\(\int \frac {\tan ^9(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [296]

Optimal result

Integrand size = 21, antiderivative size = 255 \[ \int \frac {\tan ^9(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\log (\cos (c+d x))}{a^2 d}+\frac {\left (a^2-b^2\right )^3 \left (7 a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^8 d}-\frac {2 a \left (3 a^4-8 a^2 b^2+6 b^4\right ) \sec (c+d x)}{b^7 d}+\frac {\left (5 a^4-12 a^2 b^2+6 b^4\right ) \sec ^2(c+d x)}{2 b^6 d}-\frac {4 a \left (a^2-2 b^2\right ) \sec ^3(c+d x)}{3 b^5 d}+\frac {\left (3 a^2-4 b^2\right ) \sec ^4(c+d x)}{4 b^4 d}-\frac {2 a \sec ^5(c+d x)}{5 b^3 d}+\frac {\sec ^6(c+d x)}{6 b^2 d}+\frac {\left (a^2-b^2\right )^4}{a b^8 d (a+b \sec (c+d x))} \] Output:

-ln(cos(d*x+c))/a^2/d+(a^2-b^2)^3*(7*a^2+b^2)*ln(a+b*sec(d*x+c))/a^2/b^8/d 
-2*a*(3*a^4-8*a^2*b^2+6*b^4)*sec(d*x+c)/b^7/d+1/2*(5*a^4-12*a^2*b^2+6*b^4) 
*sec(d*x+c)^2/b^6/d-4/3*a*(a^2-2*b^2)*sec(d*x+c)^3/b^5/d+1/4*(3*a^2-4*b^2) 
*sec(d*x+c)^4/b^4/d-2/5*a*sec(d*x+c)^5/b^3/d+1/6*sec(d*x+c)^6/b^2/d+(a^2-b 
^2)^4/a/b^8/d/(a+b*sec(d*x+c))
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.90 \[ \int \frac {\tan ^9(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {-\frac {b^8 \log (\cos (c+d x))}{a^2}+\frac {\left (a^2-b^2\right )^3 \left (7 a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2}-2 a b \left (3 a^4-8 a^2 b^2+6 b^4\right ) \sec (c+d x)+\frac {1}{2} b^2 \left (5 a^4-12 a^2 b^2+6 b^4\right ) \sec ^2(c+d x)-\frac {4}{3} a b^3 \left (a^2-2 b^2\right ) \sec ^3(c+d x)+\frac {1}{4} b^4 \left (3 a^2-4 b^2\right ) \sec ^4(c+d x)-\frac {2}{5} a b^5 \sec ^5(c+d x)+\frac {1}{6} b^6 \sec ^6(c+d x)+\frac {\left (a^2-b^2\right )^4}{a (a+b \sec (c+d x))}}{b^8 d} \] Input:

Integrate[Tan[c + d*x]^9/(a + b*Sec[c + d*x])^2,x]
 

Output:

(-((b^8*Log[Cos[c + d*x]])/a^2) + ((a^2 - b^2)^3*(7*a^2 + b^2)*Log[a + b*S 
ec[c + d*x]])/a^2 - 2*a*b*(3*a^4 - 8*a^2*b^2 + 6*b^4)*Sec[c + d*x] + (b^2* 
(5*a^4 - 12*a^2*b^2 + 6*b^4)*Sec[c + d*x]^2)/2 - (4*a*b^3*(a^2 - 2*b^2)*Se 
c[c + d*x]^3)/3 + (b^4*(3*a^2 - 4*b^2)*Sec[c + d*x]^4)/4 - (2*a*b^5*Sec[c 
+ d*x]^5)/5 + (b^6*Sec[c + d*x]^6)/6 + (a^2 - b^2)^4/(a*(a + b*Sec[c + d*x 
])))/(b^8*d)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 25, 4373, 522, 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.

