Integrand size = 29, antiderivative size = 148 \[ \int \frac {\sec ^5(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {5 \text {arctanh}(\sin (c+d x))}{128 a d}+\frac {\sec ^6(c+d x)}{6 a d}-\frac {\sec ^8(c+d x)}{8 a d}-\frac {5 \sec (c+d x) \tan (c+d x)}{128 a d}-\frac {5 \sec ^3(c+d x) \tan (c+d x)}{192 a d}-\frac {\sec ^5(c+d x) \tan (c+d x)}{48 a d}+\frac {\sec ^7(c+d x) \tan (c+d x)}{8 a d} \] Output:
-5/128*arctanh(sin(d*x+c))/a/d+1/6*sec(d*x+c)^6/a/d-1/8*sec(d*x+c)^8/a/d-5 /128*sec(d*x+c)*tan(d*x+c)/a/d-5/192*sec(d*x+c)^3*tan(d*x+c)/a/d-1/48*sec( d*x+c)^5*tan(d*x+c)/a/d+1/8*sec(d*x+c)^7*tan(d*x+c)/a/d
Time = 0.66 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.62 \[ \int \frac {\sec ^5(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {15 \text {arctanh}(\sin (c+d x))+\frac {4}{(-1+\sin (c+d x))^3}-\frac {3}{(-1+\sin (c+d x))^2}-\frac {3}{-1+\sin (c+d x)}+\frac {6}{(1+\sin (c+d x))^4}-\frac {6}{(1+\sin (c+d x))^2}-\frac {12}{1+\sin (c+d x)}}{384 a d} \] Input:
Integrate[(Sec[c + d*x]^5*Tan[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
-1/384*(15*ArcTanh[Sin[c + d*x]] + 4/(-1 + Sin[c + d*x])^3 - 3/(-1 + Sin[c + d*x])^2 - 3/(-1 + Sin[c + d*x]) + 6/(1 + Sin[c + d*x])^4 - 6/(1 + Sin[c + d*x])^2 - 12/(1 + Sin[c + d*x]))/(a*d)
Time = 0.91 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.01, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.552, Rules used = {3042, 3314, 3042, 3086, 25, 244, 2009, 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.
| \(\displaystyle \int \frac {\tan ^2(c+d x) \sec ^5(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2}{\cos (c+d x)^7 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3314 |
| \(\displaystyle \frac {\int \sec ^7(c+d x) \tan ^2(c+d x)dx}{a}-\frac {\int \sec ^6(c+d x) \tan ^3(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sec (c+d x)^7 \tan (c+d x)^2dx}{a}-\frac {\int \sec (c+d x)^6 \tan (c+d x)^3dx}{a}\) |
| \(\Big \downarrow \) 3086 |
| \(\displaystyle \frac {\int \sec (c+d x)^7 \tan (c+d x)^2dx}{a}-\frac {\int -\sec ^5(c+d x) \left (1-\sec ^2(c+d x)\right )d\sec (c+d x)}{a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \sec ^5(c+d x) \left (1-\sec ^2(c+d x)\right )d\sec (c+d x)}{a d}+\frac {\int \sec (c+d x)^7 \tan (c+d x)^2dx}{a}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle \frac {\int \left (\sec ^5(c+d x)-\sec ^7(c+d x)\right )d\sec (c+d x)}{a d}+\frac {\int \sec (c+d x)^7 \tan (c+d x)^2dx}{a}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\int \sec (c+d x)^7 \tan (c+d x)^2dx}{a}-\frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}-\frac {1}{8} \int \sec ^7(c+d x)dx}{a}-\frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}-\frac {1}{8} \int \csc \left (c+d x+\frac {\pi }{2}\right )^7dx}{a}-\frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {1}{8} \left (-\frac {5}{6} \int \sec ^5(c+d x)dx-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}}{a}-\frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \left (-\frac {5}{6} \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}}{a}-\frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {1}{8} \left (-\frac {5}{6} \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}}{a}-\frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \left (-\frac {5}{6} \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}}{a}-\frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {1}{8} \left (-\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}}{a}-\frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \left (-\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}}{a}-\frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{8} \left (-\frac {5}{6} \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {\tan (c+d x) \sec ^5(c+d x)}{6 d}\right )+\frac {\tan (c+d x) \sec ^7(c+d x)}{8 d}}{a}-\frac {\frac {1}{8} \sec ^8(c+d x)-\frac {1}{6} \sec ^6(c+d x)}{a d}\) |
Input:
Int[(Sec[c + d*x]^5*Tan[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
-((-1/6*Sec[c + d*x]^6 + Sec[c + d*x]^8/8)/(a*d)) + ((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*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*T an[c + d*x])/(2*d)))/4))/6)/8)/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 = 18.32 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.70
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{32+32 \sin \left (d x +c \right )}-\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{256}-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {1}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {1}{128 \sin \left (d x +c \right )-128}+\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d a}\) | \(103\) |
| default | \(\frac {-\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{32+32 \sin \left (d x +c \right )}-\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{256}-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {1}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {1}{128 \sin \left (d x +c \right )-128}+\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d a}\) | \(103\) |
| risch | \(\frac {i \left (30 i {\mathrm e}^{12 i \left (d x +c \right )}+15 \,{\mathrm e}^{13 i \left (d x +c \right )}+170 i {\mathrm e}^{10 i \left (d x +c \right )}+70 \,{\mathrm e}^{11 i \left (d x +c \right )}-1652 i {\mathrm e}^{8 i \left (d x +c \right )}+113 \,{\mathrm e}^{9 i \left (d x +c \right )}+1652 i {\mathrm e}^{6 i \left (d x +c \right )}+628 \,{\mathrm e}^{7 i \left (d x +c \right )}-170 i {\mathrm e}^{4 i \left (d x +c \right )}+113 \,{\mathrm e}^{5 i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}+70 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{192 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} d a}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 a d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 a d}\) | \(231\) |
