3.4.27 \(\int (d \sec (e+f x))^n (a-a \sec (e+f x))^{3/2} \, dx\) [327]

Optimal. Leaf size=130 \[ \frac {2 a^2 (d \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {a-a \sec (e+f x)}}+\frac {2 a^2 (1+4 n) \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};1+\sec (e+f x)\right ) (-\sec (e+f x))^{-n} (d \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {a-a \sec (e+f x)}} \]

[Out]

2*a^2*(d*sec(f*x+e))^n*tan(f*x+e)/f/(1+2*n)/(a-a*sec(f*x+e))^(1/2)+2*a^2*(1+4*n)*hypergeom([1/2, 1-n],[3/2],1+
sec(f*x+e))*(d*sec(f*x+e))^n*tan(f*x+e)/f/(1+2*n)/((-sec(f*x+e))^n)/(a-a*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3899, 21, 3891, 69, 67} \begin {gather*} \frac {2 a^2 (4 n+1) \tan (e+f x) (-\sec (e+f x))^{-n} (d \sec (e+f x))^n \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};\sec (e+f x)+1\right )}{f (2 n+1) \sqrt {a-a \sec (e+f x)}}+\frac {2 a^2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt {a-a \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^n*(a - a*Sec[e + f*x])^(3/2),x]

[Out]

(2*a^2*(d*Sec[e + f*x])^n*Tan[e + f*x])/(f*(1 + 2*n)*Sqrt[a - a*Sec[e + f*x]]) + (2*a^2*(1 + 4*n)*Hypergeometr
ic2F1[1/2, 1 - n, 3/2, 1 + Sec[e + f*x]]*(d*Sec[e + f*x])^n*Tan[e + f*x])/(f*(1 + 2*n)*(-Sec[e + f*x])^n*Sqrt[
a - a*Sec[e + f*x]])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 69

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[((-b)*(c/d))^IntPart[m]*((b*x)^FracPart[m]/(
(-d)*(x/c))^FracPart[m]), Int[((-d)*(x/c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !GtQ[c, 0] &&  !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]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 2.39, size = 377, normalized size = 2.90 \begin {gather*} \frac {2^{-\frac {3}{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 ^3\left (\frac {1}{2} (e+f x)\right ) \left (-e^{i f n x} \left (6+11 n+6 n^2+n^3\right ) \, _2F_1\left (\frac {n}{2},\frac {3}{2}+n;\frac {2+n}{2};-e^{2 i (e+f x)}\right )+3 e^{i (e+f (1+n) x)} n \left (6+5 n+n^2\right ) \, _2F_1\left (\frac {1+n}{2},\frac {3}{2}+n;\frac {3+n}{2};-e^{2 i (e+f x)}\right )+e^{2 i e} n (1+n) \left (-3 e^{i f (2+n) x} (3+n) \, _2F_1\left (\frac {3}{2}+n,\frac {2+n}{2};\frac {4+n}{2};-e^{2 i (e+f x)}\right )+e^{i (e+f (3+n) x)} (2+n) \, _2F_1\left (\frac {3}{2}+n,\frac {3+n}{2};\frac {5+n}{2};-e^{2 i (e+f x)}\right )\right )\right ) \sec ^{-\frac {3}{2}-n}(e+f x) (d \sec (e+f x))^n (a-a \sec (e+f x))^{3/2}}{f n (1+n) (2+n) (3+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^n*(a - a*Sec[e + f*x])^(3/2),x]

[Out]

(2^(-3/2 + 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 + f*x)/2]^3*(-(E^(I*f*n*x)*(6 + 11*n + 6*n^2 + n^3)*Hypergeometric2F1[n/2, 3/
2 + n, (2 + n)/2, -E^((2*I)*(e + f*x))]) + 3*E^(I*(e + f*(1 + n)*x))*n*(6 + 5*n + n^2)*Hypergeometric2F1[(1 +
n)/2, 3/2 + n, (3 + n)/2, -E^((2*I)*(e + f*x))] + E^((2*I)*e)*n*(1 + n)*(-3*E^(I*f*(2 + n)*x)*(3 + n)*Hypergeo
metric2F1[3/2 + n, (2 + n)/2, (4 + n)/2, -E^((2*I)*(e + f*x))] + E^(I*(e + f*(3 + n)*x))*(2 + n)*Hypergeometri
c2F1[3/2 + n, (3 + n)/2, (5 + n)/2, -E^((2*I)*(e + f*x))]))*Sec[e + f*x]^(-3/2 - n)*(d*Sec[e + f*x])^n*(a - a*
Sec[e + f*x])^(3/2))/(f*n*(1 + n)*(2 + n)*(3 + n))

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \left (d \sec \left (f x +e \right )\right )^{n} \left (a -a \sec \left (f x +e \right )\right )^{\frac {3}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^n*(a-a*sec(f*x+e))^(3/2),x)

[Out]

int((d*sec(f*x+e))^n*(a-a*sec(f*x+e))^(3/2),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((d*sec(f*x+e))^n*(a-a*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((-a*sec(f*x + e) + a)^(3/2)*(d*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((d*sec(f*x+e))^n*(a-a*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d \sec {\left (e + f x \right )}\right )^{n} \left (- a \left (\sec {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**n*(a-a*sec(f*x+e))**(3/2),x)

[Out]

Integral((d*sec(e + f*x))**n*(-a*(sec(e + f*x) - 1))**(3/2), x)

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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((d*sec(f*x+e))^n*(a-a*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((-a*sec(f*x + e) + a)^(3/2)*(d*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 )}^{3/2}\,{\left (\frac {d}{\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))^(3/2)*(d/cos(e + f*x))^n,x)

[Out]

int((a - a/cos(e + f*x))^(3/2)*(d/cos(e + f*x))^n, x)

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