\(\int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) [562]

Optimal result

Integrand size = 21, antiderivative size = 178 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {6 a \left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))}{b^7 d}+\frac {\left (5 a^4+9 a^2 b^2+3 b^4\right ) \tan (c+d x)}{b^6 d}-\frac {a \left (2 a^2+3 b^2\right ) \tan ^2(c+d x)}{b^5 d}+\frac {\left (a^2+b^2\right ) \tan ^3(c+d x)}{b^4 d}-\frac {a \tan ^4(c+d x)}{2 b^3 d}+\frac {\tan ^5(c+d x)}{5 b^2 d}-\frac {\left (a^2+b^2\right )^3}{b^7 d (a+b \tan (c+d x))} \] Output:

-6*a*(a^2+b^2)^2*ln(a+b*tan(d*x+c))/b^7/d+(5*a^4+9*a^2*b^2+3*b^4)*tan(d*x+ 
c)/b^6/d-a*(2*a^2+3*b^2)*tan(d*x+c)^2/b^5/d+(a^2+b^2)*tan(d*x+c)^3/b^4/d-1 
/2*a*tan(d*x+c)^4/b^3/d+1/5*tan(d*x+c)^5/b^2/d-(a^2+b^2)^3/b^7/d/(a+b*tan( 
d*x+c))
 

Mathematica [A] (verified)

Time = 1.64 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.29 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {2 b^6 \sec ^6(c+d x)+b^4 \sec ^4(c+d x) \left (a^2+4 b^2-3 a b \tan (c+d x)\right )-2 \left (8 \left (a^2+b^2\right )^3+30 a^2 \left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))+2 a b \left (-11 a^4-18 a^2 b^2-4 b^4+15 \left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))\right ) \tan (c+d x)-b^2 \left (15 a^4+29 a^2 b^2+8 b^4\right ) \tan ^2(c+d x)+a b^3 \left (5 a^2+7 b^2\right ) \tan ^3(c+d x)-2 a^2 b^4 \tan ^4(c+d x)\right )}{10 b^7 d (a+b \tan (c+d x))} \] Input:

Integrate[Sec[c + d*x]^8/(a + b*Tan[c + d*x])^2,x]
 

Output:

(2*b^6*Sec[c + d*x]^6 + b^4*Sec[c + d*x]^4*(a^2 + 4*b^2 - 3*a*b*Tan[c + d* 
x]) - 2*(8*(a^2 + b^2)^3 + 30*a^2*(a^2 + b^2)^2*Log[a + b*Tan[c + d*x]] + 
2*a*b*(-11*a^4 - 18*a^2*b^2 - 4*b^4 + 15*(a^2 + b^2)^2*Log[a + b*Tan[c + d 
*x]])*Tan[c + d*x] - b^2*(15*a^4 + 29*a^2*b^2 + 8*b^4)*Tan[c + d*x]^2 + a* 
b^3*(5*a^2 + 7*b^2)*Tan[c + d*x]^3 - 2*a^2*b^4*Tan[c + d*x]^4))/(10*b^7*d* 
(a + b*Tan[c + d*x]))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3987, 27, 476, 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.

\(\displaystyle \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^8}{(a+b \tan (c+d x))^2}dx\)

\(\Big \downarrow \) 3987

\(\displaystyle \frac {\int \frac {\left (\tan ^2(c+d x) b^2+b^2\right )^3}{b^6 (a+b \tan (c+d x))^2}d(b \tan (c+d x))}{b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (\tan ^2(c+d x) b^2+b^2\right )^3}{(a+b \tan (c+d x))^2}d(b \tan (c+d x))}{b^7 d}\)

