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

Optimal. Leaf size=27 \[ \frac{x \sqrt{\sec (c+d x)}}{b \sqrt{b \sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0027357, antiderivative size = 27, 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, 8} \[ \frac{x \sqrt{\sec (c+d x)}}{b \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[Sec[c + d*x]])/(b*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 8

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

Rubi steps

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

Mathematica [A]  time = 0.025473, size = 24, normalized size = 0.89 \[ \frac{x \sec ^{\frac{3}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.083, size = 32, normalized size = 1.2 \begin{align*}{\frac{dx+c}{d} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{3}{2}}} \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(sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.64326, size = 35, normalized size = 1.3 \begin{align*} \frac{2 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac{3}{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.93745, size = 277, normalized size = 10.26 \begin{align*} \left [-\frac{\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 )}{2 \, b^{2} d}, \frac{\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 )}{b^{\frac{3}{2}} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(-b)*log(2*sqrt(-b)*sqrt(b/cos(d*x + c))*cos(d*x + c)^(3/2)*sin(d*x + c) + 2*b*cos(d*x + c)^2 - b)/(
b^2*d), arctan(sqrt(b/cos(d*x + c))*sin(d*x + c)/(sqrt(b)*sqrt(cos(d*x + c))))/(b^(3/2)*d)]

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Sympy [A]  time = 133.827, size = 5, normalized size = 0.19 \begin{align*} \frac{x}{b^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/b**(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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