3.187 \(\int \frac {\sqrt {\cos (c+d x)}}{(b \cos (c+d x))^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 36, 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} \[ \frac {\sqrt {\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{b d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(b*d*Sqrt[b*Cos[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 {\sqrt {\cos (c+d x)}}{(b \cos (c+d x))^{3/2}} \, dx &=\frac {\sqrt {\cos (c+d x)} \int \sec (c+d x) \, dx}{b \sqrt {b \cos (c+d x)}}\\ &=\frac {\tanh ^{-1}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{b d \sqrt {b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.92 \[ \frac {\cos ^{\frac {3}{2}}(c+d x) \tanh ^{-1}(\sin (c+d x))}{d (b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^(3/2))/(d*(b*Cos[c + d*x])^(3/2))

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fricas [A]  time = 0.85, size = 116, normalized size = 3.22 \[ \left [\frac {\log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right )}{2 \, b^{\frac {3}{2}} d}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right )}{b^{2} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/2*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*sqrt(cos(d*x + c))*sin(d*x + c) - 2*b*cos(d*x + c
))/cos(d*x + c)^3)/(b^(3/2)*d), -sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x +
c))))/(b^2*d)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\cos \left (d x + c\right )}}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(cos(d*x + c))/(b*cos(d*x + c))^(3/2), x)

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maple [A]  time = 0.11, size = 42, normalized size = 1.17 \[ -\frac {2 \arctanh \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{\frac {3}{2}}\left (d x +c \right )\right )}{d \left (b \cos \left (d x +c \right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(3/2),x)

[Out]

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

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maxima [B]  time = 1.16, size = 65, normalized size = 1.81 \[ \frac {\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )}{2 \, b^{\frac {3}{2}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/2*(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))/(b^(3/2)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\sqrt {\cos \left (c+d\,x\right )}}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(1/2)/(b*cos(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^(1/2)/(b*cos(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\cos {\left (c + d x \right )}}}{\left (b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(1/2)/(b*cos(d*x+c))**(3/2),x)

[Out]

Integral(sqrt(cos(c + d*x))/(b*cos(c + d*x))**(3/2), x)

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