Integrand size = 21, antiderivative size = 130 \[ \int \frac {\tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {35 \text {arctanh}(\sin (c+d x))}{128 a d}+\frac {35 \sec (c+d x) \tan (c+d x)}{128 a d}-\frac {35 \sec (c+d x) \tan ^3(c+d x)}{192 a d}+\frac {7 \sec (c+d x) \tan ^5(c+d x)}{48 a d}-\frac {\sec (c+d x) \tan ^7(c+d x)}{8 a d}+\frac {\tan ^8(c+d x)}{8 a d} \] Output:
-35/128*arctanh(sin(d*x+c))/a/d+35/128*sec(d*x+c)*tan(d*x+c)/a/d-35/192*se c(d*x+c)*tan(d*x+c)^3/a/d+7/48*sec(d*x+c)*tan(d*x+c)^5/a/d-1/8*sec(d*x+c)* tan(d*x+c)^7/a/d+1/8*tan(d*x+c)^8/a/d
Time = 1.12 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.78 \[ \int \frac {\tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {105 \text {arctanh}(\sin (c+d x))+\frac {-48+57 \sin (c+d x)+249 \sin ^2(c+d x)-136 \sin ^3(c+d x)-424 \sin ^4(c+d x)+87 \sin ^5(c+d x)+279 \sin ^6(c+d x)}{(-1+\sin (c+d x))^3 (1+\sin (c+d x))^4}}{384 a d} \] Input:
Integrate[Tan[c + d*x]^7/(a + a*Sin[c + d*x]),x]
Output:
-1/384*(105*ArcTanh[Sin[c + d*x]] + (-48 + 57*Sin[c + d*x] + 249*Sin[c + d *x]^2 - 136*Sin[c + d*x]^3 - 424*Sin[c + d*x]^4 + 87*Sin[c + d*x]^5 + 279* Sin[c + d*x]^6)/((-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^4))/(a*d)
Time = 0.78 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3185, 3042, 3087, 15, 3091, 3042, 3091, 3042, 3091, 3042, 3091, 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.
| \(\displaystyle \int \frac {\tan ^7(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^7}{a \sin (c+d x)+a}dx\) |
| \(\Big \downarrow \) 3185 |
| \(\displaystyle \frac {\int \sec ^2(c+d x) \tan ^7(c+d x)dx}{a}-\frac {\int \sec (c+d x) \tan ^8(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sec (c+d x)^2 \tan (c+d x)^7dx}{a}-\frac {\int \sec (c+d x) \tan (c+d x)^8dx}{a}\) |
| \(\Big \downarrow \) 3087 |
| \(\displaystyle \frac {\int \tan ^7(c+d x)d\tan (c+d x)}{a d}-\frac {\int \sec (c+d x) \tan (c+d x)^8dx}{a}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {\tan ^8(c+d x)}{8 a d}-\frac {\int \sec (c+d x) \tan (c+d x)^8dx}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^8(c+d x)}{8 a d}-\frac {\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \int \sec (c+d x) \tan ^6(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^8(c+d x)}{8 a d}-\frac {\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \int \sec (c+d x) \tan (c+d x)^6dx}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^8(c+d x)}{8 a d}-\frac {\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \int \sec (c+d x) \tan ^4(c+d x)dx\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^8(c+d x)}{8 a d}-\frac {\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \int \sec (c+d x) \tan (c+d x)^4dx\right )}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^8(c+d x)}{8 a d}-\frac {\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan ^2(c+d x)dx\right )\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^8(c+d x)}{8 a d}-\frac {\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan (c+d x)^2dx\right )\right )}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^8(c+d x)}{8 a d}-\frac {\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \sec (c+d x)dx\right )\right )\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^8(c+d x)}{8 a d}-\frac {\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )\right )}{a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\tan ^8(c+d x)}{8 a d}-\frac {\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {\text {arctanh}(\sin (c+d x))}{2 d}\right )\right )\right )}{a}\) |
Input:
Int[Tan[c + d*x]^7/(a + a*Sin[c + d*x]),x]
Output:
