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

Optimal. Leaf size=216 \[ \frac {\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right )}{b^2 f (2 p+3) (2 p+5)}-\frac {(3 a-2 b (p+2)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b^2 f (2 p+3) (2 p+5)}+\frac {\tan (e+f x) \sec ^2(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b f (2 p+5)} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4146, 416, 388, 246, 245} \[ \frac {\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right )}{b^2 f (2 p+3) (2 p+5)}-\frac {(3 a-2 b (p+2)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b^2 f (2 p+3) (2 p+5)}+\frac {\tan (e+f x) \sec ^2(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b f (2 p+5)} \]

Antiderivative was successfully verified.

[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 245

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[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 388

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 416

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 4146

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*} \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx &=\frac {\operatorname {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 {\operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \, _2F_1\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*}

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Mathematica [A]  time = 2.11, size = 149, normalized size = 0.69 \[ \frac {\tan (e+f x) \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \left (a+b \sec ^2(e+f x)\right )^p \left (10 \tan ^2(e+f x) \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right )+15 \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right )+3 \tan ^4(e+f x) \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right )\right )}{15 f} \]

Antiderivative was successfully verified.

[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)

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{6}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 2.48, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{6}\left (f x +e \right )\right ) \left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p}{{\cos \left (e+f\,x\right )}^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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