3.2.72 \(\int \frac {\sqrt {\sec (c+d x)}}{(b \sec (c+d x))^{3/2}} \, dx\) [172]

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

[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, 2717} \begin {gather*} \frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{b d \sqrt {b \sec (c+d x)}} \end {gather*}

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 2717

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 = 0.05, size = 32, normalized size = 0.91 \begin {gather*} \frac {\sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (b \sec (c+d x))^{3/2}} \end {gather*}

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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Maple [A]
time = 35.29, size = 41, normalized size = 1.17

method	result	size
default	\(\frac {\sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}}{d \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right )}\)	\(41\)
risch	\(-\frac {i \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{i \left (d x +c \right )}}{2 b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {i \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{-i \left (d x +c \right )}}{2 b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(140\)

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,method=_RETURNVERBOSE)

[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 = 0.61, size = 13, normalized size = 0.37 \begin {gather*} \frac {\sin \left (d x + c\right )}{b^{\frac {3}{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))^(3/2),x, algorithm="maxima")

[Out]

sin(d*x + c)/(b^(3/2)*d)

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Fricas [A]
time = 4.14, size = 33, normalized size = 0.94 \begin {gather*} \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} \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))^(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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Sympy [A]
time = 11.70, size = 46, normalized size = 1.31 \begin {gather*} \begin {cases} \frac {\tan {\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}}}{d \left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}} & \text {for}\: d \neq 0 \\\frac {x \sqrt {\sec {\left (c \right )}}}{\left (b \sec {\left (c \right )}\right )^{\frac {3}{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))**(3/2),x)

[Out]

Piecewise((tan(c + d*x)*sqrt(sec(c + d*x))/(d*(b*sec(c + d*x))**(3/2)), Ne(d, 0)), (x*sqrt(sec(c))/(b*sec(c))*
*(3/2), 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))^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.42, size = 39, normalized size = 1.11 \begin {gather*} \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} \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))^(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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