3.9.95 \(\int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx\) [895]

3.9.95.1 Optimal result
3.9.95.2 Mathematica [A] (verified)
3.9.95.3 Rubi [A] (verified)
3.9.95.4 Maple [A] (verified)
3.9.95.5 Fricas [A] (verification not implemented)
3.9.95.6 Sympy [F(-1)]
3.9.95.7 Maxima [A] (verification not implemented)
3.9.95.8 Giac [A] (verification not implemented)
3.9.95.9 Mupad [B] (verification not implemented)

3.9.95.1 Optimal result

Integrand size = 29, antiderivative size = 264 \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {843 \log (1-\sin (c+d x))}{512 a d}-\frac {2229 \log (1+\sin (c+d x))}{512 a d}+\frac {\sin (c+d x)}{a d}-\frac {\sin ^2(c+d x)}{2 a d}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}-\frac {3 a^2}{64 d (a-a \sin (c+d x))^3}+\frac {141 a}{512 d (a-a \sin (c+d x))^2}-\frac {39}{32 d (a-a \sin (c+d x))}-\frac {a^4}{160 d (a+a \sin (c+d x))^5}+\frac {19 a^3}{256 d (a+a \sin (c+d x))^4}-\frac {53 a^2}{128 d (a+a \sin (c+d x))^3}+\frac {765 a}{512 d (a+a \sin (c+d x))^2}-\frac {1155}{256 d (a+a \sin (c+d x))} \]

output
-843/512*ln(1-sin(d*x+c))/a/d-2229/512*ln(1+sin(d*x+c))/a/d+sin(d*x+c)/a/d 
-1/2*sin(d*x+c)^2/a/d+1/256*a^3/d/(a-a*sin(d*x+c))^4-3/64*a^2/d/(a-a*sin(d 
*x+c))^3+141/512*a/d/(a-a*sin(d*x+c))^2-39/32/d/(a-a*sin(d*x+c))-1/160*a^4 
/d/(a+a*sin(d*x+c))^5+19/256*a^3/d/(a+a*sin(d*x+c))^4-53/128*a^2/d/(a+a*si 
n(d*x+c))^3+765/512*a/d/(a+a*sin(d*x+c))^2-1155/256/d/(a+a*sin(d*x+c))
 
3.9.95.2 Mathematica [A] (verified)

Time = 6.11 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.64 \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {4215 \log (1-\sin (c+d x))+11145 \log (1+\sin (c+d x))-\frac {10}{(1-\sin (c+d x))^4}+\frac {120}{(1-\sin (c+d x))^3}-\frac {705}{(1-\sin (c+d x))^2}+\frac {3120}{1-\sin (c+d x)}-2560 \sin (c+d x)+1280 \sin ^2(c+d x)+\frac {16}{(1+\sin (c+d x))^5}-\frac {190}{(1+\sin (c+d x))^4}+\frac {1060}{(1+\sin (c+d x))^3}-\frac {3825}{(1+\sin (c+d x))^2}+\frac {11550}{1+\sin (c+d x)}}{2560 a d} \]

input
Integrate[(Sin[c + d*x]^3*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]
 
output
-1/2560*(4215*Log[1 - Sin[c + d*x]] + 11145*Log[1 + Sin[c + d*x]] - 10/(1 
- Sin[c + d*x])^4 + 120/(1 - Sin[c + d*x])^3 - 705/(1 - Sin[c + d*x])^2 + 
3120/(1 - Sin[c + d*x]) - 2560*Sin[c + d*x] + 1280*Sin[c + d*x]^2 + 16/(1 
+ Sin[c + d*x])^5 - 190/(1 + Sin[c + d*x])^4 + 1060/(1 + Sin[c + d*x])^3 - 
 3825/(1 + Sin[c + d*x])^2 + 11550/(1 + Sin[c + d*x]))/(a*d)
 
3.9.95.3 Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, 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.

