3.4.34 \(\int (1-\sec (e+f x))^m (-\sec (e+f x))^n \, dx\) [334]

Optimal. Leaf size=70 \[ \frac {2^{\frac {1}{2}+m} F_1\left (\frac {1}{2};1-n,\frac {1}{2}-m;\frac {3}{2};1+\sec (e+f x),\frac {1}{2} (1+\sec (e+f x))\right ) \tan (e+f x)}{f \sqrt {1-\sec (e+f x)}} \]

[Out]

2^(1/2+m)*AppellF1(1/2,1-n,1/2-m,3/2,1+sec(f*x+e),1/2+1/2*sec(f*x+e))*tan(f*x+e)/f/(1-sec(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 70, 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 = {3910, 138} \begin {gather*} \frac {2^{m+\frac {1}{2}} \tan (e+f x) F_1\left (\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 \sqrt {1-\sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(1/2 + m)*AppellF1[1/2, 1 - n, 1/2 - m, 3/2, 1 + Sec[e + f*x], (1 + Sec[e + f*x])/2]*Tan[e + f*x])/(f*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 3910

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 - 2)*f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])), Subst[Int[(a - x)^(n
- 1)*((2*a - x)^(m - 1/2)/Sqrt[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] && GtQ[a*(d/b), 0]

Rubi steps

\begin {align*} \int (1-\sec (e+f x))^m (-\sec (e+f x))^n \, dx &=\frac {\tan (e+f x) \text {Subst}\left (\int \frac {(1-x)^{-1+n} (2-x)^{-\frac {1}{2}+m}}{\sqrt {x}} \, dx,x,1+\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}\\ &=\frac {2^{\frac {1}{2}+m} F_1\left (\frac {1}{2};1-n,\frac {1}{2}-m;\frac {3}{2};1+\sec (e+f x),\frac {1}{2} (1+\sec (e+f x))\right ) \tan (e+f x)}{f \sqrt {1-\sec (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(257\) vs. \(2(70)=140\).
time = 0.54, size = 257, normalized size = 3.67 \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 (-\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*(-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*(-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, -Ta
n[(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))

________________________________________________________________________________________

Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \left (1-\sec \left (f x +e \right )\right )^{m} \left (-\sec \left (f x +e \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (- \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*(-sec(f*x+e))**n,x)

[Out]

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

________________________________________________________________________________________

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*(-sec(f*x+e))^n,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________