3.310 \(\int \frac {\sec ^n(e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx\)
Optimal. Leaf size=61 \[ \frac {\tan (e+f x) F_1\left (\frac {1}{2};1-n,1;\frac {3}{2};1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right )}{f \sqrt {a \sec (e+f x)+a}} \]
[Out]
AppellF1(1/2,1-n,1,3/2,1-sec(f*x+e),1/2-1/2*sec(f*x+e))*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)
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Rubi [A] time = 0.13, antiderivative size = 61, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used =
{3828, 3825, 130, 429} \[ \frac {\tan (e+f x) F_1\left (\frac {1}{2};1-n,1;\frac {3}{2};1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right )}{f \sqrt {a \sec (e+f x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[Sec[e + f*x]^n/Sqrt[a + a*Sec[e + f*x]],x]
[Out]
(AppellF1[1/2, 1 - n, 1, 3/2, 1 - Sec[e + f*x], (1 - Sec[e + f*x])/2]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]
])
Rule 130
Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + (b*x^k)/e)^m*(c + (d*x^k)/e)^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]
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])
Rule 3825
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Dist[(((a*
d)/b)^n*Cot[e + f*x])/(a^(n - 2)*f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((a - x)^(n -
1)*(2*a - x)^(m - 1/2))/Sqrt[x], x], x, a - b*Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2
- b^2, 0] && !IntegerQ[m] && GtQ[a, 0] && !IntegerQ[n] && GtQ[(a*d)/b, 0]
Rule 3828
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(1 + (b*Csc[e + f*x])/a)^FracPart[m], Int[(1 + (b*Csc[e + f*x])/a)^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && !GtQ
[a, 0]
Rubi steps
\begin {align*} \int \frac {\sec ^n(e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx &=\frac {\sqrt {1+\sec (e+f x)} \int \frac {\sec ^n(e+f x)}{\sqrt {1+\sec (e+f x)}} \, dx}{\sqrt {a+a \sec (e+f x)}}\\ &=\frac {\tan (e+f x) \operatorname {Subst}\left (\int \frac {(1-x)^{-1+n}}{(2-x) \sqrt {x}} \, dx,x,1-\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(2 \tan (e+f x)) \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^{-1+n}}{2-x^2} \, dx,x,\sqrt {1-\sec (e+f x)}\right )}{f \sqrt {1-\sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {F_1\left (\frac {1}{2};1-n,1;\frac {3}{2};1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right ) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}\\ \end {align*}
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Mathematica [B] time = 6.24, size = 2964, normalized size = 48.59 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[e + f*x]^n/Sqrt[a + a*Sec[e + f*x]],x]
[Out]
(3*Sqrt[2]*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^n
*Sec[e + f*x]^(-1/2 + (-1 + 2*n)/2)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^n*Sqrt[1 + Sec[e + f*x]]*Tan[(e + f*x)/2
])/(f*Sqrt[a*(1 + Sec[e + f*x])]*(3*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^
2] + (2*(-1 + n)*AppellF1[3/2, -1/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-1 + 2*n)*App
ellF1[3/2, 1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)*((3*AppellF1[1/2
, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x]*(Sec[(e + f*x)/2]^2)^(1 + n)*(Co
s[(e + f*x)/2]^2*Sec[e + f*x])^n*Sqrt[1 + Sec[e + f*x]])/(Sqrt[2]*(3*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(
e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -1/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan
[(e + f*x)/2]^2] + (-1 + 2*n)*AppellF1[3/2, 1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan
[(e + f*x)/2]^2)) - (3*Sqrt[2]*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(S
ec[(e + f*x)/2]^2)^n*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^n*Sqrt[1 + Sec[e + f*x]]*Sin[e + f*x]*Tan[(e + f*x)/2])
/(3*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -
1/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-1 + 2*n)*AppellF1[3/2, 1/2 + n, 1 - n, 5/2,
Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) + (3*Sqrt[2]*n*AppellF1[1/2, -1/2 + n, 1 - n, 3/
2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x]*(Sec[(e + f*x)/2]^2)^n*(Cos[(e + f*x)/2]^2*Sec[e + f*
x])^n*Sqrt[1 + Sec[e + f*x]]*Tan[(e + f*x)/2]^2)/(3*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -T
an[(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -1/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]
+ (-1 + 2*n)*AppellF1[3/2, 1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)
+ (3*Sqrt[2]*Cos[e + f*x]*(Sec[(e + f*x)/2]^2)^n*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^n*Sqrt[1 + Sec[e + f*x]]*Ta
