Integrand size = 29, antiderivative size = 176 \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \text {arctanh}(\sin (c+d x))}{256 a d}-\frac {3 \sec (c+d x) \tan (c+d x)}{256 a d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{128 a d}+\frac {\sec ^5(c+d x) \tan (c+d x)}{32 a d}-\frac {\sec ^5(c+d x) \tan ^3(c+d x)}{16 a d}+\frac {\sec ^5(c+d x) \tan ^5(c+d x)}{10 a d}-\frac {\tan ^8(c+d x)}{8 a d}-\frac {\tan ^{10}(c+d x)}{10 a d} \] Output:
-3/256*arctanh(sin(d*x+c))/a/d-3/256*sec(d*x+c)*tan(d*x+c)/a/d-1/128*sec(d *x+c)^3*tan(d*x+c)/a/d+1/32*sec(d*x+c)^5*tan(d*x+c)/a/d-1/16*sec(d*x+c)^5* tan(d*x+c)^3/a/d+1/10*sec(d*x+c)^5*tan(d*x+c)^5/a/d-1/8*tan(d*x+c)^8/a/d-1 /10*tan(d*x+c)^10/a/d
Time = 2.72 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.69 \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {30 \text {arctanh}(\sin (c+d x))-\frac {2 \left (32+47 \sin (c+d x)-113 \sin ^2(c+d x)-183 \sin ^3(c+d x)+137 \sin ^4(c+d x)+265 \sin ^5(c+d x)-55 \sin ^6(c+d x)+15 \sin ^7(c+d x)+15 \sin ^8(c+d x)\right )}{(-1+\sin (c+d x))^4 (1+\sin (c+d x))^5}}{2560 a d} \] Input:
Integrate[(Sec[c + d*x]^3*Tan[c + d*x]^6)/(a + a*Sin[c + d*x]),x]
Output:
-1/2560*(30*ArcTanh[Sin[c + d*x]] - (2*(32 + 47*Sin[c + d*x] - 113*Sin[c + d*x]^2 - 183*Sin[c + d*x]^3 + 137*Sin[c + d*x]^4 + 265*Sin[c + d*x]^5 - 5 5*Sin[c + d*x]^6 + 15*Sin[c + d*x]^7 + 15*Sin[c + d*x]^8))/((-1 + Sin[c + d*x])^4*(1 + Sin[c + d*x])^5))/(a*d)
Time = 1.10 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.02, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.586, Rules used = {3042, 3314, 3042, 3087, 244, 2009, 3091, 3042, 3091, 3042, 3091, 3042, 4255, 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.
| \(\displaystyle \int \frac {\tan ^6(c+d x) \sec ^3(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^6}{\cos (c+d x)^9 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3314 |
| \(\displaystyle \frac {\int \sec ^5(c+d x) \tan ^6(c+d x)dx}{a}-\frac {\int \sec ^4(c+d x) \tan ^7(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sec (c+d x)^5 \tan (c+d x)^6dx}{a}-\frac {\int \sec (c+d x)^4 \tan (c+d x)^7dx}{a}\) |
| \(\Big \downarrow \) 3087 |
| \(\displaystyle \frac {\int \sec (c+d x)^5 \tan (c+d x)^6dx}{a}-\frac {\int \tan ^7(c+d x) \left (\tan ^2(c+d x)+1\right )d\tan (c+d x)}{a d}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle \frac {\int \sec (c+d x)^5 \tan (c+d x)^6dx}{a}-\frac {\int \left (\tan ^9(c+d x)+\tan ^7(c+d x)\right )d\tan (c+d x)}{a d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\int \sec (c+d x)^5 \tan (c+d x)^6dx}{a}-\frac {\frac {1}{10} \tan ^{10}(c+d x)+\frac {1}{8} \tan ^8(c+d x)}{a d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}-\frac {1}{2} \int \sec ^5(c+d x) \tan ^4(c+d x)dx}{a}-\frac {\frac {1}{10} \tan ^{10}(c+d x)+\frac {1}{8} \tan ^8(c+d x)}{a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}-\frac {1}{2} \int \sec (c+d x)^5 \tan (c+d x)^4dx}{a}-\frac {\frac {1}{10} \tan ^{10}(c+d x)+\frac {1}{8} \tan ^8(c+d x)}{a d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {3}{8} \int \sec ^5(c+d x) \tan ^2(c+d x)dx-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}-\frac {\frac {1}{10} \tan ^{10}(c+d x)+\frac {1}{8} \tan ^8(c+d x)}{a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {3}{8} \int \sec (c+d x)^5 \tan (c+d x)^2dx-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}-\frac {\frac {1}{10} \tan ^{10}(c+d x)+\frac {1}{8} \tan ^8(c+d x)}{a d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}-\frac {1}{6} \int \sec ^5(c+d x)dx\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}-\frac {\frac {1}{10} \tan ^{10}(c+d x)+\frac {1}{8} \tan ^8(c+d x)}{a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}-\frac {1}{6} \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}-\frac {\frac {1}{10} \tan ^{10}(c+d x)+\frac {1}{8} \tan ^8(c+d x)}{a d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \int \sec ^3(c+d x)dx-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}-\frac {\frac {1}{10} \tan ^{10}(c+d x)+\frac {1}{8} \tan ^8(c+d x)}{a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}-\frac {\frac {1}{10} \tan ^{10}(c+d x)+\frac {1}{8} \tan ^8(c+d x)}{a d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}-\frac {\frac {1}{10} \tan ^{10}(c+d x)+\frac {1}{8} \tan ^8(c+d x)}{a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}-\frac {\frac {1}{10} \tan ^{10}(c+d x)+\frac {1}{8} \tan ^8(c+d x)}{a d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}-\frac {\frac {1}{10} \tan ^{10}(c+d x)+\frac {1}{8} \tan ^8(c+d x)}{a d}\) |
