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

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Rubi [A]  time = 0.01, antiderivative size = 33, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 1.00 \[ \frac {\sqrt {b \sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt {\sec (c+d x)}} \]

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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fricas [A]  time = 0.69, size = 111, normalized size = 3.36 \[ \left [\frac {\sqrt {b} \log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{2}}\right )}{2 \, d}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b}\right )}{d}\right ] \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sec \left (d x + c\right )} \sqrt {\sec \left (d x + c\right )}\,{d x} \]

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)

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maple [A]  time = 0.98, size = 52, normalized size = 1.58 \[ -\frac {2 \arctanh \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {\frac {b}{\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )}{d} \]

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

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maxima [B]  time = 1.02, size = 65, normalized size = 1.97 \[ \frac {\sqrt {b} {\left (\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )}}{2 \, d} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sec {\left (c + d x \right )}} \sqrt {\sec {\left (c + d x \right )}}\, dx \]

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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