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

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

[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 = 0.18, 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 = {3814, 21, 3806, 67, 65} \[ \frac {2 a^2 (4 n+1) \tan (e+f x) (-\sec (e+f x))^{-n} (d \sec (e+f x))^n \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};\sec (e+f x)+1\right )}{f (2 n+1) \sqrt {a-a \sec (e+f x)}}+\frac {2 a^2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt {a-a \sec (e+f x)}} \]

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 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 67

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 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]

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 ) \operatorname {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 ) \operatorname {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]  time = 1.16, size = 346, normalized size = 2.66 \[ -\frac {2^{n-\frac {3}{2}} e^{-\frac {1}{2} i (2 n+1) (e+f x)} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{n+\frac {1}{2}} \csc ^3\left (\frac {1}{2} (e+f x)\right ) (a-a \sec (e+f x))^{3/2} \left (3 n \left (n^2+4 n+3\right ) e^{i (n+2) (e+f x)} \, _2F_1\left (1,\frac {1-n}{2};\frac {n+4}{2};-e^{2 i (e+f x)}\right )+\left (n^3+6 n^2+11 n+6\right ) e^{i n (e+f x)} \, _2F_1\left (1,\frac {1}{2} (-n-1);\frac {n+2}{2};-e^{2 i (e+f x)}\right )-n (n+2) \left ((n+1) e^{i (n+3) (e+f x)} \, _2F_1\left (1,1-\frac {n}{2};\frac {n+5}{2};-e^{2 i (e+f x)}\right )+3 (n+3) e^{i (n+1) (e+f x)} \, _2F_1\left (1,-\frac {n}{2};\frac {n+3}{2};-e^{2 i (e+f x)}\right )\right )\right ) \sec ^{-n-\frac {3}{2}}(e+f x) (d \sec (e+f x))^n}{f n (n+1) (n+2) (n+3)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((2^(-3/2 + n)*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(1/2 + n)*Csc[(e + f*x)/2]^3*(E^(I*n*(e + f*x))*(6
+ 11*n + 6*n^2 + n^3)*Hypergeometric2F1[1, (-1 - n)/2, (2 + n)/2, -E^((2*I)*(e + f*x))] + 3*E^(I*(2 + n)*(e +
f*x))*n*(3 + 4*n + n^2)*Hypergeometric2F1[1, (1 - n)/2, (4 + n)/2, -E^((2*I)*(e + f*x))] - n*(2 + n)*(E^(I*(3
+ n)*(e + f*x))*(1 + n)*Hypergeometric2F1[1, 1 - n/2, (5 + n)/2, -E^((2*I)*(e + f*x))] + 3*E^(I*(1 + n)*(e +
f*x))*(3 + n)*Hypergeometric2F1[1, -1/2*n, (3 + 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))/(E^((I/2)*(1 + 2*n)*(e + f*x))*f*n*(1 + n)*(2 + n)*(3 + n)))

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a \sec \left (f x + e\right ) - a\right )} \sqrt {-a \sec \left (f x + e\right ) + a} \left (d \sec \left (f x + e\right )\right )^{n}, x\right ) \]

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

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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maple [F]  time = 1.18, 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 \[ \int {\left (-a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (d \sec \left (f x + e\right )\right )^{n}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a-\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

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

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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