3.136 \(\int \sqrt{\sec (c+d x)} \sqrt{b \sec (c+d x)} \, dx\)

Optimal. Leaf size=33 \[ \frac{\sqrt{b \sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt{\sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.006833, 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, 3770} \[ \frac{\sqrt{b \sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sec[c + d*x]]*Sqrt[b*Sec[c + d*x]],x]

[Out]

(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 \sqrt{\sec (c+d x)} \sqrt{b \sec (c+d x)} \, dx &=\frac{\sqrt{b \sec (c+d x)} \int \sec (c+d x) \, dx}{\sqrt{\sec (c+d x)}}\\ &=\frac{\tanh ^{-1}(\sin (c+d x)) \sqrt{b \sec (c+d x)}}{d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0113776, size = 33, normalized size = 1. \[ \frac{\sqrt{b \sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sec[c + d*x]]*Sqrt[b*Sec[c + d*x]],x]

[Out]

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

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Maple [A]  time = 0.116, size = 52, normalized size = 1.6 \begin{align*} -2\,{\frac{\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}}{d}{\it Artanh} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \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)^(1/2)*(b*sec(d*x+c))^(1/2),x)

[Out]

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

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Maxima [B]  time = 2.06356, size = 88, normalized size = 2.67 \begin{align*} \frac{\sqrt{b}{\left (\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 ) - \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 )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.49578, size = 290, normalized size = 8.79 \begin{align*} \left [\frac{\sqrt{b} \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} \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(sec(d*x+c)^(1/2)*(b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/2*sqrt(b)*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)*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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec{\left (c + d x \right )}} \sqrt{\sec{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(b*sec(c + d*x))*sqrt(sec(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (d x + c\right )} \sqrt{\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)^(1/2)*(b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(d*x + c))*sqrt(sec(d*x + c)), x)