3.2.36 \(\int \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)} \, dx\) [136]

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

[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, 3855} \begin {gather*} \frac {\sqrt {b \sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt {\sec (c+d x)}} \end {gather*}

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 3855

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 = 0.02, size = 33, normalized size = 1.00 \begin {gather*} \frac {\tanh ^{-1}(\sin (c+d x)) \sqrt {b \sec (c+d x)}}{d \sqrt {\sec (c+d x)}} \end {gather*}

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 = 33.72, size = 52, normalized size = 1.58

method	result	size
default	\(-\frac {2 \cos \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {\frac {b}{\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )}{d}\)	\(52\)
risch	\(\frac {2 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \cos \left (d x +c \right )}{d}-\frac {2 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \cos \left (d x +c \right )}{d}\)	\(152\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (29) = 58\).
time = 0.62, size = 65, normalized size = 1.97 \begin {gather*} \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 {gather*}

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 = 2.89, size = 111, normalized size = 3.36 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {b \sec {\left (c + d x \right )}} \sqrt {\sec {\left (c + d x \right )}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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