3.182 \(\int \frac{\cos ^{\frac{11}{2}}(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0290272, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {17, 2635, 8} \[ \frac{3 x \sqrt{\cos (c+d x)}}{8 b \sqrt{b \cos (c+d x)}}+\frac{\sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{4 b d \sqrt{b \cos (c+d x)}}+\frac{3 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{8 b d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(11/2)/(b*Cos[c + d*x])^(3/2),x]

[Out]

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

Rule 8

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

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{11}{2}}(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx &=\frac{\sqrt{\cos (c+d x)} \int \cos ^4(c+d x) \, dx}{b \sqrt{b \cos (c+d x)}}\\ &=\frac{\cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{4 b 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 \sqrt{b \cos (c+d x)}}\\ &=\frac{3 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{8 b d \sqrt{b \cos (c+d x)}}+\frac{\cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{4 b d \sqrt{b \cos (c+d x)}}+\frac{\left (3 \sqrt{\cos (c+d x)}\right ) \int 1 \, dx}{8 b \sqrt{b \cos (c+d x)}}\\ &=\frac{3 x \sqrt{\cos (c+d x)}}{8 b \sqrt{b \cos (c+d x)}}+\frac{3 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{8 b d \sqrt{b \cos (c+d x)}}+\frac{\cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{4 b d \sqrt{b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0837066, size = 55, normalized size = 0.51 \[ \frac{(12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x))) \cos ^{\frac{3}{2}}(c+d x)}{32 d (b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(11/2)/(b*Cos[c + d*x])^(3/2),x]

[Out]

(Cos[c + d*x]^(3/2)*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(32*d*(b*Cos[c + d*x])^(3/2))

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Maple [A]  time = 0.178, size = 62, normalized size = 0.6 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +3\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +3\,dx+3\,c}{8\,d} \left ( \cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}} \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.86491, size = 66, normalized size = 0.62 \begin{align*} \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{3}{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(3/2),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^(3/2)*d)

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Fricas [A]  time = 2.32015, size = 512, normalized size = 4.79 \begin{align*} \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^{2} 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^{2} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(3/2),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^2*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^2*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(11/2)/(b*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{\frac{11}{2}}}{\left (b \cos \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(11/2)/(b*cos(d*x + c))^(3/2), x)