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

Optimal. Leaf size=69 \[ \frac {b^2 x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 69, 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, 2635, 8} \[ \frac {b^2 x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 45, normalized size = 0.65 \[ \frac {(2 (c+d x)+\sin (2 (c+d x))) (b \cos (c+d x))^{5/2}}{4 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Cos[c + d*x])^(5/2)*(2*(c + d*x) + Sin[2*(c + d*x)]))/(4*d*Cos[c + d*x]^(5/2))

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fricas [A]  time = 1.04, size = 159, normalized size = 2.30 \[ \left [\frac {2 \, \sqrt {b \cos \left (d x + c\right )} b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \sqrt {-b} b^{2} \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{4 \, d}, \frac {\sqrt {b \cos \left (d x + c\right )} b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right )}{2 \, d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*sqrt(b*cos(d*x + c))*b^2*sqrt(cos(d*x + c))*sin(d*x + c) + sqrt(-b)*b^2*log(2*b*cos(d*x + c)^2 - 2*sqr
t(b*cos(d*x + c))*sqrt(-b)*sqrt(cos(d*x + c))*sin(d*x + c) - b))/d, 1/2*(sqrt(b*cos(d*x + c))*b^2*sqrt(cos(d*x
+ c))*sin(d*x + c) + b^(5/2)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2))))/d]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 42, normalized size = 0.61 \[ \frac {\left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right ) \left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}}}{2 d \cos \left (d x +c \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.09, size = 32, normalized size = 0.46 \[ \frac {{\left (2 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} \sqrt {b}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(d*x + c)*b^2 + b^2*sin(2*d*x + 2*c))*sqrt(b)/d

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mupad [B]  time = 0.44, size = 40, normalized size = 0.58 \[ \frac {b^2\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (\sin \left (2\,c+2\,d\,x\right )+2\,d\,x\right )}{4\,d\,\sqrt {\cos \left (c+d\,x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*(b*cos(c + d*x))^(1/2)*(sin(2*c + 2*d*x) + 2*d*x))/(4*d*cos(c + d*x)^(1/2))

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

[Out]

Timed out

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