\(\int \sec ^4(e+f x) (a+b \sec ^2(e+f x))^p \, dx\) [305]

Optimal result

Integrand size = 23, antiderivative size = 131 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}-\frac {(a-2 b (1+p)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^p \left (\frac {a+b+b \tan ^2(e+f x)}{a+b}\right )^{-p}}{b f (3+2 p)} \] Output:

tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(p+1)/b/f/(3+2*p)-(a-2*b*(p+1))*hypergeom( 
[1/2, -p],[3/2],-b*tan(f*x+e)^2/(a+b))*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^p/b 
/f/(3+2*p)/(((a+b+b*tan(f*x+e)^2)/(a+b))^p)
 

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.96 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\left (a+b \sec ^2(e+f x)\right )^p \tan (e+f x) \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p} \left ((-a+2 b (1+p)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )+\left (a+b+b \tan ^2(e+f x)\right ) \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^p\right )}{b f (3+2 p)} \] Input:

Integrate[Sec[e + f*x]^4*(a + b*Sec[e + f*x]^2)^p,x]
 

Output:

((a + b*Sec[e + f*x]^2)^p*Tan[e + f*x]*((-a + 2*b*(1 + p))*Hypergeometric2 
F1[1/2, -p, 3/2, -((b*Tan[e + f*x]^2)/(a + b))] + (a + b + b*Tan[e + f*x]^ 
2)*(1 + (b*Tan[e + f*x]^2)/(a + b))^p))/(b*f*(3 + 2*p)*(1 + (b*Tan[e + f*x 
]^2)/(a + b))^p)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4634, 299, 238, 237}

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.

\(\displaystyle \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (e+f x)^4 \left (a+b \sec (e+f x)^2\right )^pdx\)

\(\Big \downarrow \) 4634

\(\displaystyle \frac {\int \left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^pd\tan (e+f x)}{f}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b (2 p+3)}-\frac {(a-2 b (p+1)) \int \left (b \tan ^2(e+f x)+a+b\right )^pd\tan (e+f x)}{b (2 p+3)}}{f}\)

\(\Big \downarrow \) 238

\(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b (2 p+3)}-\frac {(a-2 b (p+1)) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \int \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^pd\tan (e+f x)}{b (2 p+3)}}{f}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b (2 p+3)}-\frac {(a-2 b (p+1)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )}{b (2 p+3)}}{f}\)

Input:

Int[Sec[e + f*x]^4*(a + b*Sec[e + f*x]^2)^p,x]
 

Output:

((Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(1 + p))/(b*(3 + 2*p)) - ((a - 2 
*b*(1 + p))*Hypergeometric2F1[1/2, -p, 3/2, -((b*Tan[e + f*x]^2)/(a + b))] 
*Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^p)/(b*(3 + 2*p)*(1 + (b*Tan[e + f 
*x]^2)/(a + b))^p))/f
 

Defintions of rubi rules used

rule 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 3042

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

rule 4634

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

Maple [F]

\[\int \sec \left (f x +e \right )^{4} \left (a +b \sec \left (f x +e \right )^{2}\right )^{p}d x\]

Input:

int(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)
 

Output:

int(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)
 

Fricas [F]

\[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{4} \,d x } \] Input:

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="fricas")
 

Output:

integral((b*sec(f*x + e)^2 + a)^p*sec(f*x + e)^4, x)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)**4*(a+b*sec(f*x+e)**2)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{4} \,d x } \] Input:

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^p*sec(f*x + e)^4, x)
 

Giac [F]

\[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{4} \,d x } \] Input:

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^p*sec(f*x + e)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p}{{\cos \left (e+f\,x\right )}^4} \,d x \] Input:

int((a + b/cos(e + f*x)^2)^p/cos(e + f*x)^4,x)
 

Output:

int((a + b/cos(e + f*x)^2)^p/cos(e + f*x)^4, x)
 

Reduce [F]

\[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int \left (\sec \left (f x +e \right )^{2} b +a \right )^{p} \sec \left (f x +e \right )^{4}d x \] Input:

int(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)
 

Output:

int((sec(e + f*x)**2*b + a)**p*sec(e + f*x)**4,x)