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

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

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3814, 21, 3806, 67, 65} \[ \frac {2 a^2 (4 n+1) \sin (e+f x) (-\sec (e+f x))^n \sec ^{1-n}(e+f x) \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};1-\sec (e+f x)\right )}{f (2 n+1) \sqrt {a \sec (e+f x)+a}}+\frac {2 a^2 \tan (e+f x) (-\sec (e+f x))^n}{f (2 n+1) \sqrt {a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(1 + 4*n)*Hypergeometric2F1[1/2, 1 - n, 3/2, 1 - Sec[e + f*x]]*(-Sec[e + f*x])^n*Sec[e + f*x]^(1 - n)*S
in[e + f*x])/(f*(1 + 2*n)*Sqrt[a + a*Sec[e + f*x]]) + (2*a^2*(-Sec[e + f*x])^n*Tan[e + f*x])/(f*(1 + 2*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 (-\sec (e+f x))^n (a+a \sec (e+f x))^{3/2} \, dx &=\frac {2 a^2 (-\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 {(-\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 (-\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 (-\sec (e+f x))^n \sqrt {a+a \sec (e+f x)} \, dx}{1+2 n}\\ &=\frac {2 a^2 (-\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) \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 (-\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 \sec ^{1-n}(e+f x) \sin (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 (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 \sec ^{1-n}(e+f x) \sin (e+f x)}{f (1+2 n) \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 (-\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 [A]  time = 0.38, size = 88, normalized size = 0.68 \[ \frac {a \tan \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} (-\sec (e+f x))^n \left ((4 n+1) \cos ^{n+\frac {1}{2}}(e+f x) \, _2F_1\left (\frac {1}{2},n+\frac {3}{2};\frac {3}{2};2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )-1\right )}{f n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-1 + (1 + 4*n)*Cos[e + f*x]^(1/2 + n)*Hypergeometric2F1[1/2, 3/2 + n, 3/2, 2*Sin[(e + f*x)/2]^2])*(-Sec[e
+ f*x])^n*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(f*n)

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (-\sec \left (f x + e\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.95, size = 0, normalized size = 0.00 \[ \int \left (-\sec \left (f x +e \right )\right )^{n} \left (a +a \sec \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(3/2)*(-1/cos(e + f*x))^n,x)

[Out]

int((a + a/cos(e + f*x))^(3/2)*(-1/cos(e + f*x))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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