Integrand size = 29, antiderivative size = 152 \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 \text {arctanh}(\sin (c+d x))}{128 a d}+\frac {5 \sec (c+d x) \tan (c+d x)}{128 a d}-\frac {5 \sec ^3(c+d x) \tan (c+d x)}{64 a d}+\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a d}-\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a d}+\frac {\tan ^6(c+d x)}{6 a d}+\frac {\tan ^8(c+d x)}{8 a d} \] Output:
5/128*arctanh(sin(d*x+c))/a/d+5/128*sec(d*x+c)*tan(d*x+c)/a/d-5/64*sec(d*x +c)^3*tan(d*x+c)/a/d+5/48*sec(d*x+c)^3*tan(d*x+c)^3/a/d-1/8*sec(d*x+c)^3*t an(d*x+c)^5/a/d+1/6*tan(d*x+c)^6/a/d+1/8*tan(d*x+c)^8/a/d
Time = 1.15 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.61 \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {15 \text {arctanh}(\sin (c+d x))-\frac {4}{(-1+\sin (c+d x))^3}-\frac {15}{(-1+\sin (c+d x))^2}-\frac {15}{-1+\sin (c+d x)}+\frac {6}{(1+\sin (c+d x))^4}-\frac {24}{(1+\sin (c+d x))^3}+\frac {30}{(1+\sin (c+d x))^2}}{384 a d} \] Input:
Integrate[(Sec[c + d*x]^2*Tan[c + d*x]^5)/(a + a*Sin[c + d*x]),x]
Output:
(15*ArcTanh[Sin[c + d*x]] - 4/(-1 + Sin[c + d*x])^3 - 15/(-1 + Sin[c + d*x ])^2 - 15/(-1 + Sin[c + d*x]) + 6/(1 + Sin[c + d*x])^4 - 24/(1 + Sin[c + d *x])^3 + 30/(1 + Sin[c + d*x])^2)/(384*a*d)
Time = 0.95 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.01, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.517, Rules used = {3042, 3314, 3042, 3087, 244, 2009, 3091, 3042, 3091, 3042, 3091, 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.
| \(\displaystyle \int \frac {\tan ^5(c+d x) \sec ^2(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^5}{\cos (c+d x)^7 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3314 |
| \(\displaystyle \frac {\int \sec ^4(c+d x) \tan ^5(c+d x)dx}{a}-\frac {\int \sec ^3(c+d x) \tan ^6(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sec (c+d x)^4 \tan (c+d x)^5dx}{a}-\frac {\int \sec (c+d x)^3 \tan (c+d x)^6dx}{a}\) |
| \(\Big \downarrow \) 3087 |
| \(\displaystyle \frac {\int \tan ^5(c+d x) \left (\tan ^2(c+d x)+1\right )d\tan (c+d x)}{a d}-\frac {\int \sec (c+d x)^3 \tan (c+d x)^6dx}{a}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle \frac {\int \left (\tan ^7(c+d x)+\tan ^5(c+d x)\right )d\tan (c+d x)}{a d}-\frac {\int \sec (c+d x)^3 \tan (c+d x)^6dx}{a}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{8} \tan ^8(c+d x)+\frac {1}{6} \tan ^6(c+d x)}{a d}-\frac {\int \sec (c+d x)^3 \tan (c+d x)^6dx}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\frac {1}{8} \tan ^8(c+d x)+\frac {1}{6} \tan ^6(c+d x)}{a d}-\frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \int \sec ^3(c+d x) \tan ^4(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \tan ^8(c+d x)+\frac {1}{6} \tan ^6(c+d x)}{a d}-\frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \int \sec (c+d x)^3 \tan (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\frac {1}{8} \tan ^8(c+d x)+\frac {1}{6} \tan ^6(c+d x)}{a d}-\frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac {1}{2} \int \sec ^3(c+d x) \tan ^2(c+d x)dx\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \tan ^8(c+d x)+\frac {1}{6} \tan ^6(c+d x)}{a d}-\frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac {1}{2} \int \sec (c+d x)^3 \tan (c+d x)^2dx\right )}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\frac {1}{8} \tan ^8(c+d x)+\frac {1}{6} \tan ^6(c+d x)}{a d}-\frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \int \sec ^3(c+d x)dx-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \tan ^8(c+d x)+\frac {1}{6} \tan ^6(c+d x)}{a d}-\frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )}{a}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {1}{8} \tan ^8(c+d x)+\frac {1}{6} \tan ^6(c+d x)}{a d}-\frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \tan ^8(c+d x)+\frac {1}{6} \tan ^6(c+d x)}{a d}-\frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )}{a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{8} \tan ^8(c+d x)+\frac {1}{6} \tan ^6(c+d x)}{a d}-\frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )}{a}\) |
Input:
Int[(Sec[c + d*x]^2*Tan[c + d*x]^5)/(a + a*Sin[c + d*x]),x]
Output:
(Tan[c + d*x]^6/6 + Tan[c + d*x]^8/8)/(a*d) - ((Sec[c + d*x]^3*Tan[c + d*x ]^5)/(8*d) - (5*((Sec[c + d*x]^3*Tan[c + d*x]^3)/(6*d) + (-1/4*(Sec[c + d* x]^3*Tan[c + d*x])/d + (ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d))/4)/2))/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[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 = 4.22 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.68
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {5}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5}{128 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{256}+\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d a}\) | \(103\) |
| default | \(\frac {-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {5}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5}{128 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{256}+\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d a}\) | \(103\) |
