3.9.99 \(\int \frac {\sec (c+d x) \tan ^8(c+d x)}{a+a \sin (c+d x)} \, dx\) [899]

3.9.99.1 Optimal result
3.9.99.2 Mathematica [A] (verified)
3.9.99.3 Rubi [A] (verified)
3.9.99.4 Maple [A] (verified)
3.9.99.5 Fricas [A] (verification not implemented)
3.9.99.6 Sympy [F(-1)]
3.9.99.7 Maxima [A] (verification not implemented)
3.9.99.8 Giac [A] (verification not implemented)
3.9.99.9 Mupad [B] (verification not implemented)

3.9.99.1 Optimal result

Integrand size = 27, antiderivative size = 160 \[ \int \frac {\sec (c+d x) \tan ^8(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {7 \text {arctanh}(\sin (c+d x))}{256 a d}+\frac {7 \sec (c+d x) \tan (c+d x)}{256 a d}-\frac {7 \sec ^3(c+d x) \tan (c+d x)}{128 a d}+\frac {7 \sec ^3(c+d x) \tan ^3(c+d x)}{96 a d}-\frac {7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}+\frac {\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}-\frac {\tan ^{10}(c+d x)}{10 a d} \]

output
7/256*arctanh(sin(d*x+c))/a/d+7/256*sec(d*x+c)*tan(d*x+c)/a/d-7/128*sec(d* 
x+c)^3*tan(d*x+c)/a/d+7/96*sec(d*x+c)^3*tan(d*x+c)^3/a/d-7/80*sec(d*x+c)^3 
*tan(d*x+c)^5/a/d+1/10*sec(d*x+c)^3*tan(d*x+c)^7/a/d-1/10*tan(d*x+c)^10/a/ 
d
 
3.9.99.2 Mathematica [A] (verified)

Time = 1.58 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.76 \[ \int \frac {\sec (c+d x) \tan ^8(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {210 \text {arctanh}(\sin (c+d x))+\frac {-768-978 \sin (c+d x)+2862 \sin ^2(c+d x)+3842 \sin ^3(c+d x)-3838 \sin ^4(c+d x)-5630 \sin ^5(c+d x)+2050 \sin ^6(c+d x)+3630 \sin ^7(c+d x)-210 \sin ^8(c+d x)}{(-1+\sin (c+d x))^4 (1+\sin (c+d x))^5}}{7680 a d} \]

input
Integrate[(Sec[c + d*x]*Tan[c + d*x]^8)/(a + a*Sin[c + d*x]),x]
 
output
(210*ArcTanh[Sin[c + d*x]] + (-768 - 978*Sin[c + d*x] + 2862*Sin[c + d*x]^ 
2 + 3842*Sin[c + d*x]^3 - 3838*Sin[c + d*x]^4 - 5630*Sin[c + d*x]^5 + 2050 
*Sin[c + d*x]^6 + 3630*Sin[c + d*x]^7 - 210*Sin[c + d*x]^8)/((-1 + Sin[c + 
 d*x])^4*(1 + Sin[c + d*x])^5))/(7680*a*d)
 
3.9.99.3 Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.04, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.593, Rules used = {3042, 3314, 3042, 3087, 15, 3091, 3042, 3091, 3042, 3091, 3042, 3091, 3042, 4255, 3042, 4257}

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.

\(\displaystyle \int \frac {\tan ^8(c+d x) \sec (c+d x)}{a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^8}{\cos (c+d x)^9 (a \sin (c+d x)+a)}dx\)

\(\Big \downarrow \) 3314

\(\displaystyle \frac {\int \sec ^3(c+d x) \tan ^8(c+d x)dx}{a}-\frac {\int \sec ^2(c+d x) \tan ^9(c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sec (c+d x)^3 \tan (c+d x)^8dx}{a}-\frac {\int \sec (c+d x)^2 \tan (c+d x)^9dx}{a}\)

