3.348 \(\int \sqrt {d \sec (e+f x)} (a+a \sec (e+f x))^m \, dx\)
Optimal. Leaf size=96 \[ -\frac {2 \tan (e+f x) \sqrt {d \sec (e+f x)} (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\sec (e+f x),-\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)}} \]
[Out]
-2*AppellF1(1/2,1/2-m,1/2,3/2,-sec(f*x+e),sec(f*x+e))*(1+sec(f*x+e))^(-1/2-m)*(a+a*sec(f*x+e))^m*(d*sec(f*x+e)
)^(1/2)*tan(f*x+e)/f/(1-sec(f*x+e))^(1/2)
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Rubi [A] time = 0.12, 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 =
{3828, 3827, 133} \[ -\frac {2 \tan (e+f x) \sqrt {d \sec (e+f x)} (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\sec (e+f x),-\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[d*Sec[e + f*x]]*(a + a*Sec[e + f*x])^m,x]
[Out]
(-2*AppellF1[1/2, 1/2, 1/2 - m, 3/2, Sec[e + f*x], -Sec[e + f*x]]*Sqrt[d*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]])
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 \sqrt {d \sec (e+f x)} (a+a \sec (e+f x))^m \, dx &=\left ((1+\sec (e+f x))^{-m} (a+a \sec (e+f x))^m\right ) \int \sqrt {d \sec (e+f x)} (1+\sec (e+f x))^m \, 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 ) \operatorname {Subst}\left (\int \frac {(1+x)^{-\frac {1}{2}+m}}{\sqrt {1-x} \sqrt {d x}} \, 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 {3}{2};\sec (e+f x),-\sec (e+f x)\right ) \sqrt {d \sec (e+f x)} (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)}}\\ \end {align*}
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Mathematica [B] time = 14.64, size = 2225, normalized size = 23.18 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sqrt[d*Sec[e + f*x]]*(a + a*Sec[e + f*x])^m,x]
[Out]
(2^(1 + m)*AppellF1[1/2, 1/2 + m, 1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sqrt[d*Sec[e + f*x]]*(Cos
[(e + f*x)/2]^2*Sec[e + f*x])^(1/2 + m)*(a*(1 + Sec[e + f*x]))^m*Tan[(e + f*x)/2])/(f*Sqrt[Sec[(e + f*x)/2]^2]
*(AppellF1[1/2, 1/2 + m, 1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - ((AppellF1[3/2, 1/2 + m, 3/2, 5/
2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - (1 + 2*m)*AppellF1[3/2, 3/2 + m, 1/2, 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, 1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e
+ f*x)/2]^2]*Sqrt[Sec[(e + f*x)/2]^2]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(1/2 + m))/(AppellF1[1/2, 1/2 + m, 1/
2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - ((AppellF1[3/2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan
[(e + f*x)/2]^2] - (1 + 2*m)*AppellF1[3/2, 3/2 + m, 1/2, 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, 1/2, 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)/(Sqrt[Sec[(e + f*x)/2]^2]*(AppellF1[1/2, 1/2 + m, 1/2, 3/2
, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - ((AppellF1[3/2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e +
f*x)/2]^2] - (1 + 2*m)*AppellF1[3/2, 3/2 + m, 1/2, 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]*(-1/6*(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]) + ((1/2 + m)*App
ellF1[3/2, 3/2 + m, 1/2, 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)
)/(Sqrt[Sec[(e + f*x)/2]^2]*(AppellF1[1/2, 1/2 + m, 1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - ((App
ellF1[3/2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - (1 + 2*m)*AppellF1[3/2, 3/2 + m, 1/2,
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, 1/
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]*
(-1/6*(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]) + ((1/2 + m)*AppellF1[3/2, 3/2 + m, 1/2, 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 - ((AppellF1[3/2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - (1 + 2
*m)*AppellF1[3/2, 3/2 + m, 1/2, 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*((-9*AppellF1[5/2, 1/2 + m, 5/2, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*S
ec[(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, -Ta
n[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5 - (1 + 2*m)*((-3*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])/10 + (3*(3/2 + m)*AppellF1[5/2, 5/2
+ m, 1/2, 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)))/3))/(Sqrt[S
ec[(e + f*x)/2]^2]*(AppellF1[1/2, 1/2 + m, 1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - ((AppellF1[3/2
, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - (1 + 2*m)*AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)/3)^2) + (2^(1 + m)*(1/2 + m)*AppellF1[1/2, 1/2 + m,
1/2, 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]*(-(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]))/(Sqrt[
Sec[(e + f*x)/2]^2]*(AppellF1[1/2, 1/2 + m, 1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - ((AppellF1[3/
2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - (1 + 2*m)*AppellF1[3/2, 3/2 + m, 1/2, 5/2, Ta
n[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)/3))))
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fricas [F] time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \sec \left (f x + e\right )} {\left (a \sec \left (f x + e\right ) + a\right )}^{m}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^m,x, algorithm="fricas")
[Out]
integral(sqrt(d*sec(f*x + e))*(a*sec(f*x + e) + a)^m, x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d \sec \left (f x + e\right )} {\left (a \sec \left (f x + e\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^m,x, algorithm="giac")
[Out]
integrate(sqrt(d*sec(f*x + e))*(a*sec(f*x + e) + a)^m, x)
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maple [F] time = 1.18, size = 0, normalized size = 0.00 \[ \int \sqrt {d \sec \left (f x +e \right )}\, \left (a +a \sec \left (f x +e \right )\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^m,x)
[Out]
int((d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^m,x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d \sec \left (f x + e\right )} {\left (a \sec \left (f x + e\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^m,x, algorithm="maxima")
[Out]
integrate(sqrt(d*sec(f*x + e))*(a*sec(f*x + e) + a)^m, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}} \,d x \]
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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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m} \sqrt {d \sec {\left (e + f x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))**(1/2)*(a+a*sec(f*x+e))**m,x)
[Out]
Integral((a*(sec(e + f*x) + 1))**m*sqrt(d*sec(e + f*x)), x)
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