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

Optimal. Leaf size=27 \[ \frac{x \sqrt{\cos (c+d x)}}{b^2 \sqrt{b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0026813, antiderivative size = 27, 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, 8} \[ \frac{x \sqrt{\cos (c+d x)}}{b^2 \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[Cos[c + d*x]])/(b^2*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 8

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

Rubi steps

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

Mathematica [A]  time = 0.0124874, size = 24, normalized size = 0.89 \[ \frac{x \cos ^{\frac{5}{2}}(c+d x)}{(b \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.136, size = 28, normalized size = 1. \begin{align*}{\frac{dx+c}{d} \left ( \cos \left ( dx+c \right ) \right ) ^{{\frac{5}{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)^(5/2)/(b*cos(d*x+c))^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.57901, size = 35, normalized size = 1.3 \begin{align*} \frac{2 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac{5}{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.85471, size = 277, normalized size = 10.26 \begin{align*} \left [-\frac{\sqrt{-b} \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt{b \cos \left (d x + c\right )} \sqrt{-b} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{2 \, b^{3} d}, \frac{\arctan \left (\frac{\sqrt{b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt{b} \cos \left (d x + c\right )^{\frac{3}{2}}}\right )}{b^{\frac{5}{2}} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(-b)*log(2*b*cos(d*x + c)^2 + 2*sqrt(b*cos(d*x + c))*sqrt(-b)*sqrt(cos(d*x + c))*sin(d*x + c) - b)/(
b^3*d), arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))/(b^(5/2)*d)]

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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)**(5/2)/(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{\cos \left (d x + c\right )^{\frac{5}{2}}}{\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)^(5/2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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