\(\Big \downarrow \) 3087

\(\displaystyle \frac {\int \sec (c+d x)^3 \tan (c+d x)^8dx}{a}-\frac {\int \tan ^9(c+d x)d\tan (c+d x)}{a d}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {\int \sec (c+d x)^3 \tan (c+d x)^8dx}{a}-\frac {\tan ^{10}(c+d x)}{10 a d}\)

\(\Big \downarrow \) 3091

\(\displaystyle \frac {\frac {\tan ^7(c+d x) \sec ^3(c+d x)}{10 d}-\frac {7}{10} \int \sec ^3(c+d x) \tan ^6(c+d x)dx}{a}-\frac {\tan ^{10}(c+d x)}{10 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\tan ^7(c+d x) \sec ^3(c+d x)}{10 d}-\frac {7}{10} \int \sec (c+d x)^3 \tan (c+d x)^6dx}{a}-\frac {\tan ^{10}(c+d x)}{10 a d}\)

\(\Big \downarrow \) 3091

\(\displaystyle \frac {\frac {\tan ^7(c+d x) \sec ^3(c+d x)}{10 d}-\frac {7}{10} \left (\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \int \sec ^3(c+d x) \tan ^4(c+d x)dx\right )}{a}-\frac {\tan ^{10}(c+d x)}{10 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\tan ^7(c+d x) \sec ^3(c+d x)}{10 d}-\frac {7}{10} \left (\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \int \sec (c+d x)^3 \tan (c+d x)^4dx\right )}{a}-\frac {\tan ^{10}(c+d x)}{10 a d}\)

\(\Big \downarrow \) 3091

\(\displaystyle \frac {\frac {\tan ^7(c+d x) \sec ^3(c+d x)}{10 d}-\frac {7}{10} \left (\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac {1}{2} \int \sec ^3(c+d x) \tan ^2(c+d x)dx\right )\right )}{a}-\frac {\tan ^{10}(c+d x)}{10 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\tan ^7(c+d x) \sec ^3(c+d x)}{10 d}-\frac {7}{10} \left (\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac {1}{2} \int \sec (c+d x)^3 \tan (c+d x)^2dx\right )\right )}{a}-\frac {\tan ^{10}(c+d x)}{10 a d}\)

\(\Big \downarrow \) 3091

\(\displaystyle \frac {\frac {\tan ^7(c+d x) \sec ^3(c+d x)}{10 d}-\frac {7}{10} \left (\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \int \sec ^3(c+d x)dx-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )\right )}{a}-\frac {\tan ^{10}(c+d x)}{10 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\tan ^7(c+d x) \sec ^3(c+d x)}{10 d}-\frac {7}{10} \left (\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )\right )}{a}-\frac {\tan ^{10}(c+d x)}{10 a d}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\frac {\tan ^7(c+d x) \sec ^3(c+d x)}{10 d}-\frac {7}{10} \left (\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )\right )}{a}-\frac {\tan ^{10}(c+d x)}{10 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\tan ^7(c+d x) \sec ^3(c+d x)}{10 d}-\frac {7}{10} \left (\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )\right )}{a}-\frac {\tan ^{10}(c+d x)}{10 a d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {\tan ^7(c+d x) \sec ^3(c+d x)}{10 d}-\frac {7}{10} \left (\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )\right )}{a}-\frac {\tan ^{10}(c+d x)}{10 a d}\)

input
Int[(Sec[c + d*x]*Tan[c + d*x]^8)/(a + a*Sin[c + d*x]),x]
 
output
-1/10*Tan[c + d*x]^10/(a*d) + ((Sec[c + d*x]^3*Tan[c + d*x]^7)/(10*d) - (7 
*((Sec[c + d*x]^3*Tan[c + d*x]^5)/(8*d) - (5*((Sec[c + d*x]^3*Tan[c + d*x] 
^3)/(6*d) + (-1/4*(Sec[c + d*x]^3*Tan[c + d*x])/d + (ArcTanh[Sin[c + d*x]] 
/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d))/4)/2))/8))/10)/a
 

