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

Optimal. Leaf size=89 \[ \frac{\sqrt{2} \tan (e+f x) (1-\sec (e+f x))^m F_1\left (m+\frac{1}{2};1-n,\frac{1}{2};m+\frac{3}{2};1-\sec (e+f x),\frac{1}{2} (1-\sec (e+f x))\right )}{f (2 m+1) \sqrt{\sec (e+f x)+1}} \]

[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]*(1 - Sec[e + f*x])^m*T
an[e + f*x])/(f*(1 + 2*m)*Sqrt[1 + Sec[e + f*x]])

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Rubi [A]  time = 0.0701287, antiderivative size = 89, normalized size of antiderivative = 1., 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 = {3826, 133} \[ \frac{\sqrt{2} \tan (e+f x) (1-\sec (e+f x))^m F_1\left (m+\frac{1}{2};1-n,\frac{1}{2};m+\frac{3}{2};1-\sec (e+f x),\frac{1}{2} (1-\sec (e+f x))\right )}{f (2 m+1) \sqrt{\sec (e+f x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Sec[e + f*x])^m*Sec[e + f*x]^n,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]*(1 - Sec[e + f*x])^m*T
an[e + f*x])/(f*(1 + 2*m)*Sqrt[1 + Sec[e + f*x]])

Rule 3826

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 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 ^n(e+f x) \, dx &=\frac{\tan (e+f x) \operatorname{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)} \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 ) (1-\sec (e+f x))^m \tan (e+f x)}{f (1+2 m) \sqrt{1+\sec (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 2.20441, size = 255, normalized size = 2.87 \[ \frac{(2 m+3) \sin (e+f x) (1-\sec (e+f x))^m \sec ^n(e+f x) F_1\left (m+\frac{1}{2};m+n,1-n;m+\frac{3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{f (2 m+1) \left (2 \tan ^2\left (\frac{1}{2} (e+f x)\right ) \left ((n-1) F_1\left (m+\frac{3}{2};m+n,2-n;m+\frac{5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+(m+n) F_1\left (m+\frac{3}{2};m+n+1,1-n;m+\frac{5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )+(2 m+3) F_1\left (m+\frac{1}{2};m+n,1-n;m+\frac{3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )} \]

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, -Tan[(
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 = 0.702, 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 ) + 1\right )}^{m} \sec \left (f x + e\right )^{n}\,{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) + 1)^m*sec(f*x + e)^n, 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 ) + 1\right )}^{m} \sec \left (f x + e\right )^{n}, 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) + 1)^m*sec(f*x + e)^n, x)

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

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