3.222 \(\int \frac {(b \sec (c+d x))^n}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=80 \[ -\frac {2 \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (7-2 n);\frac {1}{4} (11-2 n);\cos ^2(c+d x)\right )}{d (7-2 n) \sqrt {\sin ^2(c+d x)} \sec ^{\frac {7}{2}}(c+d x)} \]

[Out]

-2*hypergeom([1/2, 7/4-1/2*n],[11/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(7-2*n)/sec(d*x+c)^(7/2
)/(sin(d*x+c)^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 80, normalized size of antiderivative = 1.00, 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} \[ -\frac {2 \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (7-2 n);\frac {1}{4} (11-2 n);\cos ^2(c+d x)\right )}{d (7-2 n) \sqrt {\sin ^2(c+d x)} \sec ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^n/Sec[c + d*x]^(5/2),x]

[Out]

(-2*Hypergeometric2F1[1/2, (7 - 2*n)/4, (11 - 2*n)/4, Cos[c + d*x]^2]*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(7 -
2*n)*Sec[c + d*x]^(7/2)*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 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]

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]

Rubi steps

\begin {align*} \int \frac {(b \sec (c+d x))^n}{\sec ^{\frac {5}{2}}(c+d x)} \, dx &=\left (\sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{-\frac {5}{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 {5}{2}-n}(c+d x) \, dx\\ &=-\frac {2 \, _2F_1\left (\frac {1}{2},\frac {1}{4} (7-2 n);\frac {1}{4} (11-2 n);\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (7-2 n) \sec ^{\frac {7}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.17, size = 81, normalized size = 1.01 \[ \frac {\sqrt {-\tan ^2(c+d x)} \csc (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} \left (n-\frac {5}{2}\right );\frac {1}{2} \left (n-\frac {1}{2}\right );\sec ^2(c+d x)\right )}{d \left (n-\frac {5}{2}\right ) \sec ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[c + d*x])^n/Sec[c + d*x]^(5/2),x]

[Out]

(Csc[c + d*x]*Hypergeometric2F1[1/2, (-5/2 + n)/2, (-1/2 + n)/2, Sec[c + d*x]^2]*(b*Sec[c + d*x])^n*Sqrt[-Tan[
c + d*x]^2])/(d*(-5/2 + n)*Sec[c + d*x]^(7/2))

________________________________________________________________________________________

fricas [F]  time = 1.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {5}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^n/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c))^n/sec(d*x + c)^(5/2), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^n/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^n/sec(d*x + c)^(5/2), x)

________________________________________________________________________________________

maple [F]  time = 0.99, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \sec \left (d x +c \right )\right )^{n}}{\sec \left (d x +c \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c))^n/sec(d*x+c)^(5/2),x)

[Out]

int((b*sec(d*x+c))^n/sec(d*x+c)^(5/2),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^n/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c))^n/sec(d*x + c)^(5/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cos(c + d*x))^n/(1/cos(c + d*x))^(5/2),x)

[Out]

int((b/cos(c + d*x))^n/(1/cos(c + d*x))^(5/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \sec {\left (c + d x \right )}\right )^{n}}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))**n/sec(d*x+c)**(5/2),x)

[Out]

Integral((b*sec(c + d*x))**n/sec(c + d*x)**(5/2), x)

________________________________________________________________________________________