3.10.1 \(\int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx\) [901]

3.10.1.1 Optimal result
3.10.1.2 Mathematica [A] (verified)
3.10.1.3 Rubi [A] (verified)
3.10.1.4 Maple [A] (verified)
3.10.1.5 Fricas [A] (verification not implemented)
3.10.1.6 Sympy [F(-1)]
3.10.1.7 Maxima [A] (verification not implemented)
3.10.1.8 Giac [A] (verification not implemented)
3.10.1.9 Mupad [B] (verification not implemented)

3.10.1.1 Optimal result

Integrand size = 29, antiderivative size = 176 \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \text {arctanh}(\sin (c+d x))}{256 a d}-\frac {3 \sec (c+d x) \tan (c+d x)}{256 a d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{128 a d}+\frac {\sec ^5(c+d x) \tan (c+d x)}{32 a d}-\frac {\sec ^5(c+d x) \tan ^3(c+d x)}{16 a d}+\frac {\sec ^5(c+d x) \tan ^5(c+d x)}{10 a d}-\frac {\tan ^8(c+d x)}{8 a d}-\frac {\tan ^{10}(c+d x)}{10 a d} \]

output
-3/256*arctanh(sin(d*x+c))/a/d-3/256*sec(d*x+c)*tan(d*x+c)/a/d-1/128*sec(d 
*x+c)^3*tan(d*x+c)/a/d+1/32*sec(d*x+c)^5*tan(d*x+c)/a/d-1/16*sec(d*x+c)^5* 
tan(d*x+c)^3/a/d+1/10*sec(d*x+c)^5*tan(d*x+c)^5/a/d-1/8*tan(d*x+c)^8/a/d-1 
/10*tan(d*x+c)^10/a/d
 
3.10.1.2 Mathematica [A] (verified)

Time = 1.72 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.69 \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {30 \text {arctanh}(\sin (c+d x))-\frac {2 \left (32+47 \sin (c+d x)-113 \sin ^2(c+d x)-183 \sin ^3(c+d x)+137 \sin ^4(c+d x)+265 \sin ^5(c+d x)-55 \sin ^6(c+d x)+15 \sin ^7(c+d x)+15 \sin ^8(c+d x)\right )}{(-1+\sin (c+d x))^4 (1+\sin (c+d x))^5}}{2560 a d} \]

input
Integrate[(Sec[c + d*x]^3*Tan[c + d*x]^6)/(a + a*Sin[c + d*x]),x]
 
output
-1/2560*(30*ArcTanh[Sin[c + d*x]] - (2*(32 + 47*Sin[c + d*x] - 113*Sin[c + 
 d*x]^2 - 183*Sin[c + d*x]^3 + 137*Sin[c + d*x]^4 + 265*Sin[c + d*x]^5 - 5 
5*Sin[c + d*x]^6 + 15*Sin[c + d*x]^7 + 15*Sin[c + d*x]^8))/((-1 + Sin[c + 
d*x])^4*(1 + Sin[c + d*x])^5))/(a*d)
 
3.10.1.3 Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.02, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.586, Rules used = {3042, 3314, 3042, 3087, 244, 2009, 3091, 3042, 3091, 3042, 3091, 3042, 4255, 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.

\(\displaystyle \int \frac {\tan ^6(c+d x) \sec ^3(c+d x)}{a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^6}{\cos (c+d x)^9 (a \sin (c+d x)+a)}dx\)

\(\Big \downarrow \) 3314

\(\displaystyle \frac {\int \sec ^5(c+d x) \tan ^6(c+d x)dx}{a}-\frac {\int \sec ^4(c+d x) \tan ^7(c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sec (c+d x)^5 \tan (c+d x)^6dx}{a}-\frac {\int \sec (c+d x)^4 \tan (c+d x)^7dx}{a}\)

\(\Big \downarrow \) 3087

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

\(\Big \downarrow \) 244

\(\displaystyle \frac {\int \sec (c+d x)^5 \tan (c+d x)^6dx}{a}-\frac {\int \left (\tan ^9(c+d x)+\tan ^7(c+d x)\right )d\tan (c+d x)}{a d}\)

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 3091

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3091

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3091

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

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

3.10.1.3.1 Defintions of rubi rules used

rule 244
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

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 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]
 
