3.4.35 \(\int (-\sec (e+f x))^n (a+a \sec (e+f x))^m \, dx\) [335]
Optimal. Leaf size=87 \[ \frac {\sqrt {2} F_1\left (\frac {1}{2}+m;1-n,\frac {1}{2};\frac {3}{2}+m;1+\sec (e+f x),\frac {1}{2} (1+\sec (e+f x))\right ) (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+2 m) \sqrt {1-\sec (e+f x)}} \]
[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))*(a+a*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 = 0.08, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3913, 3911,
138} \begin {gather*} \frac {\sqrt {2} \tan (e+f x) (a \sec (e+f x)+a)^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*(a + a*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]*(a + a*Sec[e + f*x])^m
*Tan[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]
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 (-\sec (e+f x))^n (a+a \sec (e+f x))^m \, dx &=\left ((1+\sec (e+f x))^{-m} (a+a \sec (e+f x))^m\right ) \int (-\sec (e+f x))^n (1+\sec (e+f x))^m \, dx\\ &=\frac {\left ((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)^{-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)}}\\ &=\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 ) (a+a \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. \(2250\) vs. \(2(87)=174\).
time = 6.27, size = 2250, normalized size = 25.86 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(-Sec[e + f*x])^n*(a + a*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)*(a*(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, T
an[(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, -Ta
n[(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)*(C
os[(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)*A
ppellF1[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 \[\int \left (-\sec \left (f x +e \right )\right )^{n} \left (a +a \sec \left (f x +e \right )\right )^{m}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((-sec(f*x+e))^n*(a+a*sec(f*x+e))^m,x)
[Out]
int((-sec(f*x+e))^n*(a+a*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*(a+a*sec(f*x+e))^m,x, algorithm="maxima")
[Out]
integrate((a*sec(f*x + e) + a)^m*(-sec(f*x + e))^n, 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*(a+a*sec(f*x+e))^m,x, algorithm="fricas")
[Out]
integral((a*sec(f*x + e) + a)^m*(-sec(f*x + e))^n, 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 (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-sec(f*x+e))**n*(a+a*sec(f*x+e))**m,x)
[Out]
Integral((-sec(e + f*x))**n*(a*(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*(a+a*sec(f*x+e))^m,x, algorithm="giac")
[Out]
integrate((a*sec(f*x + e) + a)^m*(-sec(f*x + e))^n, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\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((a + a/cos(e + f*x))^m*(-1/cos(e + f*x))^n,x)
[Out]
int((a + a/cos(e + f*x))^m*(-1/cos(e + f*x))^n, x)
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