3.320 \(\int (-\sec (e+f x))^n (a-a \sec (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=178 \[ \frac{2 a^3 \left (16 n^2+24 n+3\right ) \tan (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},1-n,\frac{3}{2},\sec (e+f x)+1\right )}{f (2 n+1) (2 n+3) \sqrt{a-a \sec (e+f x)}}+\frac{2 a^2 \tan (e+f x) \sqrt{a-a \sec (e+f x)} (-\sec (e+f x))^n}{f (2 n+3)}+\frac{2 a^3 (4 n+7) \tan (e+f x) (-\sec (e+f x))^n}{f (2 n+1) (2 n+3) \sqrt{a-a \sec (e+f x)}} \]

[Out]

(2*a^3*(3 + 24*n + 16*n^2)*Hypergeometric2F1[1/2, 1 - n, 3/2, 1 + Sec[e + f*x]]*Tan[e + f*x])/(f*(1 + 2*n)*(3
+ 2*n)*Sqrt[a - a*Sec[e + f*x]]) + (2*a^3*(7 + 4*n)*(-Sec[e + f*x])^n*Tan[e + f*x])/(f*(1 + 2*n)*(3 + 2*n)*Sqr
t[a - a*Sec[e + f*x]]) + (2*a^2*(-Sec[e + f*x])^n*Sqrt[a - a*Sec[e + f*x]]*Tan[e + f*x])/(f*(3 + 2*n))

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Rubi [A]  time = 0.334391, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3814, 4016, 3806, 65} \[ \frac{2 a^3 \left (16 n^2+24 n+3\right ) \tan (e+f x) \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};\sec (e+f x)+1\right )}{f (2 n+1) (2 n+3) \sqrt{a-a \sec (e+f x)}}+\frac{2 a^2 \tan (e+f x) \sqrt{a-a \sec (e+f x)} (-\sec (e+f x))^n}{f (2 n+3)}+\frac{2 a^3 (4 n+7) \tan (e+f x) (-\sec (e+f x))^n}{f (2 n+1) (2 n+3) \sqrt{a-a \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(-Sec[e + f*x])^n*(a - a*Sec[e + f*x])^(5/2),x]

[Out]

(2*a^3*(3 + 24*n + 16*n^2)*Hypergeometric2F1[1/2, 1 - n, 3/2, 1 + Sec[e + f*x]]*Tan[e + f*x])/(f*(1 + 2*n)*(3
+ 2*n)*Sqrt[a - a*Sec[e + f*x]]) + (2*a^3*(7 + 4*n)*(-Sec[e + f*x])^n*Tan[e + f*x])/(f*(1 + 2*n)*(3 + 2*n)*Sqr
t[a - a*Sec[e + f*x]]) + (2*a^2*(-Sec[e + f*x])^n*Sqrt[a - a*Sec[e + f*x]]*Tan[e + f*x])/(f*(3 + 2*n))

Rule 3814

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[b/(m + n - 1), Int[(a
+ b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]

Rule 4016

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

Rule 3806

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

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

\begin{align*} \int (-\sec (e+f x))^n (a-a \sec (e+f x))^{5/2} \, dx &=\frac{2 a^2 (-\sec (e+f x))^n \sqrt{a-a \sec (e+f x)} \tan (e+f x)}{f (3+2 n)}-\frac{(2 a) \int (-\sec (e+f x))^n \sqrt{a-a \sec (e+f x)} \left (-a \left (\frac{3}{2}+2 n\right )+a \left (\frac{7}{2}+2 n\right ) \sec (e+f x)\right ) \, dx}{3+2 n}\\ &=\frac{2 a^3 (7+4 n) (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) (3+2 n) \sqrt{a-a \sec (e+f x)}}+\frac{2 a^2 (-\sec (e+f x))^n \sqrt{a-a \sec (e+f x)} \tan (e+f x)}{f (3+2 n)}+\frac{\left (a^2 \left (3+24 n+16 n^2\right )\right ) \int (-\sec (e+f x))^n \sqrt{a-a \sec (e+f x)} \, dx}{3+8 n+4 n^2}\\ &=\frac{2 a^3 (7+4 n) (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) (3+2 n) \sqrt{a-a \sec (e+f x)}}+\frac{2 a^2 (-\sec (e+f x))^n \sqrt{a-a \sec (e+f x)} \tan (e+f x)}{f (3+2 n)}+\frac{\left (a^4 \left (3+24 n+16 n^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(-x)^{-1+n}}{\sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{f \left (3+8 n+4 n^2\right ) \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a^3 \left (3+24 n+16 n^2\right ) \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};1+\sec (e+f x)\right ) \tan (e+f x)}{f \left (3+8 n+4 n^2\right ) \sqrt{a-a \sec (e+f x)}}+\frac{2 a^3 (7+4 n) (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) (3+2 n) \sqrt{a-a \sec (e+f x)}}+\frac{2 a^2 (-\sec (e+f x))^n \sqrt{a-a \sec (e+f x)} \tan (e+f x)}{f (3+2 n)}\\ \end{align*}

