3.198 \(\int \frac{\sqrt{\cos (c+d x)}}{(b \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=35 \[ \frac{\sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0132257, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {17, 3767, 8} \[ \frac{\sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sin[c + d*x]/(b^2*d*Sqrt[Cos[c + d*x]]*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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\sqrt{\cos (c+d x)}}{(b \cos (c+d x))^{5/2}} \, dx &=\frac{\sqrt{\cos (c+d x)} \int \sec ^2(c+d x) \, dx}{b^2 \sqrt{b \cos (c+d x)}}\\ &=-\frac{\sqrt{\cos (c+d x)} \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{b^2 d \sqrt{b \cos (c+d x)}}\\ &=\frac{\sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0170582, size = 32, normalized size = 0.91 \[ \frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{d (b \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^(3/2)*Sin[c + d*x])/(d*(b*Cos[c + d*x])^(5/2))

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Maple [A]  time = 0.237, size = 29, normalized size = 0.8 \begin{align*}{\frac{\sin \left ( dx+c \right ) }{d} \left ( \cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}} \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(5/2),x)

[Out]

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

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Maxima [B]  time = 1.74199, size = 90, normalized size = 2.57 \begin{align*} \frac{2 \, \sqrt{b} \sin \left (2 \, d x + 2 \, c\right )}{{\left (b^{3} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{3} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b^{3} \cos \left (2 \, d x + 2 \, c\right ) + b^{3}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.66569, size = 84, normalized size = 2.4 \begin{align*} \frac{\sqrt{b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{b^{3} d \cos \left (d x + c\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

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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(cos(d*x+c)**(1/2)/(b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\cos \left (d x + c\right )}}{\left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(cos(d*x + c))/(b*cos(d*x + c))^(5/2), x)