\(\int (a+b \sec (c+d x))^n \tan ^5(c+d x) \, dx\) [350]

Optimal result

Integrand size = 21, antiderivative size = 178 \[ \int (a+b \sec (c+d x))^n \tan ^5(c+d x) \, dx=-\frac {a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{1+n}}{b^4 d (1+n)}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)}+\frac {\left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{2+n}}{b^4 d (2+n)}-\frac {3 a (a+b \sec (c+d x))^{3+n}}{b^4 d (3+n)}+\frac {(a+b \sec (c+d x))^{4+n}}{b^4 d (4+n)} \] Output:

-a*(a^2-2*b^2)*(a+b*sec(d*x+c))^(1+n)/b^4/d/(1+n)-hypergeom([1, 1+n],[2+n] 
,(a+b*sec(d*x+c))/a)*(a+b*sec(d*x+c))^(1+n)/a/d/(1+n)+(3*a^2-2*b^2)*(a+b*s 
ec(d*x+c))^(2+n)/b^4/d/(2+n)-3*a*(a+b*sec(d*x+c))^(3+n)/b^4/d/(3+n)+(a+b*s 
ec(d*x+c))^(4+n)/b^4/d/(4+n)
 

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.77 \[ \int (a+b \sec (c+d x))^n \tan ^5(c+d x) \, dx=\frac {(a+b \sec (c+d x))^{1+n} \left (-\frac {a^3-2 a b^2}{1+n}-\frac {b^4 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b \sec (c+d x)}{a}\right )}{a+a n}+\frac {\left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))}{2+n}-\frac {3 a (a+b \sec (c+d x))^2}{3+n}+\frac {(a+b \sec (c+d x))^3}{4+n}\right )}{b^4 d} \] Input:

Integrate[(a + b*Sec[c + d*x])^n*Tan[c + d*x]^5,x]
 

Output:

((a + b*Sec[c + d*x])^(1 + n)*(-((a^3 - 2*a*b^2)/(1 + n)) - (b^4*Hypergeom 
etric2F1[1, 1 + n, 2 + n, 1 + (b*Sec[c + d*x])/a])/(a + a*n) + ((3*a^2 - 2 
*b^2)*(a + b*Sec[c + d*x]))/(2 + n) - (3*a*(a + b*Sec[c + d*x])^2)/(3 + n) 
 + (a + b*Sec[c + d*x])^3/(4 + n)))/(b^4*d)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.90, 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.

\(\displaystyle \int \tan ^5(c+d x) (a+b \sec (c+d x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^5 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^ndx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \left (a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^ndx\)

\(\Big \downarrow \) 4373

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

\(\Big \downarrow \) 522

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{n+1}}{n+1}+\frac {\left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{n+2}}{n+2}-\frac {b^4 (a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b \sec (c+d x)}{a}+1\right )}{a (n+1)}-\frac {3 a (a+b \sec (c+d x))^{n+3}}{n+3}+\frac {(a+b \sec (c+d x))^{n+4}}{n+4}}{b^4 d}\)

Input:

Int[(a + b*Sec[c + d*x])^n*Tan[c + d*x]^5,x]
 

Output:

(-((a*(a^2 - 2*b^2)*(a + b*Sec[c + d*x])^(1 + n))/(1 + n)) - (b^4*Hypergeo 
metric2F1[1, 1 + n, 2 + n, 1 + (b*Sec[c + d*x])/a]*(a + b*Sec[c + d*x])^(1 
 + n))/(a*(1 + n)) + ((3*a^2 - 2*b^2)*(a + b*Sec[c + d*x])^(2 + n))/(2 + n 
) - (3*a*(a + b*Sec[c + d*x])^(3 + n))/(3 + n) + (a + b*Sec[c + d*x])^(4 + 
 n)/(4 + n))/(b^4*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 [F]

\[\int \left (a +b \sec \left (d x +c \right )\right )^{n} \tan \left (d x +c \right )^{5}d x\]

Input:

int((a+b*sec(d*x+c))^n*tan(d*x+c)^5,x)
 

Output:

int((a+b*sec(d*x+c))^n*tan(d*x+c)^5,x)
 

Fricas [F]

\[ \int (a+b \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5} \,d x } \] Input:

integrate((a+b*sec(d*x+c))^n*tan(d*x+c)^5,x, algorithm="fricas")
 

Output:

integral((b*sec(d*x + c) + a)^n*tan(d*x + c)^5, x)
 

