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

Optimal result

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

[Out]

-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*sin(d*x+c))^3+765/512*a/d/(a+a*sin(d
*x+c))^2-1155/256/d/(a+a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 12, 90} \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {a^4}{160 d (a \sin (c+d x)+a)^5}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}+\frac {19 a^3}{256 d (a \sin (c+d x)+a)^4}-\frac {3 a^2}{64 d (a-a \sin (c+d x))^3}-\frac {53 a^2}{128 d (a \sin (c+d x)+a)^3}-\frac {\sin ^2(c+d x)}{2 a d}+\frac {141 a}{512 d (a-a \sin (c+d x))^2}+\frac {765 a}{512 d (a \sin (c+d x)+a)^2}-\frac {39}{32 d (a-a \sin (c+d x))}-\frac {1155}{256 d (a \sin (c+d x)+a)}+\frac {\sin (c+d x)}{a d}-\frac {843 \log (1-\sin (c+d x))}{512 a d}-\frac {2229 \log (\sin (c+d x)+1)}{512 a d} \]

[In]

Int[(Sin[c + d*x]^3*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]

[Out]

(-843*Log[1 - Sin[c + d*x]])/(512*a*d) - (2229*Log[1 + Sin[c + d*x]])/(512*a*d) + Sin[c + d*x]/(a*d) - Sin[c +
d*x]^2/(2*a*d) + a^3/(256*d*(a - a*Sin[c + d*x])^4) - (3*a^2)/(64*d*(a - a*Sin[c + d*x])^3) + (141*a)/(512*d*
(a - a*Sin[c + d*x])^2) - 39/(32*d*(a - a*Sin[c + d*x])) - a^4/(160*d*(a + a*Sin[c + d*x])^5) + (19*a^3)/(256*
d*(a + a*Sin[c + d*x])^4) - (53*a^2)/(128*d*(a + a*Sin[c + d*x])^3) + (765*a)/(512*d*(a + a*Sin[c + d*x])^2) -
1155/(256*d*(a + a*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(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 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[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] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {a^9 \text {Subst}\left (\int \frac {x^{12}}{a^{12} (a-x)^5 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \frac {x^{12}}{(a-x)^5 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {\text {Subst}\left (\int \left (a+\frac {a^6}{64 (a-x)^5}-\frac {9 a^5}{64 (a-x)^4}+\frac {141 a^4}{256 (a-x)^3}-\frac {39 a^3}{32 (a-x)^2}+\frac {843 a^2}{512 (a-x)}-x+\frac {a^7}{32 (a+x)^6}-\frac {19 a^6}{64 (a+x)^5}+\frac {159 a^5}{128 (a+x)^4}-\frac {765 a^4}{256 (a+x)^3}+\frac {1155 a^3}{256 (a+x)^2}-\frac {2229 a^2}{512 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = -\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))} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

Integrate[(Sin[c + d*x]^3*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]

[Out]

-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)

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 )}\)	\(560\)

[In]

int(sec(d*x+c)^9*sin(d*x+c)^12/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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+39/3
2/(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)))

Fricas [A] (verification not implemented)

none

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

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^12/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-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)

Sympy [F(-1)]

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

[In]

integrate(sec(d*x+c)**9*sin(d*x+c)**12/(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

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

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^12/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-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*si
n(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*s
in(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

Giac [A] (verification not implemented)

none

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

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^12/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-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

Mupad [B] (verification not implemented)

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

[In]

int(sin(c + d*x)^12/(cos(c + d*x)^9*(a + a*sin(c + d*x))),x)

[Out]

((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 -
(1823*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*tan(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)^1
0 + 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*tan(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)