3.9.99.3.1 Defintions of rubi rules used

rule 15
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

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

rule 3087
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S 
ymbol] :> Simp[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 3091
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] - Simp[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] && IntegersQ[2*m, 2*n]
 

rule 3314
Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/(( 
a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a   Int[Cos[e + f 
*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d)   Int[Cos[e + f*x]^(p 
 - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & 
& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p 
+ 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
 

rule 4255
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] + Simp[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 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 
3.9.99.4 Maple [A] (verified)

Time = 1.74 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.87

method result size
derivativedivides \(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {5}{192 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {37}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {7}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {7 \ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {11}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {47}{384 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {93}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {35}{256 \left (1+\sin \left (d x +c \right )\right )}+\frac {7 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) \(139\)
default \(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {5}{192 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {37}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {7}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {7 \ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {11}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {47}{384 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {93}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {35}{256 \left (1+\sin \left (d x +c \right )\right )}+\frac {7 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) \(139\)
risch \(-\frac {i \left (2890 i {\mathrm e}^{14 i \left (d x +c \right )}+105 \,{\mathrm e}^{17 i \left (d x +c \right )}-3630 i {\mathrm e}^{16 i \left (d x +c \right )}+3260 \,{\mathrm e}^{15 i \left (d x +c \right )}-23674 i {\mathrm e}^{8 i \left (d x +c \right )}+9044 \,{\mathrm e}^{13 i \left (d x +c \right )}+25102 i {\mathrm e}^{6 i \left (d x +c \right )}+24388 \,{\mathrm e}^{11 i \left (d x +c \right )}-25102 i {\mathrm e}^{12 i \left (d x +c \right )}+24710 \,{\mathrm e}^{9 i \left (d x +c \right )}+23674 i {\mathrm e}^{10 i \left (d x +c \right )}+24388 \,{\mathrm e}^{7 i \left (d x +c \right )}-2890 i {\mathrm e}^{4 i \left (d x +c \right )}+9044 \,{\mathrm e}^{5 i \left (d x +c \right )}+3630 i {\mathrm e}^{2 i \left (d x +c \right )}+3260 \,{\mathrm e}^{3 i \left (d x +c \right )}+105 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{1920 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} d a}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 d a}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 a d}\) \(277\)
parallelrisch \(\frac {\left (-105 \cos \left (10 d x +10 c \right )-22050 \cos \left (2 d x +2 c \right )-12600 \cos \left (4 d x +4 c \right )-4725 \cos \left (6 d x +6 c \right )-1050 \cos \left (8 d x +8 c \right )-13230\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (105 \cos \left (10 d x +10 c \right )+22050 \cos \left (2 d x +2 c \right )+12600 \cos \left (4 d x +4 c \right )+4725 \cos \left (6 d x +6 c \right )+1050 \cos \left (8 d x +8 c \right )+13230\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-69720 \sin \left (3 d x +3 c \right )+23128 \sin \left (5 d x +5 c \right )-8210 \sin \left (7 d x +7 c \right )+210 \sin \left (9 d x +9 c \right )+384 \cos \left (10 d x +10 c \right )+80640 \cos \left (2 d x +2 c \right )-46080 \cos \left (4 d x +4 c \right )+17280 \cos \left (6 d x +6 c \right )-3840 \cos \left (8 d x +8 c \right )+95340 \sin \left (d x +c \right )-48384}{3840 a d \left (\cos \left (10 d x +10 c \right )+10 \cos \left (8 d x +8 c \right )+45 \cos \left (6 d x +6 c \right )+120 \cos \left (4 d x +4 c \right )+210 \cos \left (2 d x +2 c \right )+126\right )}\) \(315\)

input
int(sec(d*x+c)^9*sin(d*x+c)^8/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 
output
1/d/a*(1/256/(sin(d*x+c)-1)^4+5/192/(sin(d*x+c)-1)^3+37/512/(sin(d*x+c)-1) 
^2+7/64/(sin(d*x+c)-1)-7/512*ln(sin(d*x+c)-1)-1/160/(1+sin(d*x+c))^5+11/25 
6/(1+sin(d*x+c))^4-47/384/(1+sin(d*x+c))^3+93/512/(1+sin(d*x+c))^2-35/256/ 
(1+sin(d*x+c))+7/512*ln(1+sin(d*x+c)))
 
