Integrand size = 21, antiderivative size = 119 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {x}{a^2}-\frac {3 \text {arctanh}(\sin (c+d x))}{4 a^2 d}-\frac {\tan (c+d x)}{a^2 d}+\frac {3 \sec (c+d x) \tan (c+d x)}{4 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {\sec (c+d x) \tan ^3(c+d x)}{2 a^2 d}+\frac {\tan ^5(c+d x)}{5 a^2 d} \] Output:
x/a^2-3/4*arctanh(sin(d*x+c))/a^2/d-tan(d*x+c)/a^2/d+3/4*sec(d*x+c)*tan(d* x+c)/a^2/d+1/3*tan(d*x+c)^3/a^2/d-1/2*sec(d*x+c)*tan(d*x+c)^3/a^2/d+1/5*ta n(d*x+c)^5/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(495\) vs. \(2(119)=238\).
Time = 5.25 (sec) , antiderivative size = 495, normalized size of antiderivative = 4.16 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (240 x+\frac {180 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {180 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {\left (293 \cos \left (\frac {d x}{2}\right )+333 \cos \left (2 c+\frac {3 d x}{2}\right )+287 \cos \left (2 c+\frac {5 d x}{2}\right )+67 \cos \left (4 c+\frac {7 d x}{2}\right )+68 \cos \left (4 c+\frac {9 d x}{2}\right )\right ) \sec (c) \sec ^5(c+d x) \sin \left (\frac {d x}{2}\right )}{2 d}+\frac {36 \sin \left (\frac {c}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}-\frac {151 \sin \left (\frac {c}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {36 \sin \left (\frac {c}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}-\frac {151 \sin \left (\frac {c}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {\cos \left (\frac {c}{2}\right ) \sec (c) \sec ^4(c+d x) \left (308 \sin \left (\frac {c}{2}\right )-43 \sin \left (\frac {c}{2}+d x\right )-43 \sin \left (\frac {3 c}{2}+d x\right )-346 \sin \left (\frac {3 c}{2}+2 d x\right )+346 \sin \left (\frac {5 c}{2}+2 d x\right )+149 \sin \left (\frac {5 c}{2}+3 d x\right )+149 \sin \left (\frac {7 c}{2}+3 d x\right )\right )}{4 d}\right )}{60 a^2 (1+\sec (c+d x))^2} \] Input:
Integrate[Tan[c + d*x]^8/(a + a*Sec[c + d*x])^2,x]
Output:
(Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*(240*x + (180*Log[Cos[(c + d*x)/2] - Si n[(c + d*x)/2]])/d - (180*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d - (( 293*Cos[(d*x)/2] + 333*Cos[2*c + (3*d*x)/2] + 287*Cos[2*c + (5*d*x)/2] + 6 7*Cos[4*c + (7*d*x)/2] + 68*Cos[4*c + (9*d*x)/2])*Sec[c]*Sec[c + d*x]^5*Si n[(d*x)/2])/(2*d) + (36*Sin[c/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/ 2] - Sin[(c + d*x)/2])^4) - (151*Sin[c/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[( c + d*x)/2] - Sin[(c + d*x)/2])^2) + (36*Sin[c/2])/(d*(Cos[c/2] + Sin[c/2] )*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4) - (151*Sin[c/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (Cos[c/2]*Sec[c]*Sec [c + d*x]^4*(308*Sin[c/2] - 43*Sin[c/2 + d*x] - 43*Sin[(3*c)/2 + d*x] - 34 6*Sin[(3*c)/2 + 2*d*x] + 346*Sin[(5*c)/2 + 2*d*x] + 149*Sin[(5*c)/2 + 3*d* x] + 149*Sin[(7*c)/2 + 3*d*x]))/(4*d)))/(60*a^2*(1 + Sec[c + d*x])^2)
Time = 0.47 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4376, 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.
