3.350 \(\int \frac{(a+a \sec (e+f x))^m}{(d \sec (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=98 \[ \frac{2 \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac{1}{2}} (a \sec (e+f x)+a)^m F_1\left (-\frac{3}{2};\frac{1}{2},\frac{1}{2}-m;-\frac{1}{2};\sec (e+f x),-\sec (e+f x)\right )}{3 f \sqrt{1-\sec (e+f x)} (d \sec (e+f x))^{3/2}} \]
[Out]
(2*AppellF1[-3/2, 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])/(3*f*Sqrt[1 - Sec[e + f*x]]*(d*Sec[e + f*x])^(3/2))
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Rubi [A] time = 0.135567, antiderivative size = 98, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used =
{3828, 3827, 133} \[ \frac{2 \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac{1}{2}} (a \sec (e+f x)+a)^m F_1\left (-\frac{3}{2};\frac{1}{2},\frac{1}{2}-m;-\frac{1}{2};\sec (e+f x),-\sec (e+f x)\right )}{3 f \sqrt{1-\sec (e+f x)} (d \sec (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sec[e + f*x])^m/(d*Sec[e + f*x])^(3/2),x]
[Out]
(2*AppellF1[-3/2, 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])/(3*f*Sqrt[1 - Sec[e + f*x]]*(d*Sec[e + f*x])^(3/2))
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]
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 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])
Rubi steps
\begin{align*} \int \frac{(a+a \sec (e+f x))^m}{(d \sec (e+f x))^{3/2}} \, dx &=\left ((1+\sec (e+f x))^{-m} (a+a \sec (e+f x))^m\right ) \int \frac{(1+\sec (e+f x))^m}{(d \sec (e+f x))^{3/2}} \, 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} (d x)^{5/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{1-\sec (e+f x)}}\\ &=\frac{2 F_1\left (-\frac{3}{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)}{3 f \sqrt{1-\sec (e+f x)} (d \sec (e+f x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 19.3691, size = 3349, normalized size = 34.17 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sec[e + f*x])^m/(d*Sec[e + f*x])^(3/2),x]
[Out]
(2^(1 + m)*Sec[e + f*x]^(3/2)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(1/2 + m)*(a*(1 + Sec[e + f*x]))^m*((Cos[2*(e
+ f*x)]^3*Sqrt[Sec[e + f*x]]*(1 + Sec[e + f*x])^m)/4 + Cos[2*(e + f*x)]^2*Sqrt[Sec[e + f*x]]*((1 + Sec[e + f*x
])^m/2 + (I/4)*(1 + Sec[e + f*x])^m*Sin[2*(e + f*x)]) + Cos[2*(e + f*x)]*Sqrt[Sec[e + f*x]]*((1 + Sec[e + f*x]
)^m/4 + ((1 + Sec[e + f*x])^m*Sin[2*(e + f*x)]^2)/4) + Sqrt[Sec[e + f*x]]*((-I/4)*(1 + Sec[e + f*x])^m*Sin[2*(
e + f*x)] + ((1 + Sec[e + f*x])^m*Sin[2*(e + f*x)]^2)/2 + (I/4)*(1 + Sec[e + f*x])^m*Sin[2*(e + f*x)]^3))*Tan[
(e + f*x)/2]*(-(AppellF1[3/2, -1/2 + m, 5/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(
e + f*x)/2]^2)^(1/2 + m)*Tan[(e + f*x)/2]^2) - (9*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(
e + f*x)/2]^2]*Cos[e + f*x])/((Sec[(e + f*x)/2]^2)^(3/2)*(-3*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2
]^2, -Tan[(e + f*x)/2]^2] + (5*AppellF1[3/2, -1/2 + m, 7/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1
- 2*m)*AppellF1[3/2, 1/2 + m, 5/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))))/(3*f
*(d*Sec[e + f*x])^(3/2)*(1 + Sec[e + f*x])^m*((2^m*Sec[(e + f*x)/2]^2*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(1/2 +
m)*(-(AppellF1[3/2, -1/2 + m, 5/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/
2]^2)^(1/2 + m)*Tan[(e + f*x)/2]^2) - (9*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/
2]^2]*Cos[e + f*x])/((Sec[(e + f*x)/2]^2)^(3/2)*(-3*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^2, -Tan
[(e + f*x)/2]^2] + (5*AppellF1[3/2, -1/2 + m, 7/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 - 2*m)*A
ppellF1[3/2, 1/2 + m, 5/2, 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]*(-(AppellF1[3/2, -1/2 + m, 5/2, 5/2, Tan[(e + f
*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(1/2 + m)*Tan[(e + f*x)/2]
) - (Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(1/2 + m)*Tan[(e + f*x)/2]^2*((-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)*Ap
pellF1[3/2, -1/2 + m, 5/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(-1
/2 + m)*Tan[(e + f*x)/2]^2*(-(Sec[(e + f*x)/2]^2*Sin[e + f*x]) + Cos[e + f*x]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)
