3.4.49 \(\int \frac {(a+a \sec (e+f x))^m}{\sqrt {d \sec (e+f x)}} \, dx\) [349]
Optimal. Leaf size=96 \[ \frac {2 F_1\left (-\frac {1}{2};\frac {1}{2},\frac {1}{2}-m;\frac {1}{2};\sec (e+f x),-\sec (e+f x)\right ) (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)}{f \sqrt {1-\sec (e+f x)} \sqrt {d \sec (e+f x)}} \]
[Out]
2*AppellF1(-1/2,1/2-m,1/2,1/2,-sec(f*x+e),sec(f*x+e))*(1+sec(f*x+e))^(-1/2-m)*(a+a*sec(f*x+e))^m*tan(f*x+e)/f/
(1-sec(f*x+e))^(1/2)/(d*sec(f*x+e))^(1/2)
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Rubi [A]
time = 0.08, antiderivative size = 96, 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 = {3913, 3912,
138} \begin {gather*} \frac {2 \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m F_1\left (-\frac {1}{2};\frac {1}{2},\frac {1}{2}-m;\frac {1}{2};\sec (e+f x),-\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)} \sqrt {d \sec (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + a*Sec[e + f*x])^m/Sqrt[d*Sec[e + f*x]],x]
[Out]
(2*AppellF1[-1/2, 1/2, 1/2 - m, 1/2, Sec[e + f*x], -Sec[e + f*x]]*(1 + Sec[e + f*x])^(-1/2 - m)*(a + a*Sec[e +
f*x])^m*Tan[e + f*x])/(f*Sqrt[1 - Sec[e + f*x]]*Sqrt[d*Sec[e + f*x]])
Rule 138
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Rule 3912
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 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 {(a+a \sec (e+f x))^m}{\sqrt {d \sec (e+f x)}} \, dx &=\left ((1+\sec (e+f x))^{-m} (a+a \sec (e+f x))^m\right ) \int \frac {(1+\sec (e+f x))^m}{\sqrt {d \sec (e+f x)}} \, dx\\ &=-\frac {\left (d (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)\right ) \text {Subst}\left (\int \frac {(1+x)^{-\frac {1}{2}+m}}{\sqrt {1-x} (d x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)}}\\ &=\frac {2 F_1\left (-\frac {1}{2};\frac {1}{2},\frac {1}{2}-m;\frac {1}{2};\sec (e+f x),-\sec (e+f x)\right ) (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)}{f \sqrt {1-\sec (e+f x)} \sqrt {d \sec (e+f x)}}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 16.37, size = 2424, normalized size = 25.25 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sec[e + f*x])^m/Sqrt[d*Sec[e + f*x]],x]
[Out]
(-3*2^(1 + m)*AppellF1[1/2, -1/2 + m, 3/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sqrt[Sec[e + f*x]]*(C
os[(e + f*x)/2]^2*Sec[e + f*x])^(-1/2 + m)*(a*(1 + Sec[e + f*x]))^m*(Cos[2*(e + f*x)]*((1 + Sec[e + f*x])^m/(2
*Sqrt[Sec[e + f*x]]) - (I/2)*Sqrt[Sec[e + f*x]]*(1 + Sec[e + f*x])^m*Sin[e + f*x]) + ((1 + Sec[e + f*x])^m/2 +
(I/2)*(1 + Sec[e + f*x])^m*Sin[2*(e + f*x)])/Sqrt[Sec[e + f*x]] + Sqrt[Sec[e + f*x]]*Sin[e + f*x]*((-1/2*I)*(
1 + Sec[e + f*x])^m + ((1 + Sec[e + f*x])^m*Sin[2*(e + f*x)])/2))*Tan[(e + f*x)/2])/(f*(Sec[(e + f*x)/2]^2)^(3
/2)*Sqrt[d*Sec[e + f*x]]*(1 + Sec[e + f*x])^m*(-3*AppellF1[1/2, -1/2 + m, 3/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(
e + f*x)/2]^2] + (3*AppellF1[3/2, -1/2 + m, 5/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 - 2*m)*App
ellF1[3/2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)*((-3*2^m*AppellF1[
1/2, -1/2 + m, 3/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(-1/2 + m)
)/(Sqrt[Sec[(e + f*x)/2]^2]*(-3*AppellF1[1/2, -1/2 + m, 3/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (
3*AppellF1[3/2, -1/2 + m, 5/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 - 2*m)*AppellF1[3/2, 1/2 + m
, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)) + (9*2^m*AppellF1[1/2, -1/2 + m, 3/
2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(-1/2 + m)*Tan[(e + f*x)/2]
^2)/((Sec[(e + f*x)/2]^2)^(3/2)*(-3*AppellF1[1/2, -1/2 + m, 3/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]
+ (3*AppellF1[3/2, -1/2 + m, 5/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 - 2*m)*AppellF1[3/2, 1/2
+ m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)) - (3*2^(1 + m)*(Cos[(e + f*x)/2
