Integrand size = 21, antiderivative size = 99 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {x}{a^3}-\frac {13 \text {arctanh}(\sin (c+d x))}{8 a^3 d}-\frac {\tan (c+d x)}{a^3 d}+\frac {11 \sec (c+d x) \tan (c+d x)}{8 a^3 d}+\frac {\sec ^3(c+d x) \tan (c+d x)}{4 a^3 d}-\frac {\tan ^3(c+d x)}{a^3 d} \] Output:
x/a^3-13/8*arctanh(sin(d*x+c))/a^3/d-tan(d*x+c)/a^3/d+11/8*sec(d*x+c)*tan( d*x+c)/a^3/d+1/4*sec(d*x+c)^3*tan(d*x+c)/a^3/d-tan(d*x+c)^3/a^3/d
Leaf count is larger than twice the leaf count of optimal. \(230\) vs. \(2(99)=198\).
Time = 1.89 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.32 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\sec ^4(c+d x) \left (24 d x+39 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 \cos (2 (c+d x)) \left (8 d x+13 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-13 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos (4 (c+d x)) \left (8 d x+13 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-13 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-39 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+38 \sin (c+d x)-32 \sin (2 (c+d x))+22 \sin (3 (c+d x))\right )}{64 a^3 d} \] Input:
Integrate[Tan[c + d*x]^8/(a + a*Sec[c + d*x])^3,x]
Output:
(Sec[c + d*x]^4*(24*d*x + 39*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4* Cos[2*(c + d*x)]*(8*d*x + 13*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 13 *Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Cos[4*(c + d*x)]*(8*d*x + 13* Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 13*Log[Cos[(c + d*x)/2] + Sin[( c + d*x)/2]]) - 39*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 38*Sin[c + d *x] - 32*Sin[2*(c + d*x)] + 22*Sin[3*(c + d*x)]))/(64*a^3*d)
Time = 0.46 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4376, 25, 3042, 4374, 2009}
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.
| \(\displaystyle \int \frac {\tan ^8(c+d x)}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^8}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4376 |
| \(\displaystyle \frac {\int -(a-a \sec (c+d x))^3 \tan ^2(c+d x)dx}{a^6}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int (a-a \sec (c+d x))^3 \tan ^2(c+d x)dx}{a^6}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \cot \left (c+d x+\frac {\pi }{2}\right )^2 \left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3dx}{a^6}\) |
| \(\Big \downarrow \) 4374 |
| \(\displaystyle -\frac {\int \left (-\sec ^3(c+d x) \tan ^2(c+d x) a^3+3 \sec ^2(c+d x) \tan ^2(c+d x) a^3-3 \sec (c+d x) \tan ^2(c+d x) a^3+\tan ^2(c+d x) a^3\right )dx}{a^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {13 a^3 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 \tan ^3(c+d x)}{d}+\frac {a^3 \tan (c+d x)}{d}-\frac {a^3 \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {11 a^3 \tan (c+d x) \sec (c+d x)}{8 d}-a^3 x}{a^6}\) |
Input:
Int[Tan[c + d*x]^8/(a + a*Sec[c + d*x])^3,x]
Output:
-((-(a^3*x) + (13*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*Tan[c + d*x])/d - (11*a^3*Sec[c + d*x]*Tan[c + d*x])/(8*d) - (a^3*Sec[c + d*x]^3*Tan[c + d *x])/(4*d) + (a^3*Tan[c + d*x]^3)/d)/a^6)
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_))^(n_), x_Symbol] :> Int[ExpandIntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[ c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_))^(n_), x_Symbol] :> Simp[a^(2*n)/e^(2*n) Int[(e*Cot[c + d*x])^(m + 2* n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[a ^2 - b^2, 0] && ILtQ[n, 0]
Result contains complex when optimal does not.
