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

Optimal. Leaf size=36 \[ \frac{b^2 \sqrt{b \sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt{\sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.00772, antiderivative size = 36, 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, 3770} \[ \frac{b^2 \sqrt{b \sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*ArcTanh[Sin[c + d*x]]*Sqrt[b*Sec[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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0342288, size = 33, normalized size = 0.92 \[ \frac{(b \sec (c+d x))^{5/2} \tanh ^{-1}(\sin (c+d x))}{d \sec ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.097, size = 52, normalized size = 1.4 \begin{align*} -2\,{\frac{\cos \left ( dx+c \right ) }{d \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{3/2}}{\it Artanh} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{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)^(3/2),x)

[Out]

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

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Maxima [B]  time = 1.91586, size = 97, normalized size = 2.69 \begin{align*} \frac{{\left (b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} \sqrt{b}}{2 \, 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)^(3/2),x, algorithm="maxima")

[Out]

1/2*(b^2*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - b^2*log(cos(d*x + c)^2 + sin(d*x + c)^2 -
2*sin(d*x + c) + 1))*sqrt(b)/d

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Fricas [A]  time = 1.84508, size = 296, normalized size = 8.22 \begin{align*} \left [\frac{b^{\frac{5}{2}} \log \left (-\frac{b \cos \left (d x + c\right )^{2} - 2 \, \sqrt{b} \sqrt{\frac{b}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{2}}\right )}{2 \, d}, -\frac{\sqrt{-b} b^{2} \arctan \left (\frac{\sqrt{-b} \sqrt{\frac{b}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b}\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)^(3/2),x, algorithm="fricas")

[Out]

[1/2*b^(5/2)*log(-(b*cos(d*x + c)^2 - 2*sqrt(b)*sqrt(b/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*b)/co
s(d*x + c)^2)/d, -sqrt(-b)*b^2*arctan(sqrt(-b)*sqrt(b/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/b)/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)**(3/2),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{3}{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)^(3/2),x, algorithm="giac")

[Out]

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