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
Time = 2.48 (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)
Time = 1.04 (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
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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])
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]
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]))
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]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 3.11 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.87
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {11}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {47}{384 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {93}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {35}{256 \left (1+\sin \left (d x +c \right )\right )}+\frac {7 \ln \left (1+\sin \left (d x +c \right )\right )}{512}+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {5}{192 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {37}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {7}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {7 \ln \left (\sin \left (d x +c \right )-1\right )}{512}}{d a}\) | \(139\) |
| default | \(\frac {-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {11}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {47}{384 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {93}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {35}{256 \left (1+\sin \left (d x +c \right )\right )}+\frac {7 \ln \left (1+\sin \left (d x +c \right )\right )}{512}+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {5}{192 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {37}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {7}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {7 \ln \left (\sin \left (d x +c \right )-1\right )}{512}}{d a}\) | \(139\) |
| risch | \(-\frac {i \left (-23674 i {\mathrm e}^{8 i \left (d x +c \right )}+105 \,{\mathrm e}^{17 i \left (d x +c \right )}+2890 i {\mathrm e}^{14 i \left (d x +c \right )}+3260 \,{\mathrm e}^{15 i \left (d x +c \right )}+25102 i {\mathrm e}^{6 i \left (d x +c \right )}+9044 \,{\mathrm e}^{13 i \left (d x +c \right )}-25102 i {\mathrm e}^{12 i \left (d x +c \right )}+24388 \,{\mathrm e}^{11 i \left (d x +c \right )}-3630 i {\mathrm e}^{16 i \left (d x +c \right )}+24710 \,{\mathrm e}^{9 i \left (d x +c \right )}+3630 i {\mathrm e}^{2 i \left (d x +c \right )}+24388 \,{\mathrm e}^{7 i \left (d x +c \right )}+23674 i {\mathrm e}^{10 i \left (d x +c \right )}+9044 \,{\mathrm e}^{5 i \left (d x +c \right )}-2890 i {\mathrm e}^{4 i \left (d x +c \right )}+3260 \,{\mathrm e}^{3 i \left (d x +c \right )}+105 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{1920 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} d a}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 a d}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 a d}\) | \(277\) |
Input:
int(sec(d*x+c)*tan(d*x+c)^8/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/160/(1+sin(d*x+c))^5+11/256/(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))+ 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))
Time = 0.12 (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)*tan(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)
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)*tan(d*x+c)**8/(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (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)*tan(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
Time = 0.19 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.91 \[ \int \frac {\sec (c+d x) \tan ^8(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {7 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{512 \, a d} - \frac {7 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{512 \, a d} - \frac {105 \, \sin \left (d x + c\right )^{8} - 1815 \, \sin \left (d x + c\right )^{7} - 1025 \, \sin \left (d x + c\right )^{6} + 2815 \, \sin \left (d x + c\right )^{5} + 1919 \, \sin \left (d x + c\right )^{4} - 1921 \, \sin \left (d x + c\right )^{3} - 1431 \, \sin \left (d x + c\right )^{2} + 489 \, \sin \left (d x + c\right ) + 384}{3840 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{5} {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} \] Input:
integrate(sec(d*x+c)*tan(d*x+c)^8/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
7/512*log(abs(sin(d*x + c) + 1))/(a*d) - 7/512*log(abs(sin(d*x + c) - 1))/ (a*d) - 1/3840*(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 - 1 431*sin(d*x + c)^2 + 489*sin(d*x + c) + 384)/(a*d*(sin(d*x + c) + 1)^5*(si n(d*x + c) - 1)^4)
Time = 36.88 (sec) , antiderivative size = 496, normalized size of antiderivative = 3.10 \[ \int \frac {\sec (c+d x) \tan ^8(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(tan(c + d*x)^8/(cos(c + d*x)*(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))
Time = 0.21 (sec) , antiderivative size = 596, normalized size of antiderivative = 3.72 \[ \int \frac {\sec (c+d x) \tan ^8(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)*tan(d*x+c)^8/(a+a*sin(d*x+c)),x)
Output:
( - 105*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**9 - 105*log(tan((c + d*x)/ 2) - 1)*sin(c + d*x)**8 + 420*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7 + 420*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6 - 630*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 - 630*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 420* log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 + 420*log(tan((c + d*x)/2) - 1)* sin(c + d*x)**2 - 105*log(tan((c + d*x)/2) - 1)*sin(c + d*x) - 105*log(tan ((c + d*x)/2) - 1) + 105*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**9 + 105*l og(tan((c + d*x)/2) + 1)*sin(c + d*x)**8 - 420*log(tan((c + d*x)/2) + 1)*s in(c + d*x)**7 - 420*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6 + 630*log(t an((c + d*x)/2) + 1)*sin(c + d*x)**5 + 630*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 420*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 - 420*log(tan(( c + d*x)/2) + 1)*sin(c + d*x)**2 + 105*log(tan((c + d*x)/2) + 1)*sin(c + d *x) + 105*log(tan((c + d*x)/2) + 1) + 489*sin(c + d*x)**9 + 384*sin(c + d* x)**8 - 141*sin(c + d*x)**7 - 931*sin(c + d*x)**6 + 119*sin(c + d*x)**5 + 1015*sin(c + d*x)**4 - 35*sin(c + d*x)**3 - 525*sin(c + d*x)**2 + 105)/(38 40*a*d*(sin(c + d*x)**9 + sin(c + d*x)**8 - 4*sin(c + d*x)**7 - 4*sin(c + d*x)**6 + 6*sin(c + d*x)**5 + 6*sin(c + d*x)**4 - 4*sin(c + d*x)**3 - 4*si n(c + d*x)**2 + sin(c + d*x) + 1))