| \(\displaystyle \int \frac {\tan ^8(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^8}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4376 |
| \(\displaystyle \frac {\int (a-a \sec (c+d x))^2 \tan ^4(c+d x)dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \cot \left (c+d x+\frac {\pi }{2}\right )^4 \left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2dx}{a^4}\) |
| \(\Big \downarrow \) 4374 |
| \(\displaystyle \frac {\int \left (a^2 \tan ^4(c+d x)+a^2 \sec ^2(c+d x) \tan ^4(c+d x)-2 a^2 \sec (c+d x) \tan ^4(c+d x)\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {3 a^2 \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {a^2 \tan ^3(c+d x)}{3 d}-\frac {a^2 \tan (c+d x)}{d}-\frac {a^2 \tan ^3(c+d x) \sec (c+d x)}{2 d}+\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{4 d}+a^2 x}{a^4}\) |
Input:
Int[Tan[c + d*x]^8/(a + a*Sec[c + d*x])^2,x]
Output:
(a^2*x - (3*a^2*ArcTanh[Sin[c + d*x]])/(4*d) - (a^2*Tan[c + d*x])/d + (3*a ^2*Sec[c + d*x]*Tan[c + d*x])/(4*d) + (a^2*Tan[c + d*x]^3)/(3*d) - (a^2*Se c[c + d*x]*Tan[c + d*x]^3)/(2*d) + (a^2*Tan[c + d*x]^5)/(5*d))/a^4
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 = 0.74 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.35
| method | result | size |
| risch | \(\frac {x}{a^{2}}-\frac {i \left (75 \,{\mathrm e}^{9 i \left (d x +c \right )}+60 \,{\mathrm e}^{8 i \left (d x +c \right )}+30 \,{\mathrm e}^{7 i \left (d x +c \right )}+360 \,{\mathrm e}^{6 i \left (d x +c \right )}+320 \,{\mathrm e}^{4 i \left (d x +c \right )}-30 \,{\mathrm e}^{3 i \left (d x +c \right )}+280 \,{\mathrm e}^{2 i \left (d x +c \right )}-75 \,{\mathrm e}^{i \left (d x +c \right )}+68\right )}{30 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d \,a^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d \,a^{2}}\) | \(161\) |
| derivativedivides | \(\frac {-\frac {1}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {19}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}+2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {19}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4}}{d \,a^{2}}\) | \(200\) |
| default | \(\frac {-\frac {1}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {19}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}+2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {19}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4}}{d \,a^{2}}\) | \(200\) |
Input:
int(tan(d*x+c)^8/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
x/a^2-1/30*I*(75*exp(9*I*(d*x+c))+60*exp(8*I*(d*x+c))+30*exp(7*I*(d*x+c))+ 360*exp(6*I*(d*x+c))+320*exp(4*I*(d*x+c))-30*exp(3*I*(d*x+c))+280*exp(2*I* (d*x+c))-75*exp(I*(d*x+c))+68)/d/a^2/(exp(2*I*(d*x+c))+1)^5+3/4/d/a^2*ln(e xp(I*(d*x+c))-I)-3/4/d/a^2*ln(exp(I*(d*x+c))+I)
Time = 0.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.98 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {120 \, d x \cos \left (d x + c\right )^{5} - 45 \, \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) + 45 \, \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (68 \, \cos \left (d x + c\right )^{4} - 75 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) - 12\right )} \sin \left (d x + c\right )}{120 \, a^{2} d \cos \left (d x + c\right )^{5}} \] Input:
integrate(tan(d*x+c)^8/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
Output:
1/120*(120*d*x*cos(d*x + c)^5 - 45*cos(d*x + c)^5*log(sin(d*x + c) + 1) + 45*cos(d*x + c)^5*log(-sin(d*x + c) + 1) - 2*(68*cos(d*x + c)^4 - 75*cos(d *x + c)^3 + 4*cos(d*x + c)^2 + 30*cos(d*x + c) - 12)*sin(d*x + c))/(a^2*d* cos(d*x + c)^5)
\[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\tan ^{8}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(tan(d*x+c)**8/(a+a*sec(d*x+c))**2,x)
Output:
Integral(tan(c + d*x)**8/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2
Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (109) = 218\).
