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

Optimal result

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

[Out]

-(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+15)+sec(f*x+e)^2*tan(f*x+e)*(a+b+b*ta
n(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)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4231, 427, 396, 252, 251} \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )}{b^2 f (2 p+3) (2 p+5)}-\frac {(3 a-2 b (p+2)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b^2 f (2 p+3) (2 p+5)}+\frac {\tan (e+f x) \sec ^2(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b f (2 p+5)} \]

[In]

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

[Out]

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

Rule 251

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

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 396

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

Rule 427

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 4231

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (1+x^2\right )^2 \left (a+b+b x^2\right )^p \, dx,x,\tan (e+f x)\right )}{f} \\ & = \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 {\text {Subst}\left (\int \left (a+b+b x^2\right )^p \left (-a+2 b (2+p)-(3 a-2 b (2+p)) x^2\right ) \, dx,x,\tan (e+f x)\right )}{b f (5+2 p)} \\ & = -\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 ) \text {Subst}\left (\int \left (a+b+b x^2\right )^p \, dx,x,\tan (e+f x)\right )}{b^2 f (3+2 p) (5+2 p)} \\ & = -\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 (\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}\right ) \text {Subst}\left (\int \left (1+\frac {b x^2}{a+b}\right )^p \, dx,x,\tan (e+f x)\right )}{b^2 f (3+2 p) (5+2 p)} \\ & = -\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)} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

((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)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^p,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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