3.10.1.4 Maple [A] (verified)

Time = 1.69 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.79

method result size
derivativedivides \(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {9}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {1}{128 \left (\sin \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {7}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {5}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) \(139\)
default \(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {9}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {1}{128 \left (\sin \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {7}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {5}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) \(139\)
risch \(\frac {i \left (-2330 i {\mathrm e}^{14 i \left (d x +c \right )}+15 \,{\mathrm e}^{17 i \left (d x +c \right )}+30 i {\mathrm e}^{16 i \left (d x +c \right )}+100 \,{\mathrm e}^{15 i \left (d x +c \right )}+10698 i {\mathrm e}^{8 i \left (d x +c \right )}+1292 \,{\mathrm e}^{13 i \left (d x +c \right )}-5374 i {\mathrm e}^{6 i \left (d x +c \right )}+924 \,{\mathrm e}^{11 i \left (d x +c \right )}+5374 i {\mathrm e}^{12 i \left (d x +c \right )}+3530 \,{\mathrm e}^{9 i \left (d x +c \right )}-10698 i {\mathrm e}^{10 i \left (d x +c \right )}+924 \,{\mathrm e}^{7 i \left (d x +c \right )}+2330 i {\mathrm e}^{4 i \left (d x +c \right )}+1292 \,{\mathrm e}^{5 i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}+100 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{640 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} d a}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 d a}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 a d}\) \(277\)
parallelrisch \(\frac {\left (15 \cos \left (10 d x +10 c \right )+3150 \cos \left (2 d x +2 c \right )+1800 \cos \left (4 d x +4 c \right )+675 \cos \left (6 d x +6 c \right )+150 \cos \left (8 d x +8 c \right )+1890\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-15 \cos \left (10 d x +10 c \right )-3150 \cos \left (2 d x +2 c \right )-1800 \cos \left (4 d x +4 c \right )-675 \cos \left (6 d x +6 c \right )-150 \cos \left (8 d x +8 c \right )-1890\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-20760 \sin \left (3 d x +3 c \right )+6936 \sin \left (5 d x +5 c \right )-290 \sin \left (7 d x +7 c \right )-30 \sin \left (9 d x +9 c \right )-32 \cos \left (10 d x +10 c \right )+29120 \cos \left (2 d x +2 c \right )-14080 \cos \left (4 d x +4 c \right )+3680 \cos \left (6 d x +6 c \right )-320 \cos \left (8 d x +8 c \right )+37580 \sin \left (d x +c \right )-18368}{1280 a d \left (\cos \left (10 d x +10 c \right )+10 \cos \left (8 d x +8 c \right )+45 \cos \left (6 d x +6 c \right )+120 \cos \left (4 d x +4 c \right )+210 \cos \left (2 d x +2 c \right )+126\right )}\) \(315\)

input
int(sec(d*x+c)^9*sin(d*x+c)^6/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 
output
1/d/a*(1/256/(sin(d*x+c)-1)^4+1/64/(sin(d*x+c)-1)^3+9/512/(sin(d*x+c)-1)^2 
-1/128/(sin(d*x+c)-1)+3/512*ln(sin(d*x+c)-1)-1/160/(1+sin(d*x+c))^5+7/256/ 
(1+sin(d*x+c))^4-5/128/(1+sin(d*x+c))^3+5/512/(1+sin(d*x+c))^2+5/256/(1+si 
n(d*x+c))-3/512*ln(1+sin(d*x+c)))
 
3.10.1.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {30 \, \cos \left (d x + c\right )^{8} - 10 \, \cos \left (d x + c\right )^{6} + 124 \, \cos \left (d x + c\right )^{4} - 112 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, \cos \left (d x + c\right )^{6} - 310 \, \cos \left (d x + c\right )^{4} + 392 \, \cos \left (d x + c\right )^{2} - 144\right )} \sin \left (d x + c\right ) + 32}{2560 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \]

input
integrate(sec(d*x+c)^9*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="fricas" 
)
 
output
1/2560*(30*cos(d*x + c)^8 - 10*cos(d*x + c)^6 + 124*cos(d*x + c)^4 - 112*c 
os(d*x + c)^2 - 15*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(sin( 
d*x + c) + 1) + 15*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(-sin 
(d*x + c) + 1) - 2*(15*cos(d*x + c)^6 - 310*cos(d*x + c)^4 + 392*cos(d*x + 
 c)^2 - 144)*sin(d*x + c) + 32)/(a*d*cos(d*x + c)^8*sin(d*x + c) + a*d*cos 
(d*x + c)^8)
 