Time = 1.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.39
| method | result | size |
| risch | \(\frac {x}{a^{3}}-\frac {i \left (11 \,{\mathrm e}^{7 i \left (d x +c \right )}-16 \,{\mathrm e}^{6 i \left (d x +c \right )}+19 \,{\mathrm e}^{5 i \left (d x +c \right )}-19 \,{\mathrm e}^{3 i \left (d x +c \right )}+16 \,{\mathrm e}^{2 i \left (d x +c \right )}-11 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d \,a^{3}}-\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d \,a^{3}}\) | \(138\) |
| derivativedivides | \(\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {27}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {21}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {27}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {21}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8}}{d \,a^{3}}\) | \(170\) |
| default | \(\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {27}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {21}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {27}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {21}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8}}{d \,a^{3}}\) | \(170\) |
Input:
int(tan(d*x+c)^8/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
x/a^3-1/4*I/d/a^3/(exp(2*I*(d*x+c))+1)^4*(11*exp(7*I*(d*x+c))-16*exp(6*I*( d*x+c))+19*exp(5*I*(d*x+c))-19*exp(3*I*(d*x+c))+16*exp(2*I*(d*x+c))-11*exp (I*(d*x+c)))+13/8/d/a^3*ln(exp(I*(d*x+c))-I)-13/8/d/a^3*ln(exp(I*(d*x+c))+ I)
Time = 0.09 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.98 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {16 \, d x \cos \left (d x + c\right )^{4} - 13 \, \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + 13 \, \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (11 \, \cos \left (d x + c\right )^{2} - 8 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{16 \, a^{3} d \cos \left (d x + c\right )^{4}} \] Input:
integrate(tan(d*x+c)^8/(a+a*sec(d*x+c))^3,x, algorithm="fricas")
Output:
1/16*(16*d*x*cos(d*x + c)^4 - 13*cos(d*x + c)^4*log(sin(d*x + c) + 1) + 13 *cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(11*cos(d*x + c)^2 - 8*cos(d*x + c) + 2)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4)
\[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\tan ^{8}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(tan(d*x+c)**8/(a+a*sec(d*x+c))**3,x)
Output:
Integral(tan(c + d*x)**8/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3
Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (93) = 186\).
Time = 0.11 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.60 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {2 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {13 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {21 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{3} - \frac {4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac {16 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {13 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {13 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{8 \, d} \] Input:
integrate(tan(d*x+c)^8/(a+a*sec(d*x+c))^3,x, algorithm="maxima")
Output:
1/8*(2*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 13*sin(d*x + c)^3/(cos(d*x + c ) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 21*sin(d*x + c)^7/(cos( d*x + c) + 1)^7)/(a^3 - 4*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a^3* sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) + 16*arctan(sin(d*x + c)/ (cos(d*x + c) + 1))/a^3 - 13*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 13*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d
Time = 0.48 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.24 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {8 \, {\left (d x + c\right )}}{a^{3}} - \frac {13 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac {13 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {2 \, {\left (21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 13 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4} a^{3}}}{8 \, d} \] Input:
integrate(tan(d*x+c)^8/(a+a*sec(d*x+c))^3,x, algorithm="giac")
Output:
1/8*(8*(d*x + c)/a^3 - 13*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 + 13*log( abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 2*(21*tan(1/2*d*x + 1/2*c)^7 + 3*tan( 1/2*d*x + 1/2*c)^5 - 13*tan(1/2*d*x + 1/2*c)^3 + 5*tan(1/2*d*x + 1/2*c))/( (tan(1/2*d*x + 1/2*c)^2 - 1)^4*a^3))/d
Time = 13.56 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.49 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {x}{a^3}-\frac {13\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,a^3\,d}+\frac {\frac {21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )} \] Input:
int(tan(c + d*x)^8/(a + a/cos(c + d*x))^3,x)
Output:
x/a^3 - (13*atanh(tan(c/2 + (d*x)/2)))/(4*a^3*d) + ((5*tan(c/2 + (d*x)/2)) /4 - (13*tan(c/2 + (d*x)/2)^3)/4 + (3*tan(c/2 + (d*x)/2)^5)/4 + (21*tan(c/ 2 + (d*x)/2)^7)/4)/(d*(6*a^3*tan(c/2 + (d*x)/2)^4 - 4*a^3*tan(c/2 + (d*x)/ 2)^2 - 4*a^3*tan(c/2 + (d*x)/2)^6 + a^3*tan(c/2 + (d*x)/2)^8 + a^3))
Time = 0.16 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.09 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )+13 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}-26 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+13 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-13 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}+26 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-13 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 \sin \left (d x +c \right )^{4} d x -11 \sin \left (d x +c \right )^{3}-16 \sin \left (d x +c \right )^{2} d x +13 \sin \left (d x +c \right )+8 d x}{8 a^{3} d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(tan(d*x+c)^8/(a+a*sec(d*x+c))^3,x)
Output:
( - 8*cos(c + d*x)*sin(c + d*x) + 13*log(tan((c + d*x)/2) - 1)*sin(c + d*x )**4 - 26*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 13*log(tan((c + d*x) /2) - 1) - 13*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 + 26*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 - 13*log(tan((c + d*x)/2) + 1) + 8*sin(c + d* x)**4*d*x - 11*sin(c + d*x)**3 - 16*sin(c + d*x)**2*d*x + 13*sin(c + d*x) + 8*d*x)/(8*a**3*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))