3.9.51 \(\int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx\) [851]

3.9.51.1 Optimal result
3.9.51.2 Mathematica [A] (verified)
3.9.51.3 Rubi [A] (verified)
3.9.51.4 Maple [A] (verified)
3.9.51.5 Fricas [A] (verification not implemented)
3.9.51.6 Sympy [F(-1)]
3.9.51.7 Maxima [A] (verification not implemented)
3.9.51.8 Giac [A] (verification not implemented)
3.9.51.9 Mupad [B] (verification not implemented)

3.9.51.1 Optimal result

Integrand size = 25, antiderivative size = 133 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {39 a \log (1-\sin (c+d x))}{16 d}-\frac {9 a \log (1+\sin (c+d x))}{16 d}-\frac {a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{2 d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {5 a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a+a \sin (c+d x))} \]

output
-39/16*a*ln(1-sin(d*x+c))/d-9/16*a*ln(1+sin(d*x+c))/d-a*sin(d*x+c)/d-1/2*a 
*sin(d*x+c)^2/d+1/8*a^3/d/(a-a*sin(d*x+c))^2-5/4*a^2/d/(a-a*sin(d*x+c))-1/ 
8*a^2/d/(a+a*sin(d*x+c))
 
3.9.51.2 Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.10 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {15 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {a \left (12 \log (\cos (c+d x))+6 \sec ^2(c+d x)-\sec ^4(c+d x)+2 \sin ^2(c+d x)\right )}{4 d}+\frac {15 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {15 a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 a \sec (c+d x) \tan ^3(c+d x)}{d}-\frac {a \sin (c+d x) \tan ^4(c+d x)}{d} \]

input
Integrate[Sin[c + d*x]*(a + a*Sin[c + d*x])*Tan[c + d*x]^5,x]
 
output
(15*a*ArcTanh[Sin[c + d*x]])/(8*d) - (a*(12*Log[Cos[c + d*x]] + 6*Sec[c + 
d*x]^2 - Sec[c + d*x]^4 + 2*Sin[c + d*x]^2))/(4*d) + (15*a*Sec[c + d*x]*Ta 
n[c + d*x])/(8*d) - (15*a*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*a*Sec[c 
+ d*x]*Tan[c + d*x]^3)/d - (a*Sin[c + d*x]*Tan[c + d*x]^4)/d
 
3.9.51.3 Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.

\(\displaystyle \int \sin (c+d x) \tan ^5(c+d x) (a \sin (c+d x)+a) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3315

\(\displaystyle \frac {a^5 \int \frac {\sin ^6(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a^6 \sin ^6(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a d}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {\int \left (\frac {a^4}{4 (a-a \sin (c+d x))^3}-\frac {5 a^3}{4 (a-a \sin (c+d x))^2}+\frac {a^3}{8 (\sin (c+d x) a+a)^2}+\frac {39 a^2}{16 (a-a \sin (c+d x))}-\frac {9 a^2}{16 (\sin (c+d x) a+a)}-\sin (c+d x) a-a\right )d(a \sin (c+d x))}{a d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {a^4}{8 (a-a \sin (c+d x))^2}-\frac {5 a^3}{4 (a-a \sin (c+d x))}-\frac {a^3}{8 (a \sin (c+d x)+a)}-\frac {1}{2} a^2 \sin ^2(c+d x)-a^2 \sin (c+d x)-\frac {39}{16} a^2 \log (a-a \sin (c+d x))-\frac {9}{16} a^2 \log (a \sin (c+d x)+a)}{a d}\)

input
Int[Sin[c + d*x]*(a + a*Sin[c + d*x])*Tan[c + d*x]^5,x]
 
output
((-39*a^2*Log[a - a*Sin[c + d*x]])/16 - (9*a^2*Log[a + a*Sin[c + d*x]])/16 
 - a^2*Sin[c + d*x] - (a^2*Sin[c + d*x]^2)/2 + a^4/(8*(a - a*Sin[c + d*x]) 
^2) - (5*a^3)/(4*(a - a*Sin[c + d*x])) - a^3/(8*(a + a*Sin[c + d*x])))/(a* 
d)
 

3.9.51.3.1 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]
 
3.9.51.4 Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.26

method result size
derivativedivides \(\frac {a \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+a \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) \(167\)
default \(\frac {a \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+a \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) \(167\)
risch \(3 i a x +\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {6 i a c}{d}+\frac {i a \left (6 i {\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{5 i \left (d x +c \right )}-6 i {\mathrm e}^{2 i \left (d x +c \right )}+22 \,{\mathrm e}^{3 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d}-\frac {39 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {9 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\) \(208\)
parallelrisch \(\frac {6 \left (\left (-\frac {\sin \left (3 d x +3 c \right )}{2}-\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {13 \left (-1-\cos \left (2 d x +2 c \right )+\frac {\sin \left (d x +c \right )}{2}+\frac {\sin \left (3 d x +3 c \right )}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8}+\frac {3 \left (-1-\cos \left (2 d x +2 c \right )+\frac {\sin \left (d x +c \right )}{2}+\frac {\sin \left (3 d x +3 c \right )}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8}+\frac {\cos \left (2 d x +2 c \right )}{24}-\frac {\cos \left (4 d x +4 c \right )}{24}-\frac {\sin \left (d x +c \right )}{12}-\frac {17 \sin \left (3 d x +3 c \right )}{48}-\frac {\sin \left (5 d x +5 c \right )}{48}\right ) a}{d \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right )}\) \(229\)
norman \(\frac {\frac {12 a}{d}+\frac {12 a \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {15 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {25 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {11 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {11 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {25 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {15 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {52 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {30 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {30 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {39 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}-\frac {9 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}+\frac {3 a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) \(264\)