\(\displaystyle \int \frac {\tan ^9(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\cot \left (c+d x+\frac {\pi }{2}\right )^9}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^9}{\left (a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^2}dx\)

\(\Big \downarrow \) 4373

\(\displaystyle \frac {\int \frac {\cos (c+d x) \left (b^2-b^2 \sec ^2(c+d x)\right )^4}{b (a+b \sec (c+d x))^2}d(b \sec (c+d x))}{b^8 d}\)

\(\Big \downarrow \) 522

\(\displaystyle \frac {\int \left (\frac {\cos (c+d x) b^7}{a^2}+\sec ^5(c+d x) b^5-2 a \sec ^4(c+d x) b^4+\left (3 a^2-4 b^2\right ) \sec ^3(c+d x) b^3-4 a \left (a^2-2 b^2\right ) \sec ^2(c+d x) b^2+\left (5 a^4-12 b^2 a^2+6 b^4\right ) \sec (c+d x) b-2 a \left (3 a^4-8 b^2 a^2+6 b^4\right )+\frac {\left (a^2-b^2\right )^3 \left (7 a^2+b^2\right )}{a^2 (a+b \sec (c+d x))}-\frac {\left (a^2-b^2\right )^4}{a (a+b \sec (c+d x))^2}\right )d(b \sec (c+d x))}{b^8 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {b^8 \log (b \sec (c+d x))}{a^2}+\frac {\left (a^2-b^2\right )^4}{a (a+b \sec (c+d x))}+\frac {\left (a^2-b^2\right )^3 \left (7 a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2}+\frac {1}{4} b^4 \left (3 a^2-4 b^2\right ) \sec ^4(c+d x)-\frac {4}{3} a b^3 \left (a^2-2 b^2\right ) \sec ^3(c+d x)+\frac {1}{2} b^2 \left (5 a^4-12 a^2 b^2+6 b^4\right ) \sec ^2(c+d x)-2 a b \left (3 a^4-8 a^2 b^2+6 b^4\right ) \sec (c+d x)-\frac {2}{5} a b^5 \sec ^5(c+d x)+\frac {1}{6} b^6 \sec ^6(c+d x)}{b^8 d}\)

Input:

Int[Tan[c + d*x]^9/(a + b*Sec[c + d*x])^2,x]
 

Output:

((b^8*Log[b*Sec[c + d*x]])/a^2 + ((a^2 - b^2)^3*(7*a^2 + b^2)*Log[a + b*Se 
c[c + d*x]])/a^2 - 2*a*b*(3*a^4 - 8*a^2*b^2 + 6*b^4)*Sec[c + d*x] + (b^2*( 
5*a^4 - 12*a^2*b^2 + 6*b^4)*Sec[c + d*x]^2)/2 - (4*a*b^3*(a^2 - 2*b^2)*Sec 
[c + d*x]^3)/3 + (b^4*(3*a^2 - 4*b^2)*Sec[c + d*x]^4)/4 - (2*a*b^5*Sec[c + 
 d*x]^5)/5 + (b^6*Sec[c + d*x]^6)/6 + (a^2 - b^2)^4/(a*(a + b*Sec[c + d*x] 
)))/(b^8*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, 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 4373

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1))   Subst[Int[(b^2 - x^ 
2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, 
 d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 3.64 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.13