\(\displaystyle \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^{12}}{\cos (c+d x)^9 (a \sin (c+d x)+a)}dx\)

\(\Big \downarrow \) 3315

\(\displaystyle \frac {a^9 \int \frac {\sin ^{12}(c+d x)}{(a-a \sin (c+d x))^5 (\sin (c+d x) a+a)^6}d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a^{12} \sin ^{12}(c+d x)}{(a-a \sin (c+d x))^5 (\sin (c+d x) a+a)^6}d(a \sin (c+d x))}{a^3 d}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {\int \left (\frac {a^7}{32 (\sin (c+d x) a+a)^6}+\frac {a^6}{64 (a-a \sin (c+d x))^5}-\frac {19 a^6}{64 (\sin (c+d x) a+a)^5}-\frac {9 a^5}{64 (a-a \sin (c+d x))^4}+\frac {159 a^5}{128 (\sin (c+d x) a+a)^4}+\frac {141 a^4}{256 (a-a \sin (c+d x))^3}-\frac {765 a^4}{256 (\sin (c+d x) a+a)^3}-\frac {39 a^3}{32 (a-a \sin (c+d x))^2}+\frac {1155 a^3}{256 (\sin (c+d x) a+a)^2}+\frac {843 a^2}{512 (a-a \sin (c+d x))}-\frac {2229 a^2}{512 (\sin (c+d x) a+a)}-\sin (c+d x) a+a\right )d(a \sin (c+d x))}{a^3 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {a^7}{160 (a \sin (c+d x)+a)^5}+\frac {a^6}{256 (a-a \sin (c+d x))^4}+\frac {19 a^6}{256 (a \sin (c+d x)+a)^4}-\frac {3 a^5}{64 (a-a \sin (c+d x))^3}-\frac {53 a^5}{128 (a \sin (c+d x)+a)^3}+\frac {141 a^4}{512 (a-a \sin (c+d x))^2}+\frac {765 a^4}{512 (a \sin (c+d x)+a)^2}-\frac {39 a^3}{32 (a-a \sin (c+d x))}-\frac {1155 a^3}{256 (a \sin (c+d x)+a)}-\frac {1}{2} a^2 \sin ^2(c+d x)+a^2 \sin (c+d x)-\frac {843}{512} a^2 \log (a-a \sin (c+d x))-\frac {2229}{512} a^2 \log (a \sin (c+d x)+a)}{a^3 d}\)

input
Int[(Sin[c + d*x]^3*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]
 
output
((-843*a^2*Log[a - a*Sin[c + d*x]])/512 - (2229*a^2*Log[a + a*Sin[c + d*x] 
])/512 + a^2*Sin[c + d*x] - (a^2*Sin[c + d*x]^2)/2 + a^6/(256*(a - a*Sin[c 
 + d*x])^4) - (3*a^5)/(64*(a - a*Sin[c + d*x])^3) + (141*a^4)/(512*(a - a* 
Sin[c + d*x])^2) - (39*a^3)/(32*(a - a*Sin[c + d*x])) - a^7/(160*(a + a*Si 
n[c + d*x])^5) + (19*a^6)/(256*(a + a*Sin[c + d*x])^4) - (53*a^5)/(128*(a 
+ a*Sin[c + d*x])^3) + (765*a^4)/(512*(a + a*Sin[c + d*x])^2) - (1155*a^3) 
/(256*(a + a*Sin[c + d*x])))/(a^3*d)
 

3.9.95.3.1 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]
 