n[(e + f*x)/2]*(-1/3*((1 - n)*AppellF1[3/2, -1/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec
[(e + f*x)/2]^2*Tan[(e + f*x)/2]) + ((-1/2 + n)*AppellF1[3/2, 1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e
+ f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/3))/(3*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]
^2, -Tan[(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -1/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/
2]^2] + (-1 + 2*n)*AppellF1[3/2, 1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/
2]^2) - (3*Sqrt[2]*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x]*(
Sec[(e + f*x)/2]^2)^n*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^n*Sqrt[1 + Sec[e + f*x]]*Tan[(e + f*x)/2]*((2*(-1 + n)
*AppellF1[3/2, -1/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-1 + 2*n)*AppellF1[3/2, 1/2 +
n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2] + 3*(-1/3*((1 -
n)*AppellF1[3/2, -1/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*
x)/2]) + ((-1/2 + n)*AppellF1[3/2, 1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)
/2]^2*Tan[(e + f*x)/2])/3) + Tan[(e + f*x)/2]^2*(2*(-1 + n)*((-3*(2 - n)*AppellF1[5/2, -1/2 + n, 3 - n, 7/2, T
an[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5 + (3*(-1/2 + n)*AppellF1[5/2, 1
/2 + n, 2 - n, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5) + (-1 + 2
*n)*((-3*(1 - n)*AppellF1[5/2, 1/2 + n, 2 - n, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^
2*Tan[(e + f*x)/2])/5 + (3*(1/2 + n)*AppellF1[5/2, 3/2 + n, 1 - n, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^
2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5))))/(3*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[
(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -1/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (
-1 + 2*n)*AppellF1[3/2, 1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)^2 +
(3*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^n*(Cos[(
e + f*x)/2]^2*Sec[e + f*x])^n*Tan[(e + f*x)/2]*Tan[e + f*x])/(Sqrt[2]*Sqrt[1 + Sec[e + f*x]]*(3*AppellF1[1/2,
-1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -1/2 + n, 2 - n, 5/
2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-1 + 2*n)*AppellF1[3/2, 1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2
, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)) + (3*Sqrt[2]*n*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/
2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x]*(Sec[(e + f*x)/2]^2)^n*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(-1 + n)*Sqrt
[1 + Sec[e + f*x]]*Tan[(e + f*x)/2]*(-(Cos[(e + f*x)/2]*Sec[e + f*x]*Sin[(e + f*x)/2]) + Cos[(e + f*x)/2]^2*Se
c[e + f*x]*Tan[e + f*x]))/(3*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2
*(-1 + n)*AppellF1[3/2, -1/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-1 + 2*n)*AppellF1[3
/2, 1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)))
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fricas [F] time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sec \left (f x + e\right )^{n}}{\sqrt {a \sec \left (f x + e\right ) + a}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)^n/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
integral(sec(f*x + e)^n/sqrt(a*sec(f*x + e) + a), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )^{n}}{\sqrt {a \sec \left (f x + e\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)^n/(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate(sec(f*x + e)^n/sqrt(a*sec(f*x + e) + a), x)
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maple [F] time = 1.14, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{n}\left (f x +e \right )}{\sqrt {a +a \sec \left (f x +e \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(f*x+e)^n/(a+a*sec(f*x+e))^(1/2),x)
[Out]
int(sec(f*x+e)^n/(a+a*sec(f*x+e))^(1/2),x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )^{n}}{\sqrt {a \sec \left (f x + e\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)^n/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate(sec(f*x + e)^n/sqrt(a*sec(f*x + e) + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1/cos(e + f*x))^n/(a + a/cos(e + f*x))^(1/2),x)
[Out]
int((1/cos(e + f*x))^n/(a + a/cos(e + f*x))^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{n}{\left (e + f x \right )}}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)**n/(a+a*sec(f*x+e))**(1/2),x)
[Out]
Integral(sec(e + f*x)**n/sqrt(a*(sec(e + f*x) + 1)), x)
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