\(\int \sec ^6(c+d x) (a+b \tan (c+d x))^n \, dx\) [654]

Optimal result

Integrand size = 21, antiderivative size = 161 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^{1+n}}{b^5 d (1+n)}-\frac {4 a \left (a^2+b^2\right ) (a+b \tan (c+d x))^{2+n}}{b^5 d (2+n)}+\frac {2 \left (3 a^2+b^2\right ) (a+b \tan (c+d x))^{3+n}}{b^5 d (3+n)}-\frac {4 a (a+b \tan (c+d x))^{4+n}}{b^5 d (4+n)}+\frac {(a+b \tan (c+d x))^{5+n}}{b^5 d (5+n)} \] Output:

(a^2+b^2)^2*(a+b*tan(d*x+c))^(1+n)/b^5/d/(1+n)-4*a*(a^2+b^2)*(a+b*tan(d*x+ 
c))^(2+n)/b^5/d/(2+n)+2*(3*a^2+b^2)*(a+b*tan(d*x+c))^(3+n)/b^5/d/(3+n)-4*a 
*(a+b*tan(d*x+c))^(4+n)/b^5/d/(4+n)+(a+b*tan(d*x+c))^(5+n)/b^5/d/(5+n)
 

Mathematica [A] (verified)

Time = 2.19 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {(a+b \tan (c+d x))^{1+n} \left (b^4 \sec ^4(c+d x)+4 \left (a^2+b^2\right ) \left (\frac {a^2+b^2}{1+n}-\frac {2 a (a+b \tan (c+d x))}{2+n}+\frac {(a+b \tan (c+d x))^2}{3+n}\right )-4 a (a+b \tan (c+d x)) \left (\frac {a^2+b^2}{2+n}-\frac {2 a (a+b \tan (c+d x))}{3+n}+\frac {(a+b \tan (c+d x))^2}{4+n}\right )\right )}{b^5 d (5+n)} \] Input:

Integrate[Sec[c + d*x]^6*(a + b*Tan[c + d*x])^n,x]
 

Output:

((a + b*Tan[c + d*x])^(1 + n)*(b^4*Sec[c + d*x]^4 + 4*(a^2 + b^2)*((a^2 + 
b^2)/(1 + n) - (2*a*(a + b*Tan[c + d*x]))/(2 + n) + (a + b*Tan[c + d*x])^2 
/(3 + n)) - 4*a*(a + b*Tan[c + d*x])*((a^2 + b^2)/(2 + n) - (2*a*(a + b*Ta 
n[c + d*x]))/(3 + n) + (a + b*Tan[c + d*x])^2/(4 + n))))/(b^5*d*(5 + n))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.86, 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.

\(\displaystyle \int \sec ^6(c+d x) (a+b \tan (c+d x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (c+d x)^6 (a+b \tan (c+d x))^ndx\)

\(\Big \downarrow \) 3987

\(\displaystyle \frac {\int \frac {(a+b \tan (c+d x))^n \left (\tan ^2(c+d x) b^2+b^2\right )^2}{b^4}d(b \tan (c+d x))}{b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (a+b \tan (c+d x))^n \left (\tan ^2(c+d x) b^2+b^2\right )^2d(b \tan (c+d x))}{b^5 d}\)

\(\Big \downarrow \) 476

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

\(\Big \downarrow \) 2009

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

Input:

Int[Sec[c + d*x]^6*(a + b*Tan[c + d*x])^n,x]
 

Output:

(((a^2 + b^2)^2*(a + b*Tan[c + d*x])^(1 + n))/(1 + n) - (4*a*(a^2 + b^2)*( 
a + b*Tan[c + d*x])^(2 + n))/(2 + n) + (2*(3*a^2 + b^2)*(a + b*Tan[c + d*x 
])^(3 + n))/(3 + n) - (4*a*(a + b*Tan[c + d*x])^(4 + n))/(4 + n) + (a + b* 
Tan[c + d*x])^(5 + n)/(5 + n))/(b^5*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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(461\) vs. \(2(161)=322\).

Time = 0.33 (sec) , antiderivative size = 462, normalized size of antiderivative = 2.87

