3.4.4 \(\int \sec ^6(e+f x) (a+b \sec ^2(e+f x))^p \, dx\) [304]

3.4.4.1 Optimal result

Integrand size = 23, antiderivative size = 216 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=-\frac {(3 a-2 b (2+p)) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\sec ^2(e+f x) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{b^2 f (3+2 p) (5+2 p)} \]

output

-(3*a-2*b*(2+p))*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(p+1)/b^2/f/(4*p^2+16*p+1 
5)+sec(f*x+e)^2*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(p+1)/b/f/(5+2*p)+(3*a^2-4 
*a*b*(p+1)+4*b^2*(p^2+3*p+2))*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^2/f/(4*p^2+16*p+15)/((1+b*tan(f*x 
+e)^2/(a+b))^p)
 

3.4.4.2 Mathematica [A] (verified)

Time = 2.16 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.69 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\left (a+b \sec ^2(e+f x)\right )^p \tan (e+f x) \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p} \left (15 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )+10 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x)+3 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan ^4(e+f x)\right )}{15 f} \]

input

Integrate[Sec[e + f*x]^6*(a + b*Sec[e + f*x]^2)^p,x]
 

output

((a + b*Sec[e + f*x]^2)^p*Tan[e + f*x]*(15*Hypergeometric2F1[1/2, -p, 3/2, 
 -((b*Tan[e + f*x]^2)/(a + b))] + 10*Hypergeometric2F1[3/2, -p, 5/2, -((b* 
Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]^2 + 3*Hypergeometric2F1[5/2, -p, 7/ 
2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]^4))/(15*f*(1 + (b*Tan[e + f 
*x]^2)/(a + b))^p)
 

3.4.4.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4634, 318, 25, 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.

\(\displaystyle \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (e+f x)^6 \left (a+b \sec (e+f x)^2\right )^pdx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 318

\(\displaystyle \frac {\frac {\int -\left (b \tan ^2(e+f x)+a+b\right )^p \left ((3 a-2 b (p+2)) \tan ^2(e+f x)+a-2 b (p+2)\right )d\tan (e+f x)}{b (2 p+5)}+\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b (2 p+5)}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b (2 p+5)}-\frac {\int \left (b \tan ^2(e+f x)+a+b\right )^p \left ((3 a-2 b (p+2)) \tan ^2(e+f x)+a-2 b (p+2)\right )d\tan (e+f x)}{b (2 p+5)}}{f}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b (2 p+5)}-\frac {\frac {(3 a-2 b (p+2)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b (2 p+3)}-\frac {\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \int \left (b \tan ^2(e+f x)+a+b\right )^pd\tan (e+f x)}{b (2 p+3)}}{b (2 p+5)}}{f}\)

\(\Big \downarrow \) 238

\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b (2 p+5)}-\frac {\frac {(3 a-2 b (p+2)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b (2 p+3)}-\frac {\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \int \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^pd\tan (e+f x)}{b (2 p+3)}}{b (2 p+5)}}{f}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b (2 p+5)}-\frac {\frac {(3 a-2 b (p+2)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b (2 p+3)}-\frac {\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )}{b (2 p+3)}}{b (2 p+5)}}{f}\)

input

Int[Sec[e + f*x]^6*(a + b*Sec[e + f*x]^2)^p,x]
 

output

((Tan[e + f*x]*(1 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x]^2)^(1 + p))/(b 
*(5 + 2*p)) - (((3*a - 2*b*(2 + p))*Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2 
)^(1 + p))/(b*(3 + 2*p)) - ((3*a^2 - 4*a*b*(1 + p) + 4*b^2*(2 + 3*p + p^2) 
)*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))/(b*(5 + 2*p)))/f
 

3.4.4.3.1 Defintions of rubi rules used

rule 25

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

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 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

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]
 

3.4.4.4 Maple [F]

\[\int \sec \left (f x +e \right )^{6} \left (a +b \sec \left (f x +e \right )^{2}\right )^{p}d x\]

input

int(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^p,x)
 

output

int(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^p,x)
 

3.4.4.5 Fricas [F]

\[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{6} \,d x } \]

input

integrate(sec(f*x+e)^6*(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)^6, x)
 

3.4.4.6 Sympy [F(-1)]

Timed out. \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Timed out} \]

input

integrate(sec(f*x+e)**6*(a+b*sec(f*x+e)**2)**p,x)
 

output

Timed out
 

3.4.4.7 Maxima [F]

\[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{6} \,d x } \]

input

integrate(sec(f*x+e)^6*(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)^6, x)
 

3.4.4.8 Giac [F]

\[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{6} \,d x } \]

input

integrate(sec(f*x+e)^6*(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)^6, x)
 

3.4.4.9 Mupad [F(-1)]

Timed out. \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p}{{\cos \left (e+f\,x\right )}^6} \,d x \]

input

int((a + b/cos(e + f*x)^2)^p/cos(e + f*x)^6,x)
 

output

int((a + b/cos(e + f*x)^2)^p/cos(e + f*x)^6, x)