\(\int (a+b \sec (c+d x))^n \tan ^2(c+d x) \, dx\) [356]

Optimal result

Integrand size = 21, antiderivative size = 21 \[ \int (a+b \sec (c+d x))^n \tan ^2(c+d x) \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-n,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{1+n} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-1-n} \tan (c+d x)}{b d \sqrt {1+\sec (c+d x)}}-\frac {\sqrt {2} a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \tan (c+d x)}{b d \sqrt {1+\sec (c+d x)}}-\text {Int}\left ((a+b \sec (c+d x))^n,x\right ) \] Output:

2^(1/2)*AppellF1(1/2,-1-n,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c 
))*(a+b*sec(d*x+c))^(1+n)*((a+b*sec(d*x+c))/(a+b))^(-1-n)*tan(d*x+c)/b/d/( 
1+sec(d*x+c))^(1/2)-2^(1/2)*a*AppellF1(1/2,-n,1/2,3/2,b*(1-sec(d*x+c))/(a+ 
b),1/2-1/2*sec(d*x+c))*(a+b*sec(d*x+c))^n*tan(d*x+c)/b/d/(1+sec(d*x+c))^(1 
/2)/(((a+b*sec(d*x+c))/(a+b))^n)-Defer(Int)((a+b*sec(d*x+c))^n,x)
 

Mathematica [N/A]

Not integrable

Time = 10.54 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int (a+b \sec (c+d x))^n \tan ^2(c+d x) \, dx=\int (a+b \sec (c+d x))^n \tan ^2(c+d x) \, dx \] Input:

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

Output:

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

Rubi [N/A]

Not integrable

Time = 0.98 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {3042, 4382, 3042, 4551, 25, 3042, 4321, 156, 155, 4412, 3042, 4273, 4321, 156, 155}

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.

\(\displaystyle \int \tan ^2(c+d x) (a+b \sec (c+d x))^n \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4382

\(\displaystyle \int \left (\sec ^2(c+d x)-1\right ) (a+b \sec (c+d x))^ndx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4551

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

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{n+1}dx}{b}-\frac {\int (b+a \sec (c+d x)) (a+b \sec (c+d x))^ndx}{b}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4321

\(\displaystyle -\frac {\int \left (b+a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^ndx}{b}-\frac {\tan (c+d x) \int \frac {(a+b \sec (c+d x))^{n+1}}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}\)

\(\Big \downarrow \) 156

\(\displaystyle -\frac {\int \left (b+a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^ndx}{b}-\frac {(a+b) \tan (c+d x) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \int \frac {\left (\frac {a}{a+b}+\frac {b \sec (c+d x)}{a+b}\right )^{n+1}}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}\)

\(\Big \downarrow \) 155

\(\displaystyle \frac {\sqrt {2} (a+b) \tan (c+d x) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n-1,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1}}-\frac {\int \left (b+a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^ndx}{b}\)

\(\Big \downarrow \) 4412

\(\displaystyle \frac {\sqrt {2} (a+b) \tan (c+d x) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n-1,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1}}-\frac {b \int (a+b \sec (c+d x))^ndx+a \int \sec (c+d x) (a+b \sec (c+d x))^ndx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {2} (a+b) \tan (c+d x) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n-1,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1}}-\frac {b \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^ndx+a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^ndx}{b}\)

\(\Big \downarrow \) 4273

\(\displaystyle \frac {\sqrt {2} (a+b) \tan (c+d x) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n-1,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1}}-\frac {a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^ndx+b \int (a+b \sec (c+d x))^ndx}{b}\)

\(\Big \downarrow \) 4321

\(\displaystyle \frac {\sqrt {2} (a+b) \tan (c+d x) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n-1,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1}}-\frac {b \int (a+b \sec (c+d x))^ndx-\frac {a \tan (c+d x) \int \frac {(a+b \sec (c+d x))^n}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}}{b}\)

\(\Big \downarrow \) 156

\(\displaystyle \frac {\sqrt {2} (a+b) \tan (c+d x) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n-1,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1}}-\frac {b \int (a+b \sec (c+d x))^ndx-\frac {a \tan (c+d x) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \int \frac {\left (\frac {a}{a+b}+\frac {b \sec (c+d x)}{a+b}\right )^n}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}}{b}\)

\(\Big \downarrow \) 155

\(\displaystyle \frac {\sqrt {2} (a+b) \tan (c+d x) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n-1,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1}}-\frac {b \int (a+b \sec (c+d x))^ndx+\frac {\sqrt {2} a \tan (c+d x) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{d \sqrt {\sec (c+d x)+1}}}{b}\)

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 25

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

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

rule 3042

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

rule 4273

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Unintegrable[ 
(a + b*Csc[c + d*x])^n, x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0 
] &&  !IntegerQ[2*n]
 

rule 4321

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ 
Symbol] :> Simp[Cot[e + f*x]/(f*Sqrt[1 + Csc[e + f*x]]*Sqrt[1 - Csc[e + f*x 
]])   Subst[Int[(a + b*x)^m/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Csc[e + f*x]] 
, x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*m]
 

rule 4382

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

rule 4412

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[c   Int[(a + b*Csc[e + f*x])^m, x], x] + Sim 
p[d   Int[(a + b*Csc[e + f*x])^m*Csc[e + f*x], x], x] /; FreeQ[{a, b, c, d, 
 e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[2*m]
 

rule 4551

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. 
) + (a_))^(m_), x_Symbol] :> Simp[1/b   Int[(a + b*Csc[e + f*x])^m*(A*b - a 
*C*Csc[e + f*x]), x], x] + Simp[C/b   Int[Csc[e + f*x]*(a + b*Csc[e + f*x]) 
^(m + 1), x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && NeQ[a^2 - b^2, 0] && 
  !IntegerQ[2*m]
 

Maple [N/A]

Not integrable

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

\[\int \left (a +b \sec \left (d x +c \right )\right )^{n} \tan \left (d x +c \right )^{2}d x\]

Input:

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

Output:

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

Fricas [N/A]

Not integrable

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int (a+b \sec (c+d x))^n \tan ^2(c+d x) \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2} \,d x } \] Input:

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

Output:

integral((b*sec(d*x + c) + a)^n*tan(d*x + c)^2, x)
 

Sympy [N/A]

Not integrable

Time = 4.87 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int (a+b \sec (c+d x))^n \tan ^2(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{n} \tan ^{2}{\left (c + d x \right )}\, dx \] Input:

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

Output:

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

Maxima [N/A]

Not integrable

Time = 5.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int (a+b \sec (c+d x))^n \tan ^2(c+d x) \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2} \,d x } \] Input:

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

Output:

integrate((b*sec(d*x + c) + a)^n*tan(d*x + c)^2, x)
 

Giac [N/A]

Not integrable

Time = 0.41 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int (a+b \sec (c+d x))^n \tan ^2(c+d x) \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2} \,d x } \] Input:

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

Output:

integrate((b*sec(d*x + c) + a)^n*tan(d*x + c)^2, x)
 

Mupad [N/A]

Not integrable

Time = 11.45 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int (a+b \sec (c+d x))^n \tan ^2(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^2\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^n \,d x \] Input:

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

Output:

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

Reduce [N/A]

Not integrable

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int (a+b \sec (c+d x))^n \tan ^2(c+d x) \, dx=\int \left (\sec \left (d x +c \right ) b +a \right )^{n} \tan \left (d x +c \right )^{2}d x \] Input:

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

Output:

int((sec(c + d*x)*b + a)**n*tan(c + d*x)**2,x)