3.9.95.4 Maple [A] (verified)

Time = 4.70 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.59

method result size
derivativedivides \(\frac {-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right )+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {3}{64 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {141}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {39}{32 \left (\sin \left (d x +c \right )-1\right )}-\frac {843 \ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {19}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {53}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {765}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1155}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {2229 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) \(155\)
default \(\frac {-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right )+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {3}{64 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {141}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {39}{32 \left (\sin \left (d x +c \right )-1\right )}-\frac {843 \ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {19}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {53}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {765}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1155}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {2229 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) \(155\)
risch \(\frac {6 i x}{a}+\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 a d}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 a d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a d}+\frac {12 i c}{d a}-\frac {i \left (-31370 i {\mathrm e}^{14 i \left (d x +c \right )}+4215 \,{\mathrm e}^{17 i \left (d x +c \right )}-10770 i {\mathrm e}^{16 i \left (d x +c \right )}+40900 \,{\mathrm e}^{15 i \left (d x +c \right )}+20922 i {\mathrm e}^{8 i \left (d x +c \right )}+152108 \,{\mathrm e}^{13 i \left (d x +c \right )}+59954 i {\mathrm e}^{6 i \left (d x +c \right )}+316476 \,{\mathrm e}^{11 i \left (d x +c \right )}-59954 i {\mathrm e}^{12 i \left (d x +c \right )}+398010 \,{\mathrm e}^{9 i \left (d x +c \right )}-20922 i {\mathrm e}^{10 i \left (d x +c \right )}+316476 \,{\mathrm e}^{7 i \left (d x +c \right )}+31370 i {\mathrm e}^{4 i \left (d x +c \right )}+152108 \,{\mathrm e}^{5 i \left (d x +c \right )}+10770 i {\mathrm e}^{2 i \left (d x +c \right )}+40900 \,{\mathrm e}^{3 i \left (d x +c \right )}+4215 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{640 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} d a}-\frac {2229 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 a d}-\frac {843 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 d a}\) \(364\)
parallelrisch \(\frac {\left (215040 \sin \left (3 d x +3 c \right )+153600 \sin \left (5 d x +5 c \right )+53760 \sin \left (7 d x +7 c \right )+7680 \sin \left (9 d x +9 c \right )+860160 \cos \left (2 d x +2 c \right )+430080 \cos \left (4 d x +4 c \right )+122880 \cos \left (6 d x +6 c \right )+15360 \cos \left (8 d x +8 c \right )+107520 \sin \left (d x +c \right )+537600\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-118020 \sin \left (3 d x +3 c \right )-84300 \sin \left (5 d x +5 c \right )-29505 \sin \left (7 d x +7 c \right )-4215 \sin \left (9 d x +9 c \right )-472080 \cos \left (2 d x +2 c \right )-236040 \cos \left (4 d x +4 c \right )-67440 \cos \left (6 d x +6 c \right )-8430 \cos \left (8 d x +8 c \right )-59010 \sin \left (d x +c \right )-295050\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-312060 \sin \left (3 d x +3 c \right )-222900 \sin \left (5 d x +5 c \right )-78015 \sin \left (7 d x +7 c \right )-11145 \sin \left (9 d x +9 c \right )-1248240 \cos \left (2 d x +2 c \right )-624120 \cos \left (4 d x +4 c \right )-178320 \cos \left (6 d x +6 c \right )-22290 \cos \left (8 d x +8 c \right )-156030 \sin \left (d x +c \right )-780150\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+68316 \sin \left (3 d x +3 c \right )+73420 \sin \left (5 d x +5 c \right )+27516 \sin \left (7 d x +7 c \right )+7648 \sin \left (9 d x +9 c \right )-320 \cos \left (10 d x +10 c \right )+160 \sin \left (11 d x +11 c \right )-4856 \cos \left (2 d x +2 c \right )+5032 \cos \left (4 d x +4 c \right )+3128 \cos \left (6 d x +6 c \right )+786 \cos \left (8 d x +8 c \right )+51788 \sin \left (d x +c \right )-3770}{1280 a d \left (70+\sin \left (9 d x +9 c \right )+7 \sin \left (7 d x +7 c \right )+20 \sin \left (5 d x +5 c \right )+28 \sin \left (3 d x +3 c \right )+14 \sin \left (d x +c \right )+2 \cos \left (8 d x +8 c \right )+16 \cos \left (6 d x +6 c \right )+56 \cos \left (4 d x +4 c \right )+112 \cos \left (2 d x +2 c \right )\right )}\) \(560\)

input
int(sec(d*x+c)^9*sin(d*x+c)^12/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 
output
1/d/a*(-1/2*sin(d*x+c)^2+sin(d*x+c)+1/256/(sin(d*x+c)-1)^4+3/64/(sin(d*x+c 
)-1)^3+141/512/(sin(d*x+c)-1)^2+39/32/(sin(d*x+c)-1)-843/512*ln(sin(d*x+c) 
-1)-1/160/(1+sin(d*x+c))^5+19/256/(1+sin(d*x+c))^4-53/128/(1+sin(d*x+c))^3 
+765/512/(1+sin(d*x+c))^2-1155/256/(1+sin(d*x+c))-2229/512*ln(1+sin(d*x+c) 
))
 