\(\Big \downarrow \) 476

\(\displaystyle \frac {\int \left (5 \left (\frac {3 \left (3 a^2+b^2\right ) b^2}{5 a^4}+1\right ) a^4-2 b^3 \tan ^3(c+d x) a-2 b \left (2 a^2+3 b^2\right ) \tan (c+d x) a-\frac {6 \left (a^2+b^2\right )^2 a}{a+b \tan (c+d x)}+b^4 \tan ^4(c+d x)+3 b^2 \left (a^2+b^2\right ) \tan ^2(c+d x)+\frac {\left (a^2+b^2\right )^3}{(a+b \tan (c+d x))^2}\right )d(b \tan (c+d x))}{b^7 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-a b^2 \left (2 a^2+3 b^2\right ) \tan ^2(c+d x)-\frac {\left (a^2+b^2\right )^3}{a+b \tan (c+d x)}-6 a \left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))+b^3 \left (a^2+b^2\right ) \tan ^3(c+d x)+b \left (5 a^4+9 a^2 b^2+3 b^4\right ) \tan (c+d x)-\frac {1}{2} a b^4 \tan ^4(c+d x)+\frac {1}{5} b^5 \tan ^5(c+d x)}{b^7 d}\)

Input:

Int[Sec[c + d*x]^8/(a + b*Tan[c + d*x])^2,x]
 

Output:

(-6*a*(a^2 + b^2)^2*Log[a + b*Tan[c + d*x]] + b*(5*a^4 + 9*a^2*b^2 + 3*b^4 
)*Tan[c + d*x] - a*b^2*(2*a^2 + 3*b^2)*Tan[c + d*x]^2 + b^3*(a^2 + b^2)*Ta 
n[c + d*x]^3 - (a*b^4*Tan[c + d*x]^4)/2 + (b^5*Tan[c + d*x]^5)/5 - (a^2 + 
b^2)^3/(a + b*Tan[c + d*x]))/(b^7*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 476

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, 
 x] && IGtQ[p, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3987

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[1/(b*f)   Subst[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), 
 x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b^2, 
0] && IntegerQ[m/2]
 

Maple [A] (verified)

Time = 254.47 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.13

method	result	size
derivativedivides	\(\frac {\frac {\frac {\tan \left (d x +c \right )^{5} b^{4}}{5}-\frac {a \,b^{3} \tan \left (d x +c \right )^{4}}{2}+a^{2} b^{2} \tan \left (d x +c \right )^{3}+b^{4} \tan \left (d x +c \right )^{3}-2 a^{3} b \tan \left (d x +c \right )^{2}-3 a \,b^{3} \tan \left (d x +c \right )^{2}+5 a^{4} \tan \left (d x +c \right )+9 b^{2} a^{2} \tan \left (d x +c \right )+3 b^{4} \tan \left (d x +c \right )}{b^{6}}-\frac {a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}{b^{7} \left (a +b \tan \left (d x +c \right )\right )}-\frac {6 a \left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{7}}}{d}\)	\(201\)
default	\(\frac {\frac {\frac {\tan \left (d x +c \right )^{5} b^{4}}{5}-\frac {a \,b^{3} \tan \left (d x +c \right )^{4}}{2}+a^{2} b^{2} \tan \left (d x +c \right )^{3}+b^{4} \tan \left (d x +c \right )^{3}-2 a^{3} b \tan \left (d x +c \right )^{2}-3 a \,b^{3} \tan \left (d x +c \right )^{2}+5 a^{4} \tan \left (d x +c \right )+9 b^{2} a^{2} \tan \left (d x +c \right )+3 b^{4} \tan \left (d x +c \right )}{b^{6}}-\frac {a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}{b^{7} \left (a +b \tan \left (d x +c \right )\right )}-\frac {6 a \left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{7}}}{d}\)	\(201\)
risch	\(-\frac {4 i \left (-15 i a^{5}+8 b^{5}+25 a^{2} b^{3}+15 a^{4} b -25 i a^{3} b^{2}-8 i a \,b^{4}+32 b^{5} {\mathrm e}^{2 i \left (d x +c \right )}-15 i a^{5} {\mathrm e}^{10 i \left (d x +c \right )}+60 a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}+15 a^{4} b \,{\mathrm e}^{8 i \left (d x +c \right )}-150 i a^{5} {\mathrm e}^{6 i \left (d x +c \right )}-75 i a^{5} {\mathrm e}^{8 i \left (d x +c \right )}-75 i a^{5} {\mathrm e}^{2 i \left (d x +c \right )}+100 a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+80 a^{2} b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+90 a^{4} b \,{\mathrm e}^{4 i \left (d x +c \right )}+15 a^{2} b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+60 a^{4} b \,{\mathrm e}^{6 i \left (d x +c \right )}+140 a^{2} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-150 i a^{5} {\mathrm e}^{4 i \left (d x +c \right )}+40 b^{5} {\mathrm e}^{4 i \left (d x +c \right )}-60 i a \,b^{4} {\mathrm e}^{8 i \left (d x +c \right )}-250 i a^{3} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-80 i a \,b^{4} {\mathrm e}^{6 i \left (d x +c \right )}-240 i a^{3} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-60 i a \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-120 i a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-33 i a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-30 i a^{3} b^{2} {\mathrm e}^{10 i \left (d x +c \right )}-135 i a^{3} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-15 i a \,b^{4} {\mathrm e}^{10 i \left (d x +c \right )}\right )}{5 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right ) b^{6} d}-\frac {6 a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{7} d}-\frac {12 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{5} d}-\frac {6 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{3} d}+\frac {6 a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{7} d}+\frac {12 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{5} d}+\frac {6 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{3} d}\)	\(686\)