method	result	size
derivativedivides	\(\frac {-\frac {a^{8}-4 a^{6} b^{2}+6 a^{4} b^{4}-4 a^{2} b^{6}+b^{8}}{a^{2} b^{7} \left (b +a \cos \left (d x +c \right )\right )}+\frac {\left (7 a^{8}-20 a^{6} b^{2}+18 a^{4} b^{4}-4 a^{2} b^{6}-b^{8}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{8} a^{2}}-\frac {-3 a^{2}+4 b^{2}}{4 b^{4} \cos \left (d x +c \right )^{4}}-\frac {-5 a^{4}+12 a^{2} b^{2}-6 b^{4}}{2 b^{6} \cos \left (d x +c \right )^{2}}+\frac {\left (-7 a^{6}+20 a^{4} b^{2}-18 a^{2} b^{4}+4 b^{6}\right ) \ln \left (\cos \left (d x +c \right )\right )}{b^{8}}+\frac {1}{6 b^{2} \cos \left (d x +c \right )^{6}}-\frac {2 a}{5 b^{3} \cos \left (d x +c \right )^{5}}-\frac {4 a \left (a^{2}-2 b^{2}\right )}{3 b^{5} \cos \left (d x +c \right )^{3}}-\frac {2 a \left (3 a^{4}-8 a^{2} b^{2}+6 b^{4}\right )}{b^{7} \cos \left (d x +c \right )}}{d}\)	\(287\)
default	\(\frac {-\frac {a^{8}-4 a^{6} b^{2}+6 a^{4} b^{4}-4 a^{2} b^{6}+b^{8}}{a^{2} b^{7} \left (b +a \cos \left (d x +c \right )\right )}+\frac {\left (7 a^{8}-20 a^{6} b^{2}+18 a^{4} b^{4}-4 a^{2} b^{6}-b^{8}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{8} a^{2}}-\frac {-3 a^{2}+4 b^{2}}{4 b^{4} \cos \left (d x +c \right )^{4}}-\frac {-5 a^{4}+12 a^{2} b^{2}-6 b^{4}}{2 b^{6} \cos \left (d x +c \right )^{2}}+\frac {\left (-7 a^{6}+20 a^{4} b^{2}-18 a^{2} b^{4}+4 b^{6}\right ) \ln \left (\cos \left (d x +c \right )\right )}{b^{8}}+\frac {1}{6 b^{2} \cos \left (d x +c \right )^{6}}-\frac {2 a}{5 b^{3} \cos \left (d x +c \right )^{5}}-\frac {4 a \left (a^{2}-2 b^{2}\right )}{3 b^{5} \cos \left (d x +c \right )^{3}}-\frac {2 a \left (3 a^{4}-8 a^{2} b^{2}+6 b^{4}\right )}{b^{7} \cos \left (d x +c \right )}}{d}\)	\(287\)
risch	\(\text {Expression too large to display}\)	\(1191\)

Input:

int(tan(d*x+c)^9/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-(a^8-4*a^6*b^2+6*a^4*b^4-4*a^2*b^6+b^8)/a^2/b^7/(b+a*cos(d*x+c))+(7* 
a^8-20*a^6*b^2+18*a^4*b^4-4*a^2*b^6-b^8)/b^8/a^2*ln(b+a*cos(d*x+c))-1/4*(- 
3*a^2+4*b^2)/b^4/cos(d*x+c)^4-1/2*(-5*a^4+12*a^2*b^2-6*b^4)/b^6/cos(d*x+c) 
^2+(-7*a^6+20*a^4*b^2-18*a^2*b^4+4*b^6)/b^8*ln(cos(d*x+c))+1/6/b^2/cos(d*x 
+c)^6-2/5/b^3*a/cos(d*x+c)^5-4/3*a*(a^2-2*b^2)/b^5/cos(d*x+c)^3-2*a*(3*a^4 
-8*a^2*b^2+6*b^4)/b^7/cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.66 \[ \int \frac {\tan ^9(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {14 \, a^{3} b^{6} \cos \left (d x + c\right ) - 10 \, a^{2} b^{7} + 60 \, {\left (7 \, a^{8} b - 20 \, a^{6} b^{3} + 18 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} \cos \left (d x + c\right )^{6} + 30 \, {\left (7 \, a^{7} b^{2} - 20 \, a^{5} b^{4} + 18 \, a^{3} b^{6}\right )} \cos \left (d x + c\right )^{5} - 10 \, {\left (7 \, a^{6} b^{3} - 20 \, a^{4} b^{5} + 18 \, a^{2} b^{7}\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (7 \, a^{5} b^{4} - 20 \, a^{3} b^{6}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (7 \, a^{4} b^{5} - 20 \, a^{2} b^{7}\right )} \cos \left (d x + c\right )^{2} - 60 \, {\left ({\left (7 \, a^{9} - 20 \, a^{7} b^{2} + 18 \, a^{5} b^{4} - 4 \, a^{3} b^{6} - a b^{8}\right )} \cos \left (d x + c\right )^{7} + {\left (7 \, a^{8} b - 20 \, a^{6} b^{3} + 18 \, a^{4} b^{5} - 4 \, a^{2} b^{7} - b^{9}\right )} \cos \left (d x + c\right )^{6}\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + 60 \, {\left ({\left (7 \, a^{9} - 20 \, a^{7} b^{2} + 18 \, a^{5} b^{4} - 4 \, a^{3} b^{6}\right )} \cos \left (d x + c\right )^{7} + {\left (7 \, a^{8} b - 20 \, a^{6} b^{3} + 18 \, a^{4} b^{5} - 4 \, a^{2} b^{7}\right )} \cos \left (d x + c\right )^{6}\right )} \log \left (-\cos \left (d x + c\right )\right )}{60 \, {\left (a^{3} b^{8} d \cos \left (d x + c\right )^{7} + a^{2} b^{9} d \cos \left (d x + c\right )^{6}\right )}} \] Input:

integrate(tan(d*x+c)^9/(a+b*sec(d*x+c))^2,x, algorithm="fricas")
 