3.10.1.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

input
integrate(sec(d*x+c)**9*sin(d*x+c)**6/(a+a*sin(d*x+c)),x)
 
output
Timed out
 
3.10.1.7 Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.22 \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{8} + 15 \, \sin \left (d x + c\right )^{7} - 55 \, \sin \left (d x + c\right )^{6} + 265 \, \sin \left (d x + c\right )^{5} + 137 \, \sin \left (d x + c\right )^{4} - 183 \, \sin \left (d x + c\right )^{3} - 113 \, \sin \left (d x + c\right )^{2} + 47 \, \sin \left (d x + c\right ) + 32\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{2560 \, d} \]

input
integrate(sec(d*x+c)^9*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima" 
)
 
output
1/2560*(2*(15*sin(d*x + c)^8 + 15*sin(d*x + c)^7 - 55*sin(d*x + c)^6 + 265 
*sin(d*x + c)^5 + 137*sin(d*x + c)^4 - 183*sin(d*x + c)^3 - 113*sin(d*x + 
c)^2 + 47*sin(d*x + c) + 32)/(a*sin(d*x + c)^9 + a*sin(d*x + c)^8 - 4*a*si 
n(d*x + c)^7 - 4*a*sin(d*x + c)^6 + 6*a*sin(d*x + c)^5 + 6*a*sin(d*x + c)^ 
4 - 4*a*sin(d*x + c)^3 - 4*a*sin(d*x + c)^2 + a*sin(d*x + c) + a) - 15*log 
(sin(d*x + c) + 1)/a + 15*log(sin(d*x + c) - 1)/a)/d
 
3.10.1.8 Giac [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {5 \, {\left (25 \, \sin \left (d x + c\right )^{4} - 84 \, \sin \left (d x + c\right )^{3} + 66 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right ) - 3\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac {137 \, \sin \left (d x + c\right )^{5} + 885 \, \sin \left (d x + c\right )^{4} + 2270 \, \sin \left (d x + c\right )^{3} + 2470 \, \sin \left (d x + c\right )^{2} + 1265 \, \sin \left (d x + c\right ) + 253}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{10240 \, d} \]

input
integrate(sec(d*x+c)^9*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="giac")
 
output
-1/10240*(60*log(abs(sin(d*x + c) + 1))/a - 60*log(abs(sin(d*x + c) - 1))/ 
a + 5*(25*sin(d*x + c)^4 - 84*sin(d*x + c)^3 + 66*sin(d*x + c)^2 - 12*sin( 
d*x + c) - 3)/(a*(sin(d*x + c) - 1)^4) - (137*sin(d*x + c)^5 + 885*sin(d*x 
 + c)^4 + 2270*sin(d*x + c)^3 + 2470*sin(d*x + c)^2 + 1265*sin(d*x + c) + 
253)/(a*(sin(d*x + c) + 1)^5))/d
 
3.10.1.9 Mupad [B] (verification not implemented)

Time = 18.12 (sec) , antiderivative size = 496, normalized size of antiderivative = 2.82 \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{128}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}}{64}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{32}-\frac {23\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{64}+\frac {67\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{160}+\frac {383\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{320}+\frac {2841\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{160}+\frac {741\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{320}+\frac {1377\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{64}+\frac {741\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{320}+\frac {2841\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{160}+\frac {383\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{320}+\frac {67\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160}-\frac {23\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{64}+\frac {3\,\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 )}-\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{128\,a\,d} \]

input
int(sin(c + d*x)^6/(cos(c + d*x)^9*(a + a*sin(c + d*x))),x)
 
output
((3*tan(c/2 + (d*x)/2))/128 + (3*tan(c/2 + (d*x)/2)^2)/64 - (5*tan(c/2 + ( 
d*x)/2)^3)/32 - (23*tan(c/2 + (d*x)/2)^4)/64 + (67*tan(c/2 + (d*x)/2)^5)/1 
60 + (383*tan(c/2 + (d*x)/2)^6)/320 + (2841*tan(c/2 + (d*x)/2)^7)/160 + (7 
41*tan(c/2 + (d*x)/2)^8)/320 + (1377*tan(c/2 + (d*x)/2)^9)/64 + (741*tan(c 
/2 + (d*x)/2)^10)/320 + (2841*tan(c/2 + (d*x)/2)^11)/160 + (383*tan(c/2 + 
(d*x)/2)^12)/320 + (67*tan(c/2 + (d*x)/2)^13)/160 - (23*tan(c/2 + (d*x)/2) 
^14)/64 - (5*tan(c/2 + (d*x)/2)^15)/32 + (3*tan(c/2 + (d*x)/2)^16)/64 + (3 
*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)) - (3*atanh(tan(c/2 + (d*x)/2)))/(128*a*d)