3.337 \(\int (d \sec (e+f x))^n (1+\sec (e+f x))^m \, dx\)
Optimal. Leaf size=79 \[ -\frac {\tan (e+f x) (d \sec (e+f x))^n F_1\left (n;\frac {1}{2},\frac {1}{2}-m;n+1;\sec (e+f x),-\sec (e+f x)\right )}{f n \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}} \]
[Out]
-AppellF1(n,1/2-m,1/2,1+n,-sec(f*x+e),sec(f*x+e))*(d*sec(f*x+e))^n*tan(f*x+e)/f/n/(1-sec(f*x+e))^(1/2)/(1+sec(
f*x+e))^(1/2)
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Rubi [A] time = 0.06, antiderivative size = 79, 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 =
{3827, 133} \[ -\frac {\tan (e+f x) (d \sec (e+f x))^n F_1\left (n;\frac {1}{2},\frac {1}{2}-m;n+1;\sec (e+f x),-\sec (e+f x)\right )}{f n \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}} \]
Antiderivative was successfully verified.
[In]
Int[(d*Sec[e + f*x])^n*(1 + Sec[e + f*x])^m,x]
[Out]
-((AppellF1[n, 1/2, 1/2 - m, 1 + n, Sec[e + f*x], -Sec[e + f*x]]*(d*Sec[e + f*x])^n*Tan[e + f*x])/(f*n*Sqrt[1
- Sec[e + f*x]]*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]
Rubi steps
\begin {align*} \int (d \sec (e+f x))^n (1+\sec (e+f x))^m \, dx &=-\frac {(d \tan (e+f x)) \operatorname {Subst}\left (\int \frac {(d x)^{-1+n} (1+x)^{-\frac {1}{2}+m}}{\sqrt {1-x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}\\ &=-\frac {F_1\left (n;\frac {1}{2},\frac {1}{2}-m;1+n;\sec (e+f x),-\sec (e+f x)\right ) (d \sec (e+f x))^n \tan (e+f x)}{f n \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}\\ \end {align*}
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Mathematica [B] time = 6.21, size = 2248, normalized size = 28.46 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(d*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)*(d*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, Ta
n[(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]*Se
c[(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)*Appe
llF1[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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fricas [F] time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (d \sec \left (f x + e\right )\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{m}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^n*(1+sec(f*x+e))^m,x, algorithm="fricas")
[Out]
integral((d*sec(f*x + e))^n*(sec(f*x + e) + 1)^m, x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^n*(1+sec(f*x+e))^m,x, algorithm="giac")
[Out]
integrate((d*sec(f*x + e))^n*(sec(f*x + e) + 1)^m, x)
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maple [F] time = 2.41, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{n} \left (1+\sec \left (f x +e \right )\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*sec(f*x+e))^n*(1+sec(f*x+e))^m,x)
[Out]
int((d*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 \[ \int \left (d \sec \left (f x + e\right )\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^n*(1+sec(f*x+e))^m,x, algorithm="maxima")
[Out]
integrate((d*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 \[ \int {\left (\frac {1}{\cos \left (e+f\,x\right )}+1\right )}^m\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1/cos(e + f*x) + 1)^m*(d/cos(e + f*x))^n,x)
[Out]
int((1/cos(e + f*x) + 1)^m*(d/cos(e + f*x))^n, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec {\left (e + f x \right )}\right )^{n} \left (\sec {\left (e + f x \right )} + 1\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))**n*(1+sec(f*x+e))**m,x)
[Out]
Integral((d*sec(e + f*x))**n*(sec(e + f*x) + 1)**m, x)
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