/2]) + (9*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sin[e + f*x])/((Sec[(e +
f*x)/2]^2)^(3/2)*(-3*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (5*AppellF1[
3/2, -1/2 + m, 7/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 - 2*m)*AppellF1[3/2, 1/2 + m, 5/2, 5/2,
Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)) + (27*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x]*Tan[(e + f*x)/2])/(2*(Sec[(e + f*x)/2]^2)^(3/2)*(-3*AppellF1[1
/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (5*AppellF1[3/2, -1/2 + m, 7/2, 5/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 - 2*m)*AppellF1[3/2, 1/2 + m, 5/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e +
f*x)/2]^2])*Tan[(e + f*x)/2]^2)) - (9*Cos[e + f*x]*((-5*AppellF1[3/2, -1/2 + m, 7/2, 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])/6 + ((-1/2 + m)*AppellF1[3/2, 1/2 + m, 5/2, 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
*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (5*AppellF1[3/2, -1/2 + m, 7/2,
5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 - 2*m)*AppellF1[3/2, 1/2 + m, 5/2, 5/2, Tan[(e + f*x)/2]^2,
-Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)) + (9*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e
+ f*x)/2]^2]*Cos[e + f*x]*((5*AppellF1[3/2, -1/2 + m, 7/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1
- 2*m)*AppellF1[3/2, 1/2 + m, 5/2, 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*((-5*AppellF1[3/2, -1/2 + m, 7/2, 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])/6 + ((-1/2 + m)*AppellF1[3/2, 1/2 + m, 5/2, 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*(5*((-21*AppellF1[5/2, -1/2 + m, 9/2, 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*(-1/2 + m)*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])/5) + (1 - 2*m)*((-
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, 3/2 + 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))))/((Sec[(e + f*x)/2]^2)^(3/2)*(-3*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^
2, -Tan[(e + f*x)/2]^2] + (5*AppellF1[3/2, -1/2 + m, 7/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 -
2*m)*AppellF1[3/2, 1/2 + m, 5/2, 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)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(-1/2 + m)*Tan[(e + f*x)/2]*(-(AppellF1[3/2, -1/2 + m,
5/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(1/2 + m)*Tan[(e + f*x)/2
]^2) - (9*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x])/((Sec[(e +
f*x)/2]^2)^(3/2)*(-3*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (5*AppellF1[
3/2, -1/2 + m, 7/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 - 2*m)*AppellF1[3/2, 1/2 + m, 5/2, 5/2,
Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/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))
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Maple [F] time = 0.186, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+a\sec \left ( fx+e \right ) \right ) ^{m} \left ( d\sec \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(f*x+e))^m/(d*sec(f*x+e))^(3/2),x)
[Out]
int((a+a*sec(f*x+e))^m/(d*sec(f*x+e))^(3/2),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (f x + e\right ) + a\right )}^{m}}{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(f*x+e))^m/(d*sec(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate((a*sec(f*x + e) + a)^m/(d*sec(f*x + e))^(3/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \sec \left (f x + e\right )}{\left (a \sec \left (f x + e\right ) + a\right )}^{m}}{d^{2} \sec \left (f x + e\right )^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(f*x+e))^m/(d*sec(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
integral(sqrt(d*sec(f*x + e))*(a*sec(f*x + e) + a)^m/(d^2*sec(f*x + e)^2), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(f*x+e))**m/(d*sec(f*x+e))**(3/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (f x + e\right ) + a\right )}^{m}}{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(f*x+e))^m/(d*sec(f*x+e))^(3/2),x, algorithm="giac")
[Out]
integrate((a*sec(f*x + e) + a)^m/(d*sec(f*x + e))^(3/2), x)