3.2.81 \(\int \frac {\sqrt {\sec (c+d x)}}{(b \sec (c+d x))^{5/2}} \, dx\) [181]

Optimal. Leaf size=69 \[ \frac {x \sqrt {\sec (c+d x)}}{2 b^2 \sqrt {b \sec (c+d x)}}+\frac {\sin (c+d x)}{2 b^2 d \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}} \]

[Out]

1/2*sin(d*x+c)/b^2/d/sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+1/2*x*sec(d*x+c)^(1/2)/b^2/(b*sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {17, 2715, 8} \begin {gather*} \frac {x \sqrt {\sec (c+d x)}}{2 b^2 \sqrt {b \sec (c+d x)}}+\frac {\sin (c+d x)}{2 b^2 d \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[Sec[c + d*x]])/(2*b^2*Sqrt[b*Sec[c + d*x]]) + Sin[c + d*x]/(2*b^2*d*Sqrt[Sec[c + d*x]]*Sqrt[b*Sec[c +
d*x]])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 17

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[a^(m + 1/2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v])
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rubi steps

\begin {align*} \int \frac {\sqrt {\sec (c+d x)}}{(b \sec (c+d x))^{5/2}} \, dx &=\frac {\sqrt {\sec (c+d x)} \int \cos ^2(c+d x) \, dx}{b^2 \sqrt {b \sec (c+d x)}}\\ &=\frac {\sin (c+d x)}{2 b^2 d \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {\sqrt {\sec (c+d x)} \int 1 \, dx}{2 b^2 \sqrt {b \sec (c+d x)}}\\ &=\frac {x \sqrt {\sec (c+d x)}}{2 b^2 \sqrt {b \sec (c+d x)}}+\frac {\sin (c+d x)}{2 b^2 d \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 48, normalized size = 0.70 \begin {gather*} \frac {\sqrt {\sec (c+d x)} (2 (c+d x)+\sin (2 (c+d x)))}{4 b^2 d \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(2*(c + d*x) + Sin[2*(c + d*x)]))/(4*b^2*d*Sqrt[b*Sec[c + d*x]])

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Maple [A]
time = 32.28, size = 54, normalized size = 0.78

method	result	size
default	\(\frac {\left (\sin \left (d x +c \right ) \cos \left (d x +c \right )+d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{2 d \cos \left (d x +c \right )^{2} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}\)	\(54\)
risch	\(\frac {\sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, x}{2 b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {i \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(197\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/2/d*(sin(d*x+c)*cos(d*x+c)+d*x+c)*(1/cos(d*x+c))^(1/2)/cos(d*x+c)^2/(b/cos(d*x+c))^(5/2)

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Maxima [A]
time = 0.58, size = 25, normalized size = 0.36 \begin {gather*} \frac {2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )}{4 \, b^{\frac {5}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/4*(2*d*x + 2*c + sin(2*d*x + 2*c))/(b^(5/2)*d)

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Fricas [A]
time = 3.57, size = 165, normalized size = 2.39 \begin {gather*} \left [\frac {2 \, \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) - \sqrt {-b} \log \left (2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + 2 \, b \cos \left (d x + c\right )^{2} - b\right )}{4 \, b^{3} d}, \frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {b} \sqrt {\cos \left (d x + c\right )}}\right )}{2 \, b^{3} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(b/cos(d*x + c))*cos(d*x + c)^(3/2)*sin(d*x + c) - sqrt(-b)*log(2*sqrt(-b)*sqrt(b/cos(d*x + c))*co
s(d*x + c)^(3/2)*sin(d*x + c) + 2*b*cos(d*x + c)^2 - b))/(b^3*d), 1/2*(sqrt(b/cos(d*x + c))*cos(d*x + c)^(3/2)
*sin(d*x + c) + sqrt(b)*arctan(sqrt(b/cos(d*x + c))*sin(d*x + c)/(sqrt(b)*sqrt(cos(d*x + c)))))/(b^3*d)]

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Sympy [A]
time = 19.41, size = 107, normalized size = 1.55 \begin {gather*} \begin {cases} \frac {x \tan ^{2}{\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}}}{2 \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}} + \frac {x \sqrt {\sec {\left (c + d x \right )}}}{2 \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}} + \frac {\tan {\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}}}{2 d \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}} & \text {for}\: d \neq 0 \\\frac {x \sqrt {\sec {\left (c \right )}}}{\left (b \sec {\left (c \right )}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(1/2)/(b*sec(d*x+c))**(5/2),x)

[Out]

Piecewise((x*tan(c + d*x)**2*sqrt(sec(c + d*x))/(2*(b*sec(c + d*x))**(5/2)) + x*sqrt(sec(c + d*x))/(2*(b*sec(c
+ d*x))**(5/2)) + tan(c + d*x)*sqrt(sec(c + d*x))/(2*d*(b*sec(c + d*x))**(5/2)), Ne(d, 0)), (x*sqrt(sec(c))/(
b*sec(c))**(5/2), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.63, size = 64, normalized size = 0.93 \begin {gather*} \frac {\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,\left (\sin \left (c+d\,x\right )+\sin \left (3\,c+3\,d\,x\right )+4\,d\,x\,\cos \left (c+d\,x\right )\right )}{8\,b^3\,d\,\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b/cos(c + d*x))^(1/2)*(sin(c + d*x) + sin(3*c + 3*d*x) + 4*d*x*cos(c + d*x)))/(8*b^3*d*cos(c + d*x)*(1/cos(c
+ d*x))^(1/2))

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