Time = 0.11 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.53 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {110 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {328 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {530 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {105 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}\right )}}{a^{2} - \frac {5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {10 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {45 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} - \frac {45 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{60 \, d} \] Input:
integrate(tan(d*x+c)^8/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
Output:
-1/60*(2*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 110*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 328*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 530*sin(d*x + c)^ 7/(cos(d*x + c) + 1)^7 + 105*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/(a^2 - 5 *a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 10*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10) - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 45*log(sin(d*x + c)/(cos (d*x + c) + 1) + 1)/a^2 - 45*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2) /d
Time = 0.34 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.14 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {60 \, {\left (d x + c\right )}}{a^{2}} - \frac {45 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} + \frac {45 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, {\left (105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 530 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 328 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 110 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5} a^{2}}}{60 \, d} \] Input:
integrate(tan(d*x+c)^8/(a+a*sec(d*x+c))^2,x, algorithm="giac")
Output:
1/60*(60*(d*x + c)/a^2 - 45*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 + 45*lo g(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + 2*(105*tan(1/2*d*x + 1/2*c)^9 - 530 *tan(1/2*d*x + 1/2*c)^7 + 328*tan(1/2*d*x + 1/2*c)^5 - 110*tan(1/2*d*x + 1 /2*c)^3 + 15*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^5*a^2))/d
Time = 12.65 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.50 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {x}{a^2}-\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^2\,d}+\frac {\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}-\frac {53\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+\frac {164\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}-\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )} \] Input:
int(tan(c + d*x)^8/(a + a/cos(c + d*x))^2,x)
Output:
x/a^2 - (3*atanh(tan(c/2 + (d*x)/2)))/(2*a^2*d) + (tan(c/2 + (d*x)/2)/2 - (11*tan(c/2 + (d*x)/2)^3)/3 + (164*tan(c/2 + (d*x)/2)^5)/15 - (53*tan(c/2 + (d*x)/2)^7)/3 + (7*tan(c/2 + (d*x)/2)^9)/2)/(d*(5*a^2*tan(c/2 + (d*x)/2) ^2 - 10*a^2*tan(c/2 + (d*x)/2)^4 + 10*a^2*tan(c/2 + (d*x)/2)^6 - 5*a^2*tan (c/2 + (d*x)/2)^8 + a^2*tan(c/2 + (d*x)/2)^10 - a^2))
Time = 0.16 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.48 \[ \int \frac {\tan ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}-90 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}+90 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+60 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} d x -75 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d x +45 \cos \left (d x +c \right ) \sin \left (d x +c \right )+60 \cos \left (d x +c \right ) d x -68 \sin \left (d x +c \right )^{5}+140 \sin \left (d x +c \right )^{3}-60 \sin \left (d x +c \right )}{60 \cos \left (d x +c \right ) a^{2} d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(tan(d*x+c)^8/(a+a*sec(d*x+c))^2,x)
Output:
(45*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 - 90*cos(c + d* x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1) - 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 + 90*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 - 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1) + 60*cos(c + d*x)*sin(c + d*x)**4*d*x - 75* cos(c + d*x)*sin(c + d*x)**3 - 120*cos(c + d*x)*sin(c + d*x)**2*d*x + 45*c os(c + d*x)*sin(c + d*x) + 60*cos(c + d*x)*d*x - 68*sin(c + d*x)**5 + 140* sin(c + d*x)**3 - 60*sin(c + d*x))/(60*cos(c + d*x)*a**2*d*(sin(c + d*x)** 4 - 2*sin(c + d*x)**2 + 1))