3.2.39 \(\int \frac {\sqrt {b \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [139]

Optimal. Leaf size=63 \[ \frac {x \sqrt {b \sec (c+d x)}}{2 \sqrt {\sec (c+d x)}}+\frac {\sqrt {b \sec (c+d x)} \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

1/2*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(3/2)+1/2*x*(b*sec(d*x+c))^(1/2)/sec(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 63, 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 {b \sec (c+d x)}}{2 \sqrt {\sec (c+d x)}}+\frac {\sin (c+d x) \sqrt {b \sec (c+d x)}}{2 d \sec ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[b*Sec[c + d*x]])/(2*Sqrt[Sec[c + d*x]]) + (Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/(2*d*Sec[c + d*x]^(3/2))

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 {b \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx &=\frac {\sqrt {b \sec (c+d x)} \int \cos ^2(c+d x) \, dx}{\sqrt {\sec (c+d x)}}\\ &=\frac {\sqrt {b \sec (c+d x)} \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\sqrt {b \sec (c+d x)} \int 1 \, dx}{2 \sqrt {\sec (c+d x)}}\\ &=\frac {x \sqrt {b \sec (c+d x)}}{2 \sqrt {\sec (c+d x)}}+\frac {\sqrt {b \sec (c+d x)} \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.07, size = 45, normalized size = 0.71 \begin {gather*} \frac {\sqrt {b \sec (c+d x)} (2 (c+d x)+\sin (2 (c+d x)))}{4 d \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 33.36, size = 54, normalized size = 0.86

method	result	size
default	\(\frac {\left (\sin \left (d x +c \right ) \cos \left (d x +c \right )+d x +c \right ) \sqrt {\frac {b}{\cos \left (d x +c \right )}}}{2 d \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \cos \left (d x +c \right )^{2}}\)	\(54\)
risch	\(\frac {\sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, x}{2 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{2 i \left (d x +c \right )}}{8 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {i \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{-2 i \left (d x +c \right )}}{8 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(188\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.58, size = 25, normalized size = 0.40 \begin {gather*} \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} \sqrt {b}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 3.14, size = 158, normalized size = 2.51 \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 \, 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 \, d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^(1/2)/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))*c
os(d*x + c)^(3/2)*sin(d*x + c) + 2*b*cos(d*x + c)^2 - b))/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)))))/d]

________________________________________________________________________________________

Sympy [A]
time = 17.60, size = 107, normalized size = 1.70 \begin {gather*} \begin {cases} \frac {x \sqrt {b \sec {\left (c + d x \right )}} \tan ^{2}{\left (c + d x \right )}}{2 \sec ^{\frac {5}{2}}{\left (c + d x \right )}} + \frac {x \sqrt {b \sec {\left (c + d x \right )}}}{2 \sec ^{\frac {5}{2}}{\left (c + d x \right )}} + \frac {\sqrt {b \sec {\left (c + d x \right )}} \tan {\left (c + d x \right )}}{2 d \sec ^{\frac {5}{2}}{\left (c + d x \right )}} & \text {for}\: d \neq 0 \\\frac {x \sqrt {b \sec {\left (c \right )}}}{\sec ^{\frac {5}{2}}{\left (c \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.44, size = 41, normalized size = 0.65 \begin {gather*} \frac {\left (\sin \left (2\,c+2\,d\,x\right )+2\,d\,x\right )\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}}{4\,d\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________