3.303 \(\int (d \sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=117 \[ \frac{2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt{\sec (e+f x)+1}}-\frac{(4 n+1) \tan (e+f x) (d \sec (e+f x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},n,n+1,\sec (e+f x)\right )}{f n (2 n+1) \sqrt{1-\sec (e+f x)} \sqrt{\sec (e+f x)+1}} \]

[Out]

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

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Rubi [A]  time = 0.125207, antiderivative size = 117, normalized size of antiderivative = 1., 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, 21, 3806, 64} \[ \frac{2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt{\sec (e+f x)+1}}-\frac{(4 n+1) \tan (e+f x) (d \sec (e+f x))^n \, _2F_1\left (\frac{1}{2},n;n+1;\sec (e+f x)\right )}{f n (2 n+1) \sqrt{1-\sec (e+f x)} \sqrt{\sec (e+f x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^n*(1 + Sec[e + f*x])^(3/2),x]

[Out]

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

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 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 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 64

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

Rubi steps

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

Mathematica [A]  time = 0.365603, size = 85, normalized size = 0.73 \[ \frac{\tan \left (\frac{1}{2} (e+f x)\right ) \sqrt{\sec (e+f x)+1} (d \sec (e+f x))^n \left ((4 n+1) \cos ^{n+\frac{1}{2}}(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},n+\frac{3}{2},\frac{3}{2},2 \sin ^2\left (\frac{1}{2} (e+f x)\right )\right )-1\right )}{f n} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^n*(1 + Sec[e + f*x])^(3/2),x]

[Out]

((-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])*(d*Sec[e +
f*x])^n*Sqrt[1 + Sec[e + f*x]]*Tan[(e + f*x)/2])/(f*n)

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Maple [F]  time = 0.164, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{n} \left ( 1+\sec \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^n*(1+sec(f*x+e))^(3/2),x)

[Out]

int((d*sec(f*x+e))^n*(1+sec(f*x+e))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{n}{\left (\sec \left (f x + e\right ) + 1\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^n*(1+sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^n*(sec(f*x + e) + 1)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (d \sec \left (f x + e\right )\right )^{n}{\left (\sec \left (f x + e\right ) + 1\right )}^{\frac{3}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^n*(1+sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral((d*sec(f*x + e))^n*(sec(f*x + e) + 1)^(3/2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**n*(1+sec(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{n}{\left (\sec \left (f x + e\right ) + 1\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^n*(1+sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^n*(sec(f*x + e) + 1)^(3/2), x)