Tan[c + d*x]^8/(8*a*d) - ((Sec[c + d*x]*Tan[c + d*x]^7)/(8*d) - (7*((Sec[c + d*x]*Tan[c + d*x]^5)/(6*d) - (5*((Sec[c + d*x]*Tan[c + d*x]^3)/(4*d) - (3*(-1/2*ArcTanh[Sin[c + d*x]]/d + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4)) /6))/8)/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[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[1/a Int[Sec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x ] - Simp[1/(b*g) Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /; Fre eQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.37 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {9}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {29}{128 \left (\sin \left (d x +c \right )-1\right )}+\frac {35 \ln \left (\sin \left (d x +c \right )-1\right )}{256}+\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{48 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {19}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )}-\frac {35 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d a}\) | \(115\) |
| default | \(\frac {-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {9}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {29}{128 \left (\sin \left (d x +c \right )-1\right )}+\frac {35 \ln \left (\sin \left (d x +c \right )-1\right )}{256}+\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{48 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {19}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )}-\frac {35 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d a}\) | \(115\) |
| risch | \(-\frac {i \left (279 \,{\mathrm e}^{i \left (d x +c \right )}+279 \,{\mathrm e}^{13 i \left (d x +c \right )}+22 \,{\mathrm e}^{11 i \left (d x +c \right )}+1385 \,{\mathrm e}^{9 i \left (d x +c \right )}+1385 \,{\mathrm e}^{5 i \left (d x +c \right )}+22 \,{\mathrm e}^{3 i \left (d x +c \right )}+218 i {\mathrm e}^{10 i \left (d x +c \right )}-300 \,{\mathrm e}^{7 i \left (d x +c \right )}+174 i {\mathrm e}^{12 i \left (d x +c \right )}-174 i {\mathrm e}^{2 i \left (d x +c \right )}+300 i {\mathrm e}^{8 i \left (d x +c \right )}-300 i {\mathrm e}^{6 i \left (d x +c \right )}-218 i {\mathrm e}^{4 i \left (d x +c \right )}\right )}{192 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d a}+\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 a d}-\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 a d}\) | \(231\) |
Input:
int(tan(d*x+c)^7/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/96/(sin(d*x+c)-1)^3-9/128/(sin(d*x+c)-1)^2-29/128/(sin(d*x+c)-1) +35/256*ln(sin(d*x+c)-1)+1/64/(1+sin(d*x+c))^4-5/48/(1+sin(d*x+c))^3+19/64 /(1+sin(d*x+c))^2-1/2/(1+sin(d*x+c))-35/256*ln(1+sin(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.28 \[ \int \frac {\tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {558 \, \cos \left (d x + c\right )^{6} - 826 \, \cos \left (d x + c\right )^{4} + 476 \, \cos \left (d x + c\right )^{2} + 105 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (87 \, \cos \left (d x + c\right )^{4} - 38 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) - 112}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \] Input:
integrate(tan(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/768*(558*cos(d*x + c)^6 - 826*cos(d*x + c)^4 + 476*cos(d*x + c)^2 + 105 *(cos(d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(sin(d*x + c) + 1) - 10 5*(cos(d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(-sin(d*x + c) + 1) - 2*(87*cos(d*x + c)^4 - 38*cos(d*x + c)^2 + 8)*sin(d*x + c) - 112)/(a*d*cos (d*x + c)^6*sin(d*x + c) + a*d*cos(d*x + c)^6)
\[ \int \frac {\tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\tan ^{7}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(tan(d*x+c)**7/(a+a*sin(d*x+c)),x)
Output:
Integral(tan(c + d*x)**7/(sin(c + d*x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.35 \[ \int \frac {\tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (279 \, \sin \left (d x + c\right )^{6} + 87 \, \sin \left (d x + c\right )^{5} - 424 \, \sin \left (d x + c\right )^{4} - 136 \, \sin \left (d x + c\right )^{3} + 249 \, \sin \left (d x + c\right )^{2} + 57 \, \sin \left (d x + c\right ) - 48\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} + \frac {105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \] Input:
