)}{8 a d}-\frac {3 x}{8 a} \]

[In]

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

[Out]

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

Rule 8

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

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 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 2918

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

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

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {120 \cos (c+d x)+8 \cos (3 (c+d x))-3 \left (4 \left (3 c+3 d x+8 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+8 \sin (2 (c+d x))+\sin (4 (c+d x))\right )}{96 a d} \]

[In]

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

[Out]

(120*Cos[c + d*x] + 8*Cos[3*(c + d*x)] - 3*(4*(3*c + 3*d*x + 8*Log[Cos[(c + d*x)/2]] - 8*Log[Sin[(c + d*x)/2]]
) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(96*a*d)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {-36 d x +8 \cos \left (3 d x +3 c \right )+120 \cos \left (d x +c \right )-3 \sin \left (4 d x +4 c \right )-24 \sin \left (2 d x +2 c \right )+96 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+128}{96 d a}\)	\(68\)
risch	\(-\frac {3 x}{8 a}+\frac {5 \,{\mathrm e}^{i \left (d x +c \right )}}{8 a d}+\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}-\frac {\sin \left (4 d x +4 c \right )}{32 d a}+\frac {\cos \left (3 d x +3 c \right )}{12 a d}-\frac {\sin \left (2 d x +2 c \right )}{4 d a}\)	\(132\)
derivativedivides	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 \left (-\frac {5 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {10 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}-\frac {4}{3}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d a}\)	\(139\)
default	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 \left (-\frac {5 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {10 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}-\frac {4}{3}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d a}\)	\(139\)
norman	\(\frac {-\frac {3 x}{8 a}-\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}-\frac {15 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {15 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {15 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}-\frac {15 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}-\frac {15 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}-\frac {15 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}-\frac {15 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {15 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {3 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {3 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}+\frac {8}{3 a d}+\frac {25 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {5 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {12 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {9 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d a}+\frac {85 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}+\frac {44 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {21 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {53 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}+\frac {97 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}\)	\(436\)

[In]

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

[Out]

1/96*(-36*d*x+8*cos(3*d*x+3*c)+120*cos(d*x+c)-3*sin(4*d*x+4*c)-24*sin(2*d*x+2*c)+96*ln(tan(1/2*d*x+1/2*c))+128
)/d/a

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {8 \, \cos \left (d x + c\right )^{3} - 9 \, d x - 3 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 24 \, \cos \left (d x + c\right ) - 12 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 12 \, \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{24 \, a d} \]

[In]

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

[Out]

1/24*(8*cos(d*x + c)^3 - 9*d*x - 3*(2*cos(d*x + c)^3 + 3*cos(d*x + c))*sin(d*x + c) + 24*cos(d*x + c) - 12*log
(1/2*cos(d*x + c) + 1/2) + 12*log(-1/2*cos(d*x + c) + 1/2))/(a*d)

Sympy [F]

\[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos ^{6}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

Integral(cos(c + d*x)**6*csc(c + d*x)/(sin(c + d*x) + 1), x)/a

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (93) = 186\).

Time = 0.30 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.77 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {80 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {96 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {9 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {48 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 32}{a + \frac {4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{12 \, d} \]

[In]

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

[Out]

-1/12*((15*sin(d*x + c)/(cos(d*x + c) + 1) - 80*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 9*sin(d*x + c)^3/(cos(d*
x + c) + 1)^3 - 96*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 9*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 48*sin(d*x +
c)^6/(cos(d*x + c) + 1)^6 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 32)/(a + 4*a*sin(d*x + c)^2/(cos(d*x + c)
+ 1)^2 + 6*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8
/(cos(d*x + c) + 1)^8) + 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - 12*log(sin(d*x + c)/(cos(d*x + c) + 1))
/a)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.42 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {9 \, {\left (d x + c\right )}}{a} - \frac {24 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 96 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 80 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 32\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a}}{24 \, d} \]

[In]

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

[Out]

-1/24*(9*(d*x + c)/a - 24*log(abs(tan(1/2*d*x + 1/2*c)))/a - 2*(15*tan(1/2*d*x + 1/2*c)^7 + 48*tan(1/2*d*x + 1
/2*c)^6 - 9*tan(1/2*d*x + 1/2*c)^5 + 96*tan(1/2*d*x + 1/2*c)^4 + 9*tan(1/2*d*x + 1/2*c)^3 + 80*tan(1/2*d*x + 1
/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) + 32)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a))/d

Mupad [B] (verification not implemented)

Time = 11.58 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.23 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3\,\mathrm {atan}\left (\frac {9}{16\,\left (\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}+\frac {3}{2}\right )}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}+\frac {3}{2}\right )}\right )}{4\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {8}{3}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )} \]

[In]

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

[Out]

(3*atan(9/(16*((9*tan(c/2 + (d*x)/2))/16 + 3/2)) - (3*tan(c/2 + (d*x)/2))/(2*((9*tan(c/2 + (d*x)/2))/16 + 3/2)
)))/(4*a*d) + log(tan(c/2 + (d*x)/2))/(a*d) + ((20*tan(c/2 + (d*x)/2)^2)/3 - (5*tan(c/2 + (d*x)/2))/4 + (3*tan
(c/2 + (d*x)/2)^3)/4 + 8*tan(c/2 + (d*x)/2)^4 - (3*tan(c/2 + (d*x)/2)^5)/4 + 4*tan(c/2 + (d*x)/2)^6 + (5*tan(c
/2 + (d*x)/2)^7)/4 + 8/3)/(d*(a + 4*a*tan(c/2 + (d*x)/2)^2 + 6*a*tan(c/2 + (d*x)/2)^4 + 4*a*tan(c/2 + (d*x)/2)
^6 + a*tan(c/2 + (d*x)/2)^8))