3.148 \(\int \frac{(b \sec (c+d x))^{3/2}}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=33 \[ \frac{b \sin (c+d x) \sqrt{b \sec (c+d x)}}{d \sqrt{\sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0074591, antiderivative size = 33, 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} \[ \frac{b \sin (c+d x) \sqrt{b \sec (c+d x)}}{d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^(3/2)/Sec[c + d*x]^(5/2),x]

[Out]

(b*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{(b \sec (c+d x))^{3/2}}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{\left (b \sqrt{b \sec (c+d x)}\right ) \int \cos (c+d x) \, dx}{\sqrt{\sec (c+d x)}}\\ &=\frac{b \sqrt{b \sec (c+d x)} \sin (c+d x)}{d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0440589, size = 32, normalized size = 0.97 \[ \frac{\sin (c+d x) (b \sec (c+d x))^{3/2}}{d \sec ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[c + d*x])^(3/2)/Sec[c + d*x]^(5/2),x]

[Out]

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

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Maple [A]  time = 0.114, size = 41, normalized size = 1.2 \begin{align*}{\frac{\sin \left ( dx+c \right ) }{d\cos \left ( dx+c \right ) } \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c))^(3/2)/sec(d*x+c)^(5/2),x)

[Out]

1/d*sin(d*x+c)*(b/cos(d*x+c))^(3/2)/(1/cos(d*x+c))^(5/2)/cos(d*x+c)

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Maxima [A]  time = 2.00161, size = 18, normalized size = 0.55 \begin{align*} \frac{b^{\frac{3}{2}} \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))^(3/2)/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

b^(3/2)*sin(d*x + c)/d

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Fricas [A]  time = 1.57804, size = 78, normalized size = 2.36 \begin{align*} \frac{b \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))^(3/2)/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

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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((b*sec(d*x+c))**(3/2)/sec(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \sec \left (d x + c\right )\right )^{\frac{3}{2}}}{\sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^(3/2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^(3/2)/sec(d*x + c)^(5/2), x)