3.2.64 \(\int \frac {(b \cos (c+d x))^{5/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [164]

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

[Out]

b^2*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/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, 2717} \begin {gather*} \frac {b^2 \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[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 2717

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

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Mathematica [A]
time = 0.03, size = 32, normalized size = 0.91 \begin {gather*} \frac {(b \cos (c+d x))^{5/2} \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 29, normalized size = 0.83

method	result	size
default	\(\frac {\left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}} \sin \left (d x +c \right )}{d \cos \left (d x +c \right )^{\frac {5}{2}}}\)	\(29\)
risch	\(\frac {b^{2} \sin \left (d x +c \right ) \sqrt {b \cos \left (d x +c \right )}}{d \sqrt {\cos \left (d x +c \right )}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.59, size = 13, normalized size = 0.37 \begin {gather*} \frac {b^{\frac {5}{2}} \sin \left (d x + c\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^(5/2)*sin(d*x + c)/d

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Fricas [A]
time = 0.35, size = 31, normalized size = 0.89 \begin {gather*} \frac {\sqrt {b \cos \left (d x + c\right )} b^{2} \sin \left (d x + c\right )}{d \sqrt {\cos \left (d x + c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6189 deep

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

[Out]

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

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Mupad [B]
time = 0.32, size = 31, normalized size = 0.89 \begin {gather*} \frac {b^2\,\sin \left (c+d\,x\right )\,\sqrt {b\,\cos \left (c+d\,x\right )}}{d\,\sqrt {\cos \left (c+d\,x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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