Output:

-1/60*(14*a^3*b^6*cos(d*x + c) - 10*a^2*b^7 + 60*(7*a^8*b - 20*a^6*b^3 + 1 
8*a^4*b^5 - 4*a^2*b^7 + b^9)*cos(d*x + c)^6 + 30*(7*a^7*b^2 - 20*a^5*b^4 + 
 18*a^3*b^6)*cos(d*x + c)^5 - 10*(7*a^6*b^3 - 20*a^4*b^5 + 18*a^2*b^7)*cos 
(d*x + c)^4 + 5*(7*a^5*b^4 - 20*a^3*b^6)*cos(d*x + c)^3 - 3*(7*a^4*b^5 - 2 
0*a^2*b^7)*cos(d*x + c)^2 - 60*((7*a^9 - 20*a^7*b^2 + 18*a^5*b^4 - 4*a^3*b 
^6 - a*b^8)*cos(d*x + c)^7 + (7*a^8*b - 20*a^6*b^3 + 18*a^4*b^5 - 4*a^2*b^ 
7 - b^9)*cos(d*x + c)^6)*log(a*cos(d*x + c) + b) + 60*((7*a^9 - 20*a^7*b^2 
 + 18*a^5*b^4 - 4*a^3*b^6)*cos(d*x + c)^7 + (7*a^8*b - 20*a^6*b^3 + 18*a^4 
*b^5 - 4*a^2*b^7)*cos(d*x + c)^6)*log(-cos(d*x + c)))/(a^3*b^8*d*cos(d*x + 
 c)^7 + a^2*b^9*d*cos(d*x + c)^6)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {\tan ^9(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\tan ^{9}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \] Input:

integrate(tan(d*x+c)**9/(a+b*sec(d*x+c))**2,x)
 

Output:

Integral(tan(c + d*x)**9/(a + b*sec(c + d*x))**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.26 \[ \int \frac {\tan ^9(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {14 \, a^{3} b^{5} \cos \left (d x + c\right ) - 10 \, a^{2} b^{6} + 60 \, {\left (7 \, a^{8} - 20 \, a^{6} b^{2} + 18 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \cos \left (d x + c\right )^{6} + 30 \, {\left (7 \, a^{7} b - 20 \, a^{5} b^{3} + 18 \, a^{3} b^{5}\right )} \cos \left (d x + c\right )^{5} - 10 \, {\left (7 \, a^{6} b^{2} - 20 \, a^{4} b^{4} + 18 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (7 \, a^{5} b^{3} - 20 \, a^{3} b^{5}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (7 \, a^{4} b^{4} - 20 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )^{2}}{a^{3} b^{7} \cos \left (d x + c\right )^{7} + a^{2} b^{8} \cos \left (d x + c\right )^{6}} + \frac {60 \, {\left (7 \, a^{6} - 20 \, a^{4} b^{2} + 18 \, a^{2} b^{4} - 4 \, b^{6}\right )} \log \left (\cos \left (d x + c\right )\right )}{b^{8}} - \frac {60 \, {\left (7 \, a^{8} - 20 \, a^{6} b^{2} + 18 \, a^{4} b^{4} - 4 \, a^{2} b^{6} - b^{8}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{2} b^{8}}}{60 \, d} \] Input:

integrate(tan(d*x+c)^9/(a+b*sec(d*x+c))^2,x, algorithm="maxima")
 

Output:

