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

Optimal. Leaf size=32 \[ \frac{\sin (c+d x) \sqrt{b \sec (c+d x)}}{d \sqrt{\sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0066801, antiderivative size = 32, normalized size of antiderivative = 1., 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} \[ \frac{\sin (c+d x) \sqrt{b \sec (c+d x)}}{d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0355135, size = 32, normalized size = 1. \[ \frac{\sin (c+d x) \sqrt{b \sec (c+d x)}}{d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.127, size = 41, normalized size = 1.3 \begin{align*}{\frac{\sin \left ( dx+c \right ) }{d\cos \left ( dx+c \right ) }\sqrt{{\frac{b}{\cos \left ( dx+c \right ) }}} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.05672, size = 18, normalized size = 0.56 \begin{align*} \frac{\sqrt{b} \sin \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b)*sin(d*x + c)/d

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Fricas [A]  time = 1.40192, size = 76, normalized size = 2.38 \begin{align*} \frac{\sqrt{\frac{b}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]  time = 20.9001, size = 36, normalized size = 1.12 \begin{align*} \begin{cases} \frac{\sqrt{b} \tan{\left (c + d x \right )}}{d \sec{\left (c + d x \right )}} & \text{for}\: d \neq 0 \\\frac{x \sqrt{b \sec{\left (c \right )}}}{\sec ^{\frac{3}{2}}{\left (c \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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