Integrand size = 29, antiderivative size = 280 \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {843 \log (1-\sin (c+d x))}{512 a d}-\frac {2229 \log (1+\sin (c+d x))}{512 a d}+\frac {\sin (c+d x)}{a d}-\frac {\sin ^2(c+d x)}{2 a d}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}-\frac {3 a^2}{64 d (a-a \sin (c+d x))^3}-\frac {39}{32 d (a-a \sin (c+d x))}+\frac {19 a^3}{256 d (a+a \sin (c+d x))^4}-\frac {53 a^2}{128 d (a+a \sin (c+d x))^3}-\frac {1155}{256 d (a+a \sin (c+d x))}-\frac {a^9}{160 d \left (a^2+a^2 \sin (c+d x)\right )^5}+\frac {141 a^9}{512 d \left (a^5-a^5 \sin (c+d x)\right )^2}+\frac {765 a^9}{512 d \left (a^5+a^5 \sin (c+d x)\right )^2} \] 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))^3-39/32/d/(a-a*sin(d*x+c))+19/256*a^3/d/(a+a*sin(d*x+c))^4-53/128*a ^2/d/(a+a*sin(d*x+c))^3-1155/256/d/(a+a*sin(d*x+c))-1/160*a^9/d/(a^2+a^2*s in(d*x+c))^5+141/512*a^9/d/(a^5-a^5*sin(d*x+c))^2+765/512*a^9/d/(a^5+a^5*s in(d*x+c))^2
Time = 6.18 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.60 \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {4215 \log (1-\sin (c+d x))+11145 \log (1+\sin (c+d x))-\frac {10}{(1-\sin (c+d x))^4}+\frac {120}{(1-\sin (c+d x))^3}-\frac {705}{(1-\sin (c+d x))^2}+\frac {3120}{1-\sin (c+d x)}-2560 \sin (c+d x)+1280 \sin ^2(c+d x)+\frac {16}{(1+\sin (c+d x))^5}-\frac {190}{(1+\sin (c+d x))^4}+\frac {1060}{(1+\sin (c+d x))^3}-\frac {3825}{(1+\sin (c+d x))^2}+\frac {11550}{1+\sin (c+d x)}}{2560 a d} \] 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)
Time = 0.51 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.88, 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)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]))
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]
Time = 9.74 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.55
| method | result | size |
| derivativedivides | \(\frac {-\frac {\sin \left (d x +c \right )^{2}}{2}+\sin \left (d x +c \right )+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {3}{64 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {141}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {39}{32 \left (\sin \left (d x +c \right )-1\right )}-\frac {843 \ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {19}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {53}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {765}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1155}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {2229 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) | \(155\) |
| default | \(\frac {-\frac {\sin \left (d x +c \right )^{2}}{2}+\sin \left (d x +c \right )+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {3}{64 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {141}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {39}{32 \left (\sin \left (d x +c \right )-1\right )}-\frac {843 \ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {19}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {53}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {765}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1155}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {2229 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) | \(155\) |
| risch | \(\frac {6 i x}{a}+\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 a d}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 d a}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d a}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a d}+\frac {12 i c}{a d}-\frac {i \left (20922 i {\mathrm e}^{8 i \left (d x +c \right )}+4215 \,{\mathrm e}^{17 i \left (d x +c \right )}-31370 i {\mathrm e}^{14 i \left (d x +c \right )}+40900 \,{\mathrm e}^{15 i \left (d x +c \right )}+59954 i {\mathrm e}^{6 i \left (d x +c \right )}+152108 \,{\mathrm e}^{13 i \left (d x +c \right )}-59954 i {\mathrm e}^{12 i \left (d x +c \right )}+316476 \,{\mathrm e}^{11 i \left (d x +c \right )}-10770 i {\mathrm e}^{16 i \left (d x +c \right )}+398010 \,{\mathrm e}^{9 i \left (d x +c \right )}+10770 i {\mathrm e}^{2 i \left (d x +c \right )}+316476 \,{\mathrm e}^{7 i \left (d x +c \right )}-20922 i {\mathrm e}^{10 i \left (d x +c \right )}+152108 \,{\mathrm e}^{5 i \left (d x +c \right )}+31370 i {\mathrm e}^{4 i \left (d x +c \right )}+40900 \,{\mathrm e}^{3 i \left (d x +c \right )}+4215 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{640 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} d a}-\frac {843 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 a d}-\frac {2229 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 a d}\) | \(364\) |
Input:
int(sin(d*x+c)^3*tan(d*x+c)^9/(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) ))
