3.220 \(\int \frac{(b \sec (c+d x))^n}{\sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=80 \[ -\frac{2 \sin (c+d x) (b \sec (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{4} (3-2 n),\frac{1}{4} (7-2 n),\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt{\sin ^2(c+d x)} \sec ^{\frac{3}{2}}(c+d x)} \]

[Out]

(-2*Hypergeometric2F1[1/2, (3 - 2*n)/4, (7 - 2*n)/4, Cos[c + d*x]^2]*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(3 -
2*n)*Sec[c + d*x]^(3/2)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.0389261, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {20, 3772, 2643} \[ -\frac{2 \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (3-2 n);\frac{1}{4} (7-2 n);\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt{\sin ^2(c+d x)} \sec ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^n/Sqrt[Sec[c + d*x]],x]

[Out]

(-2*Hypergeometric2F1[1/2, (3 - 2*n)/4, (7 - 2*n)/4, Cos[c + d*x]^2]*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(3 -
2*n)*Sec[c + d*x]^(3/2)*Sqrt[Sin[c + d*x]^2])

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int \frac{(b \sec (c+d x))^n}{\sqrt{\sec (c+d x)}} \, dx &=\left (\sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{-\frac{1}{2}+n}(c+d x) \, dx\\ &=\left (\cos ^{\frac{1}{2}+n}(c+d x) \sqrt{\sec (c+d x)} (b \sec (c+d x))^n\right ) \int \cos ^{\frac{1}{2}-n}(c+d x) \, dx\\ &=-\frac{2 \, _2F_1\left (\frac{1}{2},\frac{1}{4} (3-2 n);\frac{1}{4} (7-2 n);\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) \sec ^{\frac{3}{2}}(c+d x) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.117239, size = 81, normalized size = 1.01 \[ \frac{\sqrt{-\tan ^2(c+d x)} \csc (c+d x) (b \sec (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} \left (n-\frac{1}{2}\right ),\frac{1}{2} \left (n+\frac{3}{2}\right ),\sec ^2(c+d x)\right )}{d \left (n-\frac{1}{2}\right ) \sec ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[c + d*x])^n/Sqrt[Sec[c + d*x]],x]

[Out]

(Csc[c + d*x]*Hypergeometric2F1[1/2, (-1/2 + n)/2, (3/2 + n)/2, Sec[c + d*x]^2]*(b*Sec[c + d*x])^n*Sqrt[-Tan[c
+ d*x]^2])/(d*(-1/2 + n)*Sec[c + d*x]^(3/2))

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Maple [F]  time = 0.128, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b\sec \left ( dx+c \right ) \right ) ^{n}{\frac{1}{\sqrt{\sec \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c))^n/sec(d*x+c)^(1/2),x)

[Out]

int((b*sec(d*x+c))^n/sec(d*x+c)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \sec \left (d x + c\right )\right )^{n}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^n/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c))^n/sqrt(sec(d*x + c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (b \sec \left (d x + c\right )\right )^{n}}{\sqrt{\sec \left (d x + c\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^n/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c))^n/sqrt(sec(d*x + c)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \sec{\left (c + d x \right )}\right )^{n}}{\sqrt{\sec{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))**n/sec(d*x+c)**(1/2),x)

[Out]

Integral((b*sec(c + d*x))**n/sqrt(sec(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \sec \left (d x + c\right )\right )^{n}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^n/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^n/sqrt(sec(d*x + c)), x)