Input:
int(sec(d*x+c)^5*tan(d*x+c)^2/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/64/(1+sin(d*x+c))^4+1/64/(1+sin(d*x+c))^2+1/32/(1+sin(d*x+c))-5/ 256*ln(1+sin(d*x+c))-1/96/(sin(d*x+c)-1)^3+1/128/(sin(d*x+c)-1)^2+1/128/(s in(d*x+c)-1)+5/256*ln(sin(d*x+c)-1))
Time = 0.09 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^5(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {30 \, \cos \left (d x + c\right )^{6} - 10 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{2} - 56\right )} \sin \left (d x + c\right ) + 16}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \] Input:
integrate(sec(d*x+c)^5*tan(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/768*(30*cos(d*x + c)^6 - 10*cos(d*x + c)^4 - 4*cos(d*x + c)^2 - 15*(cos( d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(sin(d*x + c) + 1) + 15*(cos( d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(-sin(d*x + c) + 1) - 2*(15*c os(d*x + c)^4 + 10*cos(d*x + c)^2 - 56)*sin(d*x + c) + 16)/(a*d*cos(d*x + c)^6*sin(d*x + c) + a*d*cos(d*x + c)^6)
\[ \int \frac {\sec ^5(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\tan ^{2}{\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(sec(d*x+c)**5*tan(d*x+c)**2/(a+a*sin(d*x+c)),x)
Output:
Integral(tan(c + d*x)**2*sec(c + d*x)**5/(sin(c + d*x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.18 \[ \int \frac {\sec ^5(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{6} + 15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{4} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )^{2} - 31 \, \sin \left (d x + c\right ) - 16\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \] Input:
integrate(sec(d*x+c)^5*tan(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/768*(2*(15*sin(d*x + c)^6 + 15*sin(d*x + c)^5 - 40*sin(d*x + c)^4 - 40*s in(d*x + c)^3 + 33*sin(d*x + c)^2 - 31*sin(d*x + c) - 16)/(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) - 15*log(sin(d*x + c) + 1)/a + 15*log(sin(d*x + c) - 1)/a)/d
Time = 0.14 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.84 \[ \int \frac {\sec ^5(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {5 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{256 \, a d} + \frac {5 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{256 \, a d} + \frac {15 \, \sin \left (d x + c\right )^{6} + 15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{4} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )^{2} - 31 \, \sin \left (d x + c\right ) - 16}{384 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{4} {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} \] Input:
integrate(sec(d*x+c)^5*tan(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-5/256*log(abs(sin(d*x + c) + 1))/(a*d) + 5/256*log(abs(sin(d*x + c) - 1)) /(a*d) + 1/384*(15*sin(d*x + c)^6 + 15*sin(d*x + c)^5 - 40*sin(d*x + c)^4 - 40*sin(d*x + c)^3 + 33*sin(d*x + c)^2 - 31*sin(d*x + c) - 16)/(a*d*(sin( d*x + c) + 1)^4*(sin(d*x + c) - 1)^3)
Time = 37.13 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.62 \[ \int \frac {\sec ^5(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{32}+\frac {221\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{96}+\frac {43\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{96}+\frac {625\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{48}+\frac {355\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{48}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{48}+\frac {625\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{192}+\frac {43\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{96}+\frac {221\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32}+\frac {5\,\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 {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a\,d} \] Input:
int(tan(c + d*x)^2/(cos(c + d*x)^5*(a + a*sin(c + d*x))),x)
Output:
((5*tan(c/2 + (d*x)/2))/64 + (5*tan(c/2 + (d*x)/2)^2)/32 + (221*tan(c/2 + (d*x)/2)^3)/96 + (43*tan(c/2 + (d*x)/2)^4)/96 + (625*tan(c/2 + (d*x)/2)^5) /192 + (35*tan(c/2 + (d*x)/2)^6)/48 + (355*tan(c/2 + (d*x)/2)^7)/48 + (35* tan(c/2 + (d*x)/2)^8)/48 + (625*tan(c/2 + (d*x)/2)^9)/192 + (43*tan(c/2 + (d*x)/2)^10)/96 + (221*tan(c/2 + (d*x)/2)^11)/96 + (5*tan(c/2 + (d*x)/2)^1 2)/32 + (5*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*tan(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*t an(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)) - (5*atanh(tan (c/2 + (d*x)/2)))/(64*a*d)
Time = 0.20 (sec) , antiderivative size = 470, normalized size of antiderivative = 3.18 \[ \int \frac {\sec ^5(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^5*tan(d*x+c)^2/(a+a*sin(d*x+c)),x)
Output:
(15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7 + 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6 - 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 - 45*log (tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 45*log(tan((c + d*x)/2) - 1)*sin( c + d*x)**3 + 45*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)**7 - 15*log(tan((c + d*x)/2) + 1)*sin(c + d* x)**6 + 45*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5 + 45*log(tan((c + d*x )/2) + 1)*sin(c + d*x)**4 - 45*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 - 45*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) - 31*sin(c + d*x)**7 - 16*s in(c + d*x)**6 + 108*sin(c + d*x)**5 + 53*sin(c + d*x)**4 - 133*sin(c + d* x)**3 - 60*sin(c + d*x)**2 + 15)/(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))