\[\frac {\tan \left (d x +c \right )^{5} {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{d \left (5+n \right )}+\frac {a \left (b^{4} n^{4}+14 b^{4} n^{3}+4 a^{2} b^{2} n^{2}+71 b^{4} n^{2}+36 a^{2} b^{2} n +154 b^{4} n +24 a^{4}+80 b^{2} a^{2}+120 b^{4}\right ) {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{b^{5} d \left (n^{5}+15 n^{4}+85 n^{3}+225 n^{2}+274 n +120\right )}+\frac {a n \tan \left (d x +c \right )^{4} {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{b d \left (n^{2}+9 n +20\right )}-\frac {2 \left (-b^{2} n^{2}+2 a^{2} n -9 b^{2} n -20 b^{2}\right ) \tan \left (d x +c \right )^{3} {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{b^{2} d \left (n^{3}+12 n^{2}+47 n +60\right )}-\frac {\left (-b^{4} n^{4}+4 a^{2} b^{2} n^{3}-14 b^{4} n^{3}+36 a^{2} b^{2} n^{2}-71 b^{4} n^{2}+24 a^{4} n +80 a^{2} b^{2} n -154 b^{4} n -120 b^{4}\right ) \tan \left (d x +c \right ) {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{b^{4} \left (n^{5}+15 n^{4}+85 n^{3}+225 n^{2}+274 n +120\right ) d}+\frac {2 \left (b^{2} n^{2}+9 b^{2} n +6 a^{2}+20 b^{2}\right ) a n \tan \left (d x +c \right )^{2} {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{b^{3} d \left (n^{4}+14 n^{3}+71 n^{2}+154 n +120\right )}\]

Input:

int(sec(d*x+c)^6*(a+b*tan(d*x+c))^n,x)
 

Output:

1/d/(5+n)*tan(d*x+c)^5*exp(n*ln(a+b*tan(d*x+c)))+a*(b^4*n^4+14*b^4*n^3+4*a 
^2*b^2*n^2+71*b^4*n^2+36*a^2*b^2*n+154*b^4*n+24*a^4+80*a^2*b^2+120*b^4)/b^ 
5/d/(n^5+15*n^4+85*n^3+225*n^2+274*n+120)*exp(n*ln(a+b*tan(d*x+c)))+a*n/b/ 
d/(n^2+9*n+20)*tan(d*x+c)^4*exp(n*ln(a+b*tan(d*x+c)))-2*(-b^2*n^2+2*a^2*n- 
9*b^2*n-20*b^2)/b^2/d/(n^3+12*n^2+47*n+60)*tan(d*x+c)^3*exp(n*ln(a+b*tan(d 
*x+c)))-(-b^4*n^4+4*a^2*b^2*n^3-14*b^4*n^3+36*a^2*b^2*n^2-71*b^4*n^2+24*a^ 
4*n+80*a^2*b^2*n-154*b^4*n-120*b^4)/b^4/(n^5+15*n^4+85*n^3+225*n^2+274*n+1 
20)/d*tan(d*x+c)*exp(n*ln(a+b*tan(d*x+c)))+2*(b^2*n^2+9*b^2*n+6*a^2+20*b^2 
)*a/b^3/d*n/(n^4+14*n^3+71*n^2+154*n+120)*tan(d*x+c)^2*exp(n*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. 420 vs. \(2 (161) = 322\).

Time = 0.13 (sec) , antiderivative size = 420, normalized size of antiderivative = 2.61 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {{\left (8 \, {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4} - {\left (a^{3} b^{2} - 3 \, a b^{4}\right )} n^{2} + 3 \, {\left (a^{3} b^{2} + 5 \, a b^{4}\right )} n\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (2 \, a b^{4} n^{3} + 3 \, {\left (a^{3} b^{2} + 3 \, a b^{4}\right )} n^{2} + {\left (3 \, a^{3} b^{2} + 7 \, a b^{4}\right )} n\right )} \cos \left (d x + c\right )^{3} + {\left (a b^{4} n^{4} + 6 \, a b^{4} n^{3} + 11 \, a b^{4} n^{2} + 6 \, a b^{4} n\right )} \cos \left (d x + c\right ) + {\left (b^{5} n^{4} + 10 \, b^{5} n^{3} + 35 \, b^{5} n^{2} + 50 \, b^{5} n + 24 \, b^{5} + 8 \, {\left (8 \, b^{5} - {\left (3 \, a^{2} b^{3} - b^{5}\right )} n^{2} - 3 \, {\left (a^{4} b + 3 \, a^{2} b^{3} - 2 \, b^{5}\right )} n\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (8 \, b^{5} - {\left (a^{2} b^{3} - b^{5}\right )} n^{3} - {\left (3 \, a^{2} b^{3} - 7 \, b^{5}\right )} n^{2} - 2 \, {\left (a^{2} b^{3} - 7 \, b^{5}\right )} n\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \left (\frac {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{n}}{{\left (b^{5} d n^{5} + 15 \, b^{5} d n^{4} + 85 \, b^{5} d n^{3} + 225 \, b^{5} d n^{2} + 274 \, b^{5} d n + 120 \, b^{5} d\right )} \cos \left (d x + c\right )^{5}} \] Input:

integrate(sec(d*x+c)^6*(a+b*tan(d*x+c))^n,x, algorithm="fricas")
 

Output:

(8*(3*a^5 + 10*a^3*b^2 + 15*a*b^4 - (a^3*b^2 - 3*a*b^4)*n^2 + 3*(a^3*b^2 + 
 5*a*b^4)*n)*cos(d*x + c)^5 + 4*(2*a*b^4*n^3 + 3*(a^3*b^2 + 3*a*b^4)*n^2 + 
 (3*a^3*b^2 + 7*a*b^4)*n)*cos(d*x + c)^3 + (a*b^4*n^4 + 6*a*b^4*n^3 + 11*a 
*b^4*n^2 + 6*a*b^4*n)*cos(d*x + c) + (b^5*n^4 + 10*b^5*n^3 + 35*b^5*n^2 + 
50*b^5*n + 24*b^5 + 8*(8*b^5 - (3*a^2*b^3 - b^5)*n^2 - 3*(a^4*b + 3*a^2*b^ 
3 - 2*b^5)*n)*cos(d*x + c)^4 + 4*(8*b^5 - (a^2*b^3 - b^5)*n^3 - (3*a^2*b^3 
 - 7*b^5)*n^2 - 2*(a^2*b^3 - 7*b^5)*n)*cos(d*x + c)^2)*sin(d*x + c))*((a*c 
os(d*x + c) + b*sin(d*x + c))/cos(d*x + c))^n/((b^5*d*n^5 + 15*b^5*d*n^4 + 
 85*b^5*d*n^3 + 225*b^5*d*n^2 + 274*b^5*d*n + 120*b^5*d)*cos(d*x + c)^5)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^n \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**6*(a+b*tan(d*x+c))**n,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.78 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {\frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{n + 1}}{b {\left (n + 1\right )}} + \frac {2 \, {\left ({\left (n^{2} + 3 \, n + 2\right )} b^{3} \tan \left (d x + c\right )^{3} + {\left (n^{2} + n\right )} a b^{2} \tan \left (d x + c\right )^{2} - 2 \, a^{2} b n \tan \left (d x + c\right ) + 2 \, a^{3}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{{\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{3}} + \frac {{\left ({\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} b^{5} \tan \left (d x + c\right )^{5} + {\left (n^{4} + 6 \, n^{3} + 11 \, n^{2} + 6 \, n\right )} a b^{4} \tan \left (d x + c\right )^{4} - 4 \, {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a^{2} b^{3} \tan \left (d x + c\right )^{3} + 12 \, {\left (n^{2} + n\right )} a^{3} b^{2} \tan \left (d x + c\right )^{2} - 24 \, a^{4} b n \tan \left (d x + c\right ) + 24 \, a^{5}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{{\left (n^{5} + 15 \, n^{4} + 85 \, n^{3} + 225 \, n^{2} + 274 \, n + 120\right )} b^{5}}}{d} \] Input:

integrate(sec(d*x+c)^6*(a+b*tan(d*x+c))^n,x, algorithm="maxima")
 

Output:

((b*tan(d*x + c) + a)^(n + 1)/(b*(n + 1)) + 2*((n^2 + 3*n + 2)*b^3*tan(d*x 
 + c)^3 + (n^2 + n)*a*b^2*tan(d*x + c)^2 - 2*a^2*b*n*tan(d*x + c) + 2*a^3) 
*(b*tan(d*x + c) + a)^n/((n^3 + 6*n^2 + 11*n + 6)*b^3) + ((n^4 + 10*n^3 + 
35*n^2 + 50*n + 24)*b^5*tan(d*x + c)^5 + (n^4 + 6*n^3 + 11*n^2 + 6*n)*a*b^ 
4*tan(d*x + c)^4 - 4*(n^3 + 3*n^2 + 2*n)*a^2*b^3*tan(d*x + c)^3 + 12*(n^2 
+ n)*a^3*b^2*tan(d*x + c)^2 - 24*a^4*b*n*tan(d*x + c) + 24*a^5)*(b*tan(d*x 
 + c) + a)^n/((n^5 + 15*n^4 + 85*n^3 + 225*n^2 + 274*n + 120)*b^5))/d
 

Giac [F]

\[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{6} \,d x } \] Input:

integrate(sec(d*x+c)^6*(a+b*tan(d*x+c))^n,x, algorithm="giac")
 

Output:

integrate((b*tan(d*x + c) + a)^n*sec(d*x + c)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^n \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n}{{\cos \left (c+d\,x\right )}^6} \,d x \] Input:

int((a + b*tan(c + d*x))^n/cos(c + d*x)^6,x)
 

Output:

int((a + b*tan(c + d*x))^n/cos(c + d*x)^6, x)
 

Reduce [F]

\[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^n \, dx=\int \left (\tan \left (d x +c \right ) b +a \right )^{n} \sec \left (d x +c \right )^{6}d x \] Input:

int(sec(d*x+c)^6*(a+b*tan(d*x+c))^n,x)
                                                                                    
                                                                                    
 

Output:

int((tan(c + d*x)*b + a)**n*sec(c + d*x)**6,x)