3.9.99.5 Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.17 \[ \int \frac {\sec (c+d x) \tan ^8(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {210 \, \cos \left (d x + c\right )^{8} + 1210 \, \cos \left (d x + c\right )^{6} - 1052 \, \cos \left (d x + c\right )^{4} + 496 \, \cos \left (d x + c\right )^{2} - 105 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (1815 \, \cos \left (d x + c\right )^{6} - 2630 \, \cos \left (d x + c\right )^{4} + 1736 \, \cos \left (d x + c\right )^{2} - 432\right )} \sin \left (d x + c\right ) - 96}{7680 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \]

input
integrate(sec(d*x+c)^9*sin(d*x+c)^8/(a+a*sin(d*x+c)),x, algorithm="fricas" 
)
 
output
-1/7680*(210*cos(d*x + c)^8 + 1210*cos(d*x + c)^6 - 1052*cos(d*x + c)^4 + 
496*cos(d*x + c)^2 - 105*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*lo 
g(sin(d*x + c) + 1) + 105*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*l 
og(-sin(d*x + c) + 1) + 2*(1815*cos(d*x + c)^6 - 2630*cos(d*x + c)^4 + 173 
6*cos(d*x + c)^2 - 432)*sin(d*x + c) - 96)/(a*d*cos(d*x + c)^8*sin(d*x + c 
) + a*d*cos(d*x + c)^8)
 
3.9.99.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x) \tan ^8(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

input
integrate(sec(d*x+c)**9*sin(d*x+c)**8/(a+a*sin(d*x+c)),x)
 
output
Timed out
 
3.9.99.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.34 \[ \int \frac {\sec (c+d x) \tan ^8(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (105 \, \sin \left (d x + c\right )^{8} - 1815 \, \sin \left (d x + c\right )^{7} - 1025 \, \sin \left (d x + c\right )^{6} + 2815 \, \sin \left (d x + c\right )^{5} + 1919 \, \sin \left (d x + c\right )^{4} - 1921 \, \sin \left (d x + c\right )^{3} - 1431 \, \sin \left (d x + c\right )^{2} + 489 \, \sin \left (d x + c\right ) + 384\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{7680 \, d} \]

input
integrate(sec(d*x+c)^9*sin(d*x+c)^8/(a+a*sin(d*x+c)),x, algorithm="maxima" 
)
 
output
-1/7680*(2*(105*sin(d*x + c)^8 - 1815*sin(d*x + c)^7 - 1025*sin(d*x + c)^6 
 + 2815*sin(d*x + c)^5 + 1919*sin(d*x + c)^4 - 1921*sin(d*x + c)^3 - 1431* 
sin(d*x + c)^2 + 489*sin(d*x + c) + 384)/(a*sin(d*x + c)^9 + a*sin(d*x + c 
)^8 - 4*a*sin(d*x + c)^7 - 4*a*sin(d*x + c)^6 + 6*a*sin(d*x + c)^5 + 6*a*s 
in(d*x + c)^4 - 4*a*sin(d*x + c)^3 - 4*a*sin(d*x + c)^2 + a*sin(d*x + c) + 
 a) - 105*log(sin(d*x + c) + 1)/a + 105*log(sin(d*x + c) - 1)/a)/d
 
3.9.99.8 Giac [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.98 \[ \int \frac {\sec (c+d x) \tan ^8(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {420 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {420 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {5 \, {\left (175 \, \sin \left (d x + c\right )^{4} - 28 \, \sin \left (d x + c\right )^{3} - 522 \, \sin \left (d x + c\right )^{2} + 588 \, \sin \left (d x + c\right ) - 189\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac {959 \, \sin \left (d x + c\right )^{5} + 8995 \, \sin \left (d x + c\right )^{4} + 20810 \, \sin \left (d x + c\right )^{3} + 21810 \, \sin \left (d x + c\right )^{2} + 11055 \, \sin \left (d x + c\right ) + 2211}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{30720 \, d} \]

