3.4.33 \(\int (-\sec (e+f x))^n (1+\sec (e+f x))^m \, dx\) [333]
Optimal. Leaf size=85 \[ \frac {\sqrt {2} F_1\left (\frac {1}{2}+m;1-n,\frac {1}{2};\frac {3}{2}+m;1+\sec (e+f x),\frac {1}{2} (1+\sec (e+f x))\right ) (1+\sec (e+f x))^m \tan (e+f x)}{f (1+2 m) \sqrt {1-\sec (e+f x)}} \]
[Out]
AppellF1(1/2+m,1-n,1/2,3/2+m,1+sec(f*x+e),1/2+1/2*sec(f*x+e))*(1+sec(f*x+e))^m*2^(1/2)*tan(f*x+e)/f/(1+2*m)/(1
-sec(f*x+e))^(1/2)
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Rubi [A]
time = 0.04, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps
used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3911, 138}
\begin {gather*} \frac {\sqrt {2} \tan (e+f x) (\sec (e+f x)+1)^m F_1\left (m+\frac {1}{2};1-n,\frac {1}{2};m+\frac {3}{2};\sec (e+f x)+1,\frac {1}{2} (\sec (e+f x)+1)\right )}{f (2 m+1) \sqrt {1-\sec (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(-Sec[e + f*x])^n*(1 + Sec[e + f*x])^m,x]
[Out]
(Sqrt[2]*AppellF1[1/2 + m, 1 - n, 1/2, 3/2 + m, 1 + Sec[e + f*x], (1 + Sec[e + f*x])/2]*(1 + Sec[e + f*x])^m*T
an[e + f*x])/(f*(1 + 2*m)*Sqrt[1 - 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 3911
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 - 1)*f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])), Subst[Int[x^(m - 1
/2)*((a - x)^(n - 1)/Sqrt[2*a - 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] && LtQ[a*(d/b), 0]
Rubi steps
\begin {align*} \int (-\sec (e+f x))^n (1+\sec (e+f x))^m \, dx &=\frac {\tan (e+f x) \text {Subst}\left (\int \frac {(1-x)^{-1+n} x^{-\frac {1}{2}+m}}{\sqrt {2-x}} \, dx,x,1+\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}\\ &=\frac {\sqrt {2} F_1\left (\frac {1}{2}+m;1-n,\frac {1}{2};\frac {3}{2}+m;1+\sec (e+f x),\frac {1}{2} (1+\sec (e+f x))\right ) (1+\sec (e+f x))^m \tan (e+f x)}{f (1+2 m) \sqrt {1-\sec (e+f x)}}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(2248\) vs. \(2(85)=170\).
time = 6.22, size = 2248, normalized size = 26.45 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(-Sec[e + f*x])^n*(1 + Sec[e + f*x])^m,x]
[Out]
(3*2^(1 + m)*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^(-
1 + n)*(-Sec[e + f*x])^n*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(m + n)*(1 + Sec[e + f*x])^m*Tan[(e + f*x)/2])/(f*(
3*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + n)*AppellF1[3/2, m + n,
2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (m + n)*AppellF1[3/2, 1 + m + n, 1 - n, 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, m + 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])^(m + n))/(3*AppellF1[1/2, m
+ n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + n)*AppellF1[3/2, m + n, 2 - n, 5/2, Tan[(
e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (m + n)*AppellF1[3/2, 1 + m + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e
+ f*x)/2]^2])*Tan[(e + f*x)/2]^2) + (3*2^(1 + m)*(-1 + n)*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2
, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^(-1 + n)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(m + n)*Tan[(e + f*x)/2
]^2)/(3*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + n)*AppellF1[3/2,
m + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (m + n)*AppellF1[3/2, 1 + m + n, 1 - n, 5/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) + (3*2^(1 + m)*(Sec[(e + f*x)/2]^2)^(-1 + n)*(Cos[(
e + f*x)/2]^2*Sec[e + f*x])^(m + n)*Tan[(e + f*x)/2]*(-1/3*((1 - n)*AppellF1[3/2, m + 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]) + ((m + n)*AppellF1[3/2, 1 + m + 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))/(3*AppellF1[1/2, m +
n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + n)*AppellF1[3/2, m + n, 2 - n, 5/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (m + n)*AppellF1[3/2, 1 + m + n, 1 - n, 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, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e +
f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^(-1 + n)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(m + n)*Tan[(e + f*x)/2]*(2*((-1 +
n)*AppellF1[3/2, m + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (m + n)*AppellF1[3/2, 1 + m +
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, m + 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
]) + ((m + n)*AppellF1[3/2, 1 + m + 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) + 2*Tan[(e + f*x)/2]^2*((-1 + n)*((-3*(2 - n)*AppellF1[5/2, m + 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*(m + n)*AppellF1[5/2, 1 + m + 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) + (m + n)*((-3*(1
- n)*AppellF1[5/2, 1 + m + 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 + m + n)*AppellF1[5/2, 2 + m + 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, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)
/2]^2] + 2*((-1 + n)*AppellF1[3/2, m + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (m + n)*Appel
lF1[3/2, 1 + m + n, 1 - n, 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)
*(m + n)*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^(-1 +
n)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(-1 + m + n)*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]))/(3*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^
2, -Tan[(e + f*x)/2]^2] + 2*((-1 + n)*AppellF1[3/2, m + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2
] + (m + n)*AppellF1[3/2, 1 + m + 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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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \left (-\sec \left (f x +e \right )\right )^{n} \left (1+\sec \left (f x +e \right )\right )^{m}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((-sec(f*x+e))^n*(1+sec(f*x+e))^m,x)
[Out]
int((-sec(f*x+e))^n*(1+sec(f*x+e))^m,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*(1+sec(f*x+e))^m,x, algorithm="maxima")
[Out]
integrate((-sec(f*x + e))^n*(sec(f*x + e) + 1)^m, 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*(1+sec(f*x+e))^m,x, algorithm="fricas")
[Out]
integral((-sec(f*x + e))^n*(sec(f*x + e) + 1)^m, x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (- \sec {\left (e + f x \right )}\right )^{n} \left (\sec {\left (e + f x \right )} + 1\right )^{m}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-sec(f*x+e))**n*(1+sec(f*x+e))**m,x)
[Out]
Integral((-sec(e + f*x))**n*(sec(e + f*x) + 1)**m, 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*(1+sec(f*x+e))^m,x, algorithm="giac")
[Out]
integrate((-sec(f*x + e))^n*(sec(f*x + e) + 1)^m, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {1}{\cos \left (e+f\,x\right )}+1\right )}^m\,{\left (-\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1/cos(e + f*x) + 1)^m*(-1/cos(e + f*x))^n,x)
[Out]
int((1/cos(e + f*x) + 1)^m*(-1/cos(e + f*x))^n, x)
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