3.334 \(\int (1-\sec (e+f x))^m (-\sec (e+f x))^n \, dx\)

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

[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]])

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Rubi [A]  time = 0.0654366, antiderivative size = 70, normalized size of antiderivative = 1., 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 = {3825, 133} \[ \frac{2^{m+\frac{1}{2}} \tan (e+f x) F_1\left (\frac{1}{2};1-n,\frac{1}{2}-m;\frac{3}{2};\sec (e+f x)+1,\frac{1}{2} (\sec (e+f x)+1)\right )}{f \sqrt{1-\sec (e+f x)}} \]

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 3825

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]

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])

Rubi steps

\begin{align*} \int (1-\sec (e+f x))^m (-\sec (e+f x))^n \, dx &=\frac{\tan (e+f x) \operatorname{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]  time = 0.297664, size = 257, normalized size = 3.67 \[ \frac{(2 m+3) \sin (e+f x) (1-\sec (e+f x))^m (-\sec (e+f x))^n F_1\left (m+\frac{1}{2};m+n,1-n;m+\frac{3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{f (2 m+1) \left (2 \tan ^2\left (\frac{1}{2} (e+f x)\right ) \left ((n-1) F_1\left (m+\frac{3}{2};m+n,2-n;m+\frac{5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+(m+n) F_1\left (m+\frac{3}{2};m+n+1,1-n;m+\frac{5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )+(2 m+3) F_1\left (m+\frac{1}{2};m+n,1-n;m+\frac{3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )} \]

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))

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Maple [F]  time = 0.724, size = 0, normalized size = 0. \begin{align*} \int \left ( 1-\sec \left ( fx+e \right ) \right ) ^{m} \left ( -\sec \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (-\sec \left (f x + e\right )\right )^{n}{\left (-\sec \left (f x + e\right ) + 1\right )}^{m}\,{d x} \end{align*}

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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (-\sec \left (f x + e\right )\right )^{n}{\left (-\sec \left (f x + e\right ) + 1\right )}^{m}, x\right ) \end{align*}

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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (- \sec{\left (e + f x \right )}\right )^{n} \left (1 - \sec{\left (e + f x \right )}\right )^{m}\, dx \end{align*}

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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (-\sec \left (f x + e\right )\right )^{n}{\left (-\sec \left (f x + e\right ) + 1\right )}^{m}\,{d x} \end{align*}

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)