Mathematica [C]  time = 24.7991, size = 429, normalized size = 2.41 \[ \frac{2^{n-\frac{5}{2}} e^{-i \left (n-\frac{1}{2}\right ) (e+f x)} \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{n-\frac{1}{2}} \csc ^5\left (\frac{e}{2}+\frac{f x}{2}\right ) (a-a \sec (e+f x))^{5/2} (-\sec (e+f x))^n \sec ^{-n-\frac{5}{2}}(e+f x) \left (\frac{e^{i n (e+f x)} \text{Hypergeometric2F1}\left (1,\frac{1}{2} (-n-3),\frac{n+2}{2},-e^{2 i (e+f x)}\right )}{n}+\frac{10 e^{i (n+2) (e+f x)} \text{Hypergeometric2F1}\left (1,\frac{1}{2} (-n-1),\frac{n+4}{2},-e^{2 i (e+f x)}\right )}{n+2}+\frac{5 e^{i (n+4) (e+f x)} \text{Hypergeometric2F1}\left (1,\frac{1-n}{2},\frac{n+6}{2},-e^{2 i (e+f x)}\right )}{n+4}-\frac{5 e^{i (n+1) (e+f x)} \text{Hypergeometric2F1}\left (1,-\frac{n}{2}-1,\frac{n+3}{2},-e^{2 i (e+f x)}\right )}{n+1}-\frac{e^{i (n+5) (e+f x)} \text{Hypergeometric2F1}\left (1,1-\frac{n}{2},\frac{n+7}{2},-e^{2 i (e+f x)}\right )}{n+5}-\frac{10 e^{i (n+3) (e+f x)} \text{Hypergeometric2F1}\left (1,-\frac{n}{2},\frac{n+5}{2},-e^{2 i (e+f x)}\right )}{n+3}\right )}{f \left (1+e^{2 i (e+f x)}\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(-Sec[e + f*x])^n*(a - a*Sec[e + f*x])^(5/2),x]

[Out]

(2^(-5/2 + n)*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(-1/2 + n)*Csc[e/2 + (f*x)/2]^5*((E^(I*n*(e + f*x))*
Hypergeometric2F1[1, (-3 - n)/2, (2 + n)/2, -E^((2*I)*(e + f*x))])/n + (10*E^(I*(2 + n)*(e + f*x))*Hypergeomet
ric2F1[1, (-1 - n)/2, (4 + n)/2, -E^((2*I)*(e + f*x))])/(2 + n) + (5*E^(I*(4 + n)*(e + f*x))*Hypergeometric2F1
[1, (1 - n)/2, (6 + n)/2, -E^((2*I)*(e + f*x))])/(4 + n) - (5*E^(I*(1 + n)*(e + f*x))*Hypergeometric2F1[1, -1
- n/2, (3 + n)/2, -E^((2*I)*(e + f*x))])/(1 + n) - (E^(I*(5 + n)*(e + f*x))*Hypergeometric2F1[1, 1 - n/2, (7 +
n)/2, -E^((2*I)*(e + f*x))])/(5 + n) - (10*E^(I*(3 + n)*(e + f*x))*Hypergeometric2F1[1, -n/2, (5 + n)/2, -E^(
(2*I)*(e + f*x))])/(3 + n))*(-Sec[e + f*x])^n*Sec[e + f*x]^(-5/2 - n)*(a - a*Sec[e + f*x])^(5/2))/(E^(I*(-1/2
+ n)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^2*f)

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Maple [F]  time = 0.183, size = 0, normalized size = 0. \begin{align*} \int \left ( -\sec \left ( fx+e \right ) \right ) ^{n} \left ( a-a\sec \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-sec(f*x+e))^n*(a-a*sec(f*x+e))^(5/2),x)

[Out]

int((-sec(f*x+e))^n*(a-a*sec(f*x+e))^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a \sec \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \left (-\sec \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sec(f*x+e))^n*(a-a*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((-a*sec(f*x + e) + a)^(5/2)*(-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 (a^{2} \sec \left (f x + e\right )^{2} - 2 \, a^{2} \sec \left (f x + e\right ) + a^{2}\right )} \sqrt{-a \sec \left (f x + e\right ) + a} \left (-\sec \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sec(f*x+e))^n*(a-a*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sec(f*x+e))**n*(a-a*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a \sec \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \left (-\sec \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sec(f*x+e))^n*(a-a*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((-a*sec(f*x + e) + a)^(5/2)*(-sec(f*x + e))^n, x)