Integrand size = 29, antiderivative size = 194 \[ \int \frac {\sec ^4(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \text {arctanh}(\sin (c+d x))}{256 a d}+\frac {\sec ^6(c+d x)}{6 a d}-\frac {\sec ^8(c+d x)}{4 a d}+\frac {\sec ^{10}(c+d x)}{10 a d}+\frac {3 \sec (c+d x) \tan (c+d x)}{256 a d}+\frac {\sec ^3(c+d x) \tan (c+d x)}{128 a d}-\frac {\sec ^5(c+d x) \tan (c+d x)}{32 a d}+\frac {\sec ^5(c+d x) \tan ^3(c+d x)}{16 a d}-\frac {\sec ^5(c+d x) \tan ^5(c+d x)}{10 a d} \] Output:
3/256*arctanh(sin(d*x+c))/a/d+1/6*sec(d*x+c)^6/a/d-1/4*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*ta n(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
Time = 5.56 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.60 \[ \int \frac {\sec ^4(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {90 \text {arctanh}(\sin (c+d x))+\frac {30}{(-1+\sin (c+d x))^4}+\frac {80}{(-1+\sin (c+d x))^3}+\frac {15}{(-1+\sin (c+d x))^2}-\frac {90}{-1+\sin (c+d x)}+\frac {48}{(1+\sin (c+d x))^5}-\frac {150}{(1+\sin (c+d x))^4}+\frac {100}{(1+\sin (c+d x))^3}+\frac {75}{(1+\sin (c+d x))^2}}{7680 a d} \] Input:
Integrate[(Sec[c + d*x]^4*Tan[c + d*x]^5)/(a + a*Sin[c + d*x]),x]
Output:
(90*ArcTanh[Sin[c + d*x]] + 30/(-1 + Sin[c + d*x])^4 + 80/(-1 + Sin[c + d* x])^3 + 15/(-1 + Sin[c + d*x])^2 - 90/(-1 + Sin[c + d*x]) + 48/(1 + Sin[c + d*x])^5 - 150/(1 + Sin[c + d*x])^4 + 100/(1 + Sin[c + d*x])^3 + 75/(1 + Sin[c + d*x])^2)/(7680*a*d)
Time = 1.10 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.01, 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, 243, 49, 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.
| \(\displaystyle \int \frac {\tan ^5(c+d x) \sec ^4(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^5}{\cos (c+d x)^9 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3314 |
| \(\displaystyle \frac {\int \sec ^6(c+d x) \tan ^5(c+d x)dx}{a}-\frac {\int \sec ^5(c+d x) \tan ^6(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sec (c+d x)^6 \tan (c+d x)^5dx}{a}-\frac {\int \sec (c+d x)^5 \tan (c+d x)^6dx}{a}\) |
| \(\Big \downarrow \) 3086 |
| \(\displaystyle \frac {\int \sec ^5(c+d x) \left (1-\sec ^2(c+d x)\right )^2d\sec (c+d x)}{a d}-\frac {\int \sec (c+d x)^5 \tan (c+d x)^6dx}{a}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\int \sec ^4(c+d x) \left (1-\sec ^2(c+d x)\right )^2d\sec ^2(c+d x)}{2 a d}-\frac {\int \sec (c+d x)^5 \tan (c+d x)^6dx}{a}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {\int \left (\sec ^8(c+d x)-2 \sec ^6(c+d x)+\sec ^4(c+d x)\right )d\sec ^2(c+d x)}{2 a d}-\frac {\int \sec (c+d x)^5 \tan (c+d x)^6dx}{a}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^{10}(c+d x)-\frac {1}{2} \sec ^8(c+d x)+\frac {1}{3} \sec ^6(c+d x)}{2 a d}-\frac {\int \sec (c+d x)^5 \tan (c+d x)^6dx}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^{10}(c+d x)-\frac {1}{2} \sec ^8(c+d x)+\frac {1}{3} \sec ^6(c+d x)}{2 a d}-\frac {\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}-\frac {1}{2} \int \sec ^5(c+d x) \tan ^4(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^{10}(c+d x)-\frac {1}{2} \sec ^8(c+d x)+\frac {1}{3} \sec ^6(c+d x)}{2 a d}-\frac {\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}-\frac {1}{2} \int \sec (c+d x)^5 \tan (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^{10}(c+d x)-\frac {1}{2} \sec ^8(c+d x)+\frac {1}{3} \sec ^6(c+d x)}{2 a d}-\frac {\frac {1}{2} \left (\frac {3}{8} \int \sec ^5(c+d x) \tan ^2(c+d x)dx-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^{10}(c+d x)-\frac {1}{2} \sec ^8(c+d x)+\frac {1}{3} \sec ^6(c+d x)}{2 a d}-\frac {\frac {1}{2} \left (\frac {3}{8} \int \sec (c+d x)^5 \tan (c+d x)^2dx-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^{10}(c+d x)-\frac {1}{2} \sec ^8(c+d x)+\frac {1}{3} \sec ^6(c+d x)}{2 a d}-\frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}-\frac {1}{6} \int \sec ^5(c+d x)dx\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^{10}(c+d x)-\frac {1}{2} \sec ^8(c+d x)+\frac {1}{3} \sec ^6(c+d x)}{2 a d}-\frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}-\frac {1}{6} \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^{10}(c+d x)-\frac {1}{2} \sec ^8(c+d x)+\frac {1}{3} \sec ^6(c+d x)}{2 a