3.9.95.5 Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.82 \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {1280 \, \cos \left (d x + c\right )^{10} + 6510 \, \cos \left (d x + c\right )^{8} + 3590 \, \cos \left (d x + c\right )^{6} - 1124 \, \cos \left (d x + c\right )^{4} + 272 \, \cos \left (d x + c\right )^{2} + 11145 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 4215 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (640 \, \cos \left (d x + c\right )^{10} + 960 \, \cos \left (d x + c\right )^{8} - 5385 \, \cos \left (d x + c\right )^{6} + 2810 \, \cos \left (d x + c\right )^{4} - 952 \, \cos \left (d x + c\right )^{2} + 144\right )} \sin \left (d x + c\right ) - 32}{2560 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \]

input
integrate(sec(d*x+c)^9*sin(d*x+c)^12/(a+a*sin(d*x+c)),x, algorithm="fricas 
")
 
output
-1/2560*(1280*cos(d*x + c)^10 + 6510*cos(d*x + c)^8 + 3590*cos(d*x + c)^6 
- 1124*cos(d*x + c)^4 + 272*cos(d*x + c)^2 + 11145*(cos(d*x + c)^8*sin(d*x 
 + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) + 4215*(cos(d*x + c)^8*sin(d 
*x + c) + cos(d*x + c)^8)*log(-sin(d*x + c) + 1) - 2*(640*cos(d*x + c)^10 
+ 960*cos(d*x + c)^8 - 5385*cos(d*x + c)^6 + 2810*cos(d*x + c)^4 - 952*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)
 
3.9.95.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

input
integrate(sec(d*x+c)**9*sin(d*x+c)**12/(a+a*sin(d*x+c)),x)
 
output
Timed out
 
3.9.95.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (4215 \, \sin \left (d x + c\right )^{8} - 5385 \, \sin \left (d x + c\right )^{7} - 18655 \, \sin \left (d x + c\right )^{6} + 13345 \, \sin \left (d x + c\right )^{5} + 30113 \, \sin \left (d x + c\right )^{4} - 11487 \, \sin \left (d x + c\right )^{3} - 21257 \, \sin \left (d x + c\right )^{2} + 3383 \, \sin \left (d x + c\right ) + 5568\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} + \frac {1280 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )}}{a} + \frac {11145 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {4215 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{2560 \, d} \]

input
integrate(sec(d*x+c)^9*sin(d*x+c)^12/(a+a*sin(d*x+c)),x, algorithm="maxima 
")
 
output
-1/2560*(2*(4215*sin(d*x + c)^8 - 5385*sin(d*x + c)^7 - 18655*sin(d*x + c) 
^6 + 13345*sin(d*x + c)^5 + 30113*sin(d*x + c)^4 - 11487*sin(d*x + c)^3 - 
21257*sin(d*x + c)^2 + 3383*sin(d*x + c) + 5568)/(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*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) + 1280*(sin(d*x + c)^2 - 2*sin(d*x + c))/a + 11145*log(sin(d*x 
 + c) + 1)/a + 4215*log(sin(d*x + c) - 1)/a)/d
 
3.9.95.8 Giac [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.69 \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {44580 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} + \frac {16860 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {5120 \, {\left (a \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )\right )}}{a^{2}} - \frac {5 \, {\left (7025 \, \sin \left (d x + c\right )^{4} - 25604 \, \sin \left (d x + c\right )^{3} + 35226 \, \sin \left (d x + c\right )^{2} - 21644 \, \sin \left (d x + c\right ) + 5005\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac {101791 \, \sin \left (d x + c\right )^{5} + 462755 \, \sin \left (d x + c\right )^{4} + 848410 \, \sin \left (d x + c\right )^{3} + 782370 \, \sin \left (d x + c\right )^{2} + 362335 \, \sin \left (d x + c\right ) + 67347}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{10240 \, d} \]

input
integrate(sec(d*x+c)^9*sin(d*x+c)^12/(a+a*sin(d*x+c)),x, algorithm="giac")
 