integrate(tan(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/768*(2*(279*sin(d*x + c)^6 + 87*sin(d*x + c)^5 - 424*sin(d*x + c)^4 - 1 36*sin(d*x + c)^3 + 249*sin(d*x + c)^2 + 57*sin(d*x + c) - 48)/(a*sin(d*x + c)^7 + a*sin(d*x + c)^6 - 3*a*sin(d*x + c)^5 - 3*a*sin(d*x + c)^4 + 3*a* sin(d*x + c)^3 + 3*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 = 0.17 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.96 \[ \int \frac {\tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {35 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{256 \, a d} + \frac {35 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{256 \, a d} - \frac {279 \, \sin \left (d x + c\right )^{6} + 87 \, \sin \left (d x + c\right )^{5} - 424 \, \sin \left (d x + c\right )^{4} - 136 \, \sin \left (d x + c\right )^{3} + 249 \, \sin \left (d x + c\right )^{2} + 57 \, \sin \left (d x + c\right ) - 48}{384 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{4} {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} \] Input:
integrate(tan(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-35/256*log(abs(sin(d*x + c) + 1))/(a*d) + 35/256*log(abs(sin(d*x + c) - 1 ))/(a*d) - 1/384*(279*sin(d*x + c)^6 + 87*sin(d*x + c)^5 - 424*sin(d*x + c )^4 - 136*sin(d*x + c)^3 + 249*sin(d*x + c)^2 + 57*sin(d*x + c) - 48)/(a*d *(sin(d*x + c) + 1)^4*(sin(d*x + c) - 1)^3)
Time = 20.81 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.98 \[ \int \frac {\tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{32}-\frac {245\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{96}-\frac {595\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{96}+\frac {791\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {231\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{16}-\frac {25\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{16}+\frac {231\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{16}+\frac {791\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{192}-\frac {595\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{96}-\frac {245\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32}+\frac {35\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+30\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-40\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+30\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-5\,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 {35\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a\,d} \] Input:
int(tan(c + d*x)^7/(a + a*sin(c + d*x)),x)
Output:
((35*tan(c/2 + (d*x)/2))/64 + (35*tan(c/2 + (d*x)/2)^2)/32 - (245*tan(c/2 + (d*x)/2)^3)/96 - (595*tan(c/2 + (d*x)/2)^4)/96 + (791*tan(c/2 + (d*x)/2) ^5)/192 + (231*tan(c/2 + (d*x)/2)^6)/16 - (25*tan(c/2 + (d*x)/2)^7)/16 + ( 231*tan(c/2 + (d*x)/2)^8)/16 + (791*tan(c/2 + (d*x)/2)^9)/192 - (595*tan(c /2 + (d*x)/2)^10)/96 - (245*tan(c/2 + (d*x)/2)^11)/96 + (35*tan(c/2 + (d*x )/2)^12)/32 + (35*tan(c/2 + (d*x)/2)^13)/64)/(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 + 9*a*tan(c/2 + ( d*x)/2)^4 + 30*a*tan(c/2 + (d*x)/2)^5 - 5*a*tan(c/2 + (d*x)/2)^6 - 40*a*ta n(c/2 + (d*x)/2)^7 - 5*a*tan(c/2 + (d*x)/2)^8 + 30*a*tan(c/2 + (d*x)/2)^9 + 9*a*tan(c/2 + (d*x)/2)^10 - 12*a*tan(c/2 + (d*x)/2)^11 - 5*a*tan(c/2 + ( d*x)/2)^12 + 2*a*tan(c/2 + (d*x)/2)^13 + a*tan(c/2 + (d*x)/2)^14)) - (35*a tanh(tan(c/2 + (d*x)/2)))/(64*a*d)
Time = 0.19 (sec) , antiderivative size = 470, normalized size of antiderivative = 3.62 \[ \int \frac {\tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(tan(d*x+c)^7/(a+a*sin(d*x+c)),x)
Output:
(105*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7 + 105*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6 - 315*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 - 315 *log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 315*log(tan((c + d*x)/2) - 1) *sin(c + d*x)**3 + 315*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)**7 - 105*log(tan((c + d*x)/2) + 1)* sin(c + d*x)**6 + 315*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5 + 315*log( tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 315*log(tan((c + d*x)/2) + 1)*sin( c + d*x)**3 - 315*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) - 57*sin(c + d*x)**7 - 336*sin(c + d*x)**6 + 84*sin(c + d*x)**5 + 595*sin(c + d*x)**4 - 35*sin(c + d*x)**3 - 420*sin(c + d*x)**2 + 105)/(384*a*d*(sin(c + d*x)* *7 + sin(c + d*x)**6 - 3*sin(c + d*x)**5 - 3*sin(c + d*x)**4 + 3*sin(c + d *x)**3 + 3*sin(c + d*x)**2 - sin(c + d*x) - 1))