input
int(sec(d*x+c)^5*sin(d*x+c)^6*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 
output
1/d*(a*(1/4*sin(d*x+c)^8/cos(d*x+c)^4-1/2*sin(d*x+c)^8/cos(d*x+c)^2-1/2*si 
n(d*x+c)^6-3/4*sin(d*x+c)^4-3/2*sin(d*x+c)^2-3*ln(cos(d*x+c)))+a*(1/4*sin( 
d*x+c)^7/cos(d*x+c)^4-3/8*sin(d*x+c)^7/cos(d*x+c)^2-3/8*sin(d*x+c)^5-5/8*s 
in(d*x+c)^3-15/8*sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c))))
 
3.9.51.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.29 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {8 \, a \cos \left (d x + c\right )^{4} + 6 \, a \cos \left (d x + c\right )^{2} - 9 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 39 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, a \cos \left (d x + c\right )^{4} + 6 \, a \cos \left (d x + c\right )^{2} - 3 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \]

input
integrate(sec(d*x+c)^5*sin(d*x+c)^6*(a+a*sin(d*x+c)),x, algorithm="fricas" 
)
 
output
1/16*(8*a*cos(d*x + c)^4 + 6*a*cos(d*x + c)^2 - 9*(a*cos(d*x + c)^2*sin(d* 
x + c) - a*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 39*(a*cos(d*x + c)^2*si 
n(d*x + c) - a*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(4*a*cos(d*x + c 
)^4 + 6*a*cos(d*x + c)^2 - 3*a)*sin(d*x + c) + 2*a)/(d*cos(d*x + c)^2*sin( 
d*x + c) - d*cos(d*x + c)^2)
 
3.9.51.6 Sympy [F(-1)]

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

input
integrate(sec(d*x+c)**5*sin(d*x+c)**6*(a+a*sin(d*x+c)),x)
 
output
Timed out
 
3.9.51.7 Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.80 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {8 \, a \sin \left (d x + c\right )^{2} + 9 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) + 39 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, a \sin \left (d x + c\right ) - \frac {2 \, {\left (9 \, a \sin \left (d x + c\right )^{2} + 3 \, a \sin \left (d x + c\right ) - 10 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \]

input
integrate(sec(d*x+c)^5*sin(d*x+c)^6*(a+a*sin(d*x+c)),x, algorithm="maxima" 
)
 
output
-1/16*(8*a*sin(d*x + c)^2 + 9*a*log(sin(d*x + c) + 1) + 39*a*log(sin(d*x + 
 c) - 1) + 16*a*sin(d*x + c) - 2*(9*a*sin(d*x + c)^2 + 3*a*sin(d*x + c) - 
10*a)/(sin(d*x + c)^3 - sin(d*x + c)^2 - sin(d*x + c) + 1))/d
 
3.9.51.8 Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.85 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {16 \, a \sin \left (d x + c\right )^{2} + 18 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 78 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 32 \, a \sin \left (d x + c\right ) - \frac {2 \, {\left (9 \, a \sin \left (d x + c\right ) + 7 \, a\right )}}{\sin \left (d x + c\right ) + 1} - \frac {117 \, a \sin \left (d x + c\right )^{2} - 194 \, a \sin \left (d x + c\right ) + 81 \, a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{32 \, d} \]

input
integrate(sec(d*x+c)^5*sin(d*x+c)^6*(a+a*sin(d*x+c)),x, algorithm="giac")
 
output
-1/32*(16*a*sin(d*x + c)^2 + 18*a*log(abs(sin(d*x + c) + 1)) + 78*a*log(ab 
s(sin(d*x + c) - 1)) + 32*a*sin(d*x + c) - 2*(9*a*sin(d*x + c) + 7*a)/(sin 
(d*x + c) + 1) - (117*a*sin(d*x + c)^2 - 194*a*sin(d*x + c) + 81*a)/(sin(d 
*x + c) - 1)^2)/d
 
3.9.51.9 Mupad [B] (verification not implemented)

Time = 9.95 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.15 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}-\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {15\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {9\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{8\,d}-\frac {39\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{8\,d}+\frac {3\,a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

input
int((sin(c + d*x)^6*(a + a*sin(c + d*x)))/cos(c + d*x)^5,x)
 
output
((3*a*tan(c/2 + (d*x)/2)^2)/2 - (15*a*tan(c/2 + (d*x)/2))/4 + 7*a*tan(c/2 
+ (d*x)/2)^3 - (7*a*tan(c/2 + (d*x)/2)^4)/2 + (11*a*tan(c/2 + (d*x)/2)^5)/ 
2 - (7*a*tan(c/2 + (d*x)/2)^6)/2 + 7*a*tan(c/2 + (d*x)/2)^7 + (3*a*tan(c/2 
 + (d*x)/2)^8)/2 - (15*a*tan(c/2 + (d*x)/2)^9)/4)/(d*(tan(c/2 + (d*x)/2)^2 
 - 2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^5 
- 2*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 - 2*tan(c/2 + (d*x)/2)^9 + 
 tan(c/2 + (d*x)/2)^10 + 1)) - (9*a*log(tan(c/2 + (d*x)/2) + 1))/(8*d) - ( 
39*a*log(tan(c/2 + (d*x)/2) - 1))/(8*d) + (3*a*log(tan(c/2 + (d*x)/2)^2 + 
1))/d