output
-1/10240*(44580*log(abs(sin(d*x + c) + 1))/a + 16860*log(abs(sin(d*x + c) 
- 1))/a + 5120*(a*sin(d*x + c)^2 - 2*a*sin(d*x + c))/a^2 - 5*(7025*sin(d*x 
 + c)^4 - 25604*sin(d*x + c)^3 + 35226*sin(d*x + c)^2 - 21644*sin(d*x + c) 
 + 5005)/(a*(sin(d*x + c) - 1)^4) - (101791*sin(d*x + c)^5 + 462755*sin(d* 
x + c)^4 + 848410*sin(d*x + c)^3 + 782370*sin(d*x + c)^2 + 362335*sin(d*x 
+ c) + 67347)/(a*(sin(d*x + c) + 1)^5))/d
 
3.9.95.9 Mupad [B] (verification not implemented)

Time = 13.56 (sec) , antiderivative size = 648, normalized size of antiderivative = 2.45 \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]

input
int(sin(c + d*x)^12/(cos(c + d*x)^9*(a + a*sin(c + d*x))),x)
 
output
((693*tan(c/2 + (d*x)/2))/128 - (75*tan(c/2 + (d*x)/2)^2)/64 - (3153*tan(c 
/2 + (d*x)/2)^3)/64 - (87*tan(c/2 + (d*x)/2)^4)/64 + (111333*tan(c/2 + (d* 
x)/2)^5)/640 + (1331*tan(c/2 + (d*x)/2)^6)/40 - (4559*tan(c/2 + (d*x)/2)^7 
)/16 - (1823*tan(c/2 + (d*x)/2)^8)/20 + (42953*tan(c/2 + (d*x)/2)^9)/320 + 
 (11713*tan(c/2 + (d*x)/2)^10)/160 + (43457*tan(c/2 + (d*x)/2)^11)/160 + ( 
11713*tan(c/2 + (d*x)/2)^12)/160 + (42953*tan(c/2 + (d*x)/2)^13)/320 - (18 
23*tan(c/2 + (d*x)/2)^14)/20 - (4559*tan(c/2 + (d*x)/2)^15)/16 + (1331*tan 
(c/2 + (d*x)/2)^16)/40 + (111333*tan(c/2 + (d*x)/2)^17)/640 - (87*tan(c/2 
+ (d*x)/2)^18)/64 - (3153*tan(c/2 + (d*x)/2)^19)/64 - (75*tan(c/2 + (d*x)/ 
2)^20)/64 + (693*tan(c/2 + (d*x)/2)^21)/128)/(d*(a + 2*a*tan(c/2 + (d*x)/2 
) - 5*a*tan(c/2 + (d*x)/2)^2 - 12*a*tan(c/2 + (d*x)/2)^3 + 7*a*tan(c/2 + ( 
d*x)/2)^4 + 26*a*tan(c/2 + (d*x)/2)^5 + 5*a*tan(c/2 + (d*x)/2)^6 - 16*a*ta 
n(c/2 + (d*x)/2)^7 - 22*a*tan(c/2 + (d*x)/2)^8 - 28*a*tan(c/2 + (d*x)/2)^9 
 + 14*a*tan(c/2 + (d*x)/2)^10 + 56*a*tan(c/2 + (d*x)/2)^11 + 14*a*tan(c/2 
+ (d*x)/2)^12 - 28*a*tan(c/2 + (d*x)/2)^13 - 22*a*tan(c/2 + (d*x)/2)^14 - 
16*a*tan(c/2 + (d*x)/2)^15 + 5*a*tan(c/2 + (d*x)/2)^16 + 26*a*tan(c/2 + (d 
*x)/2)^17 + 7*a*tan(c/2 + (d*x)/2)^18 - 12*a*tan(c/2 + (d*x)/2)^19 - 5*a*t 
an(c/2 + (d*x)/2)^20 + 2*a*tan(c/2 + (d*x)/2)^21 + a*tan(c/2 + (d*x)/2)^22 
)) - (843*log(tan(c/2 + (d*x)/2) - 1))/(256*a*d) - (2229*log(tan(c/2 + (d* 
x)/2) + 1))/(256*a*d) + (6*log(tan(c/2 + (d*x)/2)^2 + 1))/(a*d)