Sympy [F]

\[ \int (a+b \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{n} \tan ^{5}{\left (c + d x \right )}\, dx \] Input:

integrate((a+b*sec(d*x+c))**n*tan(d*x+c)**5,x)
 

Output:

Integral((a + b*sec(c + d*x))**n*tan(c + d*x)**5, x)
 

Maxima [F]

\[ \int (a+b \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5} \,d x } \] Input:

integrate((a+b*sec(d*x+c))^n*tan(d*x+c)^5,x, algorithm="maxima")
 

Output:

integrate((b*sec(d*x + c) + a)^n*tan(d*x + c)^5, x)
 

Giac [F]

\[ \int (a+b \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5} \,d x } \] Input:

integrate((a+b*sec(d*x+c))^n*tan(d*x+c)^5,x, algorithm="giac")
 

Output:

integrate((b*sec(d*x + c) + a)^n*tan(d*x + c)^5, x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+b \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^5\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^n \,d x \] Input:

int(tan(c + d*x)^5*(a + b/cos(c + d*x))^n,x)
 

Output:

int(tan(c + d*x)^5*(a + b/cos(c + d*x))^n, x)
 

Reduce [F]

\[ \int (a+b \sec (c+d x))^n \tan ^5(c+d x) \, dx =\text {Too large to display} \] Input:

int((a+b*sec(d*x+c))^n*tan(d*x+c)^5,x)
 

Output:

((sec(c + d*x)*b + a)**n*tan(c + d*x)**4*n**2 + 2*(sec(c + d*x)*b + a)**n* 
tan(c + d*x)**4*n - 4*(sec(c + d*x)*b + a)**n*tan(c + d*x)**2*n + 8*(sec(c 
 + d*x)*b + a)**n + int(((sec(c + d*x)*b + a)**n*tan(c + d*x)**5)/(sec(c + 
 d*x)*b*n**2 + 6*sec(c + d*x)*b*n + 8*sec(c + d*x)*b + a*n**2 + 6*a*n + 8* 
a),x)*a*d*n**5 + 8*int(((sec(c + d*x)*b + a)**n*tan(c + d*x)**5)/(sec(c + 
d*x)*b*n**2 + 6*sec(c + d*x)*b*n + 8*sec(c + d*x)*b + a*n**2 + 6*a*n + 8*a 
),x)*a*d*n**4 + 20*int(((sec(c + d*x)*b + a)**n*tan(c + d*x)**5)/(sec(c + 
d*x)*b*n**2 + 6*sec(c + d*x)*b*n + 8*sec(c + d*x)*b + a*n**2 + 6*a*n + 8*a 
),x)*a*d*n**3 + 16*int(((sec(c + d*x)*b + a)**n*tan(c + d*x)**5)/(sec(c + 
d*x)*b*n**2 + 6*sec(c + d*x)*b*n + 8*sec(c + d*x)*b + a*n**2 + 6*a*n + 8*a 
),x)*a*d*n**2 - 4*int(((sec(c + d*x)*b + a)**n*tan(c + d*x)**3)/(sec(c + d 
*x)*b*n**2 + 6*sec(c + d*x)*b*n + 8*sec(c + d*x)*b + a*n**2 + 6*a*n + 8*a) 
,x)*a*d*n**4 - 24*int(((sec(c + d*x)*b + a)**n*tan(c + d*x)**3)/(sec(c + d 
*x)*b*n**2 + 6*sec(c + d*x)*b*n + 8*sec(c + d*x)*b + a*n**2 + 6*a*n + 8*a) 
,x)*a*d*n**3 - 32*int(((sec(c + d*x)*b + a)**n*tan(c + d*x)**3)/(sec(c + d 
*x)*b*n**2 + 6*sec(c + d*x)*b*n + 8*sec(c + d*x)*b + a*n**2 + 6*a*n + 8*a) 
,x)*a*d*n**2 + 8*int(((sec(c + d*x)*b + a)**n*tan(c + d*x))/(sec(c + d*x)* 
b*n**2 + 6*sec(c + d*x)*b*n + 8*sec(c + d*x)*b + a*n**2 + 6*a*n + 8*a),x)* 
a*d*n**3 + 48*int(((sec(c + d*x)*b + a)**n*tan(c + d*x))/(sec(c + d*x)*b*n 
**2 + 6*sec(c + d*x)*b*n + 8*sec(c + d*x)*b + a*n**2 + 6*a*n + 8*a),x)*...