d}-\frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \int \sec ^3(c+d x)dx-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^{10}(c+d x)-\frac {1}{2} \sec ^8(c+d x)+\frac {1}{3} \sec ^6(c+d x)}{2 a d}-\frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^{10}(c+d x)-\frac {1}{2} \sec ^8(c+d x)+\frac {1}{3} \sec ^6(c+d x)}{2 a d}-\frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^{10}(c+d x)-\frac {1}{2} \sec ^8(c+d x)+\frac {1}{3} \sec ^6(c+d x)}{2 a d}-\frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^{10}(c+d x)-\frac {1}{2} \sec ^8(c+d x)+\frac {1}{3} \sec ^6(c+d x)}{2 a d}-\frac {\frac {1}{2} \left (\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{8 d}\right )+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 d}}{a}\) |
Input:
Int[(Sec[c + d*x]^4*Tan[c + d*x]^5)/(a + a*Sin[c + d*x]),x]
Output:
(Sec[c + d*x]^6/3 - Sec[c + d*x]^8/2 + Sec[c + d*x]^10/5)/(2*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*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/ (2*d)))/4)/6))/8)/2)/a
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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 = 22.44 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.65
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {1}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {3}{256 \left (\sin \left (d x +c \right )-1\right )}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{512}+\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {5}{384 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {5}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) | \(127\) |
| default | \(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {1}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {3}{256 \left (\sin \left (d x +c \right )-1\right )}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{512}+\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {5}{384 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {5}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) | \(127\) |
| risch | \(-\frac {i \left (45 \,{\mathrm e}^{17 i \left (d x +c \right )}+45 \,{\mathrm e}^{i \left (d x +c \right )}+300 \,{\mathrm e}^{15 i \left (d x +c \right )}-11484 \,{\mathrm e}^{13 i \left (d x +c \right )}+23252 \,{\mathrm e}^{11 i \left (d x +c \right )}-3746 i {\mathrm e}^{8 i \left (d x +c \right )}-40610 \,{\mathrm e}^{9 i \left (d x +c \right )}+690 i {\mathrm e}^{14 i \left (d x +c \right )}+300 \,{\mathrm e}^{3 i \left (d x +c \right )}+1798 i {\mathrm e}^{6 i \left (d x +c \right )}-1798 i {\mathrm e}^{12 i \left (d x +c \right )}+90 i {\mathrm e}^{16 i \left (d x +c \right )}-90 i {\mathrm e}^{2 i \left (d x +c \right )}+3746 i {\mathrm e}^{10 i \left (d x +c \right )}-690 i {\mathrm e}^{4 i \left (d x +c \right )}+23252 \,{\mathrm e}^{7 i \left (d x +c \right )}-11484 \,{\mathrm e}^{5 i \left (d x +c \right )}\right )}{1920 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} d a}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 a d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 a d}\) | \(277\) |
Input:
int(sec(d*x+c)^4*tan(d*x+c)^5/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/256/(sin(d*x+c)-1)^4+1/96/(sin(d*x+c)-1)^3+1/512/(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))^5-5/256/ (1+sin(d*x+c))^4+5/384/(1+sin(d*x+c))^3+5/512/(1+sin(d*x+c))^2+3/512*ln(1+ sin(d*x+c)))
Time = 0.11 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^4(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {90 \, \cos \left (d x + c\right )^{8} - 30 \, \cos \left (d x + c\right )^{6} - 1548 \, \cos \left (d x + c\right )^{4} + 2224 \, \cos \left (d x + c\right )^{2} - 45 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 45 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (45 \, \cos \left (d x + c\right )^{6} + 30 \, \cos \left (d x + c\right )^{4} - 104 \, \cos \left (d x + c\right )^{2} + 48\right )} \sin \left (d x + c\right ) - 864}{7680 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \] Input:
integrate(sec(d*x+c)^4*tan(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/7680*(90*cos(d*x + c)^8 - 30*cos(d*x + c)^6 - 1548*cos(d*x + c)^4 + 222 4*cos(d*x + c)^2 - 45*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(s in(d*x + c) + 1) + 45*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(- sin(d*x + c) + 1) - 2*(45*cos(d*x + c)^6 + 30*cos(d*x + c)^4 - 104*cos(d*x + c)^2 + 48)*sin(d*x + c) - 864)/(a*d*cos(d*x + c)^8*sin(d*x + c) + a*d*c os(d*x + c)^8)
Timed out. \[ \int \frac {\sec ^4(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**4*tan(d*x+c)**5/(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^4(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (45 \, \sin \left (d x + c\right )^{8} + 45 \, \sin \left (d x + c\right )^{7} - 165 \, \sin \left (d x + c\right )^{6} - 165 \, \sin \left (d x + c\right )^{5} - 549 \, \sin \left (d x + c\right )^{4} + 91 \, \sin \left (d x + c\right )^{3} + 301 \, \sin \left (d x + c\right )^{2} - 19 \, \sin \left (d x + c\right ) - 64\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {45 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {45 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{7680 \, d} \] Input:
integrate(sec(d*x+c)^4*tan(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/7680*(2*(45*sin(d*x + c)^8 + 45*sin(d*x + c)^7 - 165*sin(d*x + c)^6 - 1 65*sin(d*x + c)^5 - 549*sin(d*x + c)^4 + 91*sin(d*x + c)^3 + 301*sin(d*x + c)^2 - 19*sin(d*x + c) - 64)/(a*sin(d*x + c)^9 + a*sin(d*x + c)^8 - 4*a*s in(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) - 45*lo g(sin(d*x + c) + 1)/a + 45*log(sin(d*x + c) - 1)/a)/d
Time = 0.16 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.75 \[ \int \frac {\sec ^4(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{512 \, a d} - \frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{512 \, a d} - \frac {45 \, \sin \left (d x + c\right )^{8} + 45 \, \sin \left (d x + c\right )^{7} - 165 \, \sin \left (d x + c\right )^{6} - 165 \, \sin \left (d x + c\right )^{5} - 549 \, \sin \left (d x + c\right )^{4} + 91 \, \sin \left (d x + c\right )^{3} + 301 \, \sin \left (d x + c\right )^{2} - 19 \, \sin \left (d x + c\right ) - 64}{3840 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{5} {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} \] Input:
integrate(sec(d*x+c)^4*tan(d*x+c)^5/(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) - 1/3840*(45*sin(d*x + c)^8 + 45*sin(d*x + c)^7 - 165*sin(d*x + c)^6 - 165*sin(d*x + c)^5 - 549*sin(d*x + c)^4 + 91*sin(d*x + c)^3 + 301*sin(d *x + c)^2 - 19*sin(d*x + c) - 64)/(a*d*(sin(d*x + c) + 1)^5*(sin(d*x + c) - 1)^4)
Time = 38.16 (sec) , antiderivative size = 496, normalized size of antiderivative = 2.56 \[ \int \frac {\sec ^4(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(tan(c + d*x)^5/(cos(c + d*x)^4*(a + a*sin(c + d*x))),x)
Output:
(3*atanh(tan(c/2 + (d*x)/2)))/(128*a*d) + ((5*tan(c/2 + (d*x)/2)^3)/32 - ( 3*tan(c/2 + (d*x)/2)^2)/64 - (3*tan(c/2 + (d*x)/2))/128 + (23*tan(c/2 + (d *x)/2)^4)/64 - (67*tan(c/2 + (d*x)/2)^5)/160 + (9091*tan(c/2 + (d*x)/2)^6) /960 + (1717*tan(c/2 + (d*x)/2)^7)/480 + (18257*tan(c/2 + (d*x)/2)^8)/960 - (35*tan(c/2 + (d*x)/2)^9)/192 + (18257*tan(c/2 + (d*x)/2)^10)/960 + (171 7*tan(c/2 + (d*x)/2)^11)/480 + (9091*tan(c/2 + (d*x)/2)^12)/960 - (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*t an(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 = 0.20 (sec) , antiderivative size = 1266, normalized size of antiderivative = 6.53 \[ \int \frac {\sec ^4(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^4*tan(d*x+c)^5/(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 + 18 0*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*lo g(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 + 180*log(tan((c + d*x)/2) - 1)*si n(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) + 4 5*log(tan((c + d*x)/2) + 1) + 480*sec(c + d*x)**4*sin(c + d*x)**9*tan(c + d*x)**4 - 320*sec(c + d*x)**4*sin(c + d*x)**9*tan(c + d*x)**2 + 160*sec(c + d*x)**4*sin(c + d*x)**9 + 480*sec(c + d*x)**4*sin(c + d*x)**8*tan(c + d* x)**4 - 320*sec(c + d*x)**4*sin(c + d*x)**8*tan(c + d*x)**2 + 160*sec(c + d*x)**4*sin(c + d*x)**8 - 1920*sec(c + d*x)**4*sin(c + d*x)**7*tan(c + d*x )**4 + 1280*sec(c + d*x)**4*sin(c + d*x)**7*tan(c + d*x)**2 - 640*sec(c + d*x)**4*sin(c + d*x)**7 - 1920*sec(c + d*x)**4*sin(c + d*x)**6*tan(c + d*x )**4 + 1280*sec(c + d*x)**4*sin(c + d*x)**6*tan(c + d*x)**2 - 640*sec(c...