input
integrate(sec(d*x+c)^9*sin(d*x+c)^8/(a+a*sin(d*x+c)),x, algorithm="giac")
 
output
1/30720*(420*log(abs(sin(d*x + c) + 1))/a - 420*log(abs(sin(d*x + c) - 1)) 
/a + 5*(175*sin(d*x + c)^4 - 28*sin(d*x + c)^3 - 522*sin(d*x + c)^2 + 588* 
sin(d*x + c) - 189)/(a*(sin(d*x + c) - 1)^4) - (959*sin(d*x + c)^5 + 8995* 
sin(d*x + c)^4 + 20810*sin(d*x + c)^3 + 21810*sin(d*x + c)^2 + 11055*sin(d 
*x + c) + 2211)/(a*(sin(d*x + c) + 1)^5))/d
 
3.9.99.9 Mupad [B] (verification not implemented)

Time = 17.26 (sec) , antiderivative size = 496, normalized size of antiderivative = 3.10 \[ \int \frac {\sec (c+d x) \tan ^8(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {7\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{128\,a\,d}+\frac {-\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{128}-\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}}{64}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{96}+\frac {161\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{192}-\frac {469\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{480}-\frac {2681\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{960}+\frac {593\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{480}+\frac {5053\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{960}+\frac {10841\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {5053\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{960}+\frac {593\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{480}-\frac {2681\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{960}-\frac {469\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{480}+\frac {161\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{192}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}-\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{64}-\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}-7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}+20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+56\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-28\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-112\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+14\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+140\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+14\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-112\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-28\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+56\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )} \]

input
int(sin(c + d*x)^8/(cos(c + d*x)^9*(a + a*sin(c + d*x))),x)
 
output
(7*atanh(tan(c/2 + (d*x)/2)))/(128*a*d) + ((35*tan(c/2 + (d*x)/2)^3)/96 - 
(7*tan(c/2 + (d*x)/2)^2)/64 - (7*tan(c/2 + (d*x)/2))/128 + (161*tan(c/2 + 
(d*x)/2)^4)/192 - (469*tan(c/2 + (d*x)/2)^5)/480 - (2681*tan(c/2 + (d*x)/2 
)^6)/960 + (593*tan(c/2 + (d*x)/2)^7)/480 + (5053*tan(c/2 + (d*x)/2)^8)/96 
0 + (10841*tan(c/2 + (d*x)/2)^9)/192 + (5053*tan(c/2 + (d*x)/2)^10)/960 + 
(593*tan(c/2 + (d*x)/2)^11)/480 - (2681*tan(c/2 + (d*x)/2)^12)/960 - (469* 
tan(c/2 + (d*x)/2)^13)/480 + (161*tan(c/2 + (d*x)/2)^14)/192 + (35*tan(c/2 
 + (d*x)/2)^15)/96 - (7*tan(c/2 + (d*x)/2)^16)/64 - (7*tan(c/2 + (d*x)/2)^ 
17)/128)/(d*(a + 2*a*tan(c/2 + (d*x)/2) - 7*a*tan(c/2 + (d*x)/2)^2 - 16*a* 
tan(c/2 + (d*x)/2)^3 + 20*a*tan(c/2 + (d*x)/2)^4 + 56*a*tan(c/2 + (d*x)/2) 
^5 - 28*a*tan(c/2 + (d*x)/2)^6 - 112*a*tan(c/2 + (d*x)/2)^7 + 14*a*tan(c/2 
 + (d*x)/2)^8 + 140*a*tan(c/2 + (d*x)/2)^9 + 14*a*tan(c/2 + (d*x)/2)^10 - 
112*a*tan(c/2 + (d*x)/2)^11 - 28*a*tan(c/2 + (d*x)/2)^12 + 56*a*tan(c/2 + 
(d*x)/2)^13 + 20*a*tan(c/2 + (d*x)/2)^14 - 16*a*tan(c/2 + (d*x)/2)^15 - 7* 
a*tan(c/2 + (d*x)/2)^16 + 2*a*tan(c/2 + (d*x)/2)^17 + a*tan(c/2 + (d*x)/2) 
^18))