3.319 \(\int \frac {(d \sec (e+f x))^n}{(a+a \sec (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=78 \[ -\frac {\tan (e+f x) F_1\left (n;\frac {1}{2},2;n+1;\sec (e+f x),-\sec (e+f x)\right ) (d \sec (e+f x))^n}{a f n \sqrt {1-\sec (e+f x)} \sqrt {a \sec (e+f x)+a}} \]
[Out]
-AppellF1(n,2,1/2,1+n,-sec(f*x+e),sec(f*x+e))*(d*sec(f*x+e))^n*tan(f*x+e)/a/f/n/(1-sec(f*x+e))^(1/2)/(a+a*sec(
f*x+e))^(1/2)
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Rubi [A] time = 0.15, antiderivative size = 78, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used =
{3828, 3827, 133} \[ -\frac {\tan (e+f x) F_1\left (n;\frac {1}{2},2;n+1;\sec (e+f x),-\sec (e+f x)\right ) (d \sec (e+f x))^n}{a f n \sqrt {1-\sec (e+f x)} \sqrt {a \sec (e+f x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[(d*Sec[e + f*x])^n/(a + a*Sec[e + f*x])^(3/2),x]
[Out]
-((AppellF1[n, 1/2, 2, 1 + n, Sec[e + f*x], -Sec[e + f*x]]*(d*Sec[e + f*x])^n*Tan[e + f*x])/(a*f*n*Sqrt[1 - Se
c[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
Rule 133
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Rule 3827
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^2*
d*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((d*x)^(n - 1)*(a + b*x)^(m -
1/2))/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !In
tegerQ[m] && GtQ[a, 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 {(d \sec (e+f x))^n}{(a+a \sec (e+f x))^{3/2}} \, dx &=\frac {\sqrt {1+\sec (e+f x)} \int \frac {(d \sec (e+f x))^n}{(1+\sec (e+f x))^{3/2}} \, dx}{a \sqrt {a+a \sec (e+f x)}}\\ &=-\frac {(d \tan (e+f x)) \operatorname {Subst}\left (\int \frac {(d x)^{-1+n}}{\sqrt {1-x} (1+x)^2} \, dx,x,\sec (e+f x)\right )}{a f \sqrt {1-\sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=-\frac {F_1\left (n;\frac {1}{2},2;1+n;\sec (e+f x),-\sec (e+f x)\right ) (d \sec (e+f x))^n \tan (e+f x)}{a f n \sqrt {1-\sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ \end {align*}
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Mathematica [B] time = 6.22, size = 3005, normalized size = 38.53 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(d*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 - n + (-3 + 2*n)/2)*(d*Sec[e + f*x])^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)/(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*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)^(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, -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)*(-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*Co
s[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)*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 + 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]*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, 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, -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)*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 + (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*x)/2]^2*Sec[e + f*x]*Tan[e + f*x]))/(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)))
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fricas [F] time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \sec \left (f x + e\right ) + a} \left (d \sec \left (f x + e\right )\right )^{n}}{a^{2} \sec \left (f x + e\right )^{2} + 2 \, a^{2} \sec \left (f x + e\right ) + a^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*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)*(d*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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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{n}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^n/(a+a*sec(f*x+e))^(3/2),x, algorithm="giac")
[Out]
integrate((d*sec(f*x + e))^n/(a*sec(f*x + e) + a)^(3/2), x)
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maple [F] time = 0.87, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x +e \right )\right )^{n}}{\left (a +a \sec \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*sec(f*x+e))^n/(a+a*sec(f*x+e))^(3/2),x)
[Out]
int((d*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 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{n}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^n/(a+a*sec(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate((d*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 \[ \int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^n}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d/cos(e + f*x))^n/(a + a/cos(e + f*x))^(3/2),x)
[Out]
int((d/cos(e + f*x))^n/(a + a/cos(e + f*x))^(3/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec {\left (e + f x \right )}\right )^{n}}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))**n/(a+a*sec(f*x+e))**(3/2),x)
[Out]
Integral((d*sec(e + f*x))**n/(a*(sec(e + f*x) + 1))**(3/2), x)
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