Integrand size = 29, antiderivative size = 174 \[ \int \frac {\sec ^6(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \text {arctanh}(\sin (c+d x))}{256 a d}-\frac {\sec ^8(c+d x)}{8 a d}+\frac {\sec ^{10}(c+d x)}{10 a d}-\frac {3 \sec (c+d x) \tan (c+d x)}{256 a d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{128 a d}-\frac {\sec ^5(c+d x) \tan (c+d x)}{160 a d}+\frac {3 \sec ^7(c+d x) \tan (c+d x)}{80 a d}-\frac {\sec ^7(c+d x) \tan ^3(c+d x)}{10 a d} \] Output:
-3/256*arctanh(sin(d*x+c))/a/d-1/8*sec(d*x+c)^8/a/d+1/10*sec(d*x+c)^10/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-1/160*s ec(d*x+c)^5*tan(d*x+c)/a/d+3/80*sec(d*x+c)^7*tan(d*x+c)/a/d-1/10*sec(d*x+c )^7*tan(d*x+c)^3/a/d
Time = 2.87 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.60 \[ \int \frac {\sec ^6(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {30 \text {arctanh}(\sin (c+d x))-\frac {10}{(-1+\sin (c+d x))^4}+\frac {15}{(-1+\sin (c+d x))^2}-\frac {30}{-1+\sin (c+d x)}-\frac {16}{(1+\sin (c+d x))^5}+\frac {10}{(1+\sin (c+d x))^4}+\frac {20}{(1+\sin (c+d x))^3}+\frac {15}{(1+\sin (c+d x))^2}}{2560 a d} \] Input:
Integrate[(Sec[c + d*x]^6*Tan[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
-1/2560*(30*ArcTanh[Sin[c + d*x]] - 10/(-1 + Sin[c + d*x])^4 + 15/(-1 + Si n[c + d*x])^2 - 30/(-1 + Sin[c + d*x]) - 16/(1 + Sin[c + d*x])^5 + 10/(1 + Sin[c + d*x])^4 + 20/(1 + Sin[c + d*x])^3 + 15/(1 + Sin[c + d*x])^2)/(a*d )
Time = 1.07 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.02, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.621, Rules used = {3042, 3314, 3042, 3086, 25, 244, 2009, 3091, 3042, 3091, 3042, 4255, 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.
\(\displaystyle \int \frac {\tan ^3(c+d x) \sec ^6(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3}{\cos (c+d x)^9 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3314 |
\(\displaystyle \frac {\int \sec ^8(c+d x) \tan ^3(c+d x)dx}{a}-\frac {\int \sec ^7(c+d x) \tan ^4(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sec (c+d x)^8 \tan (c+d x)^3dx}{a}-\frac {\int \sec (c+d x)^7 \tan (c+d x)^4dx}{a}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\int -\sec ^7(c+d x) \left (1-\sec ^2(c+d x)\right )d\sec (c+d x)}{a d}-\frac {\int \sec (c+d x)^7 \tan (c+d x)^4dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \sec ^7(c+d x) \left (1-\sec ^2(c+d x)\right )d\sec (c+d x)}{a d}-\frac {\int \sec (c+d x)^7 \tan (c+d x)^4dx}{a}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {\int \left (\sec ^7(c+d x)-\sec ^9(c+d x)\right )d\sec (c+d x)}{a d}-\frac {\int \sec (c+d x)^7 \tan (c+d x)^4dx}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{10} \sec ^{10}(c+d x)-\frac {1}{8} \sec ^8(c+d x)}{a d}-\frac {\int \sec (c+d x)^7 \tan (c+d x)^4dx}{a}\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle \frac {\frac {1}{10} \sec ^{10}(c+d x)-\frac {1}{8} \sec ^8(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^7(c+d x)}{10 d}-\frac {3}{10} \int \sec ^7(c+d x) \tan ^2(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{10} \sec ^{10}(c+d x)-\frac {1}{8} \sec ^8(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^7(c+d x)}{10 d}-\frac {3}{10} \int \sec (c+d x)^7 \tan (c+d x)^2dx}{a}\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle \frac {\frac {1}{10} \sec ^{10}(c+d x)-\frac {1}{8} \sec ^8(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^7(c+d x)}{10 d}-\frac {3}{10} \left (\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}-\frac {1}{8} \int \sec ^7(c+d x)dx\right )}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{10} \sec ^{10}(c+d x)-\frac {1}{8} \sec ^8(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^7(c+d x)}{10 d}-\frac {3}{10} \left (\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}-\frac {1}{8} \int \csc \left (c+d x+\frac {\pi }{2}\right )^7dx\right )}{a}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {1}{10} \sec ^{10}(c+d x)-\frac {1}{8} \sec ^8(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^7(c+d x)}{10 d}-\frac {3}{10} \left (\frac {1}{8} \left (-\frac {5}{6} \int \sec ^5(c+d x)dx-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}\right )}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{10} \sec ^{10}(c+d x)-\frac {1}{8} \sec ^8(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^7(c+d x)}{10 d}-\frac {3}{10} \left (\frac {1}{8} \left (-\frac {5}{6} \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}\right )}{a}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {1}{10} \sec ^{10}(c+d x)-\frac {1}{8} \sec ^8(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^7(c+d x)}{10 d}-\frac {3}{10} \left (\frac {1}{8} \left (-\frac {5}{6} \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}\right )}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{10} \sec ^{10}(c+d x)-\frac {1}{8} \sec ^8(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^7(c+d x)}{10 d}-\frac {3}{10} \left (\frac {1}{8} \left (-\frac {5}{6} \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}\right )}{a}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {1}{10} \sec ^{10}(c+d x)-\frac {1}{8} \sec ^8(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^7(c+d x)}{10 d}-\frac {3}{10} \left (\frac {1}{8} \left (-\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}\right )}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{10} \sec ^{10}(c+d x)-\frac {1}{8} \sec ^8(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^7(c+d x)}{10 d}-\frac {3}{10} \left (\frac {1}{8} \left (-\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}\right )}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {1}{10} \sec ^{10}(c+d x)-\frac {1}{8} \sec ^8(c+d x)}{a d}-\frac {\frac {\tan ^3(c+d x) \sec ^7(c+d x)}{10 d}-\frac {3}{10} \left (\frac {1}{8} \left (-\frac {5}{6} \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}\right )}{a}\) |
Input:
Int[(Sec[c + d*x]^6*Tan[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
(-1/8*Sec[c + d*x]^8 + Sec[c + d*x]^10/10)/(a*d) - ((Sec[c + d*x]^7*Tan[c + d*x]^3)/(10*d) - (3*((Sec[c + d*x]^7*Tan[c + d*x])/(8*d) + (-1/6*(Sec[c + d*x]^5*Tan[c + d*x])/d - (5*((Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*(A rcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4))/6)/8) )/10)/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[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 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 = 55.84 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}-\frac {3}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{256 \left (\sin \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{512}+\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {1}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) | \(115\) |
default | \(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}-\frac {3}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{256 \left (\sin \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{512}+\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {1}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) | \(115\) |
risch | \(\frac {i \left (1482 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{17 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}+13770 \,{\mathrm e}^{9 i \left (d x +c \right )}+268 \,{\mathrm e}^{13 i \left (d x +c \right )}-11364 \,{\mathrm e}^{11 i \left (d x +c \right )}+100 \,{\mathrm e}^{15 i \left (d x +c \right )}+230 i {\mathrm e}^{14 i \left (d x +c \right )}+100 \,{\mathrm e}^{3 i \left (d x +c \right )}-766 i {\mathrm e}^{6 i \left (d x +c \right )}+766 i {\mathrm e}^{12 i \left (d x +c \right )}+30 i {\mathrm e}^{16 i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}-1482 i {\mathrm e}^{10 i \left (d x +c \right )}-230 i {\mathrm e}^{4 i \left (d x +c \right )}-11364 \,{\mathrm e}^{7 i \left (d x +c \right )}+268 \,{\mathrm e}^{5 i \left (d x +c \right )}\right )}{640 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} d a}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 a d}\) | \(277\) |
Input:
int(sec(d*x+c)^6*tan(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/256/(sin(d*x+c)-1)^4-3/512/(sin(d*x+c)-1)^2+3/256/(sin(d*x+c)-1)+ 3/512*ln(sin(d*x+c)-1)+1/160/(1+sin(d*x+c))^5-1/256/(1+sin(d*x+c))^4-1/128 /(1+sin(d*x+c))^3-3/512/(1+sin(d*x+c))^2-3/512*ln(1+sin(d*x+c)))
Time = 0.13 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^6(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {30 \, \cos \left (d x + c\right )^{8} - 10 \, \cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4} - 368 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, \cos \left (d x + c\right )^{6} + 10 \, \cos \left (d x + c\right )^{4} + 8 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 288}{2560 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \] Input:
integrate(sec(d*x+c)^6*tan(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/2560*(30*cos(d*x + c)^8 - 10*cos(d*x + c)^6 - 4*cos(d*x + c)^4 - 368*cos (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 + 10*cos(d*x + c)^4 + 8*cos(d*x + c)^2 - 16)*sin(d*x + c) + 288)/(a*d*cos(d*x + c)^8*sin(d*x + c) + a*d*cos(d*x + c)^8)
Timed out. \[ \int \frac {\sec ^6(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**6*tan(d*x+c)**3/(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.23 \[ \int \frac {\sec ^6(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{8} + 15 \, \sin \left (d x + c\right )^{7} - 55 \, \sin \left (d x + c\right )^{6} - 55 \, \sin \left (d x + c\right )^{5} + 73 \, \sin \left (d x + c\right )^{4} + 73 \, \sin \left (d x + c\right )^{3} + 143 \, \sin \left (d x + c\right )^{2} - 17 \, \sin \left (d x + c\right ) - 32\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{2560 \, d} \] Input:
integrate(sec(d*x+c)^6*tan(d*x+c)^3/(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 - 55* sin(d*x + c)^5 + 73*sin(d*x + c)^4 + 73*sin(d*x + c)^3 + 143*sin(d*x + c)^ 2 - 17*sin(d*x + c) - 32)/(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) - 15*log(si n(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 = 0.83 \[ \int \frac {\sec ^6(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{512 \, a d} + \frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{512 \, a d} + \frac {15 \, \sin \left (d x + c\right )^{8} + 15 \, \sin \left (d x + c\right )^{7} - 55 \, \sin \left (d x + c\right )^{6} - 55 \, \sin \left (d x + c\right )^{5} + 73 \, \sin \left (d x + c\right )^{4} + 73 \, \sin \left (d x + c\right )^{3} + 143 \, \sin \left (d x + c\right )^{2} - 17 \, \sin \left (d x + c\right ) - 32}{1280 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{5} {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} \] Input:
integrate(sec(d*x+c)^6*tan(d*x+c)^3/(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 - 55*sin(d*x + c)^5 + 73*sin(d*x + c)^4 + 73*sin(d*x + c)^3 + 143*sin(d*x + c)^2 - 17*sin(d*x + c) - 32)/(a*d*(sin(d*x + c) + 1)^5*(sin(d*x + c) - 1)^4)
Time = 37.89 (sec) , antiderivative size = 496, normalized size of antiderivative = 2.85 \[ \int \frac {\sec ^6(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(tan(c + d*x)^3/(cos(c + d*x)^6*(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 + (233*tan(c/2 + (d*x)/2)^4)/64 + (323*tan(c/2 + (d*x)/2)^5) /160 + (2687*tan(c/2 + (d*x)/2)^6)/320 - (231*tan(c/2 + (d*x)/2)^7)/160 + (5349*tan(c/2 + (d*x)/2)^8)/320 + (353*tan(c/2 + (d*x)/2)^9)/64 + (5349*ta n(c/2 + (d*x)/2)^10)/320 - (231*tan(c/2 + (d*x)/2)^11)/160 + (2687*tan(c/2 + (d*x)/2)^12)/320 + (323*tan(c/2 + (d*x)/2)^13)/160 + (233*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*t an(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 = 1016, normalized size of antiderivative = 5.84 \[ \int \frac {\sec ^6(c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^6*tan(d*x+c)^3/(a+a*sin(d*x+c)),x)
Output:
(45*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**9 + 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**8 - 180*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7 - 180*l og(tan((c + d*x)/2) - 1)*sin(c + d*x)**6 + 270*log(tan((c + d*x)/2) - 1)*s in(c + d*x)**5 + 270*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 - 180*log(t an((c + d*x)/2) - 1)*sin(c + d*x)**3 - 180*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x) + 45*log(tan((c + d *x)/2) - 1) - 45*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**9 - 45*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**8 + 180*log(tan((c + d*x)/2) + 1)*sin(c + d* x)**7 + 180*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6 - 270*log(tan((c + d *x)/2) + 1)*sin(c + d*x)**5 - 270*log(tan((c + d*x)/2) + 1)*sin(c + d*x)** 4 + 180*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 + 180*log(tan((c + d*x)/ 2) + 1)*sin(c + d*x)**2 - 45*log(tan((c + d*x)/2) + 1)*sin(c + d*x) - 45*l og(tan((c + d*x)/2) + 1) + 480*sec(c + d*x)**6*sin(c + d*x)**9*tan(c + d*x )**2 - 160*sec(c + d*x)**6*sin(c + d*x)**9 + 480*sec(c + d*x)**6*sin(c + d *x)**8*tan(c + d*x)**2 - 160*sec(c + d*x)**6*sin(c + d*x)**8 - 1920*sec(c + d*x)**6*sin(c + d*x)**7*tan(c + d*x)**2 + 640*sec(c + d*x)**6*sin(c + d* x)**7 - 1920*sec(c + d*x)**6*sin(c + d*x)**6*tan(c + d*x)**2 + 640*sec(c + d*x)**6*sin(c + d*x)**6 + 2880*sec(c + d*x)**6*sin(c + d*x)**5*tan(c + d* x)**2 - 960*sec(c + d*x)**6*sin(c + d*x)**5 + 2880*sec(c + d*x)**6*sin(c + d*x)**4*tan(c + d*x)**2 - 960*sec(c + d*x)**6*sin(c + d*x)**4 - 1920*s...