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

Optimal. Leaf size=27 \[ \frac{b^2 x \sqrt{b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0026389, 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} \[ \frac{b^2 x \sqrt{b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x*Sqrt[b*Sec[c + d*x]])/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 8

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

Rubi steps

\begin{align*} \int \frac{(b \sec (c+d x))^{5/2}}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{\left (b^2 \sqrt{b \sec (c+d x)}\right ) \int 1 \, dx}{\sqrt{\sec (c+d x)}}\\ &=\frac{b^2 x \sqrt{b \sec (c+d x)}}{\sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0194576, size = 24, normalized size = 0.89 \[ \frac{x (b \sec (c+d x))^{5/2}}{\sec ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.091, size = 32, normalized size = 1.2 \begin{align*}{\frac{dx+c}{d} \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{5}{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))^(5/2)/sec(d*x+c)^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.82461, size = 35, normalized size = 1.3 \begin{align*} \frac{2 \, b^{\frac{5}{2}} \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.97332, size = 271, normalized size = 10.04 \begin{align*} \left [\frac{\sqrt{-b} b^{2} \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 \, d}, \frac{b^{\frac{5}{2}} \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 )}{d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(-b)*b^2*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)/d, b^(5/2)*arctan(sqrt(b/cos(d*x + c))*sin(d*x + c)/(sqrt(b)*sqrt(cos(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))**(5/2)/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{5}{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))^(5/2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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