3.336 \(\int (-\sec (e+f x))^n (a-a \sec (e+f x))^m \, dx\)

Optimal. Leaf size=87 \[ \frac {2^{m+\frac {1}{2}} \tan (e+f x) (1-\sec (e+f x))^{-m-\frac {1}{2}} (a-a \sec (e+f x))^m F_1\left (\frac {1}{2};1-n,\frac {1}{2}-m;\frac {3}{2};\sec (e+f x)+1,\frac {1}{2} (\sec (e+f x)+1)\right )}{f} \]

[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))*(1-sec(f*x+e))^(-1/2-m)*(a-a*sec(f*x+e))
^m*tan(f*x+e)/f

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Rubi [A]  time = 0.12, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3828, 3825, 133} \[ \frac {2^{m+\frac {1}{2}} \tan (e+f x) (1-\sec (e+f x))^{-m-\frac {1}{2}} (a-a \sec (e+f x))^m F_1\left (\frac {1}{2};1-n,\frac {1}{2}-m;\frac {3}{2};\sec (e+f x)+1,\frac {1}{2} (\sec (e+f x)+1)\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(-Sec[e + f*x])^n*(a - a*Sec[e + f*x])^m,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]*(1 - Sec[e + f*x])^(-1
/2 - m)*(a - a*Sec[e + f*x])^m*Tan[e + f*x])/f

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 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 3828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(1 + (b*Csc[e + f*x])/a)^FracPart[m], Int[(1 + (b*Csc[e + f*x])/a)^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rubi steps

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

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Mathematica [F]  time = 0.26, size = 0, normalized size = 0.00 \[ \int (-\sec (e+f x))^n (a-a \sec (e+f x))^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-a \sec \left (f x + e\right ) + a\right )}^{m} \left (-\sec \left (f x + e\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-a \sec \left (f x + e\right ) + a\right )}^{m} \left (-\sec \left (f x + e\right )\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sec(f*x+e))^n*(a-a*sec(f*x+e))^m,x, algorithm="giac")

[Out]

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

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maple [F]  time = 2.89, size = 0, normalized size = 0.00 \[ \int \left (-\sec \left (f x +e \right )\right )^{n} \left (a -a \sec \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-a \sec \left (f x + e\right ) + a\right )}^{m} \left (-\sec \left (f x + e\right )\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a-\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (-\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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