3.173 \(\int \frac{1}{\sqrt{\sec (c+d x)} (b \sec (c+d x))^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0141337, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {18, 2635, 8} \[ \frac{x \sqrt{\sec (c+d x)}}{2 b \sqrt{b \sec (c+d x)}}+\frac{\sin (c+d x)}{2 b d \sqrt{\sec (c+d x)} \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Sec[c + d*x]]*(b*Sec[c + d*x])^(3/2)),x]

[Out]

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

Rule 18

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

Rule 2635

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] && Integer
Q[2*n]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.0659757, size = 45, normalized size = 0.65 \[ \frac{(2 (c+d x)+\sin (2 (c+d x))) \sec ^{\frac{3}{2}}(c+d x)}{4 d (b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Sec[c + d*x]]*(b*Sec[c + d*x])^(3/2)),x]

[Out]

(Sec[c + d*x]^(3/2)*(2*(c + d*x) + Sin[2*(c + d*x)]))/(4*d*(b*Sec[c + d*x])^(3/2))

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Maple [A]  time = 0.123, size = 54, normalized size = 0.8 \begin{align*}{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +dx+c}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{\frac{1}{\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}}} \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.04361, size = 34, normalized size = 0.49 \begin{align*} \frac{2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )}{4 \, b^{\frac{3}{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.97708, size = 443, normalized size = 6.42 \begin{align*} \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^{2} 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^{2} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]  time = 49.1731, size = 82, normalized size = 1.19 \begin{align*} \begin{cases} \frac{x \tan ^{2}{\left (c + d x \right )}}{2 b^{\frac{3}{2}} \sec ^{2}{\left (c + d x \right )}} + \frac{x}{2 b^{\frac{3}{2}} \sec ^{2}{\left (c + d x \right )}} + \frac{\tan{\left (c + d x \right )}}{2 b^{\frac{3}{2}} d \sec ^{2}{\left (c + d x \right )}} & \text{for}\: d \neq 0 \\\frac{x}{\left (b \sec{\left (c \right )}\right )^{\frac{3}{2}} \sqrt{\sec{\left (c \right )}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sec \left (d x + c\right )\right )^{\frac{3}{2}} \sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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