3.2.46 \(\int \frac {\sqrt {b \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [146]

Optimal. Leaf size=32 \[ \frac {\sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {17, 3852, 8} \begin {gather*} \frac {\sin (c+d x) \sqrt {b \cos (c+d x)}}{d \cos ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 3852

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {b \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx &=\frac {\sqrt {b \cos (c+d x)} \int \sec ^2(c+d x) \, dx}{\sqrt {\cos (c+d x)}}\\ &=-\frac {\sqrt {b \cos (c+d x)} \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d \sqrt {\cos (c+d x)}}\\ &=\frac {\sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 32, normalized size = 1.00 \begin {gather*} \frac {\sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.14, size = 29, normalized size = 0.91

method	result	size
default	\(\frac {\sin \left (d x +c \right ) \sqrt {b \cos \left (d x +c \right )}}{d \cos \left (d x +c \right )^{\frac {3}{2}}}\)	\(29\)
risch	\(\frac {2 i \sqrt {b \cos \left (d x +c \right )}}{\sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.57, size = 54, normalized size = 1.69 \begin {gather*} \frac {2 \, \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{{\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.39, size = 28, normalized size = 0.88 \begin {gather*} \frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {b \cos {\left (c + d x \right )}}}{\cos ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [B]
time = 0.70, size = 59, normalized size = 1.84 \begin {gather*} \frac {\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+\sin \left (2\,c+2\,d\,x\right )+1{}\mathrm {i}\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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