3.4.38 \(\int (1-\sec (e+f x))^m (d \sec (e+f x))^n \, dx\) [338]

Optimal. Leaf size=79 \[ -\frac {F_1\left (n;\frac {1}{2}-m,\frac {1}{2};1+n;\sec (e+f x),-\sec (e+f x)\right ) (d \sec (e+f x))^n \tan (e+f x)}{f n \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \]

[Out]

-AppellF1(n,1/2,1/2-m,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 = 0.04, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3912, 138} \begin {gather*} -\frac {\tan (e+f x) (d \sec (e+f x))^n F_1\left (n;\frac {1}{2}-m,\frac {1}{2};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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - Sec[e + f*x])^m*(d*Sec[e + f*x])^n,x]

[Out]

-((AppellF1[n, 1/2 - m, 1/2, 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 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 3912

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 (1-\sec (e+f x))^m (d \sec (e+f x))^n \, dx &=-\frac {(d \tan (e+f x)) \text {Subst}\left (\int \frac {(1-x)^{-\frac {1}{2}+m} (d x)^{-1+n}}{\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}-m,\frac {1}{2};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] Leaf count is larger than twice the leaf count of optimal. \(257\) vs. \(2(79)=158\).
time = 0.54, size = 257, normalized size = 3.25 \begin {gather*} \frac {(3+2 m) F_1\left (\frac {1}{2}+m;m+n,1-n;\frac {3}{2}+m;\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1-\sec (e+f x))^m (d \sec (e+f x))^n \sin (e+f x)}{f (1+2 m) \left ((3+2 m) F_1\left (\frac {1}{2}+m;m+n,1-n;\frac {3}{2}+m;\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+2 \left ((-1+n) F_1\left (\frac {3}{2}+m;m+n,2-n;\frac {5}{2}+m;\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+(m+n) F_1\left (\frac {3}{2}+m;1+m+n,1-n;\frac {5}{2}+m;\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(1 - Sec[e + f*x])^m*(d*Sec[e + f*x])^n,x]

[Out]

((3 + 2*m)*AppellF1[1/2 + m, m + n, 1 - n, 3/2 + m, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(1 - Sec[e + f*x]
)^m*(d*Sec[e + f*x])^n*Sin[e + f*x])/(f*(1 + 2*m)*((3 + 2*m)*AppellF1[1/2 + m, m + n, 1 - n, 3/2 + m, Tan[(e +
f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + n)*AppellF1[3/2 + m, m + n, 2 - n, 5/2 + m, Tan[(e + f*x)/2]^2, -T
an[(e + f*x)/2]^2] + (m + n)*AppellF1[3/2 + m, 1 + m + n, 1 - n, 5/2 + m, 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 \[\int \left (1-\sec \left (f x +e \right )\right )^{m} \left (d \sec \left (f x +e \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-sec(f*x+e))^m*(d*sec(f*x+e))^n,x)

[Out]

int((1-sec(f*x+e))^m*(d*sec(f*x+e))^n,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((1-sec(f*x+e))^m*(d*sec(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((d*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((1-sec(f*x+e))^m*(d*sec(f*x+e))^n,x, algorithm="fricas")

[Out]

integral((d*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 (d \sec {\left (e + f x \right )}\right )^{n} \left (1 - \sec {\left (e + f x \right )}\right )^{m}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-sec(f*x+e))**m*(d*sec(f*x+e))**n,x)

[Out]

Integral((d*sec(e + f*x))**n*(1 - sec(e + f*x))**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((1-sec(f*x+e))^m*(d*sec(f*x+e))^n,x, algorithm="giac")

[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 \begin {gather*} \int {\left (1-\frac {1}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (\frac {d}{\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 - 1/cos(e + f*x))^m*(d/cos(e + f*x))^n,x)

[Out]

int((1 - 1/cos(e + f*x))^m*(d/cos(e + f*x))^n, x)

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