3.162 \(\int \frac{\sec ^{\frac{5}{2}}(c+d x)}{\sqrt{b \sec (c+d x)}} \, dx\)

Optimal. Leaf size=32 \[ \frac{\sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{d \sqrt{b \sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0114377, antiderivative size = 32, 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 = {17, 3767, 8} \[ \frac{\sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{d \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(5/2)/Sqrt[b*Sec[c + d*x]],x]

[Out]

(Sec[c + d*x]^(3/2)*Sin[c + d*x])/(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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{\sec ^{\frac{5}{2}}(c+d x)}{\sqrt{b \sec (c+d x)}} \, dx &=\frac{\sqrt{\sec (c+d x)} \int \sec ^2(c+d x) \, dx}{\sqrt{b \sec (c+d x)}}\\ &=-\frac{\sqrt{\sec (c+d x)} \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d \sqrt{b \sec (c+d x)}}\\ &=\frac{\sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{d \sqrt{b \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0256689, size = 32, normalized size = 1. \[ \frac{\sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{d \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^(5/2)/Sqrt[b*Sec[c + d*x]],x]

[Out]

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

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Maple [A]  time = 0.119, size = 39, normalized size = 1.2 \begin{align*}{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{5}{2}}}{\frac{1}{\sqrt{{\frac{b}{\cos \left ( dx+c \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(5/2)/(b*sec(d*x+c))^(1/2),x)

[Out]

1/d*(1/cos(d*x+c))^(5/2)*cos(d*x+c)*sin(d*x+c)/(b/cos(d*x+c))^(1/2)

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Maxima [B]  time = 2.03861, size = 80, normalized size = 2.5 \begin{align*} \frac{2 \, \sqrt{b} \sin \left (2 \, d x + 2 \, c\right )}{{\left (b \cos \left (2 \, d x + 2 \, c\right )^{2} + b \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b \cos \left (2 \, d x + 2 \, c\right ) + b\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)/(b*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.64232, size = 81, normalized size = 2.53 \begin{align*} \frac{\sqrt{\frac{b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{b d \sqrt{\cos \left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)/(b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(5/2)/(b*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{\frac{5}{2}}}{\sqrt{b \sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)/(b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^(5/2)/sqrt(b*sec(d*x + c)), x)