3.293 \(\int \frac {\sec ^n(e+f x)}{(a+a \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=217 \[ -\frac {(3-2 n) \sin (e+f x) \sec ^{n-1}(e+f x) \, _2F_1\left (\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right )}{3 a^2 f \sqrt {\sin ^2(e+f x)}}+\frac {2 (2-n) \sin (e+f x) \sec ^n(e+f x) \, _2F_1\left (\frac {1}{2},-\frac {n}{2};\frac {2-n}{2};\cos ^2(e+f x)\right )}{3 a^2 f \sqrt {\sin ^2(e+f x)}}-\frac {2 (2-n) \sin (e+f x) \sec ^{n+1}(e+f x)}{3 a^2 f (\sec (e+f x)+1)}-\frac {\sin (e+f x) \sec ^{n+1}(e+f x)}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

-2/3*(2-n)*sec(f*x+e)^(1+n)*sin(f*x+e)/a^2/f/(1+sec(f*x+e))-1/3*sec(f*x+e)^(1+n)*sin(f*x+e)/f/(a+a*sec(f*x+e))
^2-1/3*(3-2*n)*hypergeom([1/2, 1/2-1/2*n],[3/2-1/2*n],cos(f*x+e)^2)*sec(f*x+e)^(-1+n)*sin(f*x+e)/a^2/f/(sin(f*
x+e)^2)^(1/2)+2/3*(2-n)*hypergeom([1/2, -1/2*n],[1-1/2*n],cos(f*x+e)^2)*sec(f*x+e)^n*sin(f*x+e)/a^2/f/(sin(f*x
+e)^2)^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3817, 4020, 3787, 3772, 2643} \[ -\frac {(3-2 n) \sin (e+f x) \sec ^{n-1}(e+f x) \, _2F_1\left (\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right )}{3 a^2 f \sqrt {\sin ^2(e+f x)}}+\frac {2 (2-n) \sin (e+f x) \sec ^n(e+f x) \, _2F_1\left (\frac {1}{2},-\frac {n}{2};\frac {2-n}{2};\cos ^2(e+f x)\right )}{3 a^2 f \sqrt {\sin ^2(e+f x)}}-\frac {2 (2-n) \sin (e+f x) \sec ^{n+1}(e+f x)}{3 a^2 f (\sec (e+f x)+1)}-\frac {\sin (e+f x) \sec ^{n+1}(e+f x)}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2 - n)*Sec[e + f*x]^(1 + n)*Sin[e + f*x])/(3*a^2*f*(1 + Sec[e + f*x])) - (Sec[e + f*x]^(1 + n)*Sin[e + f*
x])/(3*f*(a + a*Sec[e + f*x])^2) - ((3 - 2*n)*Hypergeometric2F1[1/2, (1 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*Sec
[e + f*x]^(-1 + n)*Sin[e + f*x])/(3*a^2*f*Sqrt[Sin[e + f*x]^2]) + (2*(2 - n)*Hypergeometric2F1[1/2, -n/2, (2 -
n)/2, Cos[e + f*x]^2]*Sec[e + f*x]^n*Sin[e + f*x])/(3*a^2*f*Sqrt[Sin[e + f*x]^2])

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3817

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

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\sec ^n(e+f x)}{(a+a \sec (e+f x))^2} \, dx &=-\frac {\sec ^{1+n}(e+f x) \sin (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {\int \frac {\sec ^n(e+f x) (a (-3+n)-a (-1+n) \sec (e+f x))}{a+a \sec (e+f x)} \, dx}{3 a^2}\\ &=-\frac {2 (2-n) \sec ^{1+n}(e+f x) \sin (e+f x)}{3 a^2 f (1+\sec (e+f x))}-\frac {\sec ^{1+n}(e+f x) \sin (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {\int \sec ^n(e+f x) \left (-a^2 (3-2 n) (1-n)-2 a^2 (2-n) n \sec (e+f x)\right ) \, dx}{3 a^4}\\ &=-\frac {2 (2-n) \sec ^{1+n}(e+f x) \sin (e+f x)}{3 a^2 f (1+\sec (e+f x))}-\frac {\sec ^{1+n}(e+f x) \sin (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {((3-2 n) (1-n)) \int \sec ^n(e+f x) \, dx}{3 a^2}+\frac {(2 (2-n) n) \int \sec ^{1+n}(e+f x) \, dx}{3 a^2}\\ &=-\frac {2 (2-n) \sec ^{1+n}(e+f x) \sin (e+f x)}{3 a^2 f (1+\sec (e+f x))}-\frac {\sec ^{1+n}(e+f x) \sin (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\left ((3-2 n) (1-n) \cos ^n(e+f x) \sec ^n(e+f x)\right ) \int \cos ^{-n}(e+f x) \, dx}{3 a^2}+\frac {\left (2 (2-n) n \cos ^n(e+f x) \sec ^n(e+f x)\right ) \int \cos ^{-1-n}(e+f x) \, dx}{3 a^2}\\ &=-\frac {2 (2-n) \sec ^{1+n}(e+f x) \sin (e+f x)}{3 a^2 f (1+\sec (e+f x))}-\frac {\sec ^{1+n}(e+f x) \sin (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {(3-2 n) \, _2F_1\left (\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right ) \sec ^{-1+n}(e+f x) \sin (e+f x)}{3 a^2 f \sqrt {\sin ^2(e+f x)}}+\frac {2 (2-n) \, _2F_1\left (\frac {1}{2},-\frac {n}{2};\frac {2-n}{2};\cos ^2(e+f x)\right ) \sec ^n(e+f x) \sin (e+f x)}{3 a^2 f \sqrt {\sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [F]  time = 10.14, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^n(e+f x)}{(a+a \sec (e+f x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sec[e + f*x]^n/(a + a*Sec[e + f*x])^2,x]

[Out]

Integrate[Sec[e + f*x]^n/(a + a*Sec[e + f*x])^2, x]

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sec \left (f x + e\right )^{n}}{a^{2} \sec \left (f x + e\right )^{2} + 2 \, a^{2} \sec \left (f x + e\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^n/(a+a*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

integral(sec(f*x + e)^n/(a^2*sec(f*x + e)^2 + 2*a^2*sec(f*x + e) + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )^{n}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^n/(a+a*sec(f*x+e))^2,x, algorithm="giac")

[Out]

integrate(sec(f*x + e)^n/(a*sec(f*x + e) + a)^2, x)

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maple [F]  time = 0.94, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{n}\left (f x +e \right )}{\left (a +a \sec \left (f x +e \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^n/(a+a*sec(f*x+e))^2,x)

[Out]

int(sec(f*x+e)^n/(a+a*sec(f*x+e))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )^{n}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^n/(a+a*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)^n/(a*sec(f*x + e) + a)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(e + f*x))^n/(a + a/cos(e + f*x))^2,x)

[Out]

int((1/cos(e + f*x))^n/(a + a/cos(e + f*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{n}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**n/(a+a*sec(f*x+e))**2,x)

[Out]

Integral(sec(e + f*x)**n/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x)/a**2

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