Input:
Int[(Sec[c + d*x]^3*Tan[c + d*x]^6)/(a + a*Sin[c + d*x]),x]
Output:
-((Tan[c + d*x]^8/8 + Tan[c + d*x]^10/10)/(a*d)) + ((Sec[c + d*x]^5*Tan[c + d*x]^5)/(10*d) + (-1/8*(Sec[c + d*x]^5*Tan[c + d*x]^3)/d + (3*((Sec[c + d*x]^5*Tan[c + d*x])/(6*d) + (-1/4*(Sec[c + d*x]^3*Tan[c + d*x])/d - (3*(A rcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4)/6))/8) /2)/a
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
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 = 15.49 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.79
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {7}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {5}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{512}+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {9}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {1}{128 \left (\sin \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{512}}{d a}\) | \(139\) |
| default | \(\frac {-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {7}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {5}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{512}+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {9}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {1}{128 \left (\sin \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{512}}{d a}\) | \(139\) |
| risch | \(\frac {i \left (10698 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{17 i \left (d x +c \right )}-2330 i {\mathrm e}^{14 i \left (d x +c \right )}+100 \,{\mathrm e}^{15 i \left (d x +c \right )}-5374 i {\mathrm e}^{6 i \left (d x +c \right )}+1292 \,{\mathrm e}^{13 i \left (d x +c \right )}+5374 i {\mathrm e}^{12 i \left (d x +c \right )}+924 \,{\mathrm e}^{11 i \left (d x +c \right )}+30 i {\mathrm e}^{16 i \left (d x +c \right )}+3530 \,{\mathrm e}^{9 i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}+924 \,{\mathrm e}^{7 i \left (d x +c \right )}-10698 i {\mathrm e}^{10 i \left (d x +c \right )}+1292 \,{\mathrm e}^{5 i \left (d x +c \right )}+2330 i {\mathrm e}^{4 i \left (d x +c \right )}+100 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{640 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} d a}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 a d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 a d}\) | \(277\) |
Input:
int(sec(d*x+c)^3*tan(d*x+c)^6/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/160/(1+sin(d*x+c))^5+7/256/(1+sin(d*x+c))^4-5/128/(1+sin(d*x+c)) ^3+5/512/(1+sin(d*x+c))^2+5/256/(1+sin(d*x+c))-3/512*ln(1+sin(d*x+c))+1/25 6/(sin(d*x+c)-1)^4+1/64/(sin(d*x+c)-1)^3+9/512/(sin(d*x+c)-1)^2-1/128/(sin (d*x+c)-1)+3/512*ln(sin(d*x+c)-1))
Time = 0.10 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {30 \, \cos \left (d x + c\right )^{8} - 10 \, \cos \left (d x + c\right )^{6} + 124 \, \cos \left (d x + c\right )^{4} - 112 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, \cos \left (d x + c\right )^{6} - 310 \, \cos \left (d x + c\right )^{4} + 392 \, \cos \left (d x + c\right )^{2} - 144\right )} \sin \left (d x + c\right ) + 32}{2560 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \] Input:
integrate(sec(d*x+c)^3*tan(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/2560*(30*cos(d*x + c)^8 - 10*cos(d*x + c)^6 + 124*cos(d*x + c)^4 - 112*c os(d*x + c)^2 - 15*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(sin( d*x + c) + 1) + 15*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(-sin (d*x + c) + 1) - 2*(15*cos(d*x + c)^6 - 310*cos(d*x + c)^4 + 392*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. \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**3*tan(d*x+c)**6/(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.22 \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{8} + 15 \, \sin \left (d x + c\right )^{7} - 55 \, \sin \left (d x + c\right )^{6} + 265 \, \sin \left (d x + c\right )^{5} + 137 \, \sin \left (d x + c\right )^{4} - 183 \, \sin \left (d x + c\right )^{3} - 113 \, \sin \left (d x + c\right )^{2} + 47 \, \sin \left (d x + c\right ) + 32\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{2560 \, d} \] Input:
