3.4.20 \(\int (-\sec (e+f x))^n (a-a \sec (e+f x))^{5/2} \, dx\) [320]

Optimal. Leaf size=178 \[ \frac {2 a^3 \left (3+24 n+16 n^2\right ) \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};1+\sec (e+f x)\right ) \tan (e+f x)}{f (1+2 n) (3+2 n) \sqrt {a-a \sec (e+f x)}}+\frac {2 a^3 (7+4 n) (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) (3+2 n) \sqrt {a-a \sec (e+f x)}}+\frac {2 a^2 (-\sec (e+f x))^n \sqrt {a-a \sec (e+f x)} \tan (e+f x)}{f (3+2 n)} \]

[Out]

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)+2*a^3*(7+4*n)*(-sec(f*x+e))^n*tan(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))^n*(a-a*
sec(f*x+e))^(1/2)*tan(f*x+e)/f/(3+2*n)

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Rubi [A]
time = 0.25, antiderivative size = 178, normalized size of antiderivative = 1.00, 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 = {3899, 4101, 3891, 67} \begin {gather*} \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^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)}}+\frac {2 a^2 \tan (e+f x) \sqrt {a-a \sec (e+f x)} (-\sec (e+f x))^n}{f (2 n+3)} \end {gather*}

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 67

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

Rule 3891

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 3899

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] /;
FreeQ[{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 4101

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 (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 ) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 26.26, size = 458, normalized size = 2.57 \begin {gather*} \frac {2^{-\frac {5}{2}+n} e^{\frac {1}{2} i (e+f (1-2 n) x)} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{-\frac {1}{2}+n} \left (1+e^{2 i (e+f x)}\right )^{-\frac {1}{2}+n} \csc ^5\left (\frac {e}{2}+\frac {f x}{2}\right ) \left (\frac {e^{i f n x} \, _2F_1\left (\frac {n}{2},\frac {5}{2}+n;\frac {2+n}{2};-e^{2 i (e+f x)}\right )}{n}-\frac {5 e^{i (e+f (1+n) x)} \, _2F_1\left (\frac {1+n}{2},\frac {5}{2}+n;\frac {3+n}{2};-e^{2 i (e+f x)}\right )}{1+n}+\frac {10 e^{i (2 e+f (2+n) x)} \, _2F_1\left (\frac {2+n}{2},\frac {5}{2}+n;\frac {4+n}{2};-e^{2 i (e+f x)}\right )}{2+n}-\frac {10 e^{i (3 e+f (3+n) x)} \, _2F_1\left (\frac {5}{2}+n,\frac {3+n}{2};\frac {5+n}{2};-e^{2 i (e+f x)}\right )}{3+n}+\frac {5 e^{i (4 e+f (4+n) x)} \, _2F_1\left (\frac {5}{2}+n,\frac {4+n}{2};\frac {6+n}{2};-e^{2 i (e+f x)}\right )}{4+n}-\frac {e^{i (5 e+f (5+n) x)} \, _2F_1\left (\frac {5}{2}+n,\frac {5+n}{2};\frac {7+n}{2};-e^{2 i (e+f x)}\right )}{5+n}\right ) (-\sec (e+f x))^n \sec ^{-\frac {5}{2}-n}(e+f x) (a-a \sec (e+f x))^{5/2}}{f} \end {gather*}

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

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \left (-\sec \left (f x +e \right )\right )^{n} \left (a -a \sec \left (f x +e \right )\right )^{\frac {5}{2}}\, dx\]

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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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