3.2.92 \(\int \frac {\cos ^{\frac {13}{2}}(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx\) [192]

Optimal. Leaf size=107 \[ \frac {3 x \sqrt {\cos (c+d x)}}{8 b^2 \sqrt {b \cos (c+d x)}}+\frac {3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 b^2 d \sqrt {b \cos (c+d x)}}+\frac {\cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 b^2 d \sqrt {b \cos (c+d x)}} \]

[Out]

3/8*cos(d*x+c)^(3/2)*sin(d*x+c)/b^2/d/(b*cos(d*x+c))^(1/2)+1/4*cos(d*x+c)^(7/2)*sin(d*x+c)/b^2/d/(b*cos(d*x+c)
)^(1/2)+3/8*x*cos(d*x+c)^(1/2)/b^2/(b*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {17, 2715, 8} \begin {gather*} \frac {3 x \sqrt {\cos (c+d x)}}{8 b^2 \sqrt {b \cos (c+d x)}}+\frac {\sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{4 b^2 d \sqrt {b \cos (c+d x)}}+\frac {3 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{8 b^2 d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(13/2)/(b*Cos[c + d*x])^(5/2),x]

[Out]

(3*x*Sqrt[Cos[c + d*x]])/(8*b^2*Sqrt[b*Cos[c + d*x]]) + (3*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(8*b^2*d*Sqrt[b*Co
s[c + d*x]]) + (Cos[c + d*x]^(7/2)*Sin[c + d*x])/(4*b^2*d*Sqrt[b*Cos[c + d*x]])

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 2715

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] && Integ
erQ[2*n]

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {13}{2}}(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx &=\frac {\sqrt {\cos (c+d x)} \int \cos ^4(c+d x) \, dx}{b^2 \sqrt {b \cos (c+d x)}}\\ &=\frac {\cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 b^2 d \sqrt {b \cos (c+d x)}}+\frac {\left (3 \sqrt {\cos (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{4 b^2 \sqrt {b \cos (c+d x)}}\\ &=\frac {3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 b^2 d \sqrt {b \cos (c+d x)}}+\frac {\cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 b^2 d \sqrt {b \cos (c+d x)}}+\frac {\left (3 \sqrt {\cos (c+d x)}\right ) \int 1 \, dx}{8 b^2 \sqrt {b \cos (c+d x)}}\\ &=\frac {3 x \sqrt {\cos (c+d x)}}{8 b^2 \sqrt {b \cos (c+d x)}}+\frac {3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 b^2 d \sqrt {b \cos (c+d x)}}+\frac {\cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 b^2 d \sqrt {b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 58, normalized size = 0.54 \begin {gather*} \frac {\sqrt {\cos (c+d x)} (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))}{32 b^2 d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(13/2)/(b*Cos[c + d*x])^(5/2),x]

[Out]

(Sqrt[Cos[c + d*x]]*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(32*b^2*d*Sqrt[b*Cos[c + d*x]])

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Maple [A]
time = 0.16, size = 62, normalized size = 0.58

method	result	size
default	\(\frac {\left (\cos ^{\frac {5}{2}}\left (d x +c \right )\right ) \left (2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 \sin \left (d x +c \right ) \cos \left (d x +c \right )+3 d x +3 c \right )}{8 d \left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}}}\)	\(62\)
risch	\(\frac {3 x \left (\sqrt {\cos }\left (d x +c \right )\right )}{8 b^{2} \sqrt {b \cos \left (d x +c \right )}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \sin \left (4 d x +4 c \right )}{32 b^{2} \sqrt {b \cos \left (d x +c \right )}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \sin \left (2 d x +2 c \right )}{4 b^{2} \sqrt {b \cos \left (d x +c \right )}\, d}\)	\(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(13/2)/(b*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/8/d*cos(d*x+c)^(5/2)*(2*sin(d*x+c)*cos(d*x+c)^3+3*sin(d*x+c)*cos(d*x+c)+3*d*x+3*c)/(b*cos(d*x+c))^(5/2)

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Maxima [A]
time = 0.58, size = 49, normalized size = 0.46 \begin {gather*} \frac {12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )}{32 \, b^{\frac {5}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(13/2)/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/32*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))))/(b^(5/2)*d)

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Fricas [A]
time = 0.44, size = 182, normalized size = 1.70 \begin {gather*} \left [\frac {2 \, \sqrt {b \cos \left (d x + c\right )} {\left (2 \, \cos \left (d x + c\right )^{2} + 3\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 3 \, \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 )}{16 \, b^{3} d}, \frac {\sqrt {b \cos \left (d x + c\right )} {\left (2 \, \cos \left (d x + c\right )^{2} + 3\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, \sqrt {b} \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 )}{8 \, b^{3} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(13/2)/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/16*(2*sqrt(b*cos(d*x + c))*(2*cos(d*x + c)^2 + 3)*sqrt(cos(d*x + c))*sin(d*x + c) - 3*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), 1/8*(sqrt(b*cos(d*
x + c))*(2*cos(d*x + c)^2 + 3)*sqrt(cos(d*x + c))*sin(d*x + c) + 3*sqrt(b)*arctan(sqrt(b*cos(d*x + c))*sin(d*x
+ c)/(sqrt(b)*cos(d*x + c)^(3/2))))/(b^3*d)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(13/2)/(b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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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(cos(d*x+c)^(13/2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(13/2)/(b*cos(d*x + c))^(5/2), x)

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Mupad [B]
time = 1.00, size = 78, normalized size = 0.73 \begin {gather*} \frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (8\,\sin \left (c+d\,x\right )+9\,\sin \left (3\,c+3\,d\,x\right )+\sin \left (5\,c+5\,d\,x\right )+24\,d\,x\,\cos \left (c+d\,x\right )\right )}{32\,b^3\,d\,\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(cos(c + d*x)^(13/2)/(b*cos(c + d*x))^(5/2),x)

[Out]

(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(8*sin(c + d*x) + 9*sin(3*c + 3*d*x) + sin(5*c + 5*d*x) + 24*d*x*co
s(c + d*x)))/(32*b^3*d*(cos(2*c + 2*d*x) + 1))

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