3.172 \(\int \frac {\sqrt {\sec (c+d x)}}{(b \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=35 \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{b d \sqrt {b \sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {17, 2637} \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{b d \sqrt {b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(b*d*Sqrt[b*Sec[c + d*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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 32, normalized size = 0.91 \[ \frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.73, size = 33, normalized size = 0.94 \[ \frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\sec \left (d x + c\right )}}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(sec(d*x + c))/(b*sec(d*x + c))^(3/2), x)

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maple [A]  time = 0.86, size = 41, normalized size = 1.17 \[ \frac {\sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{d \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.04, size = 13, normalized size = 0.37 \[ \frac {\sin \left (d x + c\right )}{b^{\frac {3}{2}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

sin(d*x + c)/(b^(3/2)*d)

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mupad [B]  time = 0.42, size = 39, normalized size = 1.11 \[ \frac {\sin \left (2\,c+2\,d\,x\right )\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{2\,b^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^(1/2)/(b/cos(c + d*x))^(3/2),x)

[Out]

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

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sympy [A]  time = 15.08, size = 36, normalized size = 1.03 \[ \begin {cases} \frac {\tan {\left (c + d x \right )}}{b^{\frac {3}{2}} d \sec {\left (c + d x \right )}} & \text {for}\: d \neq 0 \\\frac {x \sqrt {\sec {\relax (c )}}}{\left (b \sec {\relax (c )}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(1/2)/(b*sec(d*x+c))**(3/2),x)

[Out]

Piecewise((tan(c + d*x)/(b**(3/2)*d*sec(c + d*x)), Ne(d, 0)), (x*sqrt(sec(c))/(b*sec(c))**(3/2), True))

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