Time = 0.13 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {1280 \, \cos \left (d x + c\right )^{10} + 6510 \, \cos \left (d x + c\right )^{8} + 3590 \, \cos \left (d x + c\right )^{6} - 1124 \, \cos \left (d x + c\right )^{4} + 272 \, \cos \left (d x + c\right )^{2} + 11145 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 4215 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (640 \, \cos \left (d x + c\right )^{10} + 960 \, \cos \left (d x + c\right )^{8} - 5385 \, \cos \left (d x + c\right )^{6} + 2810 \, \cos \left (d x + c\right )^{4} - 952 \, \cos \left (d x + c\right )^{2} + 144\right )} \sin \left (d x + c\right ) - 32}{2560 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \] Input:
integrate(sin(d*x+c)^3*tan(d*x+c)^9/(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)
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(sin(d*x+c)**3*tan(d*x+c)**9/(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.84 \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (4215 \, \sin \left (d x + c\right )^{8} - 5385 \, \sin \left (d x + c\right )^{7} - 18655 \, \sin \left (d x + c\right )^{6} + 13345 \, \sin \left (d x + c\right )^{5} + 30113 \, \sin \left (d x + c\right )^{4} - 11487 \, \sin \left (d x + c\right )^{3} - 21257 \, \sin \left (d x + c\right )^{2} + 3383 \, \sin \left (d x + c\right ) + 5568\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} + \frac {1280 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )}}{a} + \frac {11145 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {4215 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{2560 \, d} \] Input:
integrate(sin(d*x+c)^3*tan(d*x+c)^9/(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
Time = 0.20 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.62 \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2229 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{512 \, a d} - \frac {843 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{512 \, a d} - \frac {a d \sin \left (d x + c\right )^{2} - 2 \, a d \sin \left (d x + c\right )}{2 \, a^{2} d^{2}} - \frac {4215 \, \sin \left (d x + c\right )^{8} - 5385 \, \sin \left (d x + c\right )^{7} - 18655 \, \sin \left (d x + c\right )^{6} + 13345 \, \sin \left (d x + c\right )^{5} + 30113 \, \sin \left (d x + c\right )^{4} - 11487 \, \sin \left (d x + c\right )^{3} - 21257 \, \sin \left (d x + c\right )^{2} + 3383 \, \sin \left (d x + c\right ) + 5568}{1280 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{5} {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} \] Input:
integrate(sin(d*x+c)^3*tan(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-2229/512*log(abs(sin(d*x + c) + 1))/(a*d) - 843/512*log(abs(sin(d*x + c) - 1))/(a*d) - 1/2*(a*d*sin(d*x + c)^2 - 2*a*d*sin(d*x + c))/(a^2*d^2) - 1/ 1280*(4215*sin(d*x + c)^8 - 5385*sin(d*x + c)^7 - 18655*sin(d*x + c)^6 + 1 3345*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*d*(sin(d*x + c) + 1)^5*(sin( d*x + c) - 1)^4)
Time = 34.84 (sec) , antiderivative size = 648, normalized size of antiderivative = 2.31 \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
int((sin(c + d*x)^3*tan(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)
Time = 0.19 (sec) , antiderivative size = 3758, normalized size of antiderivative = 13.42 \[ \int \frac {\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sin(d*x+c)^3*tan(d*x+c)^9/(a+a*sin(d*x+c)),x)
Output:
(160*cos(c + d*x)**2*sin(c + d*x)**9*tan(c + d*x)**6 - 1056*cos(c + d*x)** 2*sin(c + d*x)**9*tan(c + d*x)**4 + 7776*cos(c + d*x)**2*sin(c + d*x)**9*t an(c + d*x)**2 - 7776*cos(c + d*x)**2*sin(c + d*x)**9 + 160*cos(c + d*x)** 2*sin(c + d*x)**8*tan(c + d*x)**6 - 1056*cos(c + d*x)**2*sin(c + d*x)**8*t an(c + d*x)**4 + 7776*cos(c + d*x)**2*sin(c + d*x)**8*tan(c + d*x)**2 - 77 76*cos(c + d*x)**2*sin(c + d*x)**8 - 640*cos(c + d*x)**2*sin(c + d*x)**7*t an(c + d*x)**6 + 4224*cos(c + d*x)**2*sin(c + d*x)**7*tan(c + d*x)**4 - 31 104*cos(c + d*x)**2*sin(c + d*x)**7*tan(c + d*x)**2 + 31104*cos(c + d*x)** 2*sin(c + d*x)**7 - 640*cos(c + d*x)**2*sin(c + d*x)**6*tan(c + d*x)**6 + 4224*cos(c + d*x)**2*sin(c + d*x)**6*tan(c + d*x)**4 - 31104*cos(c + d*x)* *2*sin(c + d*x)**6*tan(c + d*x)**2 + 31104*cos(c + d*x)**2*sin(c + d*x)**6 + 960*cos(c + d*x)**2*sin(c + d*x)**5*tan(c + d*x)**6 - 6336*cos(c + d*x) **2*sin(c + d*x)**5*tan(c + d*x)**4 + 46656*cos(c + d*x)**2*sin(c + d*x)** 5*tan(c + d*x)**2 - 46656*cos(c + d*x)**2*sin(c + d*x)**5 + 960*cos(c + d* x)**2*sin(c + d*x)**4*tan(c + d*x)**6 - 6336*cos(c + d*x)**2*sin(c + d*x)* *4*tan(c + d*x)**4 + 46656*cos(c + d*x)**2*sin(c + d*x)**4*tan(c + d*x)**2 - 46656*cos(c + d*x)**2*sin(c + d*x)**4 - 640*cos(c + d*x)**2*sin(c + d*x )**3*tan(c + d*x)**6 + 4224*cos(c + d*x)**2*sin(c + d*x)**3*tan(c + d*x)** 4 - 31104*cos(c + d*x)**2*sin(c + d*x)**3*tan(c + d*x)**2 + 31104*cos(c + d*x)**2*sin(c + d*x)**3 - 640*cos(c + d*x)**2*sin(c + d*x)**2*tan(c + d...