3.449 \(\int (a+b \sec ^2(e+f x))^p \, dx\)
Optimal. Leaf size=83 \[ \frac{\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} F_1\left (\frac{1}{2};1,-p;\frac{3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a+b}\right )}{f} \]
[Out]
(AppellF1[1/2, 1, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]*(a + b + b*Tan[e + f*x
]^2)^p)/(f*(1 + (b*Tan[e + f*x]^2)/(a + b))^p)
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Rubi [A] time = 0.0533354, antiderivative size = 83, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used =
{4128, 430, 429} \[ \frac{\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} F_1\left (\frac{1}{2};1,-p;\frac{3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a+b}\right )}{f} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Sec[e + f*x]^2)^p,x]
[Out]
(AppellF1[1/2, 1, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]*(a + b + b*Tan[e + f*x
]^2)^p)/(f*(1 + (b*Tan[e + f*x]^2)/(a + b))^p)
Rule 4128
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]
Rule 430
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && !(IntegerQ[p] || GtQ[a, 0])
Rule 429
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Rubi steps
\begin{align*} \int \left (a+b \sec ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b+b x^2\right )^p}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\left (\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 \frac{\left (1+\frac{b x^2}{a+b}\right )^p}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{F_1\left (\frac{1}{2};1,-p;\frac{3}{2};-\tan ^2(e+f x),-\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}}{f}\\ \end{align*}
Mathematica [B] time = 6.26908, size = 2137, normalized size = 25.75 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*Sec[e + f*x]^2)^p,x]
[Out]
(3*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Cos[e + f*x]*(a + 2*b + a
*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^p*(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x])/(f*(3*(a + b)*AppellF1[1/2, -p,
1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*
x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f
*x]^2])*Tan[e + f*x]^2)*((3*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*
(a + 2*b + a*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^(-1 + p))/(3*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e +
f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e
+ f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2])*Tan[e + f*x]^2
) - (3*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*(a + 2*b + a*Cos[2*(e
+ f*x)])^p*(Sec[e + f*x]^2)^p*Sin[e + f*x]^2)/(3*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a +
b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] -
(a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2])*Tan[e + f*x]^2) + (6*(a + b
)*p*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*(a + 2*b + a*Cos[2*(e + f*x)])^p
*(Sec[e + f*x]^2)^p*Sin[e + f*x]^2)/(3*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e
+ f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*App
ellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2])*Tan[e + f*x]^2) - (6*a*(a + b)*p*Appel
lF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)
])^(-1 + p)*(Sec[e + f*x]^2)^p*Sin[e + f*x]*Sin[2*(e + f*x)])/(3*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e
+ f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan
[e + f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2])*Tan[e + f*x]
^2) + (3*(a + b)*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^p*Sin[e + f*x]*((2*b*p*AppellF
1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3*(a + b))
- (2*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/3
))/(3*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2,
1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e
+ f*x]^2)/(a + b)), -Tan[e + f*x]^2])*Tan[e + f*x]^2) - (3*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x
]^2)/(a + b)), -Tan[e + f*x]^2]*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^p*Sin[e + f*x]*
(4*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*AppellF1[3/2, -
p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2])*Sec[e + f*x]^2*Tan[e + f*x] + 3*(a + b)*((2*b*p*Ap
pellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3*(a
+ b)) - (2*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*
x])/3) + 2*Tan[e + f*x]^2*(b*p*((-6*AppellF1[5/2, 1 - p, 2, 7/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^
2]*Sec[e + f*x]^2*Tan[e + f*x])/5 - (6*b*(1 - p)*AppellF1[5/2, 2 - p, 1, 7/2, -((b*Tan[e + f*x]^2)/(a + b)), -
Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(5*(a + b))) - (a + b)*((6*b*p*AppellF1[5/2, 1 - p, 2, 7/2, -((b*
Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(5*(a + b)) - (12*AppellF1[5/2, -p, 3,
7/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/5))))/(3*(a + b)*AppellF1[1
/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan
[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Ta
n[e + f*x]^2])*Tan[e + f*x]^2)^2))
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Maple [F] time = 0.007, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*sec(f*x+e)^2)^p,x)
[Out]
int((a+b*sec(f*x+e)^2)^p,x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(f*x+e)^2)^p,x, algorithm="maxima")
[Out]
integrate((b*sec(f*x + e)^2 + a)^p, x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(f*x+e)^2)^p,x, algorithm="fricas")
[Out]
integral((b*sec(f*x + e)^2 + a)^p, x)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{p}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(f*x+e)**2)**p,x)
[Out]
Integral((a + b*sec(e + f*x)**2)**p, x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(f*x+e)^2)^p,x, algorithm="giac")
[Out]
integrate((b*sec(f*x + e)^2 + a)^p, x)