| risch | \(-\frac {i \left (625 \,{\mathrm e}^{9 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}+15 \,{\mathrm e}^{13 i \left (d x +c \right )}-442 \,{\mathrm e}^{11 i \left (d x +c \right )}+140 i {\mathrm e}^{8 i \left (d x +c \right )}-86 i {\mathrm e}^{10 i \left (d x +c \right )}-442 \,{\mathrm e}^{3 i \left (d x +c \right )}+30 i {\mathrm e}^{12 i \left (d x +c \right )}+86 i {\mathrm e}^{4 i \left (d x +c \right )}-140 i {\mathrm e}^{6 i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}-1420 \,{\mathrm e}^{7 i \left (d x +c \right )}+625 \,{\mathrm e}^{5 i \left (d x +c \right )}\right )}{192 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d a}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 a d}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 a d}\) | \(231\) |
Input:
int(sec(d*x+c)^2*tan(d*x+c)^5/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/96/(sin(d*x+c)-1)^3-5/128/(sin(d*x+c)-1)^2-5/128/(sin(d*x+c)-1)- 5/256*ln(sin(d*x+c)-1)+1/64/(1+sin(d*x+c))^4-1/16/(1+sin(d*x+c))^3+5/64/(1 +sin(d*x+c))^2+5/256*ln(1+sin(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {30 \, \cos \left (d x + c\right )^{6} - 266 \, \cos \left (d x + c\right )^{4} + 316 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 22 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) - 112}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \] Input:
integrate(sec(d*x+c)^2*tan(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/768*(30*cos(d*x + c)^6 - 266*cos(d*x + c)^4 + 316*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*cos(d*x + c)^4 - 22*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)
\[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\tan ^{5}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(sec(d*x+c)**2*tan(d*x+c)**5/(a+a*sin(d*x+c)),x)
Output:
Integral(tan(c + d*x)**5*sec(c + d*x)**2/(sin(c + d*x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{6} + 15 \, \sin \left (d x + c\right )^{5} + 88 \, \sin \left (d x + c\right )^{4} - 8 \, \sin \left (d x + c\right )^{3} - 63 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) + 16\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \] Input:
integrate(sec(d*x+c)^2*tan(d*x+c)^5/(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 + 88*sin(d*x + c)^4 - 8*s in(d*x + c)^3 - 63*sin(d*x + c)^2 + 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 = 0.16 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.81 \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{256 \, a d} - \frac {5 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{256 \, a d} - \frac {15 \, \sin \left (d x + c\right )^{6} + 15 \, \sin \left (d x + c\right )^{5} + 88 \, \sin \left (d x + c\right )^{4} - 8 \, \sin \left (d x + c\right )^{3} - 63 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) + 16}{384 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{4} {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} \] Input:
integrate(sec(d*x+c)^2*tan(d*x+c)^5/(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 + 88*sin(d*x + c)^4 - 8*sin(d*x + c)^3 - 63*sin(d*x + c)^2 + sin(d*x + c) + 16)/(a*d*(sin(d*x + c) + 1)^4*(sin(d*x + c) - 1)^3)
Time = 37.29 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.55 \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a\,d}+\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 {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{96}+\frac {85\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{96}-\frac {113\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {413\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{48}+\frac {157\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{48}+\frac {413\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{48}-\frac {113\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{192}+\frac {85\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{96}+\frac {35\,{\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 )} \] Input:
int(tan(c + d*x)^5/(cos(c + d*x)^2*(a + a*sin(c + d*x))),x)
Output:
(5*atanh(tan(c/2 + (d*x)/2)))/(64*a*d) + ((35*tan(c/2 + (d*x)/2)^3)/96 - ( 5*tan(c/2 + (d*x)/2)^2)/32 - (5*tan(c/2 + (d*x)/2))/64 + (85*tan(c/2 + (d* x)/2)^4)/96 - (113*tan(c/2 + (d*x)/2)^5)/192 + (413*tan(c/2 + (d*x)/2)^6)/ 48 + (157*tan(c/2 + (d*x)/2)^7)/48 + (413*tan(c/2 + (d*x)/2)^8)/48 - (113* tan(c/2 + (d*x)/2)^9)/192 + (85*tan(c/2 + (d*x)/2)^10)/96 + (35*tan(c/2 + (d*x)/2)^11)/96 - (5*tan(c/2 + (d*x)/2)^12)/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*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))
Time = 0.18 (sec) , antiderivative size = 1000, normalized size of antiderivative = 6.58 \[ \int \frac {\sec ^2(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^2*tan(d*x+c)^5/(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)*s in(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(ta n((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) + 64*sec(c + d*x)**2*sin (c + d*x)**7*tan(c + d*x)**4 - 64*sec(c + d*x)**2*sin(c + d*x)**7*tan(c + d*x)**2 + 64*sec(c + d*x)**2*sin(c + d*x)**7 + 64*sec(c + d*x)**2*sin(c + d*x)**6*tan(c + d*x)**4 - 64*sec(c + d*x)**2*sin(c + d*x)**6*tan(c + d*x)* *2 + 64*sec(c + d*x)**2*sin(c + d*x)**6 - 192*sec(c + d*x)**2*sin(c + d*x) **5*tan(c + d*x)**4 + 192*sec(c + d*x)**2*sin(c + d*x)**5*tan(c + d*x)**2 - 192*sec(c + d*x)**2*sin(c + d*x)**5 - 192*sec(c + d*x)**2*sin(c + d*x)** 4*tan(c + d*x)**4 + 192*sec(c + d*x)**2*sin(c + d*x)**4*tan(c + d*x)**2 - 192*sec(c + d*x)**2*sin(c + d*x)**4 + 192*sec(c + d*x)**2*sin(c + d*x)**3* tan(c + d*x)**4 - 192*sec(c + d*x)**2*sin(c + d*x)**3*tan(c + d*x)**2 + 19 2*sec(c + d*x)**2*sin(c + d*x)**3 + 192*sec(c + d*x)**2*sin(c + d*x)**2...