-1/60*((14*a^3*b^5*cos(d*x + c) - 10*a^2*b^6 + 60*(7*a^8 - 20*a^6*b^2 + 18 
*a^4*b^4 - 4*a^2*b^6 + b^8)*cos(d*x + c)^6 + 30*(7*a^7*b - 20*a^5*b^3 + 18 
*a^3*b^5)*cos(d*x + c)^5 - 10*(7*a^6*b^2 - 20*a^4*b^4 + 18*a^2*b^6)*cos(d* 
x + c)^4 + 5*(7*a^5*b^3 - 20*a^3*b^5)*cos(d*x + c)^3 - 3*(7*a^4*b^4 - 20*a 
^2*b^6)*cos(d*x + c)^2)/(a^3*b^7*cos(d*x + c)^7 + a^2*b^8*cos(d*x + c)^6) 
+ 60*(7*a^6 - 20*a^4*b^2 + 18*a^2*b^4 - 4*b^6)*log(cos(d*x + c))/b^8 - 60* 
(7*a^8 - 20*a^6*b^2 + 18*a^4*b^4 - 4*a^2*b^6 - b^8)*log(a*cos(d*x + c) + b 
)/(a^2*b^8))/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.25 \[ \int \frac {\tan ^9(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {{\left (7 \, a^{6} - 20 \, a^{4} b^{2} + 18 \, a^{2} b^{4} - 4 \, b^{6}\right )} \log \left ({\left | \cos \left (d x + c\right ) \right |}\right )}{b^{8} d} + \frac {{\left (7 \, a^{8} - 20 \, a^{6} b^{2} + 18 \, a^{4} b^{4} - 4 \, a^{2} b^{6} - b^{8}\right )} \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a^{2} b^{8} d} - \frac {14 \, a^{2} b^{6} \cos \left (d x + c\right ) - 10 \, a b^{7} + 30 \, {\left (7 \, a^{6} b^{2} - 20 \, a^{4} b^{4} + 18 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )^{5} + \frac {60 \, {\left (7 \, a^{8} b - 20 \, a^{6} b^{3} + 18 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} \cos \left (d x + c\right )^{6}}{a} - 10 \, {\left (7 \, a^{5} b^{3} - 20 \, a^{3} b^{5} + 18 \, a b^{7}\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (7 \, a^{4} b^{4} - 20 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (7 \, a^{3} b^{5} - 20 \, a b^{7}\right )} \cos \left (d x + c\right )^{2}}{60 \, {\left (a \cos \left (d x + c\right ) + b\right )} a b^{8} d \cos \left (d x + c\right )^{6}} \] Input:

integrate(tan(d*x+c)^9/(a+b*sec(d*x+c))^2,x, algorithm="giac")
 

Output:

-(7*a^6 - 20*a^4*b^2 + 18*a^2*b^4 - 4*b^6)*log(abs(cos(d*x + c)))/(b^8*d) 
+ (7*a^8 - 20*a^6*b^2 + 18*a^4*b^4 - 4*a^2*b^6 - b^8)*log(abs(a*cos(d*x + 
c) + b))/(a^2*b^8*d) - 1/60*(14*a^2*b^6*cos(d*x + c) - 10*a*b^7 + 30*(7*a^ 
6*b^2 - 20*a^4*b^4 + 18*a^2*b^6)*cos(d*x + c)^5 + 60*(7*a^8*b - 20*a^6*b^3 
 + 18*a^4*b^5 - 4*a^2*b^7 + b^9)*cos(d*x + c)^6/a - 10*(7*a^5*b^3 - 20*a^3 
*b^5 + 18*a*b^7)*cos(d*x + c)^4 + 5*(7*a^4*b^4 - 20*a^2*b^6)*cos(d*x + c)^ 
3 - 3*(7*a^3*b^5 - 20*a*b^7)*cos(d*x + c)^2)/((a*cos(d*x + c) + b)*a*b^8*d 
*cos(d*x + c)^6)
 

Mupad [B] (verification not implemented)

Time = 13.75 (sec) , antiderivative size = 760, normalized size of antiderivative = 2.98 \[ \int \frac {\tan ^9(c+d x)}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int(tan(c + d*x)^9/(a + b/cos(c + d*x))^2,x)
 

Output:

((2*(15*a*b^6 - 105*a^6*b - 105*a^7 + 15*b^7 - 191*a^2*b^5 - 191*a^3*b^4 + 
 265*a^4*b^3 + 265*a^5*b^2))/(15*a*b^7) - (2*tan(c/2 + (d*x)/2)^10*(19*a*b 
^6 - 7*a^6*b - 42*a^7 + 6*b^7 - 5*a^2*b^5 - 95*a^3*b^4 + 13*a^4*b^3 + 113* 
a^5*b^2))/(a*b^7) - (4*tan(c/2 + (d*x)/2)^6*(7*a*b^6 - 105*a^6*b - 210*a^7 
 + 30*b^7 - 145*a^2*b^5 - 362*a^3*b^4 + 244*a^4*b^3 + 523*a^5*b^2))/(3*a*b 
^7) + (2*tan(c/2 + (d*x)/2)^8*(91*a*b^6 - 105*a^6*b - 315*a^7 + 45*b^7 - 9 
9*a^2*b^5 - 613*a^3*b^4 + 223*a^4*b^3 + 809*a^5*b^2))/(3*a*b^7) + (2*tan(c 
/2 + (d*x)/2)^4*(10*a*b^6 - 350*a^6*b - 525*a^7 + 75*b^7 - 598*a^2*b^5 - 8 
62*a^3*b^4 + 860*a^4*b^3 + 1290*a^5*b^2))/(5*a*b^7) - (2*tan(c/2 + (d*x)/2 
)^2*(45*a*b^6 - 525*a^6*b - 630*a^7 + 90*b^7 - 955*a^2*b^5 - 1067*a^3*b^4 
+ 1325*a^4*b^3 + 1555*a^5*b^2))/(15*a*b^7) + (2*tan(c/2 + (d*x)/2)^12*(4*a 
*b^6 - 7*a^7 + b^7 - 18*a^3*b^4 + 20*a^5*b^2))/(a*b^7))/(d*(a + b - tan(c/ 
2 + (d*x)/2)^14*(a - b) - tan(c/2 + (d*x)/2)^2*(7*a + 5*b) + tan(c/2 + (d* 
x)/2)^12*(7*a - 5*b) + tan(c/2 + (d*x)/2)^4*(21*a + 9*b) - tan(c/2 + (d*x) 
/2)^10*(21*a - 9*b) - tan(c/2 + (d*x)/2)^6*(35*a + 5*b) + tan(c/2 + (d*x)/ 
2)^8*(35*a - 5*b))) + log(tan(c/2 + (d*x)/2)^2 + 1)/(a^2*d) - (log(tan(c/2 
 + (d*x)/2)^2 - 1)*(7*a^6 - 4*b^6 + 18*a^2*b^4 - 20*a^4*b^2))/(b^8*d) + (l 
og(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2)*(a^2 - b^2)^3* 
(7*a^2 + b^2))/(a^2*b^8*d)
 

Reduce [B] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 5249, normalized size of antiderivative = 20.58 \[ \int \frac {\tan ^9(c+d x)}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int(tan(d*x+c)^9/(a+b*sec(d*x+c))^2,x)
 

Output:

(60*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**6*a*b**8 - 180 
*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4*a*b**8 + 180*co 
s(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a*b**8 - 60*cos(c 
+ d*x)*log(tan((c + d*x)/2)**2 + 1)*a*b**8 - 420*cos(c + d*x)*log(tan((c + 
 d*x)/2) - 1)*sin(c + d*x)**6*a**9 + 1200*cos(c + d*x)*log(tan((c + d*x)/2 
) - 1)*sin(c + d*x)**6*a**7*b**2 - 1080*cos(c + d*x)*log(tan((c + d*x)/2) 
- 1)*sin(c + d*x)**6*a**5*b**4 + 240*cos(c + d*x)*log(tan((c + d*x)/2) - 1 
)*sin(c + d*x)**6*a**3*b**6 + 1260*cos(c + d*x)*log(tan((c + d*x)/2) - 1)* 
sin(c + d*x)**4*a**9 - 3600*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + 
 d*x)**4*a**7*b**2 + 3240*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d 
*x)**4*a**5*b**4 - 720*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x) 
**4*a**3*b**6 - 1260*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)** 
2*a**9 + 3600*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**7* 
b**2 - 3240*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**5*b* 
*4 + 720*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b**6 
+ 420*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**9 - 1200*cos(c + d*x)*log( 
tan((c + d*x)/2) - 1)*a**7*b**2 + 1080*cos(c + d*x)*log(tan((c + d*x)/2) - 
 1)*a**5*b**4 - 240*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**3*b**6 - 420 
*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6*a**9 + 1200*cos(c 
+ d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6*a**7*b**2 - 1080*cos(c...