3.4.11 \(\int \frac {\sec ^n(e+f x)}{(a+a \sec (e+f x))^{3/2}} \, dx\) [311]
Optimal. Leaf size=67 \[ \frac {F_1\left (\frac {1}{2};1-n,2;\frac {3}{2};1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right ) \tan (e+f x)}{2 a f \sqrt {a+a \sec (e+f x)}} \]
[Out]
1/2*AppellF1(1/2,1-n,2,3/2,1-sec(f*x+e),1/2-1/2*sec(f*x+e))*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^(1/2)
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Rubi [A]
time = 0.10, antiderivative size = 67, 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 = {3913, 3910,
129, 440} \begin {gather*} \frac {\tan (e+f x) F_1\left (\frac {1}{2};1-n,2;\frac {3}{2};1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right )}{2 a f \sqrt {a \sec (e+f x)+a}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Sec[e + f*x]^n/(a + a*Sec[e + f*x])^(3/2),x]
[Out]
(AppellF1[1/2, 1 - n, 2, 3/2, 1 - Sec[e + f*x], (1 - Sec[e + f*x])/2]*Tan[e + f*x])/(2*a*f*Sqrt[a + a*Sec[e +
f*x]])
Rule 129
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 440
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 3910
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 3913
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[a^Int
Part[m]*((a + b*Csc[e + f*x])^FracPart[m]/(1 + (b/a)*Csc[e + f*x])^FracPart[m]), Int[(1 + (b/a)*Csc[e + f*x])^
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)}{(a+a \sec (e+f x))^{3/2}} \, dx &=\frac {\sqrt {1+\sec (e+f x)} \int \frac {\sec ^n(e+f x)}{(1+\sec (e+f x))^{3/2}} \, dx}{a \sqrt {a+a \sec (e+f x)}}\\ &=\frac {\tan (e+f x) \text {Subst}\left (\int \frac {(1-x)^{-1+n}}{(2-x)^2 \sqrt {x}} \, dx,x,1-\sec (e+f x)\right )}{a f \sqrt {1-\sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(2 \tan (e+f x)) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{-1+n}}{\left (2-x^2\right )^2} \, dx,x,\sqrt {1-\sec (e+f x)}\right )}{a f \sqrt {1-\sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {F_1\left (\frac {1}{2};1-n,2;\frac {3}{2};1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right ) \tan (e+f x)}{2 a f \sqrt {a+a \sec (e+f x)}}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(2992\) vs. \(2(67)=134\).
time = 6.27, size = 2992, normalized size = 44.66 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[e + f*x]^n/(a + a*Sec[e + f*x])^(3/2),x]
[Out]
(6*AppellF1[1/2, -3/2 + 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 + (-3 + 2*n)/2)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(3/2 + n)*Tan[(e + f*x)/2]*(-1 + Tan[(e + f*x)/2]
^2)^2)/(f*(a*(1 + Sec[e + f*x]))^(3/2)*(3*AppellF1[1/2, -3/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*
x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -3/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-3 + 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)*((12*Appe
llF1[1/2, -3/2 + 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)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(3/2 + n)*Tan[(e + f*x)/2]^2*(-1 + Tan[(e + f*x)/2]^2))/(3*AppellF1[1/2
, -3/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -3/2 + n, 2 - n,
5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-3 + 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*AppellF1[1/2, -3/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -T
an[(e + f*x)/2]^2]*Cos[e + f*x]*(Sec[(e + f*x)/2]^2)^(1 + n)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(3/2 + n)*(-1 +
Tan[(e + f*x)/2]^2)^2)/(3*AppellF1[1/2, -3/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(
-1 + n)*AppellF1[3/2, -3/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-3 + 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) - (6*AppellF1[1/2, -3/2
+ 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])^(3/2 + n)*Sin[e + f*x]*Tan[(e + f*x)/2]*(-1 + Tan[(e + f*x)/2]^2)^2)/(3*AppellF1[1/2, -3/2 + n, 1 - n, 3/
2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -3/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]
^2, -Tan[(e + f*x)/2]^2] + (-3 + 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) + (6*n*AppellF1[1/2, -3/2 + 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])^(3/2 + n)*Tan[(e + f*x)/2]^2*(-1 + Tan[(
e + f*x)/2]^2)^2)/(3*AppellF1[1/2, -3/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(-1 + n
)*AppellF1[3/2, -3/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-3 + 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) + (6*Cos[e + f*x]*(Sec[(e + f*
x)/2]^2)^n*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(3/2 + n)*Tan[(e + f*x)/2]*(-1/3*((1 - n)*AppellF1[3/2, -3/2 + 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]) + ((-3/2 + n)*Appel
lF1[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 + Tan[(e + f*x)/2]^2)^2)/(3*AppellF1[1/2, -3/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]
+ (2*(-1 + n)*AppellF1[3/2, -3/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-3 + 2*n)*Appel
lF1[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) - (6*AppellF1[1/2
, -3/2 + 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])^(3/2 + n)*Tan[(e + f*x)/2]*(-1 + Tan[(e + f*x)/2]^2)^2*((2*(-1 + n)*AppellF1[3/2, -3/
2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-3 + 2*n)*AppellF1[3/2, -1/2 + n, 1 - n, 5/2, T
an[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2] + 3*(-1/3*((1 - n)*AppellF1[3/2,
-3/2 + 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]) + ((-3/2 +
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, -3/2 + n, 3 - 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*(-3/2 + 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) + (-3 + 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, 1/2 + 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, -3/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^
2] + (2*(-1 + n)*AppellF1[3/2, -3/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-3 + 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)^2 + (6*(3/2 + n
)*AppellF1[1/2, -3/2 + 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])^(1/2 + n)*Tan[(e + f*x)/2]*(-1 + Tan[(e + f*x)/2]^2)^2*(-(Cos[(e + f*x)
/2]*Sec[e + f*x]*Sin[(e + f*x)/2]) + Cos[(e + f...
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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\sec ^{n}\left (f x +e \right )}{\left (a +a \sec \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(f*x+e)^n/(a+a*sec(f*x+e))^(3/2),x)
[Out]
int(sec(f*x+e)^n/(a+a*sec(f*x+e))^(3/2),x)
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)^n/(a+a*sec(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate(sec(f*x + e)^n/(a*sec(f*x + e) + a)^(3/2), x)
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)^n/(a+a*sec(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
integral(sqrt(a*sec(f*x + e) + a)*sec(f*x + e)^n/(a^2*sec(f*x + e)^2 + 2*a^2*sec(f*x + e) + a^2), x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{n}{\left (e + f x \right )}}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)**n/(a+a*sec(f*x+e))**(3/2),x)
[Out]
Integral(sec(e + f*x)**n/(a*(sec(e + f*x) + 1))**(3/2), x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)^n/(a+a*sec(f*x+e))^(3/2),x, algorithm="giac")
[Out]
integrate(sec(f*x + e)^n/(a*sec(f*x + e) + a)^(3/2), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1/cos(e + f*x))^n/(a + a/cos(e + f*x))^(3/2),x)
[Out]
int((1/cos(e + f*x))^n/(a + a/cos(e + f*x))^(3/2), x)
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