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

Optimal. Leaf size=177 \[ \frac {2 a^3 \left (16 n^2+24 n+3\right ) \tan (e+f x) \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};1-\sec (e+f x)\right )}{f (2 n+1) (2 n+3) \sqrt {a \sec (e+f x)+a}}+\frac {2 a^3 (4 n+7) \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) (2 n+3) \sqrt {a \sec (e+f x)+a}}+\frac {2 a^2 \sin (e+f x) \sqrt {a \sec (e+f x)+a} \sec ^{n+1}(e+f x)}{f (2 n+3)} \]

[Out]

2*a^3*(7+4*n)*sec(f*x+e)^(1+n)*sin(f*x+e)/f/(4*n^2+8*n+3)/(a+a*sec(f*x+e))^(1/2)+2*a^2*sec(f*x+e)^(1+n)*sin(f*
x+e)*(a+a*sec(f*x+e))^(1/2)/f/(3+2*n)+2*a^3*(16*n^2+24*n+3)*hypergeom([1/2, 1-n],[3/2],1-sec(f*x+e))*tan(f*x+e
)/f/(4*n^2+8*n+3)/(a+a*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3814, 4016, 3806, 65} \[ \frac {2 a^3 \left (16 n^2+24 n+3\right ) \tan (e+f x) \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};1-\sec (e+f x)\right )}{f (2 n+1) (2 n+3) \sqrt {a \sec (e+f x)+a}}+\frac {2 a^2 \sin (e+f x) \sqrt {a \sec (e+f x)+a} \sec ^{n+1}(e+f x)}{f (2 n+3)}+\frac {2 a^3 (4 n+7) \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) (2 n+3) \sqrt {a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(7 + 4*n)*Sec[e + f*x]^(1 + n)*Sin[e + f*x])/(f*(1 + 2*n)*(3 + 2*n)*Sqrt[a + a*Sec[e + f*x]]) + (2*a^2*
Sec[e + f*x]^(1 + n)*Sqrt[a + a*Sec[e + f*x]]*Sin[e + f*x])/(f*(3 + 2*n)) + (2*a^3*(3 + 24*n + 16*n^2)*Hyperge
ometric2F1[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]])

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])

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 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]

Rubi steps

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

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Mathematica [C]  time = 8.12, size = 400, normalized size = 2.26 \[ -\frac {i 2^{n-\frac {5}{2}} e^{-\frac {1}{2} i (2 n+3) (e+f x)} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{n+\frac {3}{2}} \sec ^5\left (\frac {1}{2} (e+f x)\right ) (a (\sec (e+f x)+1))^{5/2} \left (\frac {10 e^{i (n+2) (e+f x)} \, _2F_1\left (1,\frac {1}{2} (-n-1);\frac {n+4}{2};-e^{2 i (e+f x)}\right )}{n+2}+\frac {5 e^{i (n+4) (e+f x)} \, _2F_1\left (1,\frac {1-n}{2};\frac {n+6}{2};-e^{2 i (e+f x)}\right )}{n+4}+\frac {e^{i n (e+f x)} \, _2F_1\left (1,-\frac {n}{2}-\frac {3}{2};\frac {n}{2}+1;-e^{2 i (e+f x)}\right )}{n}+\frac {5 e^{i (n+1) (e+f x)} \, _2F_1\left (1,-\frac {n}{2}-1;\frac {n+3}{2};-e^{2 i (e+f x)}\right )}{n+1}+\frac {e^{i (n+5) (e+f x)} \, _2F_1\left (1,1-\frac {n}{2};\frac {n+7}{2};-e^{2 i (e+f x)}\right )}{n+5}+\frac {10 e^{i (n+3) (e+f x)} \, _2F_1\left (1,-\frac {n}{2};\frac {n+5}{2};-e^{2 i (e+f x)}\right )}{n+3}\right )}{f \sec ^{\frac {5}{2}}(e+f x)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(-5/2 + n)*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(3/2 + n)*((10*E^(I*(2 + n)*(e + f*x))*Hypergeo
metric2F1[1, (-1 - n)/2, (4 + n)/2, -E^((2*I)*(e + f*x))])/(2 + n) + (5*E^(I*(4 + n)*(e + f*x))*Hypergeometric
2F1[1, (1 - n)/2, (6 + n)/2, -E^((2*I)*(e + f*x))])/(4 + n) + (E^(I*n*(e + f*x))*Hypergeometric2F1[1, -3/2 - n
/2, 1 + n/2, -E^((2*I)*(e + f*x))])/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, -1/2*n, (5 + n)/2, -E^((2*I)*(e + f*x))]
)/(3 + n))*Sec[(e + f*x)/2]^5*(a*(1 + Sec[e + f*x]))^(5/2))/(E^((I/2)*(3 + 2*n)*(e + f*x))*f*Sec[e + f*x]^(5/2
))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} \sec \left (f x + e\right )^{2} + 2 \, a^{2} \sec \left (f x + e\right ) + a^{2}\right )} \sqrt {a \sec \left (f x + e\right ) + a} \sec \left (f x + e\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^n*(a+a*sec(f*x+e))^(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sec \left (f x + e\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.19, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{n}\left (f x +e \right )\right ) \left (a +a \sec \left (f x +e \right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^n*(a+a*sec(f*x+e))^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sec \left (f x + e\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^n*(a+a*sec(f*x+e))^(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(5/2)*(1/cos(e + f*x))^n,x)

[Out]

int((a + a/cos(e + f*x))^(5/2)*(1/cos(e + f*x))^n, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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