integrate(sec(d*x+c)^3*tan(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/2560*(2*(15*sin(d*x + c)^8 + 15*sin(d*x + c)^7 - 55*sin(d*x + c)^6 + 265 *sin(d*x + c)^5 + 137*sin(d*x + c)^4 - 183*sin(d*x + c)^3 - 113*sin(d*x + c)^2 + 47*sin(d*x + c) + 32)/(a*sin(d*x + c)^9 + a*sin(d*x + c)^8 - 4*a*si n(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) - 15*log (sin(d*x + c) + 1)/a + 15*log(sin(d*x + c) - 1)/a)/d
Time = 0.16 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.82 \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{512 \, a d} + \frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{512 \, a d} + \frac {15 \, \sin \left (d x + c\right )^{8} + 15 \, \sin \left (d x + c\right )^{7} - 55 \, \sin \left (d x + c\right )^{6} + 265 \, \sin \left (d x + c\right )^{5} + 137 \, \sin \left (d x + c\right )^{4} - 183 \, \sin \left (d x + c\right )^{3} - 113 \, \sin \left (d x + c\right )^{2} + 47 \, \sin \left (d x + c\right ) + 32}{1280 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{5} {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} \] Input:
integrate(sec(d*x+c)^3*tan(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-3/512*log(abs(sin(d*x + c) + 1))/(a*d) + 3/512*log(abs(sin(d*x + c) - 1)) /(a*d) + 1/1280*(15*sin(d*x + c)^8 + 15*sin(d*x + c)^7 - 55*sin(d*x + c)^6 + 265*sin(d*x + c)^5 + 137*sin(d*x + c)^4 - 183*sin(d*x + c)^3 - 113*sin( d*x + c)^2 + 47*sin(d*x + c) + 32)/(a*d*(sin(d*x + c) + 1)^5*(sin(d*x + c) - 1)^4)
Time = 37.16 (sec) , antiderivative size = 496, normalized size of antiderivative = 2.82 \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(tan(c + d*x)^6/(cos(c + d*x)^3*(a + a*sin(c + d*x))),x)
Output:
((3*tan(c/2 + (d*x)/2))/128 + (3*tan(c/2 + (d*x)/2)^2)/64 - (5*tan(c/2 + ( d*x)/2)^3)/32 - (23*tan(c/2 + (d*x)/2)^4)/64 + (67*tan(c/2 + (d*x)/2)^5)/1 60 + (383*tan(c/2 + (d*x)/2)^6)/320 + (2841*tan(c/2 + (d*x)/2)^7)/160 + (7 41*tan(c/2 + (d*x)/2)^8)/320 + (1377*tan(c/2 + (d*x)/2)^9)/64 + (741*tan(c /2 + (d*x)/2)^10)/320 + (2841*tan(c/2 + (d*x)/2)^11)/160 + (383*tan(c/2 + (d*x)/2)^12)/320 + (67*tan(c/2 + (d*x)/2)^13)/160 - (23*tan(c/2 + (d*x)/2) ^14)/64 - (5*tan(c/2 + (d*x)/2)^15)/32 + (3*tan(c/2 + (d*x)/2)^16)/64 + (3 *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)) - (3*atanh(tan(c/2 + (d*x)/2)))/(128*a*d)
Time = 0.19 (sec) , antiderivative size = 596, normalized size of antiderivative = 3.39 \[ \int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^3*tan(d*x+c)^6/(a+a*sin(d*x+c)),x)
Output:
(15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**9 + 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**8 - 60*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7 - 60*log (tan((c + d*x)/2) - 1)*sin(c + d*x)**6 + 90*log(tan((c + d*x)/2) - 1)*sin( c + d*x)**5 + 90*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 - 60*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 - 60*log(tan((c + d*x)/2) - 1)*sin(c + d*x )**2 + 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x) + 15*log(tan((c + d*x)/2) - 1) - 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**9 - 15*log(tan((c + d*x )/2) + 1)*sin(c + d*x)**8 + 60*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**7 + 60*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6 - 90*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5 - 90*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 + 60*log (tan((c + d*x)/2) + 1)*sin(c + d*x)**3 + 60*log(tan((c + d*x)/2) + 1)*sin( c + d*x)**2 - 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x) - 15*log(tan((c + d*x)/2) + 1) - 47*sin(c + d*x)**9 - 32*sin(c + d*x)**8 + 203*sin(c + d*x)* *7 + 133*sin(c + d*x)**6 - 17*sin(c + d*x)**5 - 145*sin(c + d*x)**4 + 5*si n(c + d*x)**3 + 75*sin(c + d*x)**2 - 15)/(1280*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*sin(c + d*x)**2 + sin(c + d*x) + 1))