Input:

int(sec(d*x+c)^8/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/b^6*(1/5*tan(d*x+c)^5*b^4-1/2*a*b^3*tan(d*x+c)^4+a^2*b^2*tan(d*x+c) 
^3+b^4*tan(d*x+c)^3-2*a^3*b*tan(d*x+c)^2-3*a*b^3*tan(d*x+c)^2+5*a^4*tan(d* 
x+c)+9*b^2*a^2*tan(d*x+c)+3*b^4*tan(d*x+c))-1/b^7*(a^6+3*a^4*b^2+3*a^2*b^4 
+b^6)/(a+b*tan(d*x+c))-6*a/b^7*(a^4+2*a^2*b^2+b^4)*ln(a+b*tan(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (174) = 348\).

Time = 0.12 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.17 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {4 \, {\left (15 \, a^{4} b^{2} + 25 \, a^{2} b^{4} + 8 \, b^{6}\right )} \cos \left (d x + c\right )^{6} - 2 \, b^{6} - 2 \, {\left (15 \, a^{4} b^{2} + 25 \, a^{2} b^{4} + 8 \, b^{6}\right )} \cos \left (d x + c\right )^{4} - {\left (5 \, a^{2} b^{4} + 4 \, b^{6}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left ({\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{6} + {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{5} \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 30 \, {\left ({\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{6} + {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{5} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) + {\left (3 \, a b^{5} \cos \left (d x + c\right ) - 4 \, {\left (15 \, a^{5} b + 25 \, a^{3} b^{3} + 8 \, a b^{5}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (5 \, a^{3} b^{3} + 7 \, a b^{5}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{10 \, {\left (a b^{7} d \cos \left (d x + c\right )^{6} + b^{8} d \cos \left (d x + c\right )^{5} \sin \left (d x + c\right )\right )}} \] Input:

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c))^2,x, algorithm="fricas")
 

Output:

-1/10*(4*(15*a^4*b^2 + 25*a^2*b^4 + 8*b^6)*cos(d*x + c)^6 - 2*b^6 - 2*(15* 
a^4*b^2 + 25*a^2*b^4 + 8*b^6)*cos(d*x + c)^4 - (5*a^2*b^4 + 4*b^6)*cos(d*x 
 + c)^2 + 30*((a^6 + 2*a^4*b^2 + a^2*b^4)*cos(d*x + c)^6 + (a^5*b + 2*a^3* 
b^3 + a*b^5)*cos(d*x + c)^5*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + 
 c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - 30*((a^6 + 2*a^4*b^2 + a^2*b^4)* 
cos(d*x + c)^6 + (a^5*b + 2*a^3*b^3 + a*b^5)*cos(d*x + c)^5*sin(d*x + c))* 
log(cos(d*x + c)^2) + (3*a*b^5*cos(d*x + c) - 4*(15*a^5*b + 25*a^3*b^3 + 8 
*a*b^5)*cos(d*x + c)^5 + 2*(5*a^3*b^3 + 7*a*b^5)*cos(d*x + c)^3)*sin(d*x + 
 c))/(a*b^7*d*cos(d*x + c)^6 + b^8*d*cos(d*x + c)^5*sin(d*x + c))
 

