3.185 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx\)

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

[Out]

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 = 35, 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, 2637} \[ \frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{b d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*Sin[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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx &=\frac {\sqrt {\cos (c+d x)} \int \cos (c+d x) \, dx}{b \sqrt {b \cos (c+d x)}}\\ &=\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 32, normalized size = 0.91 \[ \frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.92, size = 31, normalized size = 0.89 \[ \frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{b^{2} d \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(5/2)/(b*cos(d*x + c))^(3/2), x)

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maple [A]  time = 0.08, size = 29, normalized size = 0.83 \[ \frac {\sin \left (d x +c \right ) \left (\cos ^{\frac {3}{2}}\left (d x +c \right )\right )}{d \left (b \cos \left (d x +c \right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x)

[Out]

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

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maxima [A]  time = 1.48, size = 13, normalized size = 0.37 \[ \frac {\sin \left (d x + c\right )}{b^{\frac {3}{2}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

sin(d*x + c)/(b^(3/2)*d)

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mupad [B]  time = 0.32, size = 47, normalized size = 1.34 \[ \frac {\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (2\,c+2\,d\,x\right )\,\sqrt {b\,\cos \left (c+d\,x\right )}}{b^2\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(5/2)/(b*cos(c + d*x))^(3/2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(5/2)/(b*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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