]^2*Sec[e + f*x])^(-1/2 + m)*Tan[(e + f*x)/2]*(-1/2*(AppellF1[3/2, -1/2 + m, 5/2, 5/2, Tan[(e + f*x)/2]^2, -Ta
n[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]) + ((-1/2 + m)*AppellF1[3/2, 1/2 + m, 3/2, 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))/((Sec[(e + f*x)/2]^2)^(3/2)*(-3*Appe
llF1[1/2, -1/2 + m, 3/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (3*AppellF1[3/2, -1/2 + m, 5/2, 5/2,
Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 - 2*m)*AppellF1[3/2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan
[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)) + (3*2^(1 + m)*AppellF1[1/2, -1/2 + m, 3/2, 3/2, Tan[(e + f*x)/2]^2, -T
an[(e + f*x)/2]^2]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(-1/2 + m)*Tan[(e + f*x)/2]*((3*AppellF1[3/2, -1/2 + m, 5
/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 - 2*m)*AppellF1[3/2, 1/2 + m, 3/2, 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*(-1/2*(AppellF1[3/2, -1/2 + m, 5/2, 5/2, Ta
n[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]) + ((-1/2 + m)*AppellF1[3/2, 1/2 +
m, 3/2, 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*(3*((-3*AppellF1[5/2, -1/2 + m, 7/2, 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])/2 + (3*(-1/2 + m)*AppellF1[5/2, 1/2 + m, 5/2, 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*m)*((-9*AppellF1[5/2, 1/2 + m, 5/2, 7/2, Tan[(e + f*x)/2]^2, -Ta
n[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/10 + (3*(1/2 + m)*AppellF1[5/2, 3/2 + m, 3/2, 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))))/((Sec[(e + f*x)/2]^2)^(3/2)*(-
3*AppellF1[1/2, -1/2 + m, 3/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (3*AppellF1[3/2, -1/2 + m, 5/2,
5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 - 2*m)*AppellF1[3/2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2
, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)^2) - (3*2^(1 + m)*(-1/2 + m)*AppellF1[1/2, -1/2 + m, 3/2, 3/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(-3/2 + m)*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*Sec[e + f*x]*Tan[e + f*x]))/((Sec[(e + f*x)/2]^2
)^(3/2)*(-3*AppellF1[1/2, -1/2 + m, 3/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (3*AppellF1[3/2, -1/2
+ m, 5/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 - 2*m)*AppellF1[3/2, 1/2 + m, 3/2, 5/2, Tan[(e +
f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))))
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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {\left (a +a \sec \left (f x +e \right )\right )^{m}}{\sqrt {d \sec \left (f x +e \right )}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(f*x+e))^m/(d*sec(f*x+e))^(1/2),x)
[Out]
int((a+a*sec(f*x+e))^m/(d*sec(f*x+e))^(1/2),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((a+a*sec(f*x+e))^m/(d*sec(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate((a*sec(f*x + e) + a)^m/sqrt(d*sec(f*x + e)), 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((a+a*sec(f*x+e))^m/(d*sec(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
integral(sqrt(d*sec(f*x + e))*(a*sec(f*x + e) + a)^m/(d*sec(f*x + e)), x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m}}{\sqrt {d \sec {\left (e + f x \right )}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(f*x+e))**m/(d*sec(f*x+e))**(1/2),x)
[Out]
Integral((a*(sec(e + f*x) + 1))**m/sqrt(d*sec(e + f*x)), 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((a+a*sec(f*x+e))^m/(d*sec(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate((a*sec(f*x + e) + a)^m/sqrt(d*sec(f*x + e)), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m}{\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a/cos(e + f*x))^m/(d/cos(e + f*x))^(1/2),x)
[Out]
int((a + a/cos(e + f*x))^m/(d/cos(e + f*x))^(1/2), x)
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