Sympy [F]

\[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\int \frac {\sec ^{8}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \] Input:

integrate(sec(d*x+c)**8/(a+b*tan(d*x+c))**2,x)
 

Output:

Integral(sec(c + d*x)**8/(a + b*tan(c + d*x))**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.04 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {10 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}}{b^{8} \tan \left (d x + c\right ) + a b^{7}} - \frac {2 \, b^{4} \tan \left (d x + c\right )^{5} - 5 \, a b^{3} \tan \left (d x + c\right )^{4} + 10 \, {\left (a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{3} - 10 \, {\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \tan \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{4} + 9 \, a^{2} b^{2} + 3 \, b^{4}\right )} \tan \left (d x + c\right )}{b^{6}} + \frac {60 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{7}}}{10 \, d} \] Input:

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c))^2,x, algorithm="maxima")
 

Output:

-1/10*(10*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/(b^8*tan(d*x + c) + a*b^7) - 
 (2*b^4*tan(d*x + c)^5 - 5*a*b^3*tan(d*x + c)^4 + 10*(a^2*b^2 + b^4)*tan(d 
*x + c)^3 - 10*(2*a^3*b + 3*a*b^3)*tan(d*x + c)^2 + 10*(5*a^4 + 9*a^2*b^2 
+ 3*b^4)*tan(d*x + c))/b^6 + 60*(a^5 + 2*a^3*b^2 + a*b^4)*log(b*tan(d*x + 
c) + a)/b^7)/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {6 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{7} d} - \frac {a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}}{{\left (b \tan \left (d x + c\right ) + a\right )} b^{7} d} + \frac {2 \, b^{8} d^{4} \tan \left (d x + c\right )^{5} - 5 \, a b^{7} d^{4} \tan \left (d x + c\right )^{4} + 10 \, a^{2} b^{6} d^{4} \tan \left (d x + c\right )^{3} + 10 \, b^{8} d^{4} \tan \left (d x + c\right )^{3} - 20 \, a^{3} b^{5} d^{4} \tan \left (d x + c\right )^{2} - 30 \, a b^{7} d^{4} \tan \left (d x + c\right )^{2} + 50 \, a^{4} b^{4} d^{4} \tan \left (d x + c\right ) + 90 \, a^{2} b^{6} d^{4} \tan \left (d x + c\right ) + 30 \, b^{8} d^{4} \tan \left (d x + c\right )}{10 \, b^{10} d^{5}} \] Input:

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

-6*(a^5 + 2*a^3*b^2 + a*b^4)*log(abs(b*tan(d*x + c) + a))/(b^7*d) - (a^6 + 
 3*a^4*b^2 + 3*a^2*b^4 + b^6)/((b*tan(d*x + c) + a)*b^7*d) + 1/10*(2*b^8*d 
^4*tan(d*x + c)^5 - 5*a*b^7*d^4*tan(d*x + c)^4 + 10*a^2*b^6*d^4*tan(d*x + 
c)^3 + 10*b^8*d^4*tan(d*x + c)^3 - 20*a^3*b^5*d^4*tan(d*x + c)^2 - 30*a*b^ 
7*d^4*tan(d*x + c)^2 + 50*a^4*b^4*d^4*tan(d*x + c) + 90*a^2*b^6*d^4*tan(d* 
x + c) + 30*b^8*d^4*tan(d*x + c))/(b^10*d^5)
 

Mupad [B] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.45 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a^3}{b^5}-\frac {a\,\left (\frac {3}{b^2}+\frac {3\,a^2}{b^4}\right )}{b}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,b^2\,d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {1}{b^2}+\frac {a^2}{b^4}\right )}{d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {a^2\,\left (\frac {3}{b^2}+\frac {3\,a^2}{b^4}\right )}{b^2}-\frac {3}{b^2}+\frac {2\,a\,\left (\frac {2\,a^3}{b^5}-\frac {2\,a\,\left (\frac {3}{b^2}+\frac {3\,a^2}{b^4}\right )}{b}\right )}{b}\right )}{d}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^4}{2\,b^3\,d}-\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (6\,a^5+12\,a^3\,b^2+6\,a\,b^4\right )}{b^7\,d}-\frac {a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}{b\,d\,\left (\mathrm {tan}\left (c+d\,x\right )\,b^7+a\,b^6\right )} \] Input:

int(1/(cos(c + d*x)^8*(a + b*tan(c + d*x))^2),x)
 

Output:

(tan(c + d*x)^2*(a^3/b^5 - (a*(3/b^2 + (3*a^2)/b^4))/b))/d + tan(c + d*x)^ 
5/(5*b^2*d) + (tan(c + d*x)^3*(1/b^2 + a^2/b^4))/d - (tan(c + d*x)*((a^2*( 
3/b^2 + (3*a^2)/b^4))/b^2 - 3/b^2 + (2*a*((2*a^3)/b^5 - (2*a*(3/b^2 + (3*a 
^2)/b^4))/b))/b))/d - (a*tan(c + d*x)^4)/(2*b^3*d) - (log(a + b*tan(c + d* 
x))*(6*a*b^4 + 6*a^5 + 12*a^3*b^2))/(b^7*d) - (a^6 + b^6 + 3*a^2*b^4 + 3*a 
^4*b^2)/(b*d*(a*b^6 + b^7*tan(c + d*x)))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 7571, normalized size of antiderivative = 42.53 \[ \int \frac {\sec ^8(c+d x)}{(a+b \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int(sec(d*x+c)^8/(a+b*tan(d*x+c))^2,x)
 

Output:

(60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7*tan(c + d*x)*a* 
*5*b**2 + 120*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7*tan(c 
 + d*x)*a**3*b**4 + 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x) 
**7*tan(c + d*x)*a*b**6 + 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c 
+ d*x)**7*a**6*b + 120*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x) 
**7*a**4*b**3 + 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7* 
a**2*b**5 - 180*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*tan 
(c + d*x)*a**5*b**2 - 360*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d 
*x)**5*tan(c + d*x)*a**3*b**4 - 180*cos(c + d*x)*log(tan((c + d*x)/2) - 1) 
*sin(c + d*x)**5*tan(c + d*x)*a*b**6 - 180*cos(c + d*x)*log(tan((c + d*x)/ 
2) - 1)*sin(c + d*x)**5*a**6*b - 360*cos(c + d*x)*log(tan((c + d*x)/2) - 1 
)*sin(c + d*x)**5*a**4*b**3 - 180*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*s 
in(c + d*x)**5*a**2*b**5 + 180*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin( 
c + d*x)**3*tan(c + d*x)*a**5*b**2 + 360*cos(c + d*x)*log(tan((c + d*x)/2) 
 - 1)*sin(c + d*x)**3*tan(c + d*x)*a**3*b**4 + 180*cos(c + d*x)*log(tan((c 
 + d*x)/2) - 1)*sin(c + d*x)**3*tan(c + d*x)*a*b**6 + 180*cos(c + d*x)*log 
(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**6*b + 360*cos(c + d*x)*log(tan(( 
c + d*x)/2) - 1)*sin(c + d*x)**3*a**4*b**3 + 180*cos(c + d*x)*log(tan((c + 
 d*x)/2) - 1)*sin(c + d*x)**3*a**2*b**5 - 60*cos(c + d*x)*log(tan((c + d*x 
)/2) - 1)*sin(c + d*x)*tan(c + d*x)*